Geometry Regents Review (June 2, 2022) - Live on YouTube

[Music] thank you [Music] hello and welcome to the emath instruction geometry regents review Live on YouTube my name is Kirk Weiler and tonight we are going to be reviewing last year&#39;s June 2022 geometry regents exam this is primarily intended for those students in New York state who are taking the geometry regents two days from now right that&#39;s going to be Tuesday I think it&#39;s Tuesday morning but make sure to check with your local school district on the exact time and location for your test right tonight what we&#39;re going to be doing is going through last June&#39;s exam Problem by problem we&#39;re going to be looking at it from many different perspectives all right we will need our calculators a little bit on this exam mostly just to do some routine calculation work I&#39;m going to be using the ti Inspire calculator I love the calculator you can really use basically any calculator you want all right just be prepared there&#39;s some messy numbers on this exam but without further Ado let&#39;s get right into it with the first problem all right here we go part one now part one on the geometry exam has got 24 questions they&#39;re all multiple choice they&#39;re all worth two points a piece and there is no partial credit so you want to take your time on these you want to read them carefully all right they&#39;re not trying to trick you in any particular problems all right although some of them are trickier than other ones and we&#39;ll try to point those out thankfully they start us off with a pretty easy one let&#39;s take a look at it problem number one triangle a prime B Prime C Prime is the image of triangle ABC after a dilation centered at the origin the coordinates of the vertices of triangle ABC are a negative two comma 1 b 2 comma four and C 2 comma negative three all right and now for the actual question here we go if the coordinates of a prime are negative 4 comma 2 the coordinates of B Prime are which of the following all right great now you know they&#39;ve taken up a lot of space with this grid that we don&#39;t particularly need but let&#39;s go back and take a look at something really quick right you know a lot about a lot of geometry is about original pre-images triangle ABC and a transformation then that Maps them to some other figure in this case triangle a prime B Prime C Prime now we are told specifically that the transformation that we&#39;re talking about here is a dilation centered at the origin now dilations are basically taking a figure and stretching it away from a point or compressing it back down towards the point so that the figure still looks the same shape as the original but has a different size now you have to love dilations that are centered at the origin because all that you need to do right when you have a dilation centered at the origin is take a point x comma y and then multiply both the X and the Y coordinates by What&#39;s called the dilation constant oftentimes that&#39;s represented by the letter k now our real task in this is figuring out what that dilation constant is but they&#39;ve really given us the key here right they&#39;ve told us that a is at negative 2 comma one so a is at negative 2 comma one and that got mapped to a prime which was at negative four comma two That was supposed to be an a there we go negative four comma two so what exactly happened here well we multiplied the X and the y coordinate by a k value equal to 2. right now the key is with the dilation when you&#39;re mapping a figure you don&#39;t change the dilation constant depending on which vertex you&#39;re you&#39;re actually dilating if you did that you wouldn&#39;t get what what are known as two similar figures so if I want to know what the coordinates of B Prime are all I need to do is go back up here unfortunately B is at 2 comma four and I just need to multiply both of its coordinates by two and I&#39;ll find that b Prime is at four comma eight so real real simple dilations that are not centered at the origin are much trickier to perform algebraically and they would be much more difficult in a problem like this to figure out but this problem started us off very easy with probably the most constant or the most common dilation constant which is k equals two all right let&#39;s keep going let&#39;s get into problem number two in the diagram below a plane intersects a square pyramid parallel to its base which two-dimensional shape describes this cross section all right so one of the things that you&#39;re supposed to learn in geometry and you also probably looked at this a little bit in eighth grade math was if you have a two dimensional or sorry a three-dimensional solid and you start slicing it with planes think about a plane as just being a flat surface that extends forever right so I&#39;m it&#39;s I&#39;m slicing this thing with a plane what two-dimensional shape am I getting well that really always is going to depend on how the plane that you&#39;re slicing is oriented in this case it&#39;s a plane intersecting a square pyramid parallel to its space so in other words I&#39;m going to intersect this pyramid parallel to its base and what I&#39;ll see when I do that is a figure that kind of looks like that now what is that well it&#39;s just another square right that&#39;s all it is and therefore we&#39;ve got this particular answer all right awesome so I think the first two start us off nice and easy let&#39;s take a look at number three this one&#39;s a little bit trickier in the diagram below Triangle C D E is the image of triangle c a b after a dilation of d e divided by a b centered at C which statement is always true and then they give us a bunch of statements that involve the three trig ratios the sine of a is CE divided by c d etc etc all right now hopefully you all know learned that the sign is the opposite over the hypotenuse the cosine is the adjacent over hypotenuse and the tangent is the opposite over adjacent sohcahtoa some teachers love that mnemonic some teachers hate that mnemonic for me it&#39;s just a memory device right the sign of a given angle is the side opposite of the angle divided by the hypotenuse we&#39;re going to be doing a fair amount of trig in this test so might as well do a little bit right now cosine is adjacent over hypotenuse and tangent is opposite over adjacent now you might say why did they give us all of this information up here well they gave us all of that information most of which we can completely ignore all right but they gave us all that information really to tell us only one thing which is that this smaller right triangle I.E Triangle C D E is similar to this larger right triangle triangle c a b okay those two right triangles are similar and what that means is that the ratios of relatively positioned sides are going to be the same all right here&#39;s what I mean by that let&#39;s take a look at Choice one just for a moment I&#39;m going to kind of move this up because quite frankly all of that wording is pretty much irrelevant at this point let&#39;s take a look at the first one sine a is equal to CE divided by c d all right well what you should have learned all right of course now I&#39;ve moved my mnemonic out of the way is that sine is opposite over hypotenuse so if you were just going to be you know trying to figure out what the sign sign of angle a was what you would do is you&#39;d say all right well you know here&#39;s that right triangle that angle a is part of and the side opposite of angle a is BC or CB either way all right and it&#39;s hypotenuse is AC now you might look at like this choice and go well that&#39;s not correct that&#39;s not CE divided by CD except it is you see because BC divided by AC is the same as c e divided by c d and that&#39;s because those triangles are similar all right so Choice a is actually correct now for a moment before we move on to problem four let&#39;s just take a look at Choice 2 which is incorrect for a moment right cosine of a right so if we were doing cosine of a we would say well cosine of a is going to be the side adjacent to a which would be a b over its hypotenuse which is again AC now in terms of the Little Triangle right which is what all the answers are put in terms right in terms of the Little Triangle right A B divided by AC would be the same as d e divided by c d right so this would be d e divided by CD but notice the choice they give us is CD divided by CE that&#39;s not even close that&#39;s that&#39;s not even one of the three ratios at all that would be actually hypotenuse divided by opposite which is a trig ratio but not one that you know of yet anyhow so this problem really combines two things it combines the idea of the three trig ratios along with the idea that two similar right triangles are going to have ratios of relatively positioned sides that are the same that&#39;s actually the basis of all of right triangle trigonometry all right let&#39;s move on to problem number four here we go a regular pentagon is rotated about its Center what is the minimum number of degrees needed to carry the Pentagon onto itself all right wonderful well you were expected to know some of the basic regular figures right a pentagon has got five sides a hexagon has got six sides seven-sided figures they come up less often because they&#39;re more messy but for the record those are called septagons of course an eight-sided figure you&#39;re supposed to know the name of that&#39;s an octagon I&#39;m sure you do a nine-sided figure also doesn&#39;t come up very often called a nonagon I love that word and then a 10-sided figure a decagon after that I don&#39;t think that you&#39;d have to know any of them but obviously we need to know what a pentagon is okay now if we&#39;ve got any regular figure and here I&#39;m going to try to do my best drawing a pentagon a regular pentagon right the idea of a regular figure which this is most certainly not but I gave it my best shot right a regular figure has got all sides the same length and all interior angles the same measure and what that means is that it&#39;s got what&#39;s known as symmetry rotational symmetry reflective symmetry you know anyhow this is talking about rotation symmetry the idea is that we&#39;ve got a center point right I&#39;m just going to kind of draw on some things that sort of look like the radii of circles here all right now if I took this thing and I rotated it either clockwise or counterclockwise it wouldn&#39;t make any difference right it would first line up with itself let&#39;s say we were rotating counterclockwise if I rotate this particular angle all right so if I rotate whatever that angle is that&#39;s going to be the minimum angle that it&#39;ll take for that vertex to end up there that vertex to end up there Etc what is that angle well I know that the entire rotation right all the way around is 360 degrees and in the Pentagon there will be five of those sort of radial angles if you will just made that up probably not a good term so I&#39;m going to do 360 divided by five I think I&#39;ll skip the calculator on this one for now and that would be 72 degrees and therefore our answer is 72 degrees now a little twist on this problem a little variation on this problem is that they might say something like which of the following angles is not an angle of rotational symmetry or which of the following angles does not map the Pentagon onto itself now you might say well I I thought 72 was the only one well now 72 would work 2 times 72 which would be 144 would work 3 times 72 which would be 216 would work etc all right so any integer multiple of this so multiply this by two by three by four by five any of those will map the figure onto itself actually you can multiply it by 1000 and that angle would would you know map it onto itself all right so let&#39;s take a look at problem number five on the set of axes below triangle ABC is congruent to Triangle a prime B Prime C Prime let me move this up a little bit so we can see the rest all right triangle ABC maps onto triangle a prime B Prime C Prime after a what all right so one thing that you learned was that if you have two figures that are congruent to each other now congruent figures are figures that are identical in every way shape or form they have the same size they have the same shape they are simply in different locations in the plane all right but if you have two figures that are congruent to one another then there is either one rigid motion and those are translations rotations or Reflections there&#39;s either one or a sequence of those that will map literally put this thing on top of this one all right so that&#39;s our idea here we&#39;re trying to figure out which one it is now let&#39;s take a look at our four choices Choice number one is a reflection over the line Y equals negative X Choice two a reflection over the line Y equals negative X plus two number three a rotation of 180 degrees centered at one comma one and four a rotation of 180 degrees centered at the origin all right now these problems are not uncommon so before we even get into which one is the correct choice I&#39;d like to point something out that will allow you to eliminate a couple of the choices let&#39;s take a look at triangle ABC but don&#39;t break to the board right in triangle ABC if I start at Point a and then I go to point B then I go to point C then I go back to point a point B Point C Etc if I do that notice right that I have to go in a clockwise manner to sort of Trace out alphabetically those letters now if I look at triangle a prime B Prime C Prime and I go from A to B to C back to a that is also in a clockwise manner notice that in both cases the order of the vertices stays in a clockwise manner that means there cannot be a reflection over a line okay or a single reflection over a line all right that is very very important all right if there is a line reflection a single line reflection then what will happen is if the letters started in a clockwise manner they will then change to a counterclockwise manner think about the idea of when you look in a mirror and you look in your reflection your hair Parts the opposite way right you know if you raise up your right hand it looks like you&#39;re raising up your left hand Etc right Reflections change the orientation of the vertices since these don&#39;t that means that both Choice One and choice 2 are immediately eliminated they&#39;re done okay now let&#39;s talk about these two 180 degree rotations and before we do that let me just kind of get rid of all this now it is very tempting to think it&#39;s a rotation of 180 degrees centered at the origin but let&#39;s talk about 180 degree rotation centered at the origin when you&#39;ve got something like this right then what happens is a point x comma y will get mapped to the point negative x negative Y in other words it&#39;ll be exactly the same point but it will have its sign changed in other words if you took something like Point C down here or sorry up here which is at one two three four five one two three four five six so C at five comma six should go to C Prime at negative 5 negative six except C Prime is actually at negative three one two three four C Prime is at negative three negative four so it really can&#39;t be a rotation of 180 degrees centered at the origin all right so that leaves us with choice three but let&#39;s really try to understand how we can see that it&#39;s 0.3 or choice three all right a rotation centered a rotation of 180 centered at one comma one rotations of 180 degrees are some of the most special rotations out there and the reason why is that when you rotate a point by 180 degrees let&#39;s say this is your original point and let&#39;s say this is your point of rotation I don&#39;t want to use C because we already have a c up here when we rotate by 180 degrees what happens is that those three points the original the pre-image point the center of rotation and the image Point are all co-linear they&#39;re all co-linear now what&#39;s cool about this is if I think or know that I have a rotation by 180 degrees I can literally connect the point and its image Point using a straight line and somewhere on that line is the center of rotation and you might go well that&#39;s not very helpful except now let me collect connect sorry what not collect connect B and B Prime and when I do that I do I get that and now when I connect C and C Prime and I do that then what happens is I can see my center of rotation right the center of rotation has to be on each one of those lines because it is a 180 degree rotation and that is specifically a point of rotation at one comma one it doesn&#39;t work if the if it&#39;s a rotation other than 180 degrees but 180 degree rotations will leave the point the center of rotation and the image Point collinear and that&#39;s really kind of a key in understanding why at least in this problem the center of rotation is not at the origin but is actually at the point one comma one all right let&#39;s keep going problem number seven I like this one in the diagram below of triangle AED and a b c d segment a b c d a e segment AE is congruent to segment d e which statement is always true all right now they&#39;re trying to just draw you in here they want you to look at this picture and they want you to make a statement like oh EB is congruent to EC well yeah sure EB looks like it&#39;s congruent to EC except I I&#39;m not told anything anything up here that would leave me to believe that in fact all I know from this statement right is two things AE is congruent to d e and I know that points a b c and d are co-linear that&#39;s what that tells me they all lie in a straight line now you might say well I can tell that just from the picture but we&#39;re trying to avoid that here okay because from the picture it sure looks like e b and e c are the same now do I know that AC is congruent to DB well again it sure looks like AC and DB are the same length you might also be thinking oh there&#39;s some kind of partitioning there some kind of addition or subtraction proof nope uh-uh didn&#39;t didn&#39;t tell me anything right how about EBA is congruent to ecd that would be this angle right here and this angle right there again nothing tells me that so it seems like it should be Choice four angle EAC is congruent to angle EDB well let&#39;s take a look at those two e a c is this angle right and e d b is this angle and you bet I know that&#39;s true and I don&#39;t know that&#39;s true because it looks like it&#39;s true I know it&#39;s true because I know that this is an isosceles triangle right a triangle with two sides that are the same length and what I know about any triangle that has two sides the same length is that the angles opposite of those two equal sides are also equal in measure and therefore the base angles of an isosceles triangle are equal and angle EAC is congruent to angle EDB cool let&#39;s keep going number eight as shown in the diagram below right triangle ABC has side lengths of 8 and 15. if the triangle is continuously rotated about segment AC the resulting figure will be Choice One a right cone with a radius of 15 and a height of 8. choice two a right cone with a radius of 8 and a height of 15. choice 3 a right cylinder with a radius of 15 and a height of 8. and choice 4 a right cylinder with a radius of 8 and a height of 15. so here you have to know obviously the difference between a cone And the cylinder and I hope you do right a cone is something that kind of looks like this right with a nice sort of like that and a cylinder is something whoops I don&#39;t know why I made that Dash something that looks like that all right now this problem gets to what&#39;s called a volume of Revolution or volume of rotation right specifically if we took this particular figure and we rotated it about AC and that&#39;s really key the idea is if I take this thing right and I rotate it around that segment right I spin that right triangle around what does it end up looking like well its Mirror Image kind of comes around here and you get these circular cross sections and what I now have is a cone right with a height of eight and a radius of 15 and that is Choice One the way you&#39;d get a cylinder by the way is by rotating a rectangle around any one of its sides okay now depending on which side you rotate it around the rectangle will have a different you know height or radius but that&#39;s the way you get a cylinder all right let&#39;s go to the next one problem number nine in the diagram below lines K and L intersect lines M and N at points a b c and d which statement is sufficient to prove that a b c d is a parallelogram awesome well right first let&#39;s just outline that thing A B C D you might be like hey that looks like a parallelogram and it certainly does but I want to know what&#39;s enough to know that it is a parallelogram now parallelograms have many properties right but at the end of the day what makes a parallelogram a parallelogram right is the idea that a b must be parallel to CD and right I have to have a d parallel to BC right what a parallelogram is is a four-sided figure with two pairs of parallel sides and you might even want to throw in two pairs of opposite parallel sides although I&#39;m pretty sure adjacent sides can&#39;t be parallel so that would be a little bit redundant all right so let&#39;s take a look you know in Choice one we tell we know that angle one is congruent to angle three all right so that would be like angle one congruent to angle three now that&#39;s awesome if I know that angle one and angle 3 are congruent to each other then I definitely know that a d is parallel to BC right so that&#39;s what choice one gets me a d is parallel to BC because these are what are known as corresponding angles okay and if corresponding angles are created that are congruent then the lines that create them are parallel but that&#39;s not enough right it only gets me a d parallel to BC so that not enough half of what I need but not enough now Choice 2 angle 4 is congruent to angle seven let&#39;s mark those angle four is congruent to angle seven well that&#39;s also kind of great these are alternate interior angles that are created by this transversal and these two lines and therefore that would be enough to know that a b is parallel to CD which is great but still not enough right that&#39;s not enough right it would only get me a trapezoid remember trapezoids only have to have one pair of parallel sides so it&#39;s it&#39;s got to be choice three or Choice four let&#39;s take a look at choice three choice three says angle two is congruent to angle five all right now that&#39;s that alternate interior angle pair again which is awesome because that again gets me you know a b parallel to C D or k parallel to L whatever whatever you want to say so I&#39;m halfway there and now 5 is congruent to seven so where&#39;s five and seven five is congruent to seven right these two being congruent to each other will get that parallel to that these two being congruent to each other will get this parallel to this and that&#39;s it I&#39;ve got it now a little side note real quick here right why not this last choice you know it seems to have just as much information as choice three right but it&#39;s not enough right because if I look at Choice four it says one is congruent to three let me kind of get rid of these again I gotta get rid of all this so one is congruent to three which is great because then that gets me m and n parallel to each other and then three is congruent to four that doesn&#39;t do anything for me three congruent four three and four are just vertical angles so they&#39;re always congruent they have nothing to do with parallel lines whatsoever so Choice four is not correct all right here comes probably my least favorite problem on the test and that&#39;s saying a lot because on the algebra 1 test if you were with me that would be weird because that means you&#39;ve watched both of these videos and you&#39;re either taking both classes or you just love watching math videos um anyway either way that that test had all sorts of problems this is one of the very few on this test that I don&#39;t particularly love but let&#39;s take a look at it which transformation does not always preserve distance does not always preserve distance now the idea of preserving distance is very simple right if I&#39;ve got two points in space right and let&#39;s connect them with a line segment just for the fun of it right you know if I take it because it&#39;s got to be fun to connect with a line segment so if I connect them with a line segment and then I transform them right and there are so many Transformations out there it&#39;s not just okay you know rotations Reflections translations dilations there&#39;s tons of things out there right sometimes when I do something right then what will happen is that the line segment I get as the image let&#39;s say this has got a length of four centimeters then the transformation preserves the distance between the two points anytime that happens always we call it a rigid body motion okay when the distance is preserved but sometimes it doesn&#39;t you know sometimes you know we do it and we get something that changes let&#39;s say that&#39;s three centimeters that would not be distance preserving all right now translations Reflections and rotations the rigid body motions will always preserve distance dilations remember those things right from problem number one dilations where we multiply the X and the y coordinate by something those don&#39;t preserve distance okay now when I look at these right these are all algebraic rules and if you&#39;ve studied or taught geometry for a long time you would look at Choice one where X comma y gets mapped to X plus two comma Y and you know oh that is just a translation two units to the right that&#39;s going to preserve distance because it&#39;s a translation now Choice two x comma y goes to negative y comma Negative X that&#39;s actually a reflection across the line Y equals negative X but it&#39;s a reflection so therefore it is also distance preserving now choice three and choice four are very strange they are transformations that you have really never seen before in any kind of algebraic way number three x comma y goes to 2 times x comma y minus one as soon as I see that multiplication by two it&#39;s not going to preserve distance it just isn&#39;t if you multiply X or Y by any number other than 1 or negative one right because here we&#39;re multiplying by negative one right if we multiply by either by anything other than one or negative one we&#39;re getting a stretching effect okay now this is kind of weird I&#39;m not going to call this a dilation because it&#39;s being stretched in the X Direction but the Y is just shifting down one unit okay but it still isn&#39;t going to preserve distance because of that multiplication here you&#39;ve got both a multiplication by negative one plus sort of like an addition of three and two and again they couldn&#39;t have made that more confusing looking this is literally a combination of both a reflection and a translation number four all right but the one that won&#39;t preserve distance in this case says choice three and really it&#39;s all about this it&#39;s all about the fact that the x is being multiplied by two it has nothing to do with the Y minus one all right let&#39;s keep going oh end of that page all right the first one that&#39;s going to require a calculator let&#39;s take a look at number 11. in the diagram below EF segment EF is parallel to segment HG EF is equal to five HG is equal to 12. f i is equal to 1.4 X plus 3 and h i is equal to 6.1 x minus 6.5 awesome then they ask us what is the length of h i and we have to be very careful on this problem all right but let&#39;s talk about what this problem is getting at as soon as they tell me that EF is parallel to HG right e f is parallel to HG what that allows me to say is that angle F and angle H are congruent and angle e and angle G are congruent now why is that good to know well as soon as I know that two angles of one triangle are congruent to two angles of another triangle then I know that this triangle and this triangle are similar to each other they&#39;re similar to each other so I can do something like this I can say 5 divided by 12. is equal to 1.4 X plus 3 divided by 6.1 x minus 6.5 all right so this is a very very common thing that we see a lot in similarity all right this is where I&#39;m going to pick up my cheat sheet because I&#39;ve worked out all the numbers here so I don&#39;t have to necessarily go to the calculator yet we&#39;ll go to the calculator when we really need to but now to solve this equation I&#39;m just going to cross multiply that&#39;s easy enough right do a little bit of this a little bit of this we&#39;ll get 5 times 6.1 x minus 6.5 is equal to 12 times 1.4 X plus 3. all right now I&#39;ll use a little distributive property to multiply through by the 12 and the 5. let me just kind of scroll up a little bit here and we will get 30.5 x minus 32.5 trying to make those decimal points a little bit bigger is equal to 16.8 X Plus 36. let me take a pause just for a minute so you can take a look at that simple enough at this point it&#39;s really mostly algebra mostly algebra so if I combine all like terms and I get all the x&#39;s and stuff to one side or another I&#39;m not going to go through all of that eventually you&#39;ll get X is equal to 5. all right now you have to be a little bit worried about this okay and here&#39;s why X is equal to 5 and so it is very tempting to immediately bubble in choice 2. the problem here right is that they&#39;re asking me what the length of h i is and that&#39;s not equal to X right h i is going to be equal to 6.1 times 5 minus 6.5 and of course you end up doing that on your calculator and you find that it&#39;s 24. all right and that&#39;s it okay but this is actually you know one of the things that I don&#39;t love about this problem is that you could know all the geometry you could know this is about similarity you could know you&#39;re supposed to set up that ratio you can cross multiply you can do every darn thing correct get x equals five bubble that choice in and you&#39;re losing two points even though you knew the geometry you knew the algebra right but because they didn&#39;t want the value of x they wanted the length of h i you get the problem wrong don&#39;t love it but it is kind of a good problem to look at in terms of make sure to give the answer they want right make sure you&#39;re really solving for it all right let&#39;s take a look at another one where we&#39;ll actually break the calculator out in a moment 12. the square pyramid below models a toy block made of maple wood each side of the base measures 4.5 centimeters and the height of the pyramid is 10 centimeters if the density of maple is 0.676 grams per centimeter cubed what is the mass of the block to the nearest tenth of a gram all right now I know that you studied density in Geometry okay but I really want to emphasize something density which is what they give us here is how much stuff is inside of stuff okay and and I know that sounds really dumb but let me kind of really like lay it out here density in this case is 0.676 grams per centimeter cubed all right to me this is no different than saying the density of eggs is 12 eggs per carton right that&#39;s how much stuff is in stuff there are 0.676 grams in a cubic centimeter of the wood that is being used to make this square pyramid so the first thing I have to do is figure out how much volume I have how many cubic centimeters do I have now this is a good time to take a not a break for lemonade although I&#39;m dying for some we will soon but this is a good time to take a look at the formula sheet at the back of the exam all Regents math exams have the same formula sheet on them I need to move that out of the way and it is literally one of the last pages that&#39;s in here oh my goodness the ice cream truck is going on outside that&#39;s amazing I would stop and go get some except I&#39;d probably get in trouble all right so let&#39;s take a look at the formula sheet now there&#39;s a lot of stuff on this formula sheet that you can use in this test you know ironically things like let&#39;s say the trig ratios aren&#39;t on it I don&#39;t know why but you know in case you&#39;re in the middle of the exam and you really need to know that I don&#39;t know one cup is equal to eight fluid ounces there it is you&#39;ll know it you&#39;ll have it anyway whatever um so one thing that is helpful for you are things like the area formula for a circle right or the circumference formula of a circle or the volume of the general prism or the volume of a cylinder or the volume of a sphere the volume of a cone the volume of a pyramid there it is that&#39;s the one I need and it says one-third times Big B times H and that Big B that Big B that sounds really weird but that Big B is the area of the base of our pyramid big b stands for the area of the base so it&#39;s one-third times the area of the base times the height of the pyramid so now let&#39;s go back to that problem let&#39;s see if we can find it here we&#39;re getting there we&#39;re getting there getting there there it is ah excellent so what do we know well we know that it&#39;s got a square base where the sides of the base are 4.5 centimeters each and the height is 10 centimeters so if you recall right that volume formula was one-third Big B times H so that&#39;s going to be one-third times the area of the base now that&#39;s 4.5 squared right that&#39;s a square base that&#39;s its area times the height which is 10. so let&#39;s figure out what the volume of our pyramid is right now and for some reason I don&#39;t have my calculator open I don&#39;t know why I thought I had it open earlier but it must have decided to just close here we go excellent let&#39;s add a calculator um and we&#39;re going to do 1 3 times 4.5 nope let&#39;s try that again 4.5 squared times 10. right and that gives us a volume of 67.5 cubic centimeters that&#39;s how much stuff we have that&#39;s the stuff that&#39;s holding the mass is this space 67.5 cubic centimeters let&#39;s put that down all right now we know that we&#39;ve got 0.676 grams per cubic centimeter I mean again if I told you you had five egg cartons and each egg carton held 12 eggs and I asked you how many eggs are there right you would take the number of cartons and multiply it by how many eggs there were per carton if I tell you that there&#39;s 0.676 grams in every cubic centimeter and you know how many cubic centimeters there are then the mass is just that many cubic centimeters times how much how many grams there are per cubic centimeter so my mass now is going to be 67.5 cubic centimeters times 0.676 grams per cubic centimeter look at that the cubic centimeters cancel let&#39;s grab my calculator again this is awesome because I can just put times .676 and that&#39;s going to give me a mass of 45.63 foreign which to the nearest tenth is 45.6 excellent I think this is as good of a time as any to take a little break have a little lemonade before we go into one of the trickier problems on the test lemonade is so good I wonder what happens if I put this right in front of the green screen does it disappear nah it doesn&#39;t disappear feels like it should it&#39;s more or less the the color of the green screen but it&#39;s just lemonade oh my goodness it gets it gets hot in here I got like lights everywhere they&#39;re beating down on me you know uh I burn hot anyway so yeah gotta have a lot of lemonade geez I feel like I drank half that thing and I&#39;m not even halfway through the test okay all right let&#39;s take a look at one of the trickier problems on the test um you know there&#39;s a lot of times when students say oh they were trying to trick me there oh they threw in a trick or something like that and I always think to myself nine times out of ten it&#39;s just not true they&#39;re not trying to trick you they&#39;re just trying to test something all right now let&#39;s I feel like I&#39;m like shiny now from sweat um okay let&#39;s take a look like you needed to know that um let&#39;s take a look at problem number 13. in the diagram below of right triangle e f g altitude FH intersects hypotenuse EG at H if FH is 9 and EF is 15 what is eg all right now there are a few things about this problem I don&#39;t particularly like all right one of the things is that they sort of indirectly tell us that triangle e f g is a right triangle all right I mean you might say well they don&#39;t indirectly tell us that they tell us this right here it would have been kind of nice had they said this particular angle is the right angle all right but let&#39;s be very clear this is a right angle right and because FH is an altitude this is also a right angle all right so immediately there&#39;s probably many of you that are saying I know this problem right this is that whole like altitude to the hypotenuse of a right triangle problem and you are kind of correct okay but here&#39;s where the tricky part comes in let&#39;s put in our our two measurements right we know that FH is equal to nine and we know that e f is equal to 15 and we want to know what EG is okay now you probably memorized some theorems or you you did hopefully your teacher told you about how this is really just all about similarity and it is all about similarity right we&#39;ve got a large right triangle we&#39;ve got a medium right triangle and we&#39;ve got a small right triangle okay I feel like it&#39;s like the you know Goldilocks and the Three Bears kind of thing all right now the problem is none of those theorems right that you may have learned about like geometric means or you know this over this is equal to this over this none of them are going to get the job done there&#39;s not enough information all right and you can try you know like you could be like well like I don&#39;t know 15 divided by 9 is equal to EG divided by f g that would actually be the case like 15 over 9 would be equal to EG divided by f g but I don&#39;t know f g so that that doesn&#39;t help okay um so we we are actually lacking enough information to use one of those theorems all right so here&#39;s what they want you to do they want you to realize that number one oh I don&#39;t have enough information to do that similarity bit okay but I can get enough information if I use the Pythagorean theorem to figure out the length of e h all right so because I know the hypotenuse 15 and this side nine I should be able to figure out the length of e h now as well if you know the famous three four five right triangle you can also figure out eh just using a multiple of that three four five right triangle but let&#39;s say for for kicks and grins you just don&#39;t recognize that so maybe you call this a right and you do a little a squared plus 9 squared is equal to 15 squared and I&#39;m not going to go through all of that because you&#39;ve seen the Pythagorean theorem done many times but you end up getting a is equal to 12. all right so you have to do that first you absolutely need to get that 12 first so this is a problem that&#39;s really testing two distinctly different things one of those things is the Pythagorean theorem and you have to do that first or special right triangle slash Pythagorean triples all right then once you have that now you&#39;re good to go and there&#39;s many different ways of doing it one way of doing it is setting up a proportion and solving for GH or HG and then adding that onto the 12 that&#39;s one good way of doing it there&#39;s another way of doing it right away right which is also doing it by similarity so now keep in mind right when you look at the three right triangles right the big one which is kind of lying on its hypotenuse and then the two smaller ones which are lying on legs of theirs right if I hone in on the middle one right I know it&#39;s hypotenuse right and I also know it&#39;s longer leg right I know it&#39;s hypotenuse and I know it&#39;s longer leg now if I look at the big right triangle I know it&#39;s longer leg and then what they&#39;re asking me for is it&#39;s hypotenuse so what I can do is I can set up a ratio like the following I can say this longer leg divided by its hypotenuse so 12 divided by 15 right so longer leg divided by the hypotenuse is now going to be equal to this longer leg 15 divided by the hypotenuse that I&#39;m looking for EG okay now some of you have maybe learned a theorem that said that this particular leg is the geometric mean between this side and the entire hypotenuse that&#39;s great that gets you this you know there&#39;s a variety of ways that you can do it but at this point in time we now cross multiply right we get 12 times e g is equal to 15 squared and then I can just do it 15 times 15 15 squared I&#39;m not going to even evaluate that I&#39;m going to say EG is 15 squared divided by 12. that&#39;s where my calculator is very handy 15 squared divided by 12 of course 15 squared divided by 12 all I really want is the decimal version and I get 18 0.75 again you have to be a little bit careful about that because if you solve for GH directly you get 6.75 right so it&#39;d be very easy to solve for GH get 6.75 and circle Choice one all right but we want all of EG and again what makes this problem to me especially tricky and a little bit dubious from the perspective of a test maker is that they&#39;re really testing two distinctly different geometric Concepts one of them being the Pythagorean theorem let&#39;s say and the other one being this similarity when the altitude is drawn to the hypotenuse of a right triangle all right let&#39;s keep going ah back to a nice easy no tricks one 14. in triangle ABC below D is a point on a b and e is a point on AC such that d e is parallel to BC which statement is always true all right so again this is one of those problems where you might think they&#39;re trying to trick you and maybe they are but what they&#39;re really trying to do is make sure that you understand what you can get out of geometric information that you&#39;re given and what you can&#39;t so let&#39;s take a look right choice one angle Ade and angle ABC are right angles in other words is there enough information in what they told us to conclude that these two are right angles now again they look like right angles they sure do but there is nothing in any of this that talks about perpendicularity at all so even though they might look like right angles I don&#39;t know that they are all right now Choice two triangle Ade is similar to triangle ABC all right so do we know that this particular triangle is similar to this one well we&#39;re told that d e is parallel to BC these two are parallel and because they&#39;re parallel we know that that angle must be congruent to that angle and this angle must be congruent to that angle those are corresponding angles being created by two transversals and two parallel lines and therefore yep triangle Ade must be try similar to triangle ABC by the angle angle theorem of similarity so that is absolutely true by angle angle now as an aside the last two choices are trying to draw you in to think that points D and E are mid points right so like d e is half of b c d e is half of BC well that would definitely be true if d and e were the midpoints of a b and a c respectively and likewise a d is congruent to DB a d congruent to DB that would be true again if D were the midpoint of a b but even though it looks like it&#39;s the midpoint there&#39;s nothing up here whatsoever that tells us it&#39;s the midpoint okay let&#39;s keep going I never know when I&#39;m really at the end of the page all right let&#39;s take a look at 15. if one exterior angle of a triangle is acute then the triangle must be which of the following all right well first let&#39;s talk about exterior angles right if you have any kind of figure it doesn&#39;t have to be a triangle any type of figure whatsoever like let me make some kind of generic quadrilateral right if I have a generic quadrilateral like that and I go to any given interior angle let me go to this nice acute angle right here right if I extend that side all right then this angle is the exterior angle to this angle&#39;s interior angle all right exterior interior all right now you might say why do you extend that one and not this one right so you know why did I extend that side length and let&#39;s say not this side length and the answer is it really doesn&#39;t matter those two angles would have the same measure and of course the key is here if you have an acute angle here right then you must have an obtuse angle out here okay so given that let&#39;s Now erase this picture and talk about the triangle that they&#39;re referring to they say if one of the exterior angles of a triangle is acute then the triangle must be what well if we&#39;ve got an acute exterior angle right let&#39;s say we got this triangle and here if I extend this right then that is acute right an acute angle is an angle that is less than 90 degrees right an acute angle is a sharp Angle an angle less than 90 degrees if we have an acute exterior angle then the adjacent interior angle must be obtuse and as soon as an as a triangle has an obtuse angle it&#39;s called an obtuse triangle now side note there is nothing called an acute triangle I&#39;ve never heard of an acute triangle whatsoever all right uh that that&#39;s just weird um there are obtuse triangles obtuse triangles or any triangle with an obtuse angle a right triangle is of course any triangle with a right angle and an equi-angular triangle also known as an equilateral triangle is a triangle with all equal angles right a scalene triangle is a triangle with all unequal angles anyway not the point the point is if you have an acute angle on the outside you must have an obtuse angle on the inside and therefore that triangle is obtuse all right ah number 16. given the information marked on the diagrams below which pair of triangles cannot always be proven congruent all right well I will say right this is a generally speaking a pretty well constructed test but this might be the one problem on the test that I believe is sort of poorly constructed it&#39;s not trying to trick you in fact I think most people will probably get this problem right now remember we&#39;re looking for something that can&#39;t be proven congruent all right let&#39;s take this moment to talk about the major theorems that prove two triangles are congruent right the big three which are side side side angle side angle and side angle side those are the big three okay those are the basis of like everything else and then there&#39;s two additional ones which is angle angle side and the very specific hypotenuse leg all right now if we can assign any one of these to these kind of diagrams given the information they tell us then we can prove that the triangles are congruent so if we take a look at the first one right and we&#39;ve got well that side&#39;s congruent to that side we know that&#39;s a 90 degree angle now here&#39;s a little bit of a problem okay all right I&#39;m gonna assume that&#39;s a 90 degree angle as well now you might say well how what do you mean assume that&#39;s a 90 degree angle well I I got to tell you man this is a flawed problem and shame on you for writing it whoever did right there is nothing in here that tells us that points a c and d are co-linear for all I know that&#39;s a right angle and that thing&#39;s like an 89 degree angle I don&#39;t know you know what I mean but the plain fact is the test writers want you to assume it&#39;s also a 90 degree angle even though they didn&#39;t tell you that and then of course BC is congruent to itself so in theory we can prove that a b c and DBC are congruent by side angle side right this is a side angle side scenario again assuming that ACD is collinear it didn&#39;t tell us that but we we have to otherwise like actually the only one that you can say are always congruent is choice three and we&#39;ll get to that in a moment let&#39;s take a look at Choice two okay in this particular problem what do we know well we&#39;ve got angle e congruent to angle H we&#39;ve got side F G congruent to side IG and again this is a little bit flawed they want us to assume that e g h and igf are also collinear that would allow us to conclude that those are vertical angles and therefore those angles are congruent now notice here the side that we know that is congruent is not between the two angles the pairs that we know are congruent but that&#39;s okay because that still falls into angle angle side okay I&#39;ve got two pairs of angles that are congruent and I&#39;ve got a side pair that&#39;s congruent that&#39;s angle angle side so we&#39;re still good all right again a little problematic because there&#39;s absolutely nothing that tells us e g h and igf are collinear we could have these little triangles being a little like like this you know what I mean then we wouldn&#39;t know if those two angles those two vertical pairs were equal or not but you got to do what you got to do you know anyway this one this one&#39;s good to go okay and the reason why is we know this side&#39;s congruent to this side that side&#39;s congruent to that side and then we have that reflexive shared side so this is a situation where the two triangles can most certainly be proven congruent by side side okay so why isn&#39;t this one correct all right well in this situation right we&#39;ve got a side congruent to a side another side congruent to a side and an angle can grow into an angle so it feels like it could be side angle side the problem is the angle pair that&#39;s congruent is not what&#39;s called included between the two congruent side pairs if we knew that angle o was congruent to angle s if we knew that angle was congruent to that one we would have side angle side but this is one of those ones that your math teacher warns you about right this is ass and ass doesn&#39;t work and it&#39;s not because it would always get Snickers from like Teenage students sitting in a geometry class although that&#39;s a good way to remember it doesn&#39;t work right it just isn&#39;t one of the theorems right you can literally have two triangles that have this kind of scenario going on but aren&#39;t congruent to each other so this is our correct choice amongst a very flawed problem all right let&#39;s keep going hello problem 17. the diagram below shows a tree growing vertically on a hillside the angle formed by the tree trunk and the hillside is 100 degrees the distance from the base of the tree to the bottom of the hill is 140 feet what is the vertical drop x to the base of the hill to the nearest Foot well it&#39;s nice that they gave us this entire diagram it&#39;s really great now look anytime you&#39;re working with a right triangle you know one of the lengths of the right triangle you&#39;re missing one of the other lengths but you&#39;ve got some kind of an angle information think about right triangle trig again let&#39;s go back to sohcahtoa let me put it up there because here comes right triangle trig again now one of the things that&#39;s a little bit unfortunate here is that the angle that they give us that hundred degree angle isn&#39;t actually in the right triangle we need in order to use sohcahtoa which you can barely read I apologize for that in order to use sohcahtoa we need to have an angle in the right triangle well if that&#39;s a 100 degree angle then the one sitting right in here is an 80 degree angle now if I&#39;m going to use that 80 degree angle to figure out the value of x what I want to do is look at the 140 the X and the 80 and ask myself which of the three trig ratios relates those three quantities well the x is adjacent to the 80 degree angle and the 140 is the hypotenuse adjacent and hypotenuse is the cosine ratio so I&#39;m going to say that the cosine of 80 degrees is the side adjacent which is x divided by the hypotenuse which is 140. now if you really just hate the cosine ratio right as as some students of mine have pointed out if I know that that&#39;s 80 degrees then I can say that that acute angle down there is 10 degrees and I could use the the sine of 10 is equal to X over 140. that would also work just fine I&#39;m going to stick with the cosine here it&#39;s real simple now I can just multiply both sides by 140. right that&#39;s going to cancel that 140. I&#39;m going to bring my calculator back out now it&#39;s really really important right when you&#39;re doing trigonometry on your calculator in a couple days from now you have to make sure it&#39;s in degree mode on the tin Spire that&#39;ll be a little deg sitting right up here if it says Rad you have a little bit of an issue you got to get it into degree mode for those using the ti Inspire it&#39;s as simple as actually clicking on that little word which I can do with my finger okay the plain fact is if you are in New York state your teachers should in theory reset your calculators and clear all of their memories before you take this test and when they do that in theory it will put your calculator back into radian mode and that&#39;s a problem because it will give you the wrong answer because as you learned in this course radians and degrees are not the same thing they&#39;re different they&#39;re like inches and centimeters right they&#39;re different units so I&#39;m going to go back to this now that I I did my little uh screed about radians and degrees right remember that what I&#39;ve got is 140 times the cosine of 80. let me bring that back up I&#39;m just going to type that in 140 times trig the cosine of 80 and that&#39;s going to be 24.31 and there it is to the nearest Foot 24 feet all right little right triangle trig second time we&#39;ve seen it right we saw it early on in that like I think it was like problem two or something or three right that kind of ratio piece with the similar right triangles here we actually had to use it to find a missing side of a right triangle right triangle trig is primarily used for two things finding missing sides when you know one side and an angle or finding missing angles when you know two or more sides a little foreshadowing I call that foreshadowing okay here we go let&#39;s take a look at problem 18. on the set of axes below triangle let and triangle L double Prime e double Prime t double Prime are graphed in the coordinate prompt plane where triangle l-e-t is congruent to Triangle L double Prime e double Prime t double Prime wow okay let&#39;s take a look which sequence of rigid motions Maps triangle let onto triangle L double Prime e double Prime t double Prime okay all right now it&#39;s a little bit weird I find these problems because it says you know number one as soon as you&#39;ve got two primes on these letters and none on these it sort of implies that there are two Transformations that are going to happen don&#39;t get me wrong it very well could be that they do that and there&#39;s still only one but that would be really disingenuous really stupid so this whole rotation of 180 degrees about the origin I got my doubts now you remember that whole like orientation of the letters things let&#39;s look at that one again and let&#39;s let&#39;s think about the word let right if I go from L to E from E to T and from T to L that&#39;s clockwise all right if I go from L to e e to sorry L double Prime e double Prime t double Prime that&#39;s counterclockwise all right now what that means given that I&#39;ve gone clockwise and I&#39;m going going counterclockwise is that there must be one and exactly one line reflection happening here there&#39;s got to be a line reflection involved there can&#39;t be two all right if there were two then it would go from clockwise to counterclockwise back to clockwise okay but there&#39;s got to be at least one line reflection and let&#39;s see if that&#39;ll help us out a little bit right Choice One says a reflection over the x-axis followed by a reflection oh sorry a reflection over the y-axis followed by a reflection over the x-axis well that can&#39;t be the case because the orientation of the letters changed and they wouldn&#39;t if there were two line Reflections Choice two a rotation of 180 degrees about the origin well again there&#39;s a couple reasons why that shouldn&#39;t be the answer one because they&#39;ve implied that there&#39;s at least two uh two rigid body motions that occur and two again a rotation wouldn&#39;t change the orientation of the letters so that one&#39;s out now the last two choice three a rotation of 90 degrees counterclockwise about the origin followed by a reflection of the y-axis and choice four a reflection over the x-axis followed by a rotation of 90 degrees clockwise about the origin well those two both include one and only one line Reflections so they&#39;re both if you will candidates all right now how do we choose between the two well here&#39;s what you want to do pick one letter just one one of the vertices and go through this set number three okay do that and see if it works and if it does work then maybe try it with Choice four and if it doesn&#39;t work then choice three is the answer so let&#39;s take a look at choice three a rotation of 90 degrees counterclockwise about the origin followed by reflection over the y-axis and I&#39;m gonna I&#39;m gonna get rid of all this stuff all right and let&#39;s just die I think I&#39;m going to hone in on E now if I take e right which is at negative 1 3 and I rotate it 90 degrees counterclockwise about the origin what&#39;s going to happen is it&#39;s going to go right here right it&#39;s going to go down to e Prime at negative three comma negative one all right now normally I would say take your piece of paper and rotate it 90 degrees counterclockwise to do that and that&#39;s fine I just can&#39;t do it myself now the second transformation right rotation of 90 degrees counterclockwise about the origin followed by a reflection over the y-axis if I now took this and reflected it over the y-axis it would in fact land on e double Prime so that&#39;s good right it seems like that is my winner right what I would maybe want to try and it is Choice 3 is the correct Choice what I might want to try is try it with either another one of the vertices and if it works with another one of the vertices great or try Choice 4 with letter e and see if that works right and if it doesn&#39;t then you know you&#39;re you&#39;re correct all right but at this point I think that we move on it is choice three all right let&#39;s take a look at the next problem number 19. diameter roq of circle o is extended through Q to point p and tangent p a is drawn if the measure of Arc r a is 100 degrees what is the measure of angle P all right great well it would be awesome if they gave us a diagram on this I mean they&#39;ve given us diagrams where they never needed to like problem number one right but this one they didn&#39;t give us a diagram for so we definitely want to draw a good picture I just wanted to get my pen icon out of the way all right so let&#39;s start drawing it now the first thing I know is that when I name a circle circle o what that means is that o is the center of the circle diameter r o q so let&#39;s let&#39;s draw that right we&#39;ve got the center at o we&#39;ve got diameter r o q all right so that&#39;s the first thing I want to do I just want I just want to draw that picture all right now it&#39;s extended through Q to point p so I&#39;m going to extend this thing from Q to point P how far it doesn&#39;t really matter I&#39;m just trying to get a reasonable picture all right and tangent p a is drawn now what&#39;s a tangent a tangent right is a line segment that is drawn that just intersects The Circle at that one location and then if it kind of kept going it would never intersected again right so there&#39;s my tangent PA drawn now they tell me that the measure of Arc r a is 100 degrees I&#39;m going to just get rid of this for a second so r a this Arc is 100 degrees okay and I want to know the measure of angle P now there are all of these sort of like you know theorems that you&#39;re supposed to learn about how angles that are created in a circle by chords and tangents and secants and things like that how they relate to the arcs that are intercepted and what I should know about this particular case is that the measure of angle p is going to be the measure of Arc r a minus the measure of Arc a q all divided by 2. so these two line segments PR and Pa intersect two arcs one Arc is Arc a r and the other Arc is Arc AQ all right and the measure of that exterior angle will be the difference between the two arcs divided by two now I know one of the arcs it&#39;s a hundred degrees but what I don&#39;t know is Arc AQ except I do right because I know Arc RQ right this entire arc Arc RQ is 180 degrees and therefore Arc AQ is equal to 80 degrees and now I have absolutely everything I need ra is going to be 100 AQ is going to be 80 all divided by 2 100 divided 100 minus 80 is 20. 20 divided by 2 is 10 and measure of angle p is 10 degrees awesome that&#39;s it okay let&#39;s take a look at problem 20. segment JM has endpoints at J negative 5 comma 1 and M 7 comma negative 9. an equation of the perpendicular bisector of JM is blank all right now this thing is testing actually a couple of different cool Concepts all right it&#39;s actually to me testing three different concepts one of them is what&#39;s known as the point-slope form of a line meaning right forget about the perpendicular bisector let&#39;s just not even worry about that meaning if I have a line right and I know it&#39;s slope let me call it m and I know some point that&#39;s on the line let me call it X1 y1 then the equation of the line can always be written as y minus y1 equals m times x minus X1 this is probably for most of you the first year where you ever saw this form of a line used up to this point you&#39;ve always used lines of the form y equals MX plus b known as the slope intercept form of a line form of a line we just need to know the slope of the line and one point that lies on the line all right now let&#39;s talk a little bit more about perpendicular bisectors now we&#39;ve got segment JM all right now I&#39;m not going to try to draw it particularly like accurately or anything like that all right but let me like like put JM here all right and one thing that we know about Jay is that it&#39;s got the coordinates negative 5 1. and M that&#39;s got the coordinate of seven comma negative nine okay I&#39;m going to switch colors really quickly I&#39;m going to go into red um bisector I want to wage enough all right so now I need to know two things about this perpendicular bisector I need to know its slope and I need to know one point that lies on the perpendicular bisector now there is one point and one point only that you absolutely positively know the perpendicular bisector goes through and that is the midpoint of that line segment right it goes through the midpoint so let me find the midpoint right now remember the midpoint formula is super duper easy all we do is we average the x coordinates and we average the Y coordinates all right I&#39;m going to have to kind of put down my final results here negative 5 plus 7 is 2 2 divided by 2 is 1. 1 plus negative 9 is negative eight negative 8 divided by 2 is negative four so that is our X1 y1 if you will all right and I&#39;m running out of room that&#39;s okay we&#39;ll we&#39;ll figure it out in a second the other thing I need to know is the slope of my perpendicular bisector and this is really cool because even though I don&#39;t know the slope of my perpendicular bisector one thing I can do is I can figure out the slope of J M I&#39;m going to go back into uh blue just to try to keep the two things separate so if I look at the slope of J M right what&#39;s that going to be it&#39;s going to be the change in Y which is negative 9 minus 1 over the change in x 7 minus negative five or seven plus five so that&#39;s going to be negative whoops sorry about that negative 10 over 12 so the slope of j m is negative 5 6. that&#39;s the slope of JM negative 5 6. now remember the slopes of two perpendicular lines are negative reciprocals to each other and if you&#39;re starting to think this is a long problem you&#39;re probably correct anyway so the slope of our segment is negative 5 6 but the slope of our perpendicular line write M perpendicular as I like to put it will be positive six-fifths so now right I&#39;ve got my slope I&#39;ve got my one point that the line goes through I&#39;m going to steal a little bit of actually I&#39;ll put it right here it&#39;s going to be y minus negative 4 which is y Plus 4. equals six-fifths times x minus 1. where is my answer right here y plus 4 is equal to six fifths times x minus one this is a question that they absolutely love love love on the geometry regions because it ties together so many important issues the midpoint formula the slope formula perpend negative reciprocals for perpendicular lines right all that comes together in this problem along with the point-slope form of a line all right let&#39;s take a look at 21. and I think we&#39;ll go back to Blue let me bring this out and then I&#39;ll read the problem for you put this away problem 21. quadrilateral e b c f and a in segment a d are drawn below such that a b c d is a parallelogram and segment EB is congruent to segment FB and segment EF is perpendicular to segment FH if the measure of angle e is 62 degrees and the measure of angle C is 51 degrees what is the measure of angle f h b all right I love these kind of problems this is what I call a fill in the angle problem and it is a Whopper so first things first let&#39;s put in the two angles that we absolutely positively know and also identify the angle that we absolutely positively want and then we&#39;ll start filling in everything else so let&#39;s see what we know we know that angle e is 62 degrees we know that angle C is 51 degrees right what else do we know we also know that EB right yeah EB here is congruent with FB here now let&#39;s immediately get something out of that because EB and FB are congruent to each other then this triangle is an isosceles triangle and that means that this must also be 62 degrees if they tell you that two segments are congruent in the middle of a random problem involving angles look for an isosceles triangle it&#39;s definitely there now the other piece is that EF is perpendicular to FH so FH and EF are perpendicular to each other that means that&#39;s a 90 degree angle right there now what&#39;s really cool about that is that that means I can also figure out this tiny little angle in there right because the 62 and whatever that angle is must add up to 90. so that angle is simply 90 minus 62 which is 28 degrees all right I&#39;m not going to try to kind of get it in there but I&#39;ll just do that now let&#39;s be very clear about what angle we&#39;re looking for angle fhb fhb so this is my winner right here right that&#39;s the angle I gotta get now if I can figure out another angle like let&#39;s say this angle right here if I can figure out angle h b f then I&#39;ve got two out of the three angles in a triangle I can add them together subtract from 180 and I&#39;ll get my final angle so the question is how can I figure that out well there&#39;s a piece of information we haven&#39;t used yet and that&#39;s the fact that a b c d is a parallelogram let me just outline that in red since I&#39;ve got the ability to do that ABCD that is a parallelogram yay okay now I know a lot about parallelograms one thing that I know is that two angles that are adjacent in parallelograms are supplementary I know opposite angles are congruent so I could say that that one&#39;s 51 as well as that one being 51 and that&#39;s not bad that that could also help you out but what I do know is that this angle in here I&#39;m going to kind of draw it like that that angle there is 180 degrees minus 51 which is equal to 129 degrees and you might be thinking to yourself well you know that that&#39;s not very helpful you know what I really need is this angle I don&#39;t need the whole one I need just that part but I know that&#39;s 62 and I know that&#39;s 62. so that will allow me to figure out what that angle is right so just this little angle in here this one is going to be 180 minus 2 times the 62. which is 180 minus 124 which is 56 degrees so this is 56 degrees right I know that that whole angle is 129 degrees without going to my calculator I don&#39;t want to have to deal with all that so I can now figure out this angle which is 129 minus 56 and that&#39;s going to be 73 degrees and I&#39;ve just about gotten it right because now I&#39;ve got that is 73 and this is 28 so now I can do 73 plus 28 and get 101 degrees and now I will just simply do 180 minus 101 and that&#39;s 79 degrees now I know that when there are this many angles that I&#39;m filling in there&#39;s probably three or four different ways to do this problem unquestionably so if you&#39;re like no I got 79 by using the fact that the four angles of any quadrilateral add up to 360. awesome okay there are many many different ways to typically do any math problem and one like this where you&#39;re filling in angles all over the place hey anything that you need to do to get over to the angle that you want right do it use any principle you have to all right I think we&#39;re almost done with the multiple choice questions in fact we&#39;ve only got two more let&#39;s take a look 22. Point P divides the directed line segment from point A at negative four negative one to point B six comma four in the ratio two to three coordinates of Point P are which of the following all right this is a standard problem on almost every geometry regents exam sometimes it&#39;s multiple choice sometimes it&#39;s free response let&#39;s talk about the idea here okay we&#39;ve got a line segment we&#39;ll talk about what it means to be a directed line segment in a second but just for a moment let me put a at negative 4 1 and B at 6 comma four okay now they tell me that I&#39;ve got Point P somewhere on this line and it divides it into a ratio of two to three now what that means right let me just put Point p on here is that if I divided this thing up into five equal chunks y5 because two plus three is five that&#39;s it right that there would be two of them here and there would be three of them here all right that&#39;s what that means right two to three now a directed line segment means it&#39;s a line segment that goes from A to B see most line segments are just they&#39;re just hanging out oh here they are right you know but sometimes we want to think of it almost like motion like I&#39;m starting at Point a and I&#39;m walking to point B and two-fifths of the way along we come to point P all right now that&#39;s really key right p is two-fifths of the way from a to B two fifths not two-thirds not three halves not three-fifths right two-fifths right two plus three five and we are two of those five now key right now we&#39;re gonna think about how far we traveled in the X Direction and how far we traveled in the y direction and I really wish this problem had made one of those two negative it didn&#39;t right just look at the X&#39;s here I&#39;m going from negative four to six right so I&#39;m traveling a change in X of ten units right I&#39;m increasing X by 10 units to go from an x equals negative four to an x equals six now I&#39;m going from a y equals negative one to a y equals four so my change in y is equal to five all right so how do I find Point P that&#39;s really cool let me first find the x coordinate find the x coordinate of Point P what I&#39;m going to do is I&#39;m going to take the x coordinate I start at and I&#39;m starting at an x coordinate of negative 4. and then what I&#39;m going to do is I&#39;m going to add to it two-fifths of my change in X right so I&#39;m going to have negative 4 plus 2 fifths of how much my X has changed now you could totally do that on your calculator but two-fifths of 10 is actually Four so my x coordinate of p is 0. now immediately what that tells you is that the correct choice is 2. but let&#39;s pretend that this is a free response problem right and get the y coordinate as well the y coordinate of p is going to be the y coordinate we start at which is negative 1. plus two-fifths of my change in y and my change in y is 5 units and two-fifths times five is just going to be 2 negative 1 plus 2 is 1. so that&#39;s it right we use the ratio to figure out what fraction of the journey we&#39;ve sort of traveled to get to our point that partitions our line segment and then we use that fraction to add on the fraction of the change that we have in both directions X and Y to our starting coordinates all right absolutely key and obviously those changes can be negative again in this problem both changes were positive which is disappointing to me all right let&#39;s take a look at problem 23 and then maybe we&#39;ll have a little lemonade break 23. a line is dilated by a scale factor of one-third centered at a point on the line which statement is correct about the image of the line Choice One its slope is changed by a factor of one-third two its y-intercept is changed by a factor of one-third three its slope and y-intercept are changed by a factor of one-third and four the image of the line and the pre-image are the same all right now this is also a problem that they like a lot on the Regions exam so let&#39;s take a little bit of time to discuss it all right so let&#39;s say that we have a line all right maybe it&#39;s line a b let&#39;s call it line a b and we&#39;ve got some point C okay and let&#39;s say that we dilate this line and let&#39;s forget about a factor of one-third just forget about it let&#39;s say that we actually stretch this thing to make it bigger we could we could compress it to make it smaller whatever but right the idea is you know we got this dilation and maybe we stretch a out to here and we stretch B out to here and then we run into the other problem and it gets confusing this right we stretch B out to here right and we get our new image line and our new image line is a prime B Prime all right if the center of dilation is not on the line then our image line that is produced is parallel to the original line and therefore should have the same slope okay let me say that again when we dilate a line using a Center Point that&#39;s not on the line the image line that we get should have the same slope as the original line but it&#39;ll definitely have a different y-intercept okay you know if I like oriented this on I don&#39;t know the coordinate plane or something like that right then we we definitely see that the y-intercept of this line and this line are different but the slopes would be the same so this what first one that says its slope is changed by a scale factor of one-third no now what happens if the point is on the line well in that case okay let&#39;s say we&#39;ve got this line a b and our Center Point is on the line well now what happens is when we either stretch or compress based on that Center Line we get exactly the same line and that&#39;s Choice four the image line and the pre-image line are the same line and again it&#39;s kind of irritating because most of the time when you dilate a line the center of dilation is not on the line that you&#39;re dilating so it&#39;s totally different line we&#39;ll have the same slope right but it&#39;ll tend to have a different y-intercept exactly what the y-intercept is that&#39;s harder to say and less the center is at the origin then it&#39;s easy enough then you just take the y-intercept and you multiply it by whatever that number is and that&#39;s your new y-intercept okay but the slope will stay the same if the point is actually on the line that you&#39;re dilating then the the image line and the original line are exactly the same all right let&#39;s take a look that&#39;s it oh no one more multiple choice problem that apparently they just could not fit on that other page then we&#39;ll go into the free response but we&#39;ll take a lemonade break in between problem number 24. in the diagram below of circle o tangent a b is drawn from external point B and secant b c o e and D diameter aod are drawn if measure of angle o b a is equal to 36 degrees and OC is equal to 10 what is the area of shaded sector d o e all right man boy and again this is a problem with multiple things that they&#39;re testing here okay but ultimately ultimately we want to figure out what the area of this shaded sector of the circle is now what you should have learned about areas of shaded sectors is that basically it all boils down to knowing what the area of the whole circle is and knowing what that central angle is and then setting up kind of a proportion to figure out what the final area is and we&#39;ll get there okay but what&#39;s interesting is we don&#39;t know what that central angle is we don&#39;t know what angle e o d is we really need to know angle EOD so what do we know well we know the angle Oba Oba which is this thing sitting right out here this thing is 36 degrees wow okay well what else do we know well because ba is a tangent to a radius tangents and radii will be perpendicular to each other all right so we know that&#39;s 36 we know that&#39;s 90 which allows me to say that angle AOC right the measure of angle AOC is going to be 90 minus 36 and that&#39;s going to be 54 degrees so 54 degrees all right now because that angle and this angle are vertical angles I can now say that is 54 degrees and that took me a lot of work to just get over here now how do we figure out the area of the Shaded sector well what I always tell students first is figure out the total area of the circle now we know that the radius I&#39;m sorry we know that OC is 10. which means that&#39;s 10 and that&#39;s 10 Etc so the area of the total Circle which is pi r squared will be pi times 10 squared or 100 pi okay now that&#39;s the area of everything but we don&#39;t have the whole circle we only have 54 degrees out of 360 degrees of it but we can now set up a little proportion that goes something like this the area of the sector divided by the total area is equal to the central angle of the sector divided by the total angle 360 degrees and now if I want the area of the sector I&#39;m just going to multiply both sides by 100 Pi I&#39;m not going to even cross multiply I don&#39;t need to do that all right and now I&#39;ve got to figure out what all of this is in terms of Pi this is where the calculator can be very very handy okay so I&#39;m just going to take this thing out and I&#39;m not I&#39;m not going to even worry about the uh the pie right now what I&#39;m going to do is I&#39;m going to take my calculator and I&#39;m going to ask it what is 54 divided by 360 what is that fraction times 100 then I&#39;ll bring the pi in for in a second right so 54 over 360 times 100 is 15. so so all of this then is just equal to 15. pi all right we just needed to have that central angle that was it ah excellent okay we&#39;ve made it to the free response problems excellent I&#39;m going to take just a moment and hydrate gotta hydrate got to stay hydrated okay can you believe it&#39;s been over an hour and a half that&#39;s amazing I&#39;m like oh we&#39;re gonna we&#39;re just booking through this we&#39;re gonna get done in like like two hours and I&#39;m gonna take the last hour and just do my stand-up comedy routine um I know you think I&#39;ve been doing that all along but uh math is no joke math is no joke um that wasn&#39;t that wasn&#39;t even funny uh anyway so yeah um we&#39;ve now made it through 24 two point multiple choice problems okay and again they&#39;re absolutely key if you nail them and do really well on the multiple choice problems then you&#39;re guaranteed to pass the exam that&#39;s just all there is to it now granted the curve on the geometry exam is not nearly as nice and forgiving as it is on the algebra one exam it&#39;s pretty nice it&#39;s pretty forgiving but it&#39;s not nearly as nice and forgiving okay so you want to nail the multiple choice 100 right but you also have to get the free response now part two we&#39;ve got seven questions in this part and each one of them&#39;s worth two credits okay so just like the multiple choice problems they&#39;re worth exactly the same we had 24 two pointers that we just went through now we&#39;ve got seven two pointers except there are no choices to help us out so let&#39;s get right into them make sure you&#39;re sure that we show proper work that we pay attention to our rounding and our notation and all of that so that we get two out of two on every one of these problems let&#39;s take a look at our first one problem 25. the Leaning Tower of Pisa in Italy is known for its slant which occurred after its construction began the angle of the slant is 86.03 degrees from the ground the low side of the tower reaches a height of 183.27 feet from the ground so nice of them to give us this beautiful diagram even though it kind of looks like the top portion of the the Leaning Tower of Pisa was photoshopped from a different picture from the bottom version I don&#39;t know you know maybe I&#39;m crazy about that but it just definitely looks like this is different than that anyway like I&#39;m one to talk anyway then it says determine and state the slant height X of the low side of the tower to the nearest hundredth of a foot I love all of that context that they give us uh-oh um they give us all of this context right for what is really essentially just a right triangle trig problem right that&#39;s all it is now granted it might be a little bit of a right triangle trig problem that&#39;s hard to look at but ignore the Leaning Tower of Pisa just look at this thing right we know the angle 86.03 degrees we know the side that is opposite of the angle even though that&#39;s a little bit hard to write and that&#39;s 183.27 feet and we want X which happens to be the hypotenuse right we want the hypotenuse so when we think about our trig ratios for the third but not last time of this test foreshadowing um right the one that we&#39;re going to use the one that has opposite and hypotenuse in it right is the sine ratio and so what we&#39;re going to say is we&#39;re going to say the sine of 86.03 degrees you don&#39;t have to put the degree symbol there but I can&#39;t help myself is 183.27 divided by X now this is one of those ones where the thing that you&#39;re solving for is in the denominator so it&#39;s not as easy as just sort of you know like multiply this side by this side and you&#39;re done kind of thing all right if I cross multiply I&#39;ll get x times the sine of 86.03 is equal to 1 times 100 now I gotta I really don&#39;t want this here is equal to 183.27 and then I&#39;ll divide by the sine of 86.03 and divide by the sine of 86 point 0 3 and X will be 183.27 divided by the sign of 86.03 so let&#39;s take a look at what that is on the calculator again it&#39;s easy enough to do this um let&#39;s just do maybe we&#39;ll even go with the fraction bar 183 0.27 divided by trig sine of 86.03 enter 183.71 notice how close that is right um actually let me put it down here X oh boy that looks like a funny X um x equals 183.7108 Etc really watch your rounding on here right this is a two-point problem if I round this to 183.7 I&#39;ve done all the trig right I&#39;ve done all the geometry right I&#39;m going to lose a point right I&#39;m going to lose a point because I didn&#39;t round to the nearest hundredth and rounded to the nearest tenth and boy how horrible would it be just to be 184. I mean you&#39;d still only lose one point but still right so X equals 183 .71 and by the way you don&#39;t even have to put down foot I would I can&#39;t even help it so I&#39;m going to put down foot but the plain fact is they&#39;ve actually put the units down for you because they don&#39;t want to deduct for you not putting down foot all right but watch your rounding all right let&#39;s take a look at the next one now every single free response problem they&#39;re going to give you at least one sheet of paper all right so you&#39;re probably not going to use almost any of it on these two pointers that&#39;s just the way it is let&#39;s take a look at another Circle problem in 26. in the diagram below quadrilateral ABCD is inscribed in circle o the measure of Arc CD well sorry the ratio of the measure of Arc CV to the measure of Arc D A to the measure of Arc a b to the measure of Arc BC is two to three to five to five all right State sorry determine and state the measure of angle B I love the fact that they say determine and state the measure of angle B like you&#39;re going to determine the measure of angle B but then you&#39;re going to keep it to yourself do you know what I mean like they couldn&#39;t have just said determine the measure of angle B whatever let&#39;s take a look at measure of angle B now measure of angle B right angle B is right here now one thing that you should have learned in your work with inscribed angles angle B is what&#39;s called an inscribed angle is that it enters its measure right the measure of angle B will be one half the measure of the arc that it intercepts it enter intercepts Arc a c all right so I need to know what the measure of Arc AC is like if I can figure that out then I&#39;ll just find one half of that measure and I&#39;ll be good to go so what do we know well we know CD to D A to a b to BC is two to three to five to five all right and there&#39;s a common trick we do with this all right and this is the trick we do we know right that CD plus d a plus a B plus b c is 360. so what I&#39;m going to do is I&#39;m going to call CD 2x I&#39;m going to call ad3x I&#39;m going to call a b 5x and I&#39;m going to call BC also 5x right and then I&#39;m going to do the following I&#39;m going to say 2x plus 3x plus 5x plus 5x is equal to 100 whoop no it&#39;s equal to 360 Degrees all right now I&#39;m going to solve I&#39;m going to have 15x is equal to 360 degrees and I&#39;m going to divide by 15 on both sides and I&#39;m going to find the X is equal to 24 degrees all right now because I know X is 24 degrees I can now say that 3x is 3 times 24 degrees so this is going to be 72 degrees Arc a d this is going to be 2 times 24 degrees so that&#39;s going to be 48 degrees and now I can say that the measure of angle B is going to be one half of 72 plus 48 which is going to be one half of 120 degrees and that is a 60 degree angle and that&#39;s it that&#39;s quite a bit of work for two points but that is what it is right all right let&#39;s keep going here we go 27. in the diagram below right circular cone has a diameter of 10 and a slant height of 13. determine and state the volume of the cone in terms of Pi okay well again if you don&#39;t know the volume formula for a cone this is when you would go back to those formulas and find it okay but let&#39;s just save ourselves the time the volume formula for a cone is one-third pi r squared times h now that&#39;s what it&#39;s going to look like if you look at the formula sheet now all right the key in this problem is knowing what H and R represent most people don&#39;t have any problem with r all right R is the radius of the base circle of the cone so our R is five no problem the real problem here is the height the height is not 13. 13 is what&#39;s known as a slant height and I hate that piece of terminology hate it hate it hate it hate it it&#39;s not a height it&#39;s a length Okay a height is how far you are above the ground okay the 13 is not how far you are above the ground it is a length Okay it is not the H and if you use it as the H you likely lose zero you likely lose both points okay the H is this distance right that&#39;s H all right the height is perpendicular to the base always for everything cone cylinder pyramid I don&#39;t care it is always perpendicular to the base and you don&#39;t know what it is but with the Pythagorean theorem you can find it right and you can just kind of over here you can just do a little H squared plus 5 squared is equal to 13 squared H squared plus 25 is equal to 169 H squared would be 144 if we subtract 25 from both sides take the square root and we have our famous 5 12 13 right triangle and our height is 12. all right now we have absolutely everything we need our volume is going to be 1 3 pi times r squared times h now notice that they want our answer in terms of Pi meaning that there&#39;s got to be a pi in the answer so what I do at this point is I tell students look don&#39;t worry about the pie leave it out for a moment take your calculator and do the following here we go right I&#39;m going to put my my nice one-third in can&#39;t forget the one-third definitely can&#39;t make it a one-half I&#39;ve Got 5 squared times 12. that is the entire formula without the pi okay I hit it right and I get a hundred all right so the correct answer is not there the correct answer then is one whoops 100 and again it&#39;s so weird that they don&#39;t put any units down you know like a 100 Pi what cubic centimeters cubic inches cubic miles I don&#39;t know cubic Kirks I don&#39;t know what the units are here but that&#39;s it now again there&#39;s a few things that are tricky about this problem the trickiest thing of all is the fact that they give us not the height but that slant height right which means we have to use either the Pythagorean theorem or Pythagorean triple to come up with the actual height and then we use that to get our volume all right let&#39;s take a look at problem 28. in the diagram below parallelogram E F G H is mapped onto parallelogram i j k h after a reflection over line L use the properties of rigid motions to explain why parallelogram efgh is congruent to parallelogram i j k h I tell you I don&#39;t love this problem because theoretically you should just be able to say it&#39;s a rigid body motion and therefore they&#39;re congruent done like that is the definition of congruence at this level but they would not accept that they would probably give you 0 out of two so it&#39;s this simple what do rigid bodies motions do what do Reflections translations and rotations do they do two things they preserve distance and they preserve angle they preserve distance and they preserve angle right and that&#39;s all you really have to say so what do I I want to say right um I&#39;m going to say a reflection is a rigid body motion so it preserves or if you don&#39;t like the word preserves which I totally would get you know instead of maybe saying preserves maybe you say so it doesn&#39;t change right it doesn&#39;t change so it preserves both distance and angle and that&#39;s it nothing more all right if you have any short problem especially like a two-point free response problem that&#39;s asking you in some way to justify that two figures are congruent based on a rigid body motion or a sequence of rigid body motions I want you to mention that those things preserve or don&#39;t change distance and angle all right and that&#39;s it moving on okay let&#39;s get into some more messy math problem 29. Izzy is making homemade clay pendants in the shape of a solid hemisphere as modeled below each pendant has a radius of 2.8 centimeters how much clay to the nearest cubic centimeter does Izzy need to make 100 pendants all right now this is key okay we&#39;re obviously going to find out how much Clay is he needs for one pendant and then we&#39;re going to multiply by a hundred and it would be very easy to you know do all the other stuff you know figure out how much Clay is he needs for one pendant and then not multiply by a hundred now if we&#39;re talking about how much we&#39;re talking about volume or mass but in here we&#39;ve got no density right we&#39;re just talking about how much clay to the nearest cubic centimeter right cubic centimeter right that means we&#39;re talking about volume all right now it&#39;s a hemisphere which is half of a sphere and if we go back to our formula sheet what we&#39;ll see is that the volume of any sphere is four thirds pi R cubed four-thirds pi R cubed now that&#39;s any sphere right the globe your soccer ball a basketball Etc none of those are actually true spheres although the basketball is probably the closest one all right but you understand what I&#39;m I&#39;m getting at right so the volume of a pendant foreign half times four thirds times pi times 2.8 cubed cubed cubed cubed not squared I have taught geometry countless numbers of years and the number one error people will make in these kind of situations is that they&#39;ll do squaring because they&#39;re so used to pi r squared and I get it that&#39;s the area of a circle we&#39;ve already seen pi r squared like multiple times right here it&#39;s four thirds pi R cubed let&#39;s see what that number looks like I think you&#39;ll like that number actually all right let&#39;s do it we&#39;ve got um one half times four thirds whoops I didn&#39;t quite get the four but I&#39;ll get it now uh times and now I&#39;m actually going to bring the pi in for the first time times 2.8 cubed 2.8 to the third I don&#39;t really need the parentheses there but that&#39;s okay sorry about the little cursor sitting there that&#39;s a little bit better 45 .97616 this isn&#39;t the one I thought it was we&#39;re going to eventually see one where the answer is literally Pi all right now that is the volume of a single pendant all right and let&#39;s let&#39;s write that down really quick now I&#39;m not going to round this all right I&#39;m definitely not going to round it to the nearest integer that&#39;s 45.97616 Etc big big decimal point there that&#39;s one pennant and of course it would be so easy to get that answer and then just move on right so easy all right but what we want is the total which is going to be 100 times maybe I&#39;ll just do V sub p and that&#39;s easy enough right we&#39;ll just take this and multiply it by 100. and we will get four thousand 598 I is there a way for me to there well that that&#39;s not exactly what I wanted to do but and actually it&#39;s kind of cool I didn&#39;t know I could do that anyway I just wanted to get the cursor out of the way this will be 4 598 cubic centimeters now think about how easy by the way it would be to lose two points in this problem first way you could do it not putting the one half there right that would lose you one point putting squared here instead of cubed that would lose you the other point or maybe you remember the one-half but do squared there and forget to multiply by 100 right that would lose you two points there&#39;s a lot of ways that you can kind of know what you&#39;re doing on this problem and go over two so again be very careful make sure if it&#39;s a half a sphere you involve that one half if there&#39;s more than one of them make sure you multiply at the end all right words to live by all right let&#39;s take a look at another one number 30. determine and state the coordinates again determine and state determine and state the coordinates of the center and the length of the radius of the circle whose equation is x squared plus y squared plus six x equals six y plus 63. all right well what you should have learned in this course was that if you have something that looks like this it&#39;s supposed to be on minus if you&#39;ve got something that looks like that then you&#39;ve got the equation of a circle and you can just get it Center by looking at the two things that are subtracted from the X and the Y and you can figure out its radius by taking the square root of whatever&#39;s sitting over there all right unfortunately this thing doesn&#39;t look at all like this but it&#39;s definitely a circle mostly because they told us it was a circle but also because we have an x squared term and a y squared term so how are we going to make it look like this well we are going to do what&#39;s called completing the square and we&#39;re going to do it twice now the first thing I&#39;m going to do though is I&#39;m going to move things around a little bit number one I&#39;m going to put this 6X next to the x squared I&#39;m also going to take that 6y which is on this side of the equation and bring it over to this side of the equation but make it at negative 6y so I&#39;m going to start by rearranging some things a little bit and having this I&#39;m going to leave that 63 exactly where it is okay now what I&#39;m going to do is I&#39;m going to complete the square on my X terms and on my y terms now remember how completing the square Works we&#39;re going to take half of this number we&#39;re going to square it and we&#39;re going to add it to this binomial and to this side and then we&#39;ll do the same here it&#39;s a little bit annoying because in both cases right when we take half of 6 we get 3 and when we Square it we get 9. so we&#39;re going to get x squared plus 6X Plus 9. plus y squared minus 6y plus 9 equals 63 plus 9 plus 9. all right so we&#39;re doing completing the square twice now when you do completing the square correctly what happens is you end up with trinomials that are perfect squares in other words they are a binomial quantity squared specifically this one is X plus three squared this one specifically is y minus 3 squared and when we add all these together we get 81. I guess by now I have determined what the center and the radius are but I haven&#39;t stated them I don&#39;t know maybe maybe not anyway so now when I kind of compare these two forms what I can say is that my Center is that negative three comma 3 if you will you sort of look in the parentheses and you take the opposite of what you see so if it&#39;s X plus 3 The Centers at negative 3 and Y minus 3 the center is at positive three so there&#39;s our Center and our radius is going to be the square root of 81. which is just equal to 9. so I have determined and stated the center and the radius awesome Okay so we&#39;ve kind of come to the first one where I&#39;m going to break out of this program and look at and go into another program where I have the ability to use a compass and a straight edge so real quick before I do that I&#39;m just going to pop in here change my line width go into my full screen mode Let&#39;s Take a look at problem 31. use a compass and straight edge to construct a line parallel to a b through Point C shown below leave all construction marks all right you know constructions are a tricky business because there&#39;s a ton of them that you need to know and I&#39;m going to go over just this one and it&#39;s unfortunate because the odds are this is not the one you&#39;re going to see tomorrow and they are not tomorrow sorry the odds are this is not the one that you&#39;re going to see on Tuesday but you will see one on Tuesday still it&#39;s what it is all right so let&#39;s learn how to create a line that is parallel to another line that goes through a particular point so here&#39;s how I&#39;m going to do it the first thing I&#39;m going to do is I&#39;m going to take a straight edge and I&#39;m not going to rotate it I&#39;m just going to bring it down like this and I am going to draw a line that passes through the point that I&#39;d like to draw the parallel line through and passes through the line that I am trying to make it parallel to all right so just then it could be anything it could be it could be like that it could be straight up and down whatever now to give you a little preview right I&#39;m going to try to copy this angle right here up here I&#39;m going to just try to copy that angle all right we&#39;ll explain why this works in a little bit in order to copy that angle I&#39;m going to bring my compass and I&#39;m going to put the sharp end down at that intersection and I&#39;m going to draw an arc okay the arc does not have to go through Point C and in fact I wouldn&#39;t particularly make it go through Point C all right now without changing the radius on the arc I&#39;m going to move my compass up to point C and I am going to draw an arc with the same radius now it won&#39;t be exactly the same Arc but it should have the same radius okay so I&#39;ve got you know this angle with this Arc drawn in and an arc centered at Point C with the same radius now right I kind of want to think about it as this point is equivalent to point C and this point is equivalent to the point up here what I really need to do is find the equivalent of this point along this Arc the way I&#39;m going to do that is I&#39;m going to bring my compass over here I&#39;m going to put it down on that point and I&#39;m going to then adjust the compass to a new length just so it passes through that point now you should be able to be a little bit more accurate than I am I can only be as accurate as the computer allows me to be but then oh that&#39;s actually not too bad I draw that Arc now again without changing the radius of my compass which I realize is a challenge right especially if you&#39;ve got bad school issued compasses and I know that most of you do right I now draw the same Arc but centered at that other point and then I get rid of my compass it&#39;s done its job okay I have literally just repeated up here exactly what I did down here now I&#39;m going to take my straight edge and I am going to draw a line that passes through C and through the intersection of those two arcs and that is my parallel line and again it&#39;s you know it&#39;s unfortunate because there are just so many different you know you could be bisecting an angle a couple days from now you could be uh drawing a perpendicular from a point down to a line there&#39;s a ton that you could do we could spend three hours just going over well maybe not three hours we could spend a good hour going over all the constructions that you should know for the exam even though they&#39;ll probably be only one of them on there in fact I&#39;ve actually never seen an exam with two so there&#39;s that all right let&#39;s head back into this program and we are up to part three all right let&#39;s take just a little break we are just over the two hour mark at this point and now we&#39;re headed into some of the longer problems but here&#39;s the good news we only have four problems left four problems what do you mean four problems we saw part three and part four left well part three is three questions each one of them is worth Four Points so they&#39;re beefy questions you know what I mean and oftentimes unlike algebra one oftentimes in Geometry when there is a four-point question it&#39;s not like part A is two points and Part B is two points it&#39;s just a four pointer and we&#39;re gonna see that right away in the first problem we do once we get to our three-part uh three questions worth Four Points each then we&#39;ll get to that one part four question that&#39;s worth six points but let&#39;s jump right into question number 32. as modeled below a projector mounted on a ceiling is 3.74 meters from a wall where a white board is displayed the vertical distance from the ceiling to the top of the white board is 0.41 meters and the height of the Whiteboard is 1.17 meters determine and state the projection angle Theta I got to keep coming back to that to the nearest tenth of a degree all right cool now this kind of a very open problem for four points right and we&#39;re trying to find this particular angle okay now whenever you&#39;re looking for an angle in geometry it&#39;s either going to be a fill in the angle kind of game right like that one huge multiple choice question where we were like 61 61 39 58 etc etc right it&#39;s either one of those or it&#39;s right triangle trig now it&#39;s interesting so far you know we&#39;ve seen right triangle trig twice okay once was like a very theoretical part one problem then we had that Leaning Tower oh actually we&#39;ve seen it three times right we saw the the one with the tree on the hillside that&#39;s where we used it to find a length we saw the Leaning Tower of Pisa we used it to find a length there and then we saw that that early on multiple choice problem this is one where we&#39;re going to use right triangle trig to find an angle now one of the things that&#39;s very tricky tricky is that angle Theta is not part of a right triangle and right now the only kind of trig you know how to do takes place within right triangles so here&#39;s the key there are two right triangles in this problem whoops let&#39;s make that a bit bigger to make it much bigger right we&#39;ve got this right triangle the smaller one and let me kind of draw that right here all right we got this right triangle and this is 3.74 comma on decimal and 0.41 and then even though this is probably not going to help very much oh maybe I&#39;ll go highlighter um yeah I&#39;ll go highlighter and I&#39;ll go purple and then we&#39;ve got this right triangle and now I&#39;m going to go back to pen because why wouldn&#39;t I and let&#39;s draw that one a little bit bigger because it is bigger and that is 3.74 and it&#39;s the sum of those two one point five eight all right so let&#39;s talk about this a little bit right so we&#39;ve got the smaller right triangle 3.74 by 0.41 and then the larger one 3.74 by 1.58 and again that 1.58 just came from adding those two links does this relate to Theta well if I call this let&#39;s say angle X and call this say angle y right this would be X and this would be y then what I should be able to see is that Theta is going to be angle y minus angle X angle y minus angle X so I just have to find these two angles and I have to go back to sohcahtoa right so let&#39;s first take a look at the small triangle right here angle X if I look at the 0.41 that&#39;s opposite of X and the 3.74 that&#39;s adjacent to angle X and therefore if I&#39;m looking for opposite and adjacent we&#39;re talking about tangent now specifically what I can now write is that the tangent of angle X is equal to the side opposite of X 0.41 divided by the side adjacent to X 3.74 now whatever you do don&#39;t lose a point and put equals some angle okay the equal sign does not mean give me the answer never has and it never will equals means that the two numbers are the same and if you tell me that 0.41 divided by 3.74 is equal to 22 degrees well then I&#39;m going to wonder if you really understand what fractions mean and I&#39;m going to deduct a point if I&#39;m the greater all right this is not equal to the angle what is equal to the angle is What&#39;s called the inverse tangent of that ratio 0.41 divided by 3.74 why don&#39;t we figure that out all right we want the inverse tangent let me get to my notes Here just so I can I have them in front of me bring this whoops nope that&#39;s not what I want bring this back up bring this back up here we go I just want to get out of here so I&#39;m going to go into trig and I&#39;m going to go to What&#39;s called inverse tangent this is literally telling me the angle that has a tangent of 0.41 divided by 3.74 all right that&#39;s all I want to know right and that&#39;s an angle small angle 6.26 let&#39;s call it 6.26 now notice I&#39;m looking for this to the nearest tenth of a degree I&#39;m going to take that angle out one decimal place for farther now what&#39;s really great is this bigger triangle it&#39;s the same ratio well it&#39;s not the same ratio but it&#39;s the same trig function right we&#39;ve got opposite we&#39;ve got adjacent so again it&#39;s going to be the tangent of angle Y is equal to 1.58 divided by 3.74 so Y is equal to the inverse tangent don&#39;t use x for both of these of that particular fraction and Y I&#39;m going to just tell you what this is because we&#39;ve already done that on the calculator 22.90 degrees to the nearest hundredth and now Theta right Theta is just going to be y minus X and when I do that subtraction I will get 16 .6 degrees and that one&#39;s kind of tricky right I mean if they had asked you just for you know a single angle right and all you had to do was sort of set up a single ratio do the inverse trig whatever it is inverse tangent inverse sine inverse cosine that&#39;s simple enough here it&#39;s way more challenging because you actually have to think well I&#39;ve got to calculate two angles subtract and get the final answer all right let&#39;s keep going oh I&#39;m very excited about the next problem oh and this one I have to go back over to my other display uh let&#39;s do it let me get uh get the right line Style uh why am I excited because it&#39;s the proof or at least it&#39;s the euclidean geometry proof we&#39;ll see another proof in a moment let&#39;s take a look given parallelogram p q r s QT perpendicular to PS Su perpendicular to QR prove that PT segment PT is congruent to segment r u all right now the first thing I always want to do whenever I&#39;m doing a geometry proof whether it&#39;s worth four points or six points and it&#39;s going to be one of those two is I want to really sort of work my way backwards so let&#39;s let&#39;s first highlight the two things that we want to prove congruent and boy if you&#39;ve got a highlighter it&#39;s not the worst thing to be using on here right so I wanna I wanna like somehow prove that PT and r u are congruent to each other okay so so far so good now almost always not always but almost always when I&#39;m looking to prove that two segments or two angles are congruent to each other I&#39;m going to be using what&#39;s known as a cpctc proof corresponding parts of congruent triangles are congruent and what I&#39;d really love to do is I&#39;d love to prove that this triangle here RSU is congruent to this triangle here p q t I&#39;d love to prove that those two are congruent all right because if I can then it&#39;s going to be very easy to say that PT is congruent to Ru so let&#39;s see what we know well we know we&#39;ve got a parallelogram right parallelogram pqrs and that&#39;s pretty awesome because with parallelograms we know that opposite sides are congruent so I know that PQ is congruent to Sr or RS and I also know that opposite angles are congruent so I know that angle R is congruent to angle p this is good because I&#39;m almost there right then and that&#39;s just all from that first piece of information parallelogram pqrs I already have like a ton of information that&#39;s going to allow me to prove that those two triangles are congruent now I know that QT is perpendicular to PS which tells me that that&#39;s a right angle okay and I know that s u is perpendicular to QR that tells me that that is a right angle this is awesome because now I can prove that triangle r u s is congruent to Triangle p t q by angle angle side right and once I have those two congruent by angle angle side I can then state that PT is congruent to Ru by cpctc let&#39;s do it all right here we go classic now some of you may have done these things before doing like like flow diagram proofs or paragraph proofs all of that is fine the majority of people do it with t table proofs so that&#39;s what I&#39;m going with all right so first things first is I&#39;m just going to write down that first given um parallelogram p q r s and you might say why aren&#39;t you writing down all the Givens right away well you can absolutely I just like to use them when I need them so parallelogram pqrs that&#39;s given now what did I know from that well I knew from that that RS is congruent to um PQ and why do I know that because opposite sides of a parallelogram are congruent now this is very dangerous you do not want to write down definition of a parallelogram a parallelogram does is not defined as having opposite sides that are congruent it&#39;s defined as having opposite sides that are parallel all right but anyway I got that now let&#39;s get those two angles um let me say angle r is congruent to angle p and it&#39;s going to be very similar opposite angles of a parallelogram [Music] R congruent sweet all right now I really need to get those two right angles congruent but the first thing I need to do is get those two right angles now those angles aren&#39;t as convenient as angle R and angle P right and I can certainly call them by their three letter names I could call them angle Rus and angle ptq or qtp and Sur Etc like that but this is actually a great example of a case where I could also put like a little one down here and a little two down here oops and apparently Circle it in red now before I do either one of those I got to get the given in there so I&#39;m going to say QT is perpendicular to PS and Su is predict perpendicular to QR and that&#39;s a given all right now I&#39;m going to say that angle one and angle two are right angles and as my reason I&#39;m going to say perpendicular lines form right angles now I can say that the two angles are congruent angle one is congruent to angle two and my reason is that all right angles are congruent whoo wow and now I have enough to state that the two triangles are congruent to each other line seven now make sure to get your vertices correct here triangle r u s is congruent triangle our us is congruent to Triangle ptq and that is by the angle angle side theorem and we&#39;re there line eight extend that a little bit farther let me do this PT and r u p t is congruent to R U by CP CTC you probably spent a third of your school year so that you could do this one problem right all of that proof that you did for this eight lines and that&#39;s what you should expect in a four point proof is around eight lines if this was a six pointer then we might see a 10 line proof 11 12 line proof that&#39;s it but thankfully that&#39;s not the problem that we were facing all right let&#39;s take a look at the next one here we go this is another messy one 34. a concrete footing is a cylinder that&#39;s placed in the ground to support a building structure the cylinder is four feet tall and 12 inches in diameter a contractor is installing 10 footings if a bag of concrete mix makes two-thirds of a cubic foot of concrete determine and state the minimum number of bags of concrete mix needed to make all 10 footings this is at that different of a problem than that pendant one in fact one could argue that it&#39;s actually like they could have had this one as a two-pointer as well all right look we&#39;ve got this is all about volume right we know bags of concrete have two-thirds of a cubic foot of concrete they would never ever say that on a bag two-thirds of cubic foot right they&#39;d say 0.67 cubic feet or something like that anyway not the point it&#39;s just weird that they did that so what we need to do is we need to figure out the volume all right and let&#39;s first begin by finding the volume of a single cylinder now the volume of a single cylinder oops make that a little bit a little bit thicker the volume of a single cylinder is going to be pi r squared times H and they did throw in a little bit of a loop here right so the height is four feet and the diameter is 12 inches all right and we can&#39;t mix units can&#39;t have feet and inches especially not when we&#39;re talking about the bags having a volume of two-thirds of a cubic foot everything&#39;s got to be in terms of feet right now everyone should know at this point that 12 inches is one foot all right so at least they did that at least they didn&#39;t say 14 inches or 18 inches or something like that but right keep in mind that still means that the radius R is 0.5 feet and man this is critical right if you use six as the radius you&#39;re going to get this problem completely wrong now you won&#39;t lose all the points on it but come on you know what I mean you just you just can&#39;t you can&#39;t do that right make it all the same unit and the unit you really want it to all be in is feet anyway let&#39;s figure out now the volume of just one of the single ones of these pilings okay pi times the radius squared times H all right let&#39;s do it let&#39;s see what pi times 0.5 squared times H is here we go my Pi there it is times 0.5 squared got to make sure to square it times 4 enter and as many people will recognize this is actually just equal to PI right like and again we could figure out why that was but it&#39;s not worth our time and we&#39;re actually getting a little bit short on time we&#39;ve now been going for two hours and 25 minutes and I&#39;m not going past three hours because that&#39;s all the time we get for this exam anyway so what I know now is that a single one of these is 3.1415 ETC now of course we need to make 10 footings right so I need 10 times V which is going to be 31.415 Etc cubic feet all right and of course on the calculator we can always do that by just multiplying by 10. so 34 31.145926 Etc now this is another tricky part we&#39;re actually trying to find the minimum number of concrete bags that we need to buy and we were told that each one holds 2 2 3 of a cubic foot now if they said each one of them held 1.5 cubic feet I bet you&#39;d take this number and divide by 1.5 if they said certainly if they said each one of them contained five cubic feet by the way that would be a heck of a lot of concrete right if they said each one of them contained you know one point you know like two cubic feet you divide all this by two we need to take this answer and divide by two-thirds and that&#39;s really pretty easy on most modern calculators we can literally say divide by two-thirds right and that&#39;s how many bags we need right we need 47.123 Etc now let me just do this really quick all right 10v divided by two thirds let me just skip over here really quick is 47.123 ETC now last part of this problem that&#39;s a bit tricky right what is the minimum number of bags of concrete needed to make all 10 footings it would be very tempting right now to say 47 bags because definitely if they just said round to the nearest bag or round to the nearest whole number then definitely the answer is 47. the problem is 47 bags wouldn&#39;t be enough right it wouldn&#39;t be enough because we need 47.123 bags which means we need 48 bags of concrete all right and that&#39;s that and that was three part four questions which means with just a little over a half an hour left we are on our last question so let&#39;s just take a moment all right a little bit more lemonade and then we&#39;ll do our last problem [Applause] okay so part four questions on the geometry exam one part for a question it&#39;s worth six points sometimes unfortunately it can be a euclidean geometry proof all right but we already did that one sometimes it can be a really involved volume density problem but we actually had a volume density problem that was actually a multiple choice problem all right and it&#39;s not one of those what it is is what&#39;s called a coordinate geometry proof problem and just like some of the rest of these things I am going to go back over to this presentation software and do it here let&#39;s take a look at our last problem the coordinates of the vertices of triangle a b c are a negative two comma four B negative seven comma negative 1 and C negative three comma negative 3. prove that triangle a b c is isosceles parentheses the use of the set of axes on the next page is optional all right so I you know I kept you know repeating this set of axes so I wasn&#39;t like flipping back and forth and stuff like that your set is on the next sheet and that&#39;s fine anyway so we don&#39;t actually have to plot this triangle but I would always suggest doing so okay because if you do it it will give you a sense of what&#39;s going on now for me I&#39;m just going to make it sort of magically appear okay so there&#39;s my triangle ABC and it asks me to prove that ABC is isosceles so the first thing that you have to know is what an isosceles triangle is and if you don&#39;t you&#39;re going to be sunk in the water all right that&#39;s just the plain fact if they say prove it&#39;s a parallelogram and you don&#39;t know what a parallelogram is it&#39;s going to be game over same thing with a rectangle a rhombus a trapezoid a square anything it&#39;s just going to be a problem all right but what we should know because we&#39;ve discussed it repeatedly tonight is that an isosceles triangle is any triangle with at least two sides that are equal in length now one of the great things about plotting the thing is it makes it fairly easy to see that the two equal lengths if they are equal are side a b and AC and if I want to prove that they have the same length I&#39;m going to use the distance formula all right so I want to find the length of a b and I want to find the length of AC now remember the distance formula basically says to find the distance between any two coordinate points we do we subtract our x coordinates Square subtract their Y coordinates and square add those two together and take the square root now all this is is the Pythagorean theorem okay but in the coordinate plane so I want to find the length of a b and I want to find the length of AC so a b equals all right so if I look at A and B right I&#39;m going to have negative 7 minus negative 2 which will be negative seven plus two plus 4 minus negative 1 which will be four plus one all right so a b will be the square root of negative 5. squared plus 5 squared that will be the square root of 25 plus 25 and that will be the square root of 50. now the square root of 50 most certainly simplifies it simplifies as 5 times the square root of 2. but there&#39;s absolutely no need for you to simplify that we just want to prove that the length of a b and the length of AC are the same and it doesn&#39;t matter what form we have it in all right I would try to avoid putting things down like equals and then putting down some kind of rounded decimal that could be problematic because you&#39;re going to be stating that the length of a b is something rounded and that&#39;s a little bit questionable anyway let&#39;s take a look at AC all right now AC right that&#39;s a little bit harder but I&#39;m going to like I&#39;ll do this kind of underline those two that will be the square root now of negative 2 minus negative three lots of negatives in this problem so squared and then 4 minus negative 3 which will be 4 plus 3 squared so AC will be the square root of 1 squared plus 7 squared which will be the square root of 1 plus 49 which will be the square root of 50. all right now I need to have a little conclusion here I can now say therefore three dots I love using the three dots for therefore triangle ABC is isosceles because it has two sides that are the same length there we go now you might say but what about BC don&#39;t you have to calculate the length of that and the answer is absolutely not an isosceles triangle is a triangle with at least two sides that are the same length I have shown that two sides are the same length and therefore this triangle is isosceles done all right let&#39;s take a look at the second part of this problem which is not a coordinate geometry proof and should be very easy for you let&#39;s take a look State the coordinates of triangle a prime B Prime C Prime the image of triangle ABC after a translation five units to the right and five units down all right I think they threw this one in there to kind of make it a six point problem we just want to take these three points right and we want to move them five units to the right and five units down now keep in mind that the way translations work is if you&#39;re moving to the right that means you&#39;re going to add 5 to each x coordinate and if you&#39;re moving down by five you&#39;re going to subtract 5 from each y coordinate so if you will that is our algebraic rule for this particular translation so this should be really easy a at negative two comma four will get mapped to a prime negative 2 plus 5 is 3. and four minus 5 is negative one so I&#39;m just applying this rule to each one of those coordinate points B which is at negative seven negative one will get mapped to B Prime so now I want to add 5 to the x coordinate that&#39;ll be negative 2. I&#39;m going to subtract 5 from the y coordinate and that&#39;ll be negative 6. and finally C which is at negative 3 comma negative 3. we&#39;ll go to C Prime and again adding 5 to the x coordinate we&#39;ll go to 2 and subtracting 5 from the y coordinate will go to negative 8. and that&#39;s it now don&#39;t get me wrong in a six point problem my bet is that this is worth two points this is worth one point and what&#39;s still yet to come is worth the other three points all right now what&#39;s to come is on the last sheet of the exam but I&#39;ve just kind of put it all on here so the final part of this problem and the final part of this test prove that quadrilateral a a Prime C Prime C is a rhombus the use of the set of axes below is optional now one of the things that&#39;s a little bit unfortunate is that they only give you one grid for the whole thing right and so that can be a little bit tricky you do have extra graph paper on the test so you could go to that all right but again really really worthwhile to take a look at it now if I plot points a a Prime C Prime and C I&#39;m going to get something that looks like this okay now normally have we not done the first part of this problem normally I would say okay everybody I want you to think about what a rhombus is and what a rhombus is is it&#39;s any quadrilateral that has four sides that are the same length four sides the same length so then I would say hey everybody let&#39;s calculate the length of a a prime a Prime C Prime C Prime C and AC and let&#39;s just show that they&#39;re all equal and that is a great way of doing this there is nothing wrong with doing that again using the distance formula to calculate 1 2 3 4 distances actually you&#39;d only have to do three because you&#39;ve already figured out the length of AC earlier in this problem but then we would just be doing a lot more distance formulas so I&#39;d like to look with the limited time that we have left and a different way of doing this problem but again if your initial reaction is hey man I&#39;m just going to do the distance formula three times yes yes yes yes yes okay but again let&#39;s look at a different way all right there&#39;s almost always more than one way to prove these particular figures so let&#39;s say I drew in the diagonals of this rhombus now a rhombus is a parallelogram okay and the diagonals of all parallelograms have this cool property is in that they bisect each other meaning that they have the same midpoint so one way of proving that a four-sided figure is a parallelogram forget about a rhombus one way of proving it&#39;s a parallelogram is showing that the two diagonals a c Prime and a Prime C have the same midpoint let me do that really quick so I&#39;m going to take a look at a c Prime all right I&#39;m going to find their midpoint okay and the midpoint remember we looked at this earlier is the average of the two x&#39;s and the average of the two y&#39;s all right negative two plus two is zero and zero divided by two is zero four plus negative 8 is negative 4 and negative 4 divided by 2 is negative 2. so the midpoint of a c Prime is 0 negative two which you can kind of see here on the graph right that&#39;s zero negative two and now let&#39;s take a look at the midpoint of a Prime C foreign C that&#39;s going to be 3 plus negative 3 divided by 2 and negative 1 plus negative 3 divided by 2. 3 plus negative three is zero zero divided by two is zero and negative 1 plus negative three is negative four negative four divided by two is negative two all right so what has this done for us you know finding the two midpoints of the two diagonals and seeing they&#39;re the same well what that tells me is that a a Prime C Prime C is a parallelogram because it&#39;s diagonals bisect each other or have the same midpoint either way yeah you&#39;re probably saying to yourself but wait wait a second I need to prove it&#39;s a rhombus well the way you can prove something is a rhombus is by first proving it&#39;s a parallelogram and then showing that its diagonals are perpendicular to each other all right so all parallelograms have diagonals that bisect each other but we needed to do this just to prove that it was in fact a parallelogram now let&#39;s prove that the two diagonals are perpendicular to each other and we&#39;re going to do that using slopes so now I&#39;m just kind of pan down a little bit here so now I&#39;m going to go a c Prime and I&#39;m going to calculate its slope all right so I&#39;m going to do negative 8 minus 4 Change in y all divided by 2 minus negative 2 change in X that&#39;s going to be negative 12 divided by 4. and that&#39;s going to be a slope of negative 3 . now let me figure out the slope of a Prime C I&#39;m trying to do a colon there let&#39;s see m is negative 3 minus 1 change in X or sorry change in y divided by the change in x negative 3 minus 3 and that&#39;s going to be negative four nope that&#39;s something I did something wrong there um hold on negative three minus oh there it is that&#39;s negative 3 plus 1. that&#39;s not working negative 2 over negative 6 and that we&#39;ll simplify to one-third right so I&#39;ve got one slope that&#39;s negative three one slope that&#39;s positive one-third so I can now say that a Prime C A C Prime is perpendicular to ah come on this is making it very hard to finish this exam a c Prime is perpendicular to a Prime C because slopes are negative reciprocals and finally I can say therefore oh here we go a a Prime C Prime C is a rhombus let me see if I can spell it right a rhombus because it is a parallelogram with with perpendicular diagonals all right now again don&#39;t get me wrong if your initial sort of urge here is to do four distance calculations or three distance calculations because you&#39;ve already done the one for AC you&#39;d find that all four lengths were the square root of 50 just like in the first part of the problem all right and you&#39;d conclude it was a rhombus because it was a quadrilateral with four sides that were the same length or four congruent sides okay and that is fine it&#39;s good there&#39;s nothing wrong with it awesome 100 good job all right I just wanted to do these two because they could easily come up in a problem a couple days from now all right in other words if you&#39;re asked to prove that a four-sided figure is a parallelogram one great way to do that is to find the midpoint of both diagonals and if the midpoint is the same You can conclude it&#39;s a DOT it&#39;s a parallelogram because the diagonals bisect each other then if you already know it&#39;s a parallelogram this part right whoops almost fell then you can calculate the slopes of the two diagonals and if those slopes are negative reciprocals then you know the two diagonals are perpendicular to one another and you can say it&#39;s a rhombus because it&#39;s a parallelogram with perpendicular diagonals all right let&#39;s get out of here and go back to the test itself because that is the last problem right that&#39;s it let me go all the way back up to the beginning whoo that took us two hours and 45 minutes to go through now you probably aren&#39;t going to take that long on the exam because you&#39;re not going to be explaining it to somebody as you work through it then again you will have also not worked through it two or three different times like I had before even coming in here tonight all right so you know it&#39;s taking it right the plain fact is if you&#39;ve really been present in your geometry class all throughout the year right there&#39;s a lot of stuff you should have learned right you know and It&#39;s tricky because unlike algebra in Geometry there is a lot of memorization granted it&#39;s it&#39;s kind of tactile right it&#39;s Figures it&#39;s parallelism it&#39;s perpendicularity but there&#39;s a lot of stuff you have to know you have to know the definitions of a parallelogram a rectangle a rhombus a square a trapezoid an isosceles triangle a right triangle you need to know the three trig ratios apparently right especially on this exam they came up I think four times which seems like quite a bit to me you need to know the Pythagorean theorem and when to apply it you have to feel comfortable with volume formulas for pyramids cones cylinders spheres hemispheres right there&#39;s a lot of content but you&#39;ve got three hours on this exam right it won&#39;t be nearly as calculated as intensive as either the algebra 1 exam or eventually the algebra 2 exam will be for you but it will be a lot of thinking about the problem understanding what information they give you and what it tells you and what it doesn&#39;t tell you right think about those problems where they they tried to get you to say that something was true even when they didn&#39;t give you the information to really know that it was true all right so read the questions carefully take your time on the exam and you are going to do great on it okay and you&#39;re going to be all set up to take Algebra 2 next year all right well I think I&#39;ve talked your ear off tonight so I&#39;m going to just say good night and good luck a couple days from now my name is Kirk Weiler this has been the emath instruction live geometry regents review good luck in a couple days keep thinking and keep solving problems

[Music] thank you [Music] hello and welcome to the emath instruction geometry regents review Live on YouTube my name is Kirk Weiler and tonight we are going to be reviewing last year&amp;#39;s June 2022 geometry regents exam this is primarily intended for those students in New York state who are taking the geometry regents two days from now right that&amp;#39;s going to be Tuesday I think it&amp;#39;s Tuesday morning but make sure to check with your local school district on the exact time and location for your test right tonight what we&amp;#39;re going to be doing is going through last June&amp;#39;s exam Problem by problem we&amp;#39;re going to be looking at it from many different perspectives all right we will need our calculators a little bit on this exam mostly just to do some routine calculation work I&amp;#39;m going to be using the ti Inspire calculator I love the calculator you can really use basically any calculator you want all right just be prepared there&amp;#39;s some messy numbers on this exam but without further Ado let&amp;#39;s get right into it with the first problem all right here we go part one now part one on the geometry exam has got 24 questions they&amp;#39;re all multiple choice they&amp;#39;re all worth two points a piece and there is no partial credit so you want to take your time on these you want to read them carefully all right they&amp;#39;re not trying to trick you in any particular problems all right although some of them are trickier than other ones and we&amp;#39;ll try to point those out thankfully they start us off with a pretty easy one let&amp;#39;s take a look at it problem number one triangle a prime B Prime C Prime is the image of triangle ABC after a dilation centered at the origin the coordinates of the vertices of triangle ABC are a negative two comma 1 b 2 comma four and C 2 comma negative three all right and now for the actual question here we go if the coordinates of a prime are negative 4 comma 2 the coordinates of B Prime are which of the following all right great now you know they&amp;#39;ve taken up a lot of space with this grid that we don&amp;#39;t particularly need but let&amp;#39;s go back and take a look at something really quick right you know a lot about a lot of geometry is about original pre-images triangle ABC and a transformation then that Maps them to some other figure in this case triangle a prime B Prime C Prime now we are told specifically that the transformation that we&amp;#39;re talking about here is a dilation centered at the origin now dilations are basically taking a figure and stretching it away from a point or compressing it back down towards the point so that the figure still looks the same shape as the original but has a different size now you have to love dilations that are centered at the origin because all that you need to do right when you have a dilation centered at the origin is take a point x comma y and then multiply both the X and the Y coordinates by What&amp;#39;s called the dilation constant oftentimes that&amp;#39;s represented by the letter k now our real task in this is figuring out what that dilation constant is but they&amp;#39;ve really given us the key here right they&amp;#39;ve told us that a is at negative 2 comma one so a is at negative 2 comma one and that got mapped to a prime which was at negative four comma two That was supposed to be an a there we go negative four comma two so what exactly happened here well we multiplied the X and the y coordinate by a k value equal to 2. right now the key is with the dilation when you&amp;#39;re mapping a figure you don&amp;#39;t change the dilation constant depending on which vertex you&amp;#39;re you&amp;#39;re actually dilating if you did that you wouldn&amp;#39;t get what what are known as two similar figures so if I want to know what the coordinates of B Prime are all I need to do is go back up here unfortunately B is at 2 comma four and I just need to multiply both of its coordinates by two and I&amp;#39;ll find that b Prime is at four comma eight so real real simple dilations that are not centered at the origin are much trickier to perform algebraically and they would be much more difficult in a problem like this to figure out but this problem started us off very easy with probably the most constant or the most common dilation constant which is k equals two all right let&amp;#39;s keep going let&amp;#39;s get into problem number two in the diagram below a plane intersects a square pyramid parallel to its base which two-dimensional shape describes this cross section all right so one of the things that you&amp;#39;re supposed to learn in geometry and you also probably looked at this a little bit in eighth grade math was if you have a two dimensional or sorry a three-dimensional solid and you start slicing it with planes think about a plane as just being a flat surface that extends forever right so I&amp;#39;m it&amp;#39;s I&amp;#39;m slicing this thing with a plane what two-dimensional shape am I getting well that really always is going to depend on how the plane that you&amp;#39;re slicing is oriented in this case it&amp;#39;s a plane intersecting a square pyramid parallel to its space so in other words I&amp;#39;m going to intersect this pyramid parallel to its base and what I&amp;#39;ll see when I do that is a figure that kind of looks like that now what is that well it&amp;#39;s just another square right that&amp;#39;s all it is and therefore we&amp;#39;ve got this particular answer all right awesome so I think the first two start us off nice and easy let&amp;#39;s take a look at number three this one&amp;#39;s a little bit trickier in the diagram below Triangle C D E is the image of triangle c a b after a dilation of d e divided by a b centered at C which statement is always true and then they give us a bunch of statements that involve the three trig ratios the sine of a is CE divided by c d etc etc all right now hopefully you all know learned that the sign is the opposite over the hypotenuse the cosine is the adjacent over hypotenuse and the tangent is the opposite over adjacent sohcahtoa some teachers love that mnemonic some teachers hate that mnemonic for me it&amp;#39;s just a memory device right the sign of a given angle is the side opposite of the angle divided by the hypotenuse we&amp;#39;re going to be doing a fair amount of trig in this test so might as well do a little bit right now cosine is adjacent over hypotenuse and tangent is opposite over adjacent now you might say why did they give us all of this information up here well they gave us all of that information most of which we can completely ignore all right but they gave us all that information really to tell us only one thing which is that this smaller right triangle I.E Triangle C D E is similar to this larger right triangle triangle c a b okay those two right triangles are similar and what that means is that the ratios of relatively positioned sides are going to be the same all right here&amp;#39;s what I mean by that let&amp;#39;s take a look at Choice one just for a moment I&amp;#39;m going to kind of move this up because quite frankly all of that wording is pretty much irrelevant at this point let&amp;#39;s take a look at the first one sine a is equal to CE divided by c d all right well what you should have learned all right of course now I&amp;#39;ve moved my mnemonic out of the way is that sine is opposite over hypotenuse so if you were just going to be you know trying to figure out what the sign sign of angle a was what you would do is you&amp;#39;d say all right well you know here&amp;#39;s that right triangle that angle a is part of and the side opposite of angle a is BC or CB either way all right and it&amp;#39;s hypotenuse is AC now you might look at like this choice and go well that&amp;#39;s not correct that&amp;#39;s not CE divided by CD except it is you see because BC divided by AC is the same as c e divided by c d and that&amp;#39;s because those triangles are similar all right so Choice a is actually correct now for a moment before we move on to problem four let&amp;#39;s just take a look at Choice 2 which is incorrect for a moment right cosine of a right so if we were doing cosine of a we would say well cosine of a is going to be the side adjacent to a which would be a b over its hypotenuse which is again AC now in terms of the Little Triangle right which is what all the answers are put in terms right in terms of the Little Triangle right A B divided by AC would be the same as d e divided by c d right so this would be d e divided by CD but notice the choice they give us is CD divided by CE that&amp;#39;s not even close that&amp;#39;s that&amp;#39;s not even one of the three ratios at all that would be actually hypotenuse divided by opposite which is a trig ratio but not one that you know of yet anyhow so this problem really combines two things it combines the idea of the three trig ratios along with the idea that two similar right triangles are going to have ratios of relatively positioned sides that are the same that&amp;#39;s actually the basis of all of right triangle trigonometry all right let&amp;#39;s move on to problem number four here we go a regular pentagon is rotated about its Center what is the minimum number of degrees needed to carry the Pentagon onto itself all right wonderful well you were expected to know some of the basic regular figures right a pentagon has got five sides a hexagon has got six sides seven-sided figures they come up less often because they&amp;#39;re more messy but for the record those are called septagons of course an eight-sided figure you&amp;#39;re supposed to know the name of that&amp;#39;s an octagon I&amp;#39;m sure you do a nine-sided figure also doesn&amp;#39;t come up very often called a nonagon I love that word and then a 10-sided figure a decagon after that I don&amp;#39;t think that you&amp;#39;d have to know any of them but obviously we need to know what a pentagon is okay now if we&amp;#39;ve got any regular figure and here I&amp;#39;m going to try to do my best drawing a pentagon a regular pentagon right the idea of a regular figure which this is most certainly not but I gave it my best shot right a regular figure has got all sides the same length and all interior angles the same measure and what that means is that it&amp;#39;s got what&amp;#39;s known as symmetry rotational symmetry reflective symmetry you know anyhow this is talking about rotation symmetry the idea is that we&amp;#39;ve got a center point right I&amp;#39;m just going to kind of draw on some things that sort of look like the radii of circles here all right now if I took this thing and I rotated it either clockwise or counterclockwise it wouldn&amp;#39;t make any difference right it would first line up with itself let&amp;#39;s say we were rotating counterclockwise if I rotate this particular angle all right so if I rotate whatever that angle is that&amp;#39;s going to be the minimum angle that it&amp;#39;ll take for that vertex to end up there that vertex to end up there Etc what is that angle well I know that the entire rotation right all the way around is 360 degrees and in the Pentagon there will be five of those sort of radial angles if you will just made that up probably not a good term so I&amp;#39;m going to do 360 divided by five I think I&amp;#39;ll skip the calculator on this one for now and that would be 72 degrees and therefore our answer is 72 degrees now a little twist on this problem a little variation on this problem is that they might say something like which of the following angles is not an angle of rotational symmetry or which of the following angles does not map the Pentagon onto itself now you might say well I I thought 72 was the only one well now 72 would work 2 times 72 which would be 144 would work 3 times 72 which would be 216 would work etc all right so any integer multiple of this so multiply this by two by three by four by five any of those will map the figure onto itself actually you can multiply it by 1000 and that angle would would you know map it onto itself all right so let&amp;#39;s take a look at problem number five on the set of axes below triangle ABC is congruent to Triangle a prime B Prime C Prime let me move this up a little bit so we can see the rest all right triangle ABC maps onto triangle a prime B Prime C Prime after a what all right so one thing that you learned was that if you have two figures that are congruent to each other now congruent figures are figures that are identical in every way shape or form they have the same size they have the same shape they are simply in different locations in the plane all right but if you have two figures that are congruent to one another then there is either one rigid motion and those are translations rotations or Reflections there&amp;#39;s either one or a sequence of those that will map literally put this thing on top of this one all right so that&amp;#39;s our idea here we&amp;#39;re trying to figure out which one it is now let&amp;#39;s take a look at our four choices Choice number one is a reflection over the line Y equals negative X Choice two a reflection over the line Y equals negative X plus two number three a rotation of 180 degrees centered at one comma one and four a rotation of 180 degrees centered at the origin all right now these problems are not uncommon so before we even get into which one is the correct choice I&amp;#39;d like to point something out that will allow you to eliminate a couple of the choices let&amp;#39;s take a look at triangle ABC but don&amp;#39;t break to the board right in triangle ABC if I start at Point a and then I go to point B then I go to point C then I go back to point a point B Point C Etc if I do that notice right that I have to go in a clockwise manner to sort of Trace out alphabetically those letters now if I look at triangle a prime B Prime C Prime and I go from A to B to C back to a that is also in a clockwise manner notice that in both cases the order of the vertices stays in a clockwise manner that means there cannot be a reflection over a line okay or a single reflection over a line all right that is very very important all right if there is a line reflection a single line reflection then what will happen is if the letters started in a clockwise manner they will then change to a counterclockwise manner think about the idea of when you look in a mirror and you look in your reflection your hair Parts the opposite way right you know if you raise up your right hand it looks like you&amp;#39;re raising up your left hand Etc right Reflections change the orientation of the vertices since these don&amp;#39;t that means that both Choice One and choice 2 are immediately eliminated they&amp;#39;re done okay now let&amp;#39;s talk about these two 180 degree rotations and before we do that let me just kind of get rid of all this now it is very tempting to think it&amp;#39;s a rotation of 180 degrees centered at the origin but let&amp;#39;s talk about 180 degree rotation centered at the origin when you&amp;#39;ve got something like this right then what happens is a point x comma y will get mapped to the point negative x negative Y in other words it&amp;#39;ll be exactly the same point but it will have its sign changed in other words if you took something like Point C down here or sorry up here which is at one two three four five one two three four five six so C at five comma six should go to C Prime at negative 5 negative six except C Prime is actually at negative three one two three four C Prime is at negative three negative four so it really can&amp;#39;t be a rotation of 180 degrees centered at the origin all right so that leaves us with choice three but let&amp;#39;s really try to understand how we can see that it&amp;#39;s 0.3 or choice three all right a rotation centered a rotation of 180 centered at one comma one rotations of 180 degrees are some of the most special rotations out there and the reason why is that when you rotate a point by 180 degrees let&amp;#39;s say this is your original point and let&amp;#39;s say this is your point of rotation I don&amp;#39;t want to use C because we already have a c up here when we rotate by 180 degrees what happens is that those three points the original the pre-image point the center of rotation and the image Point are all co-linear they&amp;#39;re all co-linear now what&amp;#39;s cool about this is if I think or know that I have a rotation by 180 degrees I can literally connect the point and its image Point using a straight line and somewhere on that line is the center of rotation and you might go well that&amp;#39;s not very helpful except now let me collect connect sorry what not collect connect B and B Prime and when I do that I do I get that and now when I connect C and C Prime and I do that then what happens is I can see my center of rotation right the center of rotation has to be on each one of those lines because it is a 180 degree rotation and that is specifically a point of rotation at one comma one it doesn&amp;#39;t work if the if it&amp;#39;s a rotation other than 180 degrees but 180 degree rotations will leave the point the center of rotation and the image Point collinear and that&amp;#39;s really kind of a key in understanding why at least in this problem the center of rotation is not at the origin but is actually at the point one comma one all right let&amp;#39;s keep going problem number seven I like this one in the diagram below of triangle AED and a b c d segment a b c d a e segment AE is congruent to segment d e which statement is always true all right now they&amp;#39;re trying to just draw you in here they want you to look at this picture and they want you to make a statement like oh EB is congruent to EC well yeah sure EB looks like it&amp;#39;s congruent to EC except I I&amp;#39;m not told anything anything up here that would leave me to believe that in fact all I know from this statement right is two things AE is congruent to d e and I know that points a b c and d are co-linear that&amp;#39;s what that tells me they all lie in a straight line now you might say well I can tell that just from the picture but we&amp;#39;re trying to avoid that here okay because from the picture it sure looks like e b and e c are the same now do I know that AC is congruent to DB well again it sure looks like AC and DB are the same length you might also be thinking oh there&amp;#39;s some kind of partitioning there some kind of addition or subtraction proof nope uh-uh didn&amp;#39;t didn&amp;#39;t tell me anything right how about EBA is congruent to ecd that would be this angle right here and this angle right there again nothing tells me that so it seems like it should be Choice four angle EAC is congruent to angle EDB well let&amp;#39;s take a look at those two e a c is this angle right and e d b is this angle and you bet I know that&amp;#39;s true and I don&amp;#39;t know that&amp;#39;s true because it looks like it&amp;#39;s true I know it&amp;#39;s true because I know that this is an isosceles triangle right a triangle with two sides that are the same length and what I know about any triangle that has two sides the same length is that the angles opposite of those two equal sides are also equal in measure and therefore the base angles of an isosceles triangle are equal and angle EAC is congruent to angle EDB cool let&amp;#39;s keep going number eight as shown in the diagram below right triangle ABC has side lengths of 8 and 15. if the triangle is continuously rotated about segment AC the resulting figure will be Choice One a right cone with a radius of 15 and a height of 8. choice two a right cone with a radius of 8 and a height of 15. choice 3 a right cylinder with a radius of 15 and a height of 8. and choice 4 a right cylinder with a radius of 8 and a height of 15. so here you have to know obviously the difference between a cone And the cylinder and I hope you do right a cone is something that kind of looks like this right with a nice sort of like that and a cylinder is something whoops I don&amp;#39;t know why I made that Dash something that looks like that all right now this problem gets to what&amp;#39;s called a volume of Revolution or volume of rotation right specifically if we took this particular figure and we rotated it about AC and that&amp;#39;s really key the idea is if I take this thing right and I rotate it around that segment right I spin that right triangle around what does it end up looking like well its Mirror Image kind of comes around here and you get these circular cross sections and what I now have is a cone right with a height of eight and a radius of 15 and that is Choice One the way you&amp;#39;d get a cylinder by the way is by rotating a rectangle around any one of its sides okay now depending on which side you rotate it around the rectangle will have a different you know height or radius but that&amp;#39;s the way you get a cylinder all right let&amp;#39;s go to the next one problem number nine in the diagram below lines K and L intersect lines M and N at points a b c and d which statement is sufficient to prove that a b c d is a parallelogram awesome well right first let&amp;#39;s just outline that thing A B C D you might be like hey that looks like a parallelogram and it certainly does but I want to know what&amp;#39;s enough to know that it is a parallelogram now parallelograms have many properties right but at the end of the day what makes a parallelogram a parallelogram right is the idea that a b must be parallel to CD and right I have to have a d parallel to BC right what a parallelogram is is a four-sided figure with two pairs of parallel sides and you might even want to throw in two pairs of opposite parallel sides although I&amp;#39;m pretty sure adjacent sides can&amp;#39;t be parallel so that would be a little bit redundant all right so let&amp;#39;s take a look you know in Choice one we tell we know that angle one is congruent to angle three all right so that would be like angle one congruent to angle three now that&amp;#39;s awesome if I know that angle one and angle 3 are congruent to each other then I definitely know that a d is parallel to BC right so that&amp;#39;s what choice one gets me a d is parallel to BC because these are what are known as corresponding angles okay and if corresponding angles are created that are congruent then the lines that create them are parallel but that&amp;#39;s not enough right it only gets me a d parallel to BC so that not enough half of what I need but not enough now Choice 2 angle 4 is congruent to angle seven let&amp;#39;s mark those angle four is congruent to angle seven well that&amp;#39;s also kind of great these are alternate interior angles that are created by this transversal and these two lines and therefore that would be enough to know that a b is parallel to CD which is great but still not enough right that&amp;#39;s not enough right it would only get me a trapezoid remember trapezoids only have to have one pair of parallel sides so it&amp;#39;s it&amp;#39;s got to be choice three or Choice four let&amp;#39;s take a look at choice three choice three says angle two is congruent to angle five all right now that&amp;#39;s that alternate interior angle pair again which is awesome because that again gets me you know a b parallel to C D or k parallel to L whatever whatever you want to say so I&amp;#39;m halfway there and now 5 is congruent to seven so where&amp;#39;s five and seven five is congruent to seven right these two being congruent to each other will get that parallel to that these two being congruent to each other will get this parallel to this and that&amp;#39;s it I&amp;#39;ve got it now a little side note real quick here right why not this last choice you know it seems to have just as much information as choice three right but it&amp;#39;s not enough right because if I look at Choice four it says one is congruent to three let me kind of get rid of these again I gotta get rid of all this so one is congruent to three which is great because then that gets me m and n parallel to each other and then three is congruent to four that doesn&amp;#39;t do anything for me three congruent four three and four are just vertical angles so they&amp;#39;re always congruent they have nothing to do with parallel lines whatsoever so Choice four is not correct all right here comes probably my least favorite problem on the test and that&amp;#39;s saying a lot because on the algebra 1 test if you were with me that would be weird because that means you&amp;#39;ve watched both of these videos and you&amp;#39;re either taking both classes or you just love watching math videos um anyway either way that that test had all sorts of problems this is one of the very few on this test that I don&amp;#39;t particularly love but let&amp;#39;s take a look at it which transformation does not always preserve distance does not always preserve distance now the idea of preserving distance is very simple right if I&amp;#39;ve got two points in space right and let&amp;#39;s connect them with a line segment just for the fun of it right you know if I take it because it&amp;#39;s got to be fun to connect with a line segment so if I connect them with a line segment and then I transform them right and there are so many Transformations out there it&amp;#39;s not just okay you know rotations Reflections translations dilations there&amp;#39;s tons of things out there right sometimes when I do something right then what will happen is that the line segment I get as the image let&amp;#39;s say this has got a length of four centimeters then the transformation preserves the distance between the two points anytime that happens always we call it a rigid body motion okay when the distance is preserved but sometimes it doesn&amp;#39;t you know sometimes you know we do it and we get something that changes let&amp;#39;s say that&amp;#39;s three centimeters that would not be distance preserving all right now translations Reflections and rotations the rigid body motions will always preserve distance dilations remember those things right from problem number one dilations where we multiply the X and the y coordinate by something those don&amp;#39;t preserve distance okay now when I look at these right these are all algebraic rules and if you&amp;#39;ve studied or taught geometry for a long time you would look at Choice one where X comma y gets mapped to X plus two comma Y and you know oh that is just a translation two units to the right that&amp;#39;s going to preserve distance because it&amp;#39;s a translation now Choice two x comma y goes to negative y comma Negative X that&amp;#39;s actually a reflection across the line Y equals negative X but it&amp;#39;s a reflection so therefore it is also distance preserving now choice three and choice four are very strange they are transformations that you have really never seen before in any kind of algebraic way number three x comma y goes to 2 times x comma y minus one as soon as I see that multiplication by two it&amp;#39;s not going to preserve distance it just isn&amp;#39;t if you multiply X or Y by any number other than 1 or negative one right because here we&amp;#39;re multiplying by negative one right if we multiply by either by anything other than one or negative one we&amp;#39;re getting a stretching effect okay now this is kind of weird I&amp;#39;m not going to call this a dilation because it&amp;#39;s being stretched in the X Direction but the Y is just shifting down one unit okay but it still isn&amp;#39;t going to preserve distance because of that multiplication here you&amp;#39;ve got both a multiplication by negative one plus sort of like an addition of three and two and again they couldn&amp;#39;t have made that more confusing looking this is literally a combination of both a reflection and a translation number four all right but the one that won&amp;#39;t preserve distance in this case says choice three and really it&amp;#39;s all about this it&amp;#39;s all about the fact that the x is being multiplied by two it has nothing to do with the Y minus one all right let&amp;#39;s keep going oh end of that page all right the first one that&amp;#39;s going to require a calculator let&amp;#39;s take a look at number 11. in the diagram below EF segment EF is parallel to segment HG EF is equal to five HG is equal to 12. f i is equal to 1.4 X plus 3 and h i is equal to 6.1 x minus 6.5 awesome then they ask us what is the length of h i and we have to be very careful on this problem all right but let&amp;#39;s talk about what this problem is getting at as soon as they tell me that EF is parallel to HG right e f is parallel to HG what that allows me to say is that angle F and angle H are congruent and angle e and angle G are congruent now why is that good to know well as soon as I know that two angles of one triangle are congruent to two angles of another triangle then I know that this triangle and this triangle are similar to each other they&amp;#39;re similar to each other so I can do something like this I can say 5 divided by 12. is equal to 1.4 X plus 3 divided by 6.1 x minus 6.5 all right so this is a very very common thing that we see a lot in similarity all right this is where I&amp;#39;m going to pick up my cheat sheet because I&amp;#39;ve worked out all the numbers here so I don&amp;#39;t have to necessarily go to the calculator yet we&amp;#39;ll go to the calculator when we really need to but now to solve this equation I&amp;#39;m just going to cross multiply that&amp;#39;s easy enough right do a little bit of this a little bit of this we&amp;#39;ll get 5 times 6.1 x minus 6.5 is equal to 12 times 1.4 X plus 3. all right now I&amp;#39;ll use a little distributive property to multiply through by the 12 and the 5. let me just kind of scroll up a little bit here and we will get 30.5 x minus 32.5 trying to make those decimal points a little bit bigger is equal to 16.8 X Plus 36. let me take a pause just for a minute so you can take a look at that simple enough at this point it&amp;#39;s really mostly algebra mostly algebra so if I combine all like terms and I get all the x&amp;#39;s and stuff to one side or another I&amp;#39;m not going to go through all of that eventually you&amp;#39;ll get X is equal to 5. all right now you have to be a little bit worried about this okay and here&amp;#39;s why X is equal to 5 and so it is very tempting to immediately bubble in choice 2. the problem here right is that they&amp;#39;re asking me what the length of h i is and that&amp;#39;s not equal to X right h i is going to be equal to 6.1 times 5 minus 6.5 and of course you end up doing that on your calculator and you find that it&amp;#39;s 24. all right and that&amp;#39;s it okay but this is actually you know one of the things that I don&amp;#39;t love about this problem is that you could know all the geometry you could know this is about similarity you could know you&amp;#39;re supposed to set up that ratio you can cross multiply you can do every darn thing correct get x equals five bubble that choice in and you&amp;#39;re losing two points even though you knew the geometry you knew the algebra right but because they didn&amp;#39;t want the value of x they wanted the length of h i you get the problem wrong don&amp;#39;t love it but it is kind of a good problem to look at in terms of make sure to give the answer they want right make sure you&amp;#39;re really solving for it all right let&amp;#39;s take a look at another one where we&amp;#39;ll actually break the calculator out in a moment 12. the square pyramid below models a toy block made of maple wood each side of the base measures 4.5 centimeters and the height of the pyramid is 10 centimeters if the density of maple is 0.676 grams per centimeter cubed what is the mass of the block to the nearest tenth of a gram all right now I know that you studied density in Geometry okay but I really want to emphasize something density which is what they give us here is how much stuff is inside of stuff okay and and I know that sounds really dumb but let me kind of really like lay it out here density in this case is 0.676 grams per centimeter cubed all right to me this is no different than saying the density of eggs is 12 eggs per carton right that&amp;#39;s how much stuff is in stuff there are 0.676 grams in a cubic centimeter of the wood that is being used to make this square pyramid so the first thing I have to do is figure out how much volume I have how many cubic centimeters do I have now this is a good time to take a not a break for lemonade although I&amp;#39;m dying for some we will soon but this is a good time to take a look at the formula sheet at the back of the exam all Regents math exams have the same formula sheet on them I need to move that out of the way and it is literally one of the last pages that&amp;#39;s in here oh my goodness the ice cream truck is going on outside that&amp;#39;s amazing I would stop and go get some except I&amp;#39;d probably get in trouble all right so let&amp;#39;s take a look at the formula sheet now there&amp;#39;s a lot of stuff on this formula sheet that you can use in this test you know ironically things like let&amp;#39;s say the trig ratios aren&amp;#39;t on it I don&amp;#39;t know why but you know in case you&amp;#39;re in the middle of the exam and you really need to know that I don&amp;#39;t know one cup is equal to eight fluid ounces there it is you&amp;#39;ll know it you&amp;#39;ll have it anyway whatever um so one thing that is helpful for you are things like the area formula for a circle right or the circumference formula of a circle or the volume of the general prism or the volume of a cylinder or the volume of a sphere the volume of a cone the volume of a pyramid there it is that&amp;#39;s the one I need and it says one-third times Big B times H and that Big B that Big B that sounds really weird but that Big B is the area of the base of our pyramid big b stands for the area of the base so it&amp;#39;s one-third times the area of the base times the height of the pyramid so now let&amp;#39;s go back to that problem let&amp;#39;s see if we can find it here we&amp;#39;re getting there we&amp;#39;re getting there getting there there it is ah excellent so what do we know well we know that it&amp;#39;s got a square base where the sides of the base are 4.5 centimeters each and the height is 10 centimeters so if you recall right that volume formula was one-third Big B times H so that&amp;#39;s going to be one-third times the area of the base now that&amp;#39;s 4.5 squared right that&amp;#39;s a square base that&amp;#39;s its area times the height which is 10. so let&amp;#39;s figure out what the volume of our pyramid is right now and for some reason I don&amp;#39;t have my calculator open I don&amp;#39;t know why I thought I had it open earlier but it must have decided to just close here we go excellent let&amp;#39;s add a calculator um and we&amp;#39;re going to do 1 3 times 4.5 nope let&amp;#39;s try that again 4.5 squared times 10. right and that gives us a volume of 67.5 cubic centimeters that&amp;#39;s how much stuff we have that&amp;#39;s the stuff that&amp;#39;s holding the mass is this space 67.5 cubic centimeters let&amp;#39;s put that down all right now we know that we&amp;#39;ve got 0.676 grams per cubic centimeter I mean again if I told you you had five egg cartons and each egg carton held 12 eggs and I asked you how many eggs are there right you would take the number of cartons and multiply it by how many eggs there were per carton if I tell you that there&amp;#39;s 0.676 grams in every cubic centimeter and you know how many cubic centimeters there are then the mass is just that many cubic centimeters times how much how many grams there are per cubic centimeter so my mass now is going to be 67.5 cubic centimeters times 0.676 grams per cubic centimeter look at that the cubic centimeters cancel let&amp;#39;s grab my calculator again this is awesome because I can just put times .676 and that&amp;#39;s going to give me a mass of 45.63 foreign which to the nearest tenth is 45.6 excellent I think this is as good of a time as any to take a little break have a little lemonade before we go into one of the trickier problems on the test lemonade is so good I wonder what happens if I put this right in front of the green screen does it disappear nah it doesn&amp;#39;t disappear feels like it should it&amp;#39;s more or less the the color of the green screen but it&amp;#39;s just lemonade oh my goodness it gets it gets hot in here I got like lights everywhere they&amp;#39;re beating down on me you know uh I burn hot anyway so yeah gotta have a lot of lemonade geez I feel like I drank half that thing and I&amp;#39;m not even halfway through the test okay all right let&amp;#39;s take a look at one of the trickier problems on the test um you know there&amp;#39;s a lot of times when students say oh they were trying to trick me there oh they threw in a trick or something like that and I always think to myself nine times out of ten it&amp;#39;s just not true they&amp;#39;re not trying to trick you they&amp;#39;re just trying to test something all right now let&amp;#39;s I feel like I&amp;#39;m like shiny now from sweat um okay let&amp;#39;s take a look like you needed to know that um let&amp;#39;s take a look at problem number 13. in the diagram below of right triangle e f g altitude FH intersects hypotenuse EG at H if FH is 9 and EF is 15 what is eg all right now there are a few things about this problem I don&amp;#39;t particularly like all right one of the things is that they sort of indirectly tell us that triangle e f g is a right triangle all right I mean you might say well they don&amp;#39;t indirectly tell us that they tell us this right here it would have been kind of nice had they said this particular angle is the right angle all right but let&amp;#39;s be very clear this is a right angle right and because FH is an altitude this is also a right angle all right so immediately there&amp;#39;s probably many of you that are saying I know this problem right this is that whole like altitude to the hypotenuse of a right triangle problem and you are kind of correct okay but here&amp;#39;s where the tricky part comes in let&amp;#39;s put in our our two measurements right we know that FH is equal to nine and we know that e f is equal to 15 and we want to know what EG is okay now you probably memorized some theorems or you you did hopefully your teacher told you about how this is really just all about similarity and it is all about similarity right we&amp;#39;ve got a large right triangle we&amp;#39;ve got a medium right triangle and we&amp;#39;ve got a small right triangle okay I feel like it&amp;#39;s like the you know Goldilocks and the Three Bears kind of thing all right now the problem is none of those theorems right that you may have learned about like geometric means or you know this over this is equal to this over this none of them are going to get the job done there&amp;#39;s not enough information all right and you can try you know like you could be like well like I don&amp;#39;t know 15 divided by 9 is equal to EG divided by f g that would actually be the case like 15 over 9 would be equal to EG divided by f g but I don&amp;#39;t know f g so that that doesn&amp;#39;t help okay um so we we are actually lacking enough information to use one of those theorems all right so here&amp;#39;s what they want you to do they want you to realize that number one oh I don&amp;#39;t have enough information to do that similarity bit okay but I can get enough information if I use the Pythagorean theorem to figure out the length of e h all right so because I know the hypotenuse 15 and this side nine I should be able to figure out the length of e h now as well if you know the famous three four five right triangle you can also figure out eh just using a multiple of that three four five right triangle but let&amp;#39;s say for for kicks and grins you just don&amp;#39;t recognize that so maybe you call this a right and you do a little a squared plus 9 squared is equal to 15 squared and I&amp;#39;m not going to go through all of that because you&amp;#39;ve seen the Pythagorean theorem done many times but you end up getting a is equal to 12. all right so you have to do that first you absolutely need to get that 12 first so this is a problem that&amp;#39;s really testing two distinctly different things one of those things is the Pythagorean theorem and you have to do that first or special right triangle slash Pythagorean triples all right then once you have that now you&amp;#39;re good to go and there&amp;#39;s many different ways of doing it one way of doing it is setting up a proportion and solving for GH or HG and then adding that onto the 12 that&amp;#39;s one good way of doing it there&amp;#39;s another way of doing it right away right which is also doing it by similarity so now keep in mind right when you look at the three right triangles right the big one which is kind of lying on its hypotenuse and then the two smaller ones which are lying on legs of theirs right if I hone in on the middle one right I know it&amp;#39;s hypotenuse right and I also know it&amp;#39;s longer leg right I know it&amp;#39;s hypotenuse and I know it&amp;#39;s longer leg now if I look at the big right triangle I know it&amp;#39;s longer leg and then what they&amp;#39;re asking me for is it&amp;#39;s hypotenuse so what I can do is I can set up a ratio like the following I can say this longer leg divided by its hypotenuse so 12 divided by 15 right so longer leg divided by the hypotenuse is now going to be equal to this longer leg 15 divided by the hypotenuse that I&amp;#39;m looking for EG okay now some of you have maybe learned a theorem that said that this particular leg is the geometric mean between this side and the entire hypotenuse that&amp;#39;s great that gets you this you know there&amp;#39;s a variety of ways that you can do it but at this point in time we now cross multiply right we get 12 times e g is equal to 15 squared and then I can just do it 15 times 15 15 squared I&amp;#39;m not going to even evaluate that I&amp;#39;m going to say EG is 15 squared divided by 12. that&amp;#39;s where my calculator is very handy 15 squared divided by 12 of course 15 squared divided by 12 all I really want is the decimal version and I get 18 0.75 again you have to be a little bit careful about that because if you solve for GH directly you get 6.75 right so it&amp;#39;d be very easy to solve for GH get 6.75 and circle Choice one all right but we want all of EG and again what makes this problem to me especially tricky and a little bit dubious from the perspective of a test maker is that they&amp;#39;re really testing two distinctly different geometric Concepts one of them being the Pythagorean theorem let&amp;#39;s say and the other one being this similarity when the altitude is drawn to the hypotenuse of a right triangle all right let&amp;#39;s keep going ah back to a nice easy no tricks one 14. in triangle ABC below D is a point on a b and e is a point on AC such that d e is parallel to BC which statement is always true all right so again this is one of those problems where you might think they&amp;#39;re trying to trick you and maybe they are but what they&amp;#39;re really trying to do is make sure that you understand what you can get out of geometric information that you&amp;#39;re given and what you can&amp;#39;t so let&amp;#39;s take a look right choice one angle Ade and angle ABC are right angles in other words is there enough information in what they told us to conclude that these two are right angles now again they look like right angles they sure do but there is nothing in any of this that talks about perpendicularity at all so even though they might look like right angles I don&amp;#39;t know that they are all right now Choice two triangle Ade is similar to triangle ABC all right so do we know that this particular triangle is similar to this one well we&amp;#39;re told that d e is parallel to BC these two are parallel and because they&amp;#39;re parallel we know that that angle must be congruent to that angle and this angle must be congruent to that angle those are corresponding angles being created by two transversals and two parallel lines and therefore yep triangle Ade must be try similar to triangle ABC by the angle angle theorem of similarity so that is absolutely true by angle angle now as an aside the last two choices are trying to draw you in to think that points D and E are mid points right so like d e is half of b c d e is half of BC well that would definitely be true if d and e were the midpoints of a b and a c respectively and likewise a d is congruent to DB a d congruent to DB that would be true again if D were the midpoint of a b but even though it looks like it&amp;#39;s the midpoint there&amp;#39;s nothing up here whatsoever that tells us it&amp;#39;s the midpoint okay let&amp;#39;s keep going I never know when I&amp;#39;m really at the end of the page all right let&amp;#39;s take a look at 15. if one exterior angle of a triangle is acute then the triangle must be which of the following all right well first let&amp;#39;s talk about exterior angles right if you have any kind of figure it doesn&amp;#39;t have to be a triangle any type of figure whatsoever like let me make some kind of generic quadrilateral right if I have a generic quadrilateral like that and I go to any given interior angle let me go to this nice acute angle right here right if I extend that side all right then this angle is the exterior angle to this angle&amp;#39;s interior angle all right exterior interior all right now you might say why do you extend that one and not this one right so you know why did I extend that side length and let&amp;#39;s say not this side length and the answer is it really doesn&amp;#39;t matter those two angles would have the same measure and of course the key is here if you have an acute angle here right then you must have an obtuse angle out here okay so given that let&amp;#39;s Now erase this picture and talk about the triangle that they&amp;#39;re referring to they say if one of the exterior angles of a triangle is acute then the triangle must be what well if we&amp;#39;ve got an acute exterior angle right let&amp;#39;s say we got this triangle and here if I extend this right then that is acute right an acute angle is an angle that is less than 90 degrees right an acute angle is a sharp Angle an angle less than 90 degrees if we have an acute exterior angle then the adjacent interior angle must be obtuse and as soon as an as a triangle has an obtuse angle it&amp;#39;s called an obtuse triangle now side note there is nothing called an acute triangle I&amp;#39;ve never heard of an acute triangle whatsoever all right uh that that&amp;#39;s just weird um there are obtuse triangles obtuse triangles or any triangle with an obtuse angle a right triangle is of course any triangle with a right angle and an equi-angular triangle also known as an equilateral triangle is a triangle with all equal angles right a scalene triangle is a triangle with all unequal angles anyway not the point the point is if you have an acute angle on the outside you must have an obtuse angle on the inside and therefore that triangle is obtuse all right ah number 16. given the information marked on the diagrams below which pair of triangles cannot always be proven congruent all right well I will say right this is a generally speaking a pretty well constructed test but this might be the one problem on the test that I believe is sort of poorly constructed it&amp;#39;s not trying to trick you in fact I think most people will probably get this problem right now remember we&amp;#39;re looking for something that can&amp;#39;t be proven congruent all right let&amp;#39;s take this moment to talk about the major theorems that prove two triangles are congruent right the big three which are side side side angle side angle and side angle side those are the big three okay those are the basis of like everything else and then there&amp;#39;s two additional ones which is angle angle side and the very specific hypotenuse leg all right now if we can assign any one of these to these kind of diagrams given the information they tell us then we can prove that the triangles are congruent so if we take a look at the first one right and we&amp;#39;ve got well that side&amp;#39;s congruent to that side we know that&amp;#39;s a 90 degree angle now here&amp;#39;s a little bit of a problem okay all right I&amp;#39;m gonna assume that&amp;#39;s a 90 degree angle as well now you might say well how what do you mean assume that&amp;#39;s a 90 degree angle well I I got to tell you man this is a flawed problem and shame on you for writing it whoever did right there is nothing in here that tells us that points a c and d are co-linear for all I know that&amp;#39;s a right angle and that thing&amp;#39;s like an 89 degree angle I don&amp;#39;t know you know what I mean but the plain fact is the test writers want you to assume it&amp;#39;s also a 90 degree angle even though they didn&amp;#39;t tell you that and then of course BC is congruent to itself so in theory we can prove that a b c and DBC are congruent by side angle side right this is a side angle side scenario again assuming that ACD is collinear it didn&amp;#39;t tell us that but we we have to otherwise like actually the only one that you can say are always congruent is choice three and we&amp;#39;ll get to that in a moment let&amp;#39;s take a look at Choice two okay in this particular problem what do we know well we&amp;#39;ve got angle e congruent to angle H we&amp;#39;ve got side F G congruent to side IG and again this is a little bit flawed they want us to assume that e g h and igf are also collinear that would allow us to conclude that those are vertical angles and therefore those angles are congruent now notice here the side that we know that is congruent is not between the two angles the pairs that we know are congruent but that&amp;#39;s okay because that still falls into angle angle side okay I&amp;#39;ve got two pairs of angles that are congruent and I&amp;#39;ve got a side pair that&amp;#39;s congruent that&amp;#39;s angle angle side so we&amp;#39;re still good all right again a little problematic because there&amp;#39;s absolutely nothing that tells us e g h and igf are collinear we could have these little triangles being a little like like this you know what I mean then we wouldn&amp;#39;t know if those two angles those two vertical pairs were equal or not but you got to do what you got to do you know anyway this one this one&amp;#39;s good to go okay and the reason why is we know this side&amp;#39;s congruent to this side that side&amp;#39;s congruent to that side and then we have that reflexive shared side so this is a situation where the two triangles can most certainly be proven congruent by side side okay so why isn&amp;#39;t this one correct all right well in this situation right we&amp;#39;ve got a side congruent to a side another side congruent to a side and an angle can grow into an angle so it feels like it could be side angle side the problem is the angle pair that&amp;#39;s congruent is not what&amp;#39;s called included between the two congruent side pairs if we knew that angle o was congruent to angle s if we knew that angle was congruent to that one we would have side angle side but this is one of those ones that your math teacher warns you about right this is ass and ass doesn&amp;#39;t work and it&amp;#39;s not because it would always get Snickers from like Teenage students sitting in a geometry class although that&amp;#39;s a good way to remember it doesn&amp;#39;t work right it just isn&amp;#39;t one of the theorems right you can literally have two triangles that have this kind of scenario going on but aren&amp;#39;t congruent to each other so this is our correct choice amongst a very flawed problem all right let&amp;#39;s keep going hello problem 17. the diagram below shows a tree growing vertically on a hillside the angle formed by the tree trunk and the hillside is 100 degrees the distance from the base of the tree to the bottom of the hill is 140 feet what is the vertical drop x to the base of the hill to the nearest Foot well it&amp;#39;s nice that they gave us this entire diagram it&amp;#39;s really great now look anytime you&amp;#39;re working with a right triangle you know one of the lengths of the right triangle you&amp;#39;re missing one of the other lengths but you&amp;#39;ve got some kind of an angle information think about right triangle trig again let&amp;#39;s go back to sohcahtoa let me put it up there because here comes right triangle trig again now one of the things that&amp;#39;s a little bit unfortunate here is that the angle that they give us that hundred degree angle isn&amp;#39;t actually in the right triangle we need in order to use sohcahtoa which you can barely read I apologize for that in order to use sohcahtoa we need to have an angle in the right triangle well if that&amp;#39;s a 100 degree angle then the one sitting right in here is an 80 degree angle now if I&amp;#39;m going to use that 80 degree angle to figure out the value of x what I want to do is look at the 140 the X and the 80 and ask myself which of the three trig ratios relates those three quantities well the x is adjacent to the 80 degree angle and the 140 is the hypotenuse adjacent and hypotenuse is the cosine ratio so I&amp;#39;m going to say that the cosine of 80 degrees is the side adjacent which is x divided by the hypotenuse which is 140. now if you really just hate the cosine ratio right as as some students of mine have pointed out if I know that that&amp;#39;s 80 degrees then I can say that that acute angle down there is 10 degrees and I could use the the sine of 10 is equal to X over 140. that would also work just fine I&amp;#39;m going to stick with the cosine here it&amp;#39;s real simple now I can just multiply both sides by 140. right that&amp;#39;s going to cancel that 140. I&amp;#39;m going to bring my calculator back out now it&amp;#39;s really really important right when you&amp;#39;re doing trigonometry on your calculator in a couple days from now you have to make sure it&amp;#39;s in degree mode on the tin Spire that&amp;#39;ll be a little deg sitting right up here if it says Rad you have a little bit of an issue you got to get it into degree mode for those using the ti Inspire it&amp;#39;s as simple as actually clicking on that little word which I can do with my finger okay the plain fact is if you are in New York state your teachers should in theory reset your calculators and clear all of their memories before you take this test and when they do that in theory it will put your calculator back into radian mode and that&amp;#39;s a problem because it will give you the wrong answer because as you learned in this course radians and degrees are not the same thing they&amp;#39;re different they&amp;#39;re like inches and centimeters right they&amp;#39;re different units so I&amp;#39;m going to go back to this now that I I did my little uh screed about radians and degrees right remember that what I&amp;#39;ve got is 140 times the cosine of 80. let me bring that back up I&amp;#39;m just going to type that in 140 times trig the cosine of 80 and that&amp;#39;s going to be 24.31 and there it is to the nearest Foot 24 feet all right little right triangle trig second time we&amp;#39;ve seen it right we saw it early on in that like I think it was like problem two or something or three right that kind of ratio piece with the similar right triangles here we actually had to use it to find a missing side of a right triangle right triangle trig is primarily used for two things finding missing sides when you know one side and an angle or finding missing angles when you know two or more sides a little foreshadowing I call that foreshadowing okay here we go let&amp;#39;s take a look at problem 18. on the set of axes below triangle let and triangle L double Prime e double Prime t double Prime are graphed in the coordinate prompt plane where triangle l-e-t is congruent to Triangle L double Prime e double Prime t double Prime wow okay let&amp;#39;s take a look which sequence of rigid motions Maps triangle let onto triangle L double Prime e double Prime t double Prime okay all right now it&amp;#39;s a little bit weird I find these problems because it says you know number one as soon as you&amp;#39;ve got two primes on these letters and none on these it sort of implies that there are two Transformations that are going to happen don&amp;#39;t get me wrong it very well could be that they do that and there&amp;#39;s still only one but that would be really disingenuous really stupid so this whole rotation of 180 degrees about the origin I got my doubts now you remember that whole like orientation of the letters things let&amp;#39;s look at that one again and let&amp;#39;s let&amp;#39;s think about the word let right if I go from L to E from E to T and from T to L that&amp;#39;s clockwise all right if I go from L to e e to sorry L double Prime e double Prime t double Prime that&amp;#39;s counterclockwise all right now what that means given that I&amp;#39;ve gone clockwise and I&amp;#39;m going going counterclockwise is that there must be one and exactly one line reflection happening here there&amp;#39;s got to be a line reflection involved there can&amp;#39;t be two all right if there were two then it would go from clockwise to counterclockwise back to clockwise okay but there&amp;#39;s got to be at least one line reflection and let&amp;#39;s see if that&amp;#39;ll help us out a little bit right Choice One says a reflection over the x-axis followed by a reflection oh sorry a reflection over the y-axis followed by a reflection over the x-axis well that can&amp;#39;t be the case because the orientation of the letters changed and they wouldn&amp;#39;t if there were two line Reflections Choice two a rotation of 180 degrees about the origin well again there&amp;#39;s a couple reasons why that shouldn&amp;#39;t be the answer one because they&amp;#39;ve implied that there&amp;#39;s at least two uh two rigid body motions that occur and two again a rotation wouldn&amp;#39;t change the orientation of the letters so that one&amp;#39;s out now the last two choice three a rotation of 90 degrees counterclockwise about the origin followed by a reflection of the y-axis and choice four a reflection over the x-axis followed by a rotation of 90 degrees clockwise about the origin well those two both include one and only one line Reflections so they&amp;#39;re both if you will candidates all right now how do we choose between the two well here&amp;#39;s what you want to do pick one letter just one one of the vertices and go through this set number three okay do that and see if it works and if it does work then maybe try it with Choice four and if it doesn&amp;#39;t work then choice three is the answer so let&amp;#39;s take a look at choice three a rotation of 90 degrees counterclockwise about the origin followed by reflection over the y-axis and I&amp;#39;m gonna I&amp;#39;m gonna get rid of all this stuff all right and let&amp;#39;s just die I think I&amp;#39;m going to hone in on E now if I take e right which is at negative 1 3 and I rotate it 90 degrees counterclockwise about the origin what&amp;#39;s going to happen is it&amp;#39;s going to go right here right it&amp;#39;s going to go down to e Prime at negative three comma negative one all right now normally I would say take your piece of paper and rotate it 90 degrees counterclockwise to do that and that&amp;#39;s fine I just can&amp;#39;t do it myself now the second transformation right rotation of 90 degrees counterclockwise about the origin followed by a reflection over the y-axis if I now took this and reflected it over the y-axis it would in fact land on e double Prime so that&amp;#39;s good right it seems like that is my winner right what I would maybe want to try and it is Choice 3 is the correct Choice what I might want to try is try it with either another one of the vertices and if it works with another one of the vertices great or try Choice 4 with letter e and see if that works right and if it doesn&amp;#39;t then you know you&amp;#39;re you&amp;#39;re correct all right but at this point I think that we move on it is choice three all right let&amp;#39;s take a look at the next problem number 19. diameter roq of circle o is extended through Q to point p and tangent p a is drawn if the measure of Arc r a is 100 degrees what is the measure of angle P all right great well it would be awesome if they gave us a diagram on this I mean they&amp;#39;ve given us diagrams where they never needed to like problem number one right but this one they didn&amp;#39;t give us a diagram for so we definitely want to draw a good picture I just wanted to get my pen icon out of the way all right so let&amp;#39;s start drawing it now the first thing I know is that when I name a circle circle o what that means is that o is the center of the circle diameter r o q so let&amp;#39;s let&amp;#39;s draw that right we&amp;#39;ve got the center at o we&amp;#39;ve got diameter r o q all right so that&amp;#39;s the first thing I want to do I just want I just want to draw that picture all right now it&amp;#39;s extended through Q to point p so I&amp;#39;m going to extend this thing from Q to point P how far it doesn&amp;#39;t really matter I&amp;#39;m just trying to get a reasonable picture all right and tangent p a is drawn now what&amp;#39;s a tangent a tangent right is a line segment that is drawn that just intersects The Circle at that one location and then if it kind of kept going it would never intersected again right so there&amp;#39;s my tangent PA drawn now they tell me that the measure of Arc r a is 100 degrees I&amp;#39;m going to just get rid of this for a second so r a this Arc is 100 degrees okay and I want to know the measure of angle P now there are all of these sort of like you know theorems that you&amp;#39;re supposed to learn about how angles that are created in a circle by chords and tangents and secants and things like that how they relate to the arcs that are intercepted and what I should know about this particular case is that the measure of angle p is going to be the measure of Arc r a minus the measure of Arc a q all divided by 2. so these two line segments PR and Pa intersect two arcs one Arc is Arc a r and the other Arc is Arc AQ all right and the measure of that exterior angle will be the difference between the two arcs divided by two now I know one of the arcs it&amp;#39;s a hundred degrees but what I don&amp;#39;t know is Arc AQ except I do right because I know Arc RQ right this entire arc Arc RQ is 180 degrees and therefore Arc AQ is equal to 80 degrees and now I have absolutely everything I need ra is going to be 100 AQ is going to be 80 all divided by 2 100 divided 100 minus 80 is 20. 20 divided by 2 is 10 and measure of angle p is 10 degrees awesome that&amp;#39;s it okay let&amp;#39;s take a look at problem 20. segment JM has endpoints at J negative 5 comma 1 and M 7 comma negative 9. an equation of the perpendicular bisector of JM is blank all right now this thing is testing actually a couple of different cool Concepts all right it&amp;#39;s actually to me testing three different concepts one of them is what&amp;#39;s known as the point-slope form of a line meaning right forget about the perpendicular bisector let&amp;#39;s just not even worry about that meaning if I have a line right and I know it&amp;#39;s slope let me call it m and I know some point that&amp;#39;s on the line let me call it X1 y1 then the equation of the line can always be written as y minus y1 equals m times x minus X1 this is probably for most of you the first year where you ever saw this form of a line used up to this point you&amp;#39;ve always used lines of the form y equals MX plus b known as the slope intercept form of a line form of a line we just need to know the slope of the line and one point that lies on the line all right now let&amp;#39;s talk a little bit more about perpendicular bisectors now we&amp;#39;ve got segment JM all right now I&amp;#39;m not going to try to draw it particularly like accurately or anything like that all right but let me like like put JM here all right and one thing that we know about Jay is that it&amp;#39;s got the coordinates negative 5 1. and M that&amp;#39;s got the coordinate of seven comma negative nine okay I&amp;#39;m going to switch colors really quickly I&amp;#39;m going to go into red um bisector I want to wage enough all right so now I need to know two things about this perpendicular bisector I need to know its slope and I need to know one point that lies on the perpendicular bisector now there is one point and one point only that you absolutely positively know the perpendicular bisector goes through and that is the midpoint of that line segment right it goes through the midpoint so let me find the midpoint right now remember the midpoint formula is super duper easy all we do is we average the x coordinates and we average the Y coordinates all right I&amp;#39;m going to have to kind of put down my final results here negative 5 plus 7 is 2 2 divided by 2 is 1. 1 plus negative 9 is negative eight negative 8 divided by 2 is negative four so that is our X1 y1 if you will all right and I&amp;#39;m running out of room that&amp;#39;s okay we&amp;#39;ll we&amp;#39;ll figure it out in a second the other thing I need to know is the slope of my perpendicular bisector and this is really cool because even though I don&amp;#39;t know the slope of my perpendicular bisector one thing I can do is I can figure out the slope of J M I&amp;#39;m going to go back into uh blue just to try to keep the two things separate so if I look at the slope of J M right what&amp;#39;s that going to be it&amp;#39;s going to be the change in Y which is negative 9 minus 1 over the change in x 7 minus negative five or seven plus five so that&amp;#39;s going to be negative whoops sorry about that negative 10 over 12 so the slope of j m is negative 5 6. that&amp;#39;s the slope of JM negative 5 6. now remember the slopes of two perpendicular lines are negative reciprocals to each other and if you&amp;#39;re starting to think this is a long problem you&amp;#39;re probably correct anyway so the slope of our segment is negative 5 6 but the slope of our perpendicular line write M perpendicular as I like to put it will be positive six-fifths so now right I&amp;#39;ve got my slope I&amp;#39;ve got my one point that the line goes through I&amp;#39;m going to steal a little bit of actually I&amp;#39;ll put it right here it&amp;#39;s going to be y minus negative 4 which is y Plus 4. equals six-fifths times x minus 1. where is my answer right here y plus 4 is equal to six fifths times x minus one this is a question that they absolutely love love love on the geometry regions because it ties together so many important issues the midpoint formula the slope formula perpend negative reciprocals for perpendicular lines right all that comes together in this problem along with the point-slope form of a line all right let&amp;#39;s take a look at 21. and I think we&amp;#39;ll go back to Blue let me bring this out and then I&amp;#39;ll read the problem for you put this away problem 21. quadrilateral e b c f and a in segment a d are drawn below such that a b c d is a parallelogram and segment EB is congruent to segment FB and segment EF is perpendicular to segment FH if the measure of angle e is 62 degrees and the measure of angle C is 51 degrees what is the measure of angle f h b all right I love these kind of problems this is what I call a fill in the angle problem and it is a Whopper so first things first let&amp;#39;s put in the two angles that we absolutely positively know and also identify the angle that we absolutely positively want and then we&amp;#39;ll start filling in everything else so let&amp;#39;s see what we know we know that angle e is 62 degrees we know that angle C is 51 degrees right what else do we know we also know that EB right yeah EB here is congruent with FB here now let&amp;#39;s immediately get something out of that because EB and FB are congruent to each other then this triangle is an isosceles triangle and that means that this must also be 62 degrees if they tell you that two segments are congruent in the middle of a random problem involving angles look for an isosceles triangle it&amp;#39;s definitely there now the other piece is that EF is perpendicular to FH so FH and EF are perpendicular to each other that means that&amp;#39;s a 90 degree angle right there now what&amp;#39;s really cool about that is that that means I can also figure out this tiny little angle in there right because the 62 and whatever that angle is must add up to 90. so that angle is simply 90 minus 62 which is 28 degrees all right I&amp;#39;m not going to try to kind of get it in there but I&amp;#39;ll just do that now let&amp;#39;s be very clear about what angle we&amp;#39;re looking for angle fhb fhb so this is my winner right here right that&amp;#39;s the angle I gotta get now if I can figure out another angle like let&amp;#39;s say this angle right here if I can figure out angle h b f then I&amp;#39;ve got two out of the three angles in a triangle I can add them together subtract from 180 and I&amp;#39;ll get my final angle so the question is how can I figure that out well there&amp;#39;s a piece of information we haven&amp;#39;t used yet and that&amp;#39;s the fact that a b c d is a parallelogram let me just outline that in red since I&amp;#39;ve got the ability to do that ABCD that is a parallelogram yay okay now I know a lot about parallelograms one thing that I know is that two angles that are adjacent in parallelograms are supplementary I know opposite angles are congruent so I could say that that one&amp;#39;s 51 as well as that one being 51 and that&amp;#39;s not bad that that could also help you out but what I do know is that this angle in here I&amp;#39;m going to kind of draw it like that that angle there is 180 degrees minus 51 which is equal to 129 degrees and you might be thinking to yourself well you know that that&amp;#39;s not very helpful you know what I really need is this angle I don&amp;#39;t need the whole one I need just that part but I know that&amp;#39;s 62 and I know that&amp;#39;s 62. so that will allow me to figure out what that angle is right so just this little angle in here this one is going to be 180 minus 2 times the 62. which is 180 minus 124 which is 56 degrees so this is 56 degrees right I know that that whole angle is 129 degrees without going to my calculator I don&amp;#39;t want to have to deal with all that so I can now figure out this angle which is 129 minus 56 and that&amp;#39;s going to be 73 degrees and I&amp;#39;ve just about gotten it right because now I&amp;#39;ve got that is 73 and this is 28 so now I can do 73 plus 28 and get 101 degrees and now I will just simply do 180 minus 101 and that&amp;#39;s 79 degrees now I know that when there are this many angles that I&amp;#39;m filling in there&amp;#39;s probably three or four different ways to do this problem unquestionably so if you&amp;#39;re like no I got 79 by using the fact that the four angles of any quadrilateral add up to 360. awesome okay there are many many different ways to typically do any math problem and one like this where you&amp;#39;re filling in angles all over the place hey anything that you need to do to get over to the angle that you want right do it use any principle you have to all right I think we&amp;#39;re almost done with the multiple choice questions in fact we&amp;#39;ve only got two more let&amp;#39;s take a look 22. Point P divides the directed line segment from point A at negative four negative one to point B six comma four in the ratio two to three coordinates of Point P are which of the following all right this is a standard problem on almost every geometry regents exam sometimes it&amp;#39;s multiple choice sometimes it&amp;#39;s free response let&amp;#39;s talk about the idea here okay we&amp;#39;ve got a line segment we&amp;#39;ll talk about what it means to be a directed line segment in a second but just for a moment let me put a at negative 4 1 and B at 6 comma four okay now they tell me that I&amp;#39;ve got Point P somewhere on this line and it divides it into a ratio of two to three now what that means right let me just put Point p on here is that if I divided this thing up into five equal chunks y5 because two plus three is five that&amp;#39;s it right that there would be two of them here and there would be three of them here all right that&amp;#39;s what that means right two to three now a directed line segment means it&amp;#39;s a line segment that goes from A to B see most line segments are just they&amp;#39;re just hanging out oh here they are right you know but sometimes we want to think of it almost like motion like I&amp;#39;m starting at Point a and I&amp;#39;m walking to point B and two-fifths of the way along we come to point P all right now that&amp;#39;s really key right p is two-fifths of the way from a to B two fifths not two-thirds not three halves not three-fifths right two-fifths right two plus three five and we are two of those five now key right now we&amp;#39;re gonna think about how far we traveled in the X Direction and how far we traveled in the y direction and I really wish this problem had made one of those two negative it didn&amp;#39;t right just look at the X&amp;#39;s here I&amp;#39;m going from negative four to six right so I&amp;#39;m traveling a change in X of ten units right I&amp;#39;m increasing X by 10 units to go from an x equals negative four to an x equals six now I&amp;#39;m going from a y equals negative one to a y equals four so my change in y is equal to five all right so how do I find Point P that&amp;#39;s really cool let me first find the x coordinate find the x coordinate of Point P what I&amp;#39;m going to do is I&amp;#39;m going to take the x coordinate I start at and I&amp;#39;m starting at an x coordinate of negative 4. and then what I&amp;#39;m going to do is I&amp;#39;m going to add to it two-fifths of my change in X right so I&amp;#39;m going to have negative 4 plus 2 fifths of how much my X has changed now you could totally do that on your calculator but two-fifths of 10 is actually Four so my x coordinate of p is 0. now immediately what that tells you is that the correct choice is 2. but let&amp;#39;s pretend that this is a free response problem right and get the y coordinate as well the y coordinate of p is going to be the y coordinate we start at which is negative 1. plus two-fifths of my change in y and my change in y is 5 units and two-fifths times five is just going to be 2 negative 1 plus 2 is 1. so that&amp;#39;s it right we use the ratio to figure out what fraction of the journey we&amp;#39;ve sort of traveled to get to our point that partitions our line segment and then we use that fraction to add on the fraction of the change that we have in both directions X and Y to our starting coordinates all right absolutely key and obviously those changes can be negative again in this problem both changes were positive which is disappointing to me all right let&amp;#39;s take a look at problem 23 and then maybe we&amp;#39;ll have a little lemonade break 23. a line is dilated by a scale factor of one-third centered at a point on the line which statement is correct about the image of the line Choice One its slope is changed by a factor of one-third two its y-intercept is changed by a factor of one-third three its slope and y-intercept are changed by a factor of one-third and four the image of the line and the pre-image are the same all right now this is also a problem that they like a lot on the Regions exam so let&amp;#39;s take a little bit of time to discuss it all right so let&amp;#39;s say that we have a line all right maybe it&amp;#39;s line a b let&amp;#39;s call it line a b and we&amp;#39;ve got some point C okay and let&amp;#39;s say that we dilate this line and let&amp;#39;s forget about a factor of one-third just forget about it let&amp;#39;s say that we actually stretch this thing to make it bigger we could we could compress it to make it smaller whatever but right the idea is you know we got this dilation and maybe we stretch a out to here and we stretch B out to here and then we run into the other problem and it gets confusing this right we stretch B out to here right and we get our new image line and our new image line is a prime B Prime all right if the center of dilation is not on the line then our image line that is produced is parallel to the original line and therefore should have the same slope okay let me say that again when we dilate a line using a Center Point that&amp;#39;s not on the line the image line that we get should have the same slope as the original line but it&amp;#39;ll definitely have a different y-intercept okay you know if I like oriented this on I don&amp;#39;t know the coordinate plane or something like that right then we we definitely see that the y-intercept of this line and this line are different but the slopes would be the same so this what first one that says its slope is changed by a scale factor of one-third no now what happens if the point is on the line well in that case okay let&amp;#39;s say we&amp;#39;ve got this line a b and our Center Point is on the line well now what happens is when we either stretch or compress based on that Center Line we get exactly the same line and that&amp;#39;s Choice four the image line and the pre-image line are the same line and again it&amp;#39;s kind of irritating because most of the time when you dilate a line the center of dilation is not on the line that you&amp;#39;re dilating so it&amp;#39;s totally different line we&amp;#39;ll have the same slope right but it&amp;#39;ll tend to have a different y-intercept exactly what the y-intercept is that&amp;#39;s harder to say and less the center is at the origin then it&amp;#39;s easy enough then you just take the y-intercept and you multiply it by whatever that number is and that&amp;#39;s your new y-intercept okay but the slope will stay the same if the point is actually on the line that you&amp;#39;re dilating then the the image line and the original line are exactly the same all right let&amp;#39;s take a look that&amp;#39;s it oh no one more multiple choice problem that apparently they just could not fit on that other page then we&amp;#39;ll go into the free response but we&amp;#39;ll take a lemonade break in between problem number 24. in the diagram below of circle o tangent a b is drawn from external point B and secant b c o e and D diameter aod are drawn if measure of angle o b a is equal to 36 degrees and OC is equal to 10 what is the area of shaded sector d o e all right man boy and again this is a problem with multiple things that they&amp;#39;re testing here okay but ultimately ultimately we want to figure out what the area of this shaded sector of the circle is now what you should have learned about areas of shaded sectors is that basically it all boils down to knowing what the area of the whole circle is and knowing what that central angle is and then setting up kind of a proportion to figure out what the final area is and we&amp;#39;ll get there okay but what&amp;#39;s interesting is we don&amp;#39;t know what that central angle is we don&amp;#39;t know what angle e o d is we really need to know angle EOD so what do we know well we know the angle Oba Oba which is this thing sitting right out here this thing is 36 degrees wow okay well what else do we know well because ba is a tangent to a radius tangents and radii will be perpendicular to each other all right so we know that&amp;#39;s 36 we know that&amp;#39;s 90 which allows me to say that angle AOC right the measure of angle AOC is going to be 90 minus 36 and that&amp;#39;s going to be 54 degrees so 54 degrees all right now because that angle and this angle are vertical angles I can now say that is 54 degrees and that took me a lot of work to just get over here now how do we figure out the area of the Shaded sector well what I always tell students first is figure out the total area of the circle now we know that the radius I&amp;#39;m sorry we know that OC is 10. which means that&amp;#39;s 10 and that&amp;#39;s 10 Etc so the area of the total Circle which is pi r squared will be pi times 10 squared or 100 pi okay now that&amp;#39;s the area of everything but we don&amp;#39;t have the whole circle we only have 54 degrees out of 360 degrees of it but we can now set up a little proportion that goes something like this the area of the sector divided by the total area is equal to the central angle of the sector divided by the total angle 360 degrees and now if I want the area of the sector I&amp;#39;m just going to multiply both sides by 100 Pi I&amp;#39;m not going to even cross multiply I don&amp;#39;t need to do that all right and now I&amp;#39;ve got to figure out what all of this is in terms of Pi this is where the calculator can be very very handy okay so I&amp;#39;m just going to take this thing out and I&amp;#39;m not I&amp;#39;m not going to even worry about the uh the pie right now what I&amp;#39;m going to do is I&amp;#39;m going to take my calculator and I&amp;#39;m going to ask it what is 54 divided by 360 what is that fraction times 100 then I&amp;#39;ll bring the pi in for in a second right so 54 over 360 times 100 is 15. so so all of this then is just equal to 15. pi all right we just needed to have that central angle that was it ah excellent okay we&amp;#39;ve made it to the free response problems excellent I&amp;#39;m going to take just a moment and hydrate gotta hydrate got to stay hydrated okay can you believe it&amp;#39;s been over an hour and a half that&amp;#39;s amazing I&amp;#39;m like oh we&amp;#39;re gonna we&amp;#39;re just booking through this we&amp;#39;re gonna get done in like like two hours and I&amp;#39;m gonna take the last hour and just do my stand-up comedy routine um I know you think I&amp;#39;ve been doing that all along but uh math is no joke math is no joke um that wasn&amp;#39;t that wasn&amp;#39;t even funny uh anyway so yeah um we&amp;#39;ve now made it through 24 two point multiple choice problems okay and again they&amp;#39;re absolutely key if you nail them and do really well on the multiple choice problems then you&amp;#39;re guaranteed to pass the exam that&amp;#39;s just all there is to it now granted the curve on the geometry exam is not nearly as nice and forgiving as it is on the algebra one exam it&amp;#39;s pretty nice it&amp;#39;s pretty forgiving but it&amp;#39;s not nearly as nice and forgiving okay so you want to nail the multiple choice 100 right but you also have to get the free response now part two we&amp;#39;ve got seven questions in this part and each one of them&amp;#39;s worth two credits okay so just like the multiple choice problems they&amp;#39;re worth exactly the same we had 24 two pointers that we just went through now we&amp;#39;ve got seven two pointers except there are no choices to help us out so let&amp;#39;s get right into them make sure you&amp;#39;re sure that we show proper work that we pay attention to our rounding and our notation and all of that so that we get two out of two on every one of these problems let&amp;#39;s take a look at our first one problem 25. the Leaning Tower of Pisa in Italy is known for its slant which occurred after its construction began the angle of the slant is 86.03 degrees from the ground the low side of the tower reaches a height of 183.27 feet from the ground so nice of them to give us this beautiful diagram even though it kind of looks like the top portion of the the Leaning Tower of Pisa was photoshopped from a different picture from the bottom version I don&amp;#39;t know you know maybe I&amp;#39;m crazy about that but it just definitely looks like this is different than that anyway like I&amp;#39;m one to talk anyway then it says determine and state the slant height X of the low side of the tower to the nearest hundredth of a foot I love all of that context that they give us uh-oh um they give us all of this context right for what is really essentially just a right triangle trig problem right that&amp;#39;s all it is now granted it might be a little bit of a right triangle trig problem that&amp;#39;s hard to look at but ignore the Leaning Tower of Pisa just look at this thing right we know the angle 86.03 degrees we know the side that is opposite of the angle even though that&amp;#39;s a little bit hard to write and that&amp;#39;s 183.27 feet and we want X which happens to be the hypotenuse right we want the hypotenuse so when we think about our trig ratios for the third but not last time of this test foreshadowing um right the one that we&amp;#39;re going to use the one that has opposite and hypotenuse in it right is the sine ratio and so what we&amp;#39;re going to say is we&amp;#39;re going to say the sine of 86.03 degrees you don&amp;#39;t have to put the degree symbol there but I can&amp;#39;t help myself is 183.27 divided by X now this is one of those ones where the thing that you&amp;#39;re solving for is in the denominator so it&amp;#39;s not as easy as just sort of you know like multiply this side by this side and you&amp;#39;re done kind of thing all right if I cross multiply I&amp;#39;ll get x times the sine of 86.03 is equal to 1 times 100 now I gotta I really don&amp;#39;t want this here is equal to 183.27 and then I&amp;#39;ll divide by the sine of 86.03 and divide by the sine of 86 point 0 3 and X will be 183.27 divided by the sign of 86.03 so let&amp;#39;s take a look at what that is on the calculator again it&amp;#39;s easy enough to do this um let&amp;#39;s just do maybe we&amp;#39;ll even go with the fraction bar 183 0.27 divided by trig sine of 86.03 enter 183.71 notice how close that is right um actually let me put it down here X oh boy that looks like a funny X um x equals 183.7108 Etc really watch your rounding on here right this is a two-point problem if I round this to 183.7 I&amp;#39;ve done all the trig right I&amp;#39;ve done all the geometry right I&amp;#39;m going to lose a point right I&amp;#39;m going to lose a point because I didn&amp;#39;t round to the nearest hundredth and rounded to the nearest tenth and boy how horrible would it be just to be 184. I mean you&amp;#39;d still only lose one point but still right so X equals 183 .71 and by the way you don&amp;#39;t even have to put down foot I would I can&amp;#39;t even help it so I&amp;#39;m going to put down foot but the plain fact is they&amp;#39;ve actually put the units down for you because they don&amp;#39;t want to deduct for you not putting down foot all right but watch your rounding all right let&amp;#39;s take a look at the next one now every single free response problem they&amp;#39;re going to give you at least one sheet of paper all right so you&amp;#39;re probably not going to use almost any of it on these two pointers that&amp;#39;s just the way it is let&amp;#39;s take a look at another Circle problem in 26. in the diagram below quadrilateral ABCD is inscribed in circle o the measure of Arc CD well sorry the ratio of the measure of Arc CV to the measure of Arc D A to the measure of Arc a b to the measure of Arc BC is two to three to five to five all right State sorry determine and state the measure of angle B I love the fact that they say determine and state the measure of angle B like you&amp;#39;re going to determine the measure of angle B but then you&amp;#39;re going to keep it to yourself do you know what I mean like they couldn&amp;#39;t have just said determine the measure of angle B whatever let&amp;#39;s take a look at measure of angle B now measure of angle B right angle B is right here now one thing that you should have learned in your work with inscribed angles angle B is what&amp;#39;s called an inscribed angle is that it enters its measure right the measure of angle B will be one half the measure of the arc that it intercepts it enter intercepts Arc a c all right so I need to know what the measure of Arc AC is like if I can figure that out then I&amp;#39;ll just find one half of that measure and I&amp;#39;ll be good to go so what do we know well we know CD to D A to a b to BC is two to three to five to five all right and there&amp;#39;s a common trick we do with this all right and this is the trick we do we know right that CD plus d a plus a B plus b c is 360. so what I&amp;#39;m going to do is I&amp;#39;m going to call CD 2x I&amp;#39;m going to call ad3x I&amp;#39;m going to call a b 5x and I&amp;#39;m going to call BC also 5x right and then I&amp;#39;m going to do the following I&amp;#39;m going to say 2x plus 3x plus 5x plus 5x is equal to 100 whoop no it&amp;#39;s equal to 360 Degrees all right now I&amp;#39;m going to solve I&amp;#39;m going to have 15x is equal to 360 degrees and I&amp;#39;m going to divide by 15 on both sides and I&amp;#39;m going to find the X is equal to 24 degrees all right now because I know X is 24 degrees I can now say that 3x is 3 times 24 degrees so this is going to be 72 degrees Arc a d this is going to be 2 times 24 degrees so that&amp;#39;s going to be 48 degrees and now I can say that the measure of angle B is going to be one half of 72 plus 48 which is going to be one half of 120 degrees and that is a 60 degree angle and that&amp;#39;s it that&amp;#39;s quite a bit of work for two points but that is what it is right all right let&amp;#39;s keep going here we go 27. in the diagram below right circular cone has a diameter of 10 and a slant height of 13. determine and state the volume of the cone in terms of Pi okay well again if you don&amp;#39;t know the volume formula for a cone this is when you would go back to those formulas and find it okay but let&amp;#39;s just save ourselves the time the volume formula for a cone is one-third pi r squared times h now that&amp;#39;s what it&amp;#39;s going to look like if you look at the formula sheet now all right the key in this problem is knowing what H and R represent most people don&amp;#39;t have any problem with r all right R is the radius of the base circle of the cone so our R is five no problem the real problem here is the height the height is not 13. 13 is what&amp;#39;s known as a slant height and I hate that piece of terminology hate it hate it hate it hate it it&amp;#39;s not a height it&amp;#39;s a length Okay a height is how far you are above the ground okay the 13 is not how far you are above the ground it is a length Okay it is not the H and if you use it as the H you likely lose zero you likely lose both points okay the H is this distance right that&amp;#39;s H all right the height is perpendicular to the base always for everything cone cylinder pyramid I don&amp;#39;t care it is always perpendicular to the base and you don&amp;#39;t know what it is but with the Pythagorean theorem you can find it right and you can just kind of over here you can just do a little H squared plus 5 squared is equal to 13 squared H squared plus 25 is equal to 169 H squared would be 144 if we subtract 25 from both sides take the square root and we have our famous 5 12 13 right triangle and our height is 12. all right now we have absolutely everything we need our volume is going to be 1 3 pi times r squared times h now notice that they want our answer in terms of Pi meaning that there&amp;#39;s got to be a pi in the answer so what I do at this point is I tell students look don&amp;#39;t worry about the pie leave it out for a moment take your calculator and do the following here we go right I&amp;#39;m going to put my my nice one-third in can&amp;#39;t forget the one-third definitely can&amp;#39;t make it a one-half I&amp;#39;ve Got 5 squared times 12. that is the entire formula without the pi okay I hit it right and I get a hundred all right so the correct answer is not there the correct answer then is one whoops 100 and again it&amp;#39;s so weird that they don&amp;#39;t put any units down you know like a 100 Pi what cubic centimeters cubic inches cubic miles I don&amp;#39;t know cubic Kirks I don&amp;#39;t know what the units are here but that&amp;#39;s it now again there&amp;#39;s a few things that are tricky about this problem the trickiest thing of all is the fact that they give us not the height but that slant height right which means we have to use either the Pythagorean theorem or Pythagorean triple to come up with the actual height and then we use that to get our volume all right let&amp;#39;s take a look at problem 28. in the diagram below parallelogram E F G H is mapped onto parallelogram i j k h after a reflection over line L use the properties of rigid motions to explain why parallelogram efgh is congruent to parallelogram i j k h I tell you I don&amp;#39;t love this problem because theoretically you should just be able to say it&amp;#39;s a rigid body motion and therefore they&amp;#39;re congruent done like that is the definition of congruence at this level but they would not accept that they would probably give you 0 out of two so it&amp;#39;s this simple what do rigid bodies motions do what do Reflections translations and rotations do they do two things they preserve distance and they preserve angle they preserve distance and they preserve angle right and that&amp;#39;s all you really have to say so what do I I want to say right um I&amp;#39;m going to say a reflection is a rigid body motion so it preserves or if you don&amp;#39;t like the word preserves which I totally would get you know instead of maybe saying preserves maybe you say so it doesn&amp;#39;t change right it doesn&amp;#39;t change so it preserves both distance and angle and that&amp;#39;s it nothing more all right if you have any short problem especially like a two-point free response problem that&amp;#39;s asking you in some way to justify that two figures are congruent based on a rigid body motion or a sequence of rigid body motions I want you to mention that those things preserve or don&amp;#39;t change distance and angle all right and that&amp;#39;s it moving on okay let&amp;#39;s get into some more messy math problem 29. Izzy is making homemade clay pendants in the shape of a solid hemisphere as modeled below each pendant has a radius of 2.8 centimeters how much clay to the nearest cubic centimeter does Izzy need to make 100 pendants all right now this is key okay we&amp;#39;re obviously going to find out how much Clay is he needs for one pendant and then we&amp;#39;re going to multiply by a hundred and it would be very easy to you know do all the other stuff you know figure out how much Clay is he needs for one pendant and then not multiply by a hundred now if we&amp;#39;re talking about how much we&amp;#39;re talking about volume or mass but in here we&amp;#39;ve got no density right we&amp;#39;re just talking about how much clay to the nearest cubic centimeter right cubic centimeter right that means we&amp;#39;re talking about volume all right now it&amp;#39;s a hemisphere which is half of a sphere and if we go back to our formula sheet what we&amp;#39;ll see is that the volume of any sphere is four thirds pi R cubed four-thirds pi R cubed now that&amp;#39;s any sphere right the globe your soccer ball a basketball Etc none of those are actually true spheres although the basketball is probably the closest one all right but you understand what I&amp;#39;m I&amp;#39;m getting at right so the volume of a pendant foreign half times four thirds times pi times 2.8 cubed cubed cubed cubed not squared I have taught geometry countless numbers of years and the number one error people will make in these kind of situations is that they&amp;#39;ll do squaring because they&amp;#39;re so used to pi r squared and I get it that&amp;#39;s the area of a circle we&amp;#39;ve already seen pi r squared like multiple times right here it&amp;#39;s four thirds pi R cubed let&amp;#39;s see what that number looks like I think you&amp;#39;ll like that number actually all right let&amp;#39;s do it we&amp;#39;ve got um one half times four thirds whoops I didn&amp;#39;t quite get the four but I&amp;#39;ll get it now uh times and now I&amp;#39;m actually going to bring the pi in for the first time times 2.8 cubed 2.8 to the third I don&amp;#39;t really need the parentheses there but that&amp;#39;s okay sorry about the little cursor sitting there that&amp;#39;s a little bit better 45 .97616 this isn&amp;#39;t the one I thought it was we&amp;#39;re going to eventually see one where the answer is literally Pi all right now that is the volume of a single pendant all right and let&amp;#39;s let&amp;#39;s write that down really quick now I&amp;#39;m not going to round this all right I&amp;#39;m definitely not going to round it to the nearest integer that&amp;#39;s 45.97616 Etc big big decimal point there that&amp;#39;s one pennant and of course it would be so easy to get that answer and then just move on right so easy all right but what we want is the total which is going to be 100 times maybe I&amp;#39;ll just do V sub p and that&amp;#39;s easy enough right we&amp;#39;ll just take this and multiply it by 100. and we will get four thousand 598 I is there a way for me to there well that that&amp;#39;s not exactly what I wanted to do but and actually it&amp;#39;s kind of cool I didn&amp;#39;t know I could do that anyway I just wanted to get the cursor out of the way this will be 4 598 cubic centimeters now think about how easy by the way it would be to lose two points in this problem first way you could do it not putting the one half there right that would lose you one point putting squared here instead of cubed that would lose you the other point or maybe you remember the one-half but do squared there and forget to multiply by 100 right that would lose you two points there&amp;#39;s a lot of ways that you can kind of know what you&amp;#39;re doing on this problem and go over two so again be very careful make sure if it&amp;#39;s a half a sphere you involve that one half if there&amp;#39;s more than one of them make sure you multiply at the end all right words to live by all right let&amp;#39;s take a look at another one number 30. determine and state the coordinates again determine and state determine and state the coordinates of the center and the length of the radius of the circle whose equation is x squared plus y squared plus six x equals six y plus 63. all right well what you should have learned in this course was that if you have something that looks like this it&amp;#39;s supposed to be on minus if you&amp;#39;ve got something that looks like that then you&amp;#39;ve got the equation of a circle and you can just get it Center by looking at the two things that are subtracted from the X and the Y and you can figure out its radius by taking the square root of whatever&amp;#39;s sitting over there all right unfortunately this thing doesn&amp;#39;t look at all like this but it&amp;#39;s definitely a circle mostly because they told us it was a circle but also because we have an x squared term and a y squared term so how are we going to make it look like this well we are going to do what&amp;#39;s called completing the square and we&amp;#39;re going to do it twice now the first thing I&amp;#39;m going to do though is I&amp;#39;m going to move things around a little bit number one I&amp;#39;m going to put this 6X next to the x squared I&amp;#39;m also going to take that 6y which is on this side of the equation and bring it over to this side of the equation but make it at negative 6y so I&amp;#39;m going to start by rearranging some things a little bit and having this I&amp;#39;m going to leave that 63 exactly where it is okay now what I&amp;#39;m going to do is I&amp;#39;m going to complete the square on my X terms and on my y terms now remember how completing the square Works we&amp;#39;re going to take half of this number we&amp;#39;re going to square it and we&amp;#39;re going to add it to this binomial and to this side and then we&amp;#39;ll do the same here it&amp;#39;s a little bit annoying because in both cases right when we take half of 6 we get 3 and when we Square it we get 9. so we&amp;#39;re going to get x squared plus 6X Plus 9. plus y squared minus 6y plus 9 equals 63 plus 9 plus 9. all right so we&amp;#39;re doing completing the square twice now when you do completing the square correctly what happens is you end up with trinomials that are perfect squares in other words they are a binomial quantity squared specifically this one is X plus three squared this one specifically is y minus 3 squared and when we add all these together we get 81. I guess by now I have determined what the center and the radius are but I haven&amp;#39;t stated them I don&amp;#39;t know maybe maybe not anyway so now when I kind of compare these two forms what I can say is that my Center is that negative three comma 3 if you will you sort of look in the parentheses and you take the opposite of what you see so if it&amp;#39;s X plus 3 The Centers at negative 3 and Y minus 3 the center is at positive three so there&amp;#39;s our Center and our radius is going to be the square root of 81. which is just equal to 9. so I have determined and stated the center and the radius awesome Okay so we&amp;#39;ve kind of come to the first one where I&amp;#39;m going to break out of this program and look at and go into another program where I have the ability to use a compass and a straight edge so real quick before I do that I&amp;#39;m just going to pop in here change my line width go into my full screen mode Let&amp;#39;s Take a look at problem 31. use a compass and straight edge to construct a line parallel to a b through Point C shown below leave all construction marks all right you know constructions are a tricky business because there&amp;#39;s a ton of them that you need to know and I&amp;#39;m going to go over just this one and it&amp;#39;s unfortunate because the odds are this is not the one you&amp;#39;re going to see tomorrow and they are not tomorrow sorry the odds are this is not the one that you&amp;#39;re going to see on Tuesday but you will see one on Tuesday still it&amp;#39;s what it is all right so let&amp;#39;s learn how to create a line that is parallel to another line that goes through a particular point so here&amp;#39;s how I&amp;#39;m going to do it the first thing I&amp;#39;m going to do is I&amp;#39;m going to take a straight edge and I&amp;#39;m not going to rotate it I&amp;#39;m just going to bring it down like this and I am going to draw a line that passes through the point that I&amp;#39;d like to draw the parallel line through and passes through the line that I am trying to make it parallel to all right so just then it could be anything it could be it could be like that it could be straight up and down whatever now to give you a little preview right I&amp;#39;m going to try to copy this angle right here up here I&amp;#39;m going to just try to copy that angle all right we&amp;#39;ll explain why this works in a little bit in order to copy that angle I&amp;#39;m going to bring my compass and I&amp;#39;m going to put the sharp end down at that intersection and I&amp;#39;m going to draw an arc okay the arc does not have to go through Point C and in fact I wouldn&amp;#39;t particularly make it go through Point C all right now without changing the radius on the arc I&amp;#39;m going to move my compass up to point C and I am going to draw an arc with the same radius now it won&amp;#39;t be exactly the same Arc but it should have the same radius okay so I&amp;#39;ve got you know this angle with this Arc drawn in and an arc centered at Point C with the same radius now right I kind of want to think about it as this point is equivalent to point C and this point is equivalent to the point up here what I really need to do is find the equivalent of this point along this Arc the way I&amp;#39;m going to do that is I&amp;#39;m going to bring my compass over here I&amp;#39;m going to put it down on that point and I&amp;#39;m going to then adjust the compass to a new length just so it passes through that point now you should be able to be a little bit more accurate than I am I can only be as accurate as the computer allows me to be but then oh that&amp;#39;s actually not too bad I draw that Arc now again without changing the radius of my compass which I realize is a challenge right especially if you&amp;#39;ve got bad school issued compasses and I know that most of you do right I now draw the same Arc but centered at that other point and then I get rid of my compass it&amp;#39;s done its job okay I have literally just repeated up here exactly what I did down here now I&amp;#39;m going to take my straight edge and I am going to draw a line that passes through C and through the intersection of those two arcs and that is my parallel line and again it&amp;#39;s you know it&amp;#39;s unfortunate because there are just so many different you know you could be bisecting an angle a couple days from now you could be uh drawing a perpendicular from a point down to a line there&amp;#39;s a ton that you could do we could spend three hours just going over well maybe not three hours we could spend a good hour going over all the constructions that you should know for the exam even though they&amp;#39;ll probably be only one of them on there in fact I&amp;#39;ve actually never seen an exam with two so there&amp;#39;s that all right let&amp;#39;s head back into this program and we are up to part three all right let&amp;#39;s take just a little break we are just over the two hour mark at this point and now we&amp;#39;re headed into some of the longer problems but here&amp;#39;s the good news we only have four problems left four problems what do you mean four problems we saw part three and part four left well part three is three questions each one of them is worth Four Points so they&amp;#39;re beefy questions you know what I mean and oftentimes unlike algebra one oftentimes in Geometry when there is a four-point question it&amp;#39;s not like part A is two points and Part B is two points it&amp;#39;s just a four pointer and we&amp;#39;re gonna see that right away in the first problem we do once we get to our three-part uh three questions worth Four Points each then we&amp;#39;ll get to that one part four question that&amp;#39;s worth six points but let&amp;#39;s jump right into question number 32. as modeled below a projector mounted on a ceiling is 3.74 meters from a wall where a white board is displayed the vertical distance from the ceiling to the top of the white board is 0.41 meters and the height of the Whiteboard is 1.17 meters determine and state the projection angle Theta I got to keep coming back to that to the nearest tenth of a degree all right cool now this kind of a very open problem for four points right and we&amp;#39;re trying to find this particular angle okay now whenever you&amp;#39;re looking for an angle in geometry it&amp;#39;s either going to be a fill in the angle kind of game right like that one huge multiple choice question where we were like 61 61 39 58 etc etc right it&amp;#39;s either one of those or it&amp;#39;s right triangle trig now it&amp;#39;s interesting so far you know we&amp;#39;ve seen right triangle trig twice okay once was like a very theoretical part one problem then we had that Leaning Tower oh actually we&amp;#39;ve seen it three times right we saw the the one with the tree on the hillside that&amp;#39;s where we used it to find a length we saw the Leaning Tower of Pisa we used it to find a length there and then we saw that that early on multiple choice problem this is one where we&amp;#39;re going to use right triangle trig to find an angle now one of the things that&amp;#39;s very tricky tricky is that angle Theta is not part of a right triangle and right now the only kind of trig you know how to do takes place within right triangles so here&amp;#39;s the key there are two right triangles in this problem whoops let&amp;#39;s make that a bit bigger to make it much bigger right we&amp;#39;ve got this right triangle the smaller one and let me kind of draw that right here all right we got this right triangle and this is 3.74 comma on decimal and 0.41 and then even though this is probably not going to help very much oh maybe I&amp;#39;ll go highlighter um yeah I&amp;#39;ll go highlighter and I&amp;#39;ll go purple and then we&amp;#39;ve got this right triangle and now I&amp;#39;m going to go back to pen because why wouldn&amp;#39;t I and let&amp;#39;s draw that one a little bit bigger because it is bigger and that is 3.74 and it&amp;#39;s the sum of those two one point five eight all right so let&amp;#39;s talk about this a little bit right so we&amp;#39;ve got the smaller right triangle 3.74 by 0.41 and then the larger one 3.74 by 1.58 and again that 1.58 just came from adding those two links does this relate to Theta well if I call this let&amp;#39;s say angle X and call this say angle y right this would be X and this would be y then what I should be able to see is that Theta is going to be angle y minus angle X angle y minus angle X so I just have to find these two angles and I have to go back to sohcahtoa right so let&amp;#39;s first take a look at the small triangle right here angle X if I look at the 0.41 that&amp;#39;s opposite of X and the 3.74 that&amp;#39;s adjacent to angle X and therefore if I&amp;#39;m looking for opposite and adjacent we&amp;#39;re talking about tangent now specifically what I can now write is that the tangent of angle X is equal to the side opposite of X 0.41 divided by the side adjacent to X 3.74 now whatever you do don&amp;#39;t lose a point and put equals some angle okay the equal sign does not mean give me the answer never has and it never will equals means that the two numbers are the same and if you tell me that 0.41 divided by 3.74 is equal to 22 degrees well then I&amp;#39;m going to wonder if you really understand what fractions mean and I&amp;#39;m going to deduct a point if I&amp;#39;m the greater all right this is not equal to the angle what is equal to the angle is What&amp;#39;s called the inverse tangent of that ratio 0.41 divided by 3.74 why don&amp;#39;t we figure that out all right we want the inverse tangent let me get to my notes Here just so I can I have them in front of me bring this whoops nope that&amp;#39;s not what I want bring this back up bring this back up here we go I just want to get out of here so I&amp;#39;m going to go into trig and I&amp;#39;m going to go to What&amp;#39;s called inverse tangent this is literally telling me the angle that has a tangent of 0.41 divided by 3.74 all right that&amp;#39;s all I want to know right and that&amp;#39;s an angle small angle 6.26 let&amp;#39;s call it 6.26 now notice I&amp;#39;m looking for this to the nearest tenth of a degree I&amp;#39;m going to take that angle out one decimal place for farther now what&amp;#39;s really great is this bigger triangle it&amp;#39;s the same ratio well it&amp;#39;s not the same ratio but it&amp;#39;s the same trig function right we&amp;#39;ve got opposite we&amp;#39;ve got adjacent so again it&amp;#39;s going to be the tangent of angle Y is equal to 1.58 divided by 3.74 so Y is equal to the inverse tangent don&amp;#39;t use x for both of these of that particular fraction and Y I&amp;#39;m going to just tell you what this is because we&amp;#39;ve already done that on the calculator 22.90 degrees to the nearest hundredth and now Theta right Theta is just going to be y minus X and when I do that subtraction I will get 16 .6 degrees and that one&amp;#39;s kind of tricky right I mean if they had asked you just for you know a single angle right and all you had to do was sort of set up a single ratio do the inverse trig whatever it is inverse tangent inverse sine inverse cosine that&amp;#39;s simple enough here it&amp;#39;s way more challenging because you actually have to think well I&amp;#39;ve got to calculate two angles subtract and get the final answer all right let&amp;#39;s keep going oh I&amp;#39;m very excited about the next problem oh and this one I have to go back over to my other display uh let&amp;#39;s do it let me get uh get the right line Style uh why am I excited because it&amp;#39;s the proof or at least it&amp;#39;s the euclidean geometry proof we&amp;#39;ll see another proof in a moment let&amp;#39;s take a look given parallelogram p q r s QT perpendicular to PS Su perpendicular to QR prove that PT segment PT is congruent to segment r u all right now the first thing I always want to do whenever I&amp;#39;m doing a geometry proof whether it&amp;#39;s worth four points or six points and it&amp;#39;s going to be one of those two is I want to really sort of work my way backwards so let&amp;#39;s let&amp;#39;s first highlight the two things that we want to prove congruent and boy if you&amp;#39;ve got a highlighter it&amp;#39;s not the worst thing to be using on here right so I wanna I wanna like somehow prove that PT and r u are congruent to each other okay so so far so good now almost always not always but almost always when I&amp;#39;m looking to prove that two segments or two angles are congruent to each other I&amp;#39;m going to be using what&amp;#39;s known as a cpctc proof corresponding parts of congruent triangles are congruent and what I&amp;#39;d really love to do is I&amp;#39;d love to prove that this triangle here RSU is congruent to this triangle here p q t I&amp;#39;d love to prove that those two are congruent all right because if I can then it&amp;#39;s going to be very easy to say that PT is congruent to Ru so let&amp;#39;s see what we know well we know we&amp;#39;ve got a parallelogram right parallelogram pqrs and that&amp;#39;s pretty awesome because with parallelograms we know that opposite sides are congruent so I know that PQ is congruent to Sr or RS and I also know that opposite angles are congruent so I know that angle R is congruent to angle p this is good because I&amp;#39;m almost there right then and that&amp;#39;s just all from that first piece of information parallelogram pqrs I already have like a ton of information that&amp;#39;s going to allow me to prove that those two triangles are congruent now I know that QT is perpendicular to PS which tells me that that&amp;#39;s a right angle okay and I know that s u is perpendicular to QR that tells me that that is a right angle this is awesome because now I can prove that triangle r u s is congruent to Triangle p t q by angle angle side right and once I have those two congruent by angle angle side I can then state that PT is congruent to Ru by cpctc let&amp;#39;s do it all right here we go classic now some of you may have done these things before doing like like flow diagram proofs or paragraph proofs all of that is fine the majority of people do it with t table proofs so that&amp;#39;s what I&amp;#39;m going with all right so first things first is I&amp;#39;m just going to write down that first given um parallelogram p q r s and you might say why aren&amp;#39;t you writing down all the Givens right away well you can absolutely I just like to use them when I need them so parallelogram pqrs that&amp;#39;s given now what did I know from that well I knew from that that RS is congruent to um PQ and why do I know that because opposite sides of a parallelogram are congruent now this is very dangerous you do not want to write down definition of a parallelogram a parallelogram does is not defined as having opposite sides that are congruent it&amp;#39;s defined as having opposite sides that are parallel all right but anyway I got that now let&amp;#39;s get those two angles um let me say angle r is congruent to angle p and it&amp;#39;s going to be very similar opposite angles of a parallelogram [Music] R congruent sweet all right now I really need to get those two right angles congruent but the first thing I need to do is get those two right angles now those angles aren&amp;#39;t as convenient as angle R and angle P right and I can certainly call them by their three letter names I could call them angle Rus and angle ptq or qtp and Sur Etc like that but this is actually a great example of a case where I could also put like a little one down here and a little two down here oops and apparently Circle it in red now before I do either one of those I got to get the given in there so I&amp;#39;m going to say QT is perpendicular to PS and Su is predict perpendicular to QR and that&amp;#39;s a given all right now I&amp;#39;m going to say that angle one and angle two are right angles and as my reason I&amp;#39;m going to say perpendicular lines form right angles now I can say that the two angles are congruent angle one is congruent to angle two and my reason is that all right angles are congruent whoo wow and now I have enough to state that the two triangles are congruent to each other line seven now make sure to get your vertices correct here triangle r u s is congruent triangle our us is congruent to Triangle ptq and that is by the angle angle side theorem and we&amp;#39;re there line eight extend that a little bit farther let me do this PT and r u p t is congruent to R U by CP CTC you probably spent a third of your school year so that you could do this one problem right all of that proof that you did for this eight lines and that&amp;#39;s what you should expect in a four point proof is around eight lines if this was a six pointer then we might see a 10 line proof 11 12 line proof that&amp;#39;s it but thankfully that&amp;#39;s not the problem that we were facing all right let&amp;#39;s take a look at the next one here we go this is another messy one 34. a concrete footing is a cylinder that&amp;#39;s placed in the ground to support a building structure the cylinder is four feet tall and 12 inches in diameter a contractor is installing 10 footings if a bag of concrete mix makes two-thirds of a cubic foot of concrete determine and state the minimum number of bags of concrete mix needed to make all 10 footings this is at that different of a problem than that pendant one in fact one could argue that it&amp;#39;s actually like they could have had this one as a two-pointer as well all right look we&amp;#39;ve got this is all about volume right we know bags of concrete have two-thirds of a cubic foot of concrete they would never ever say that on a bag two-thirds of cubic foot right they&amp;#39;d say 0.67 cubic feet or something like that anyway not the point it&amp;#39;s just weird that they did that so what we need to do is we need to figure out the volume all right and let&amp;#39;s first begin by finding the volume of a single cylinder now the volume of a single cylinder oops make that a little bit a little bit thicker the volume of a single cylinder is going to be pi r squared times H and they did throw in a little bit of a loop here right so the height is four feet and the diameter is 12 inches all right and we can&amp;#39;t mix units can&amp;#39;t have feet and inches especially not when we&amp;#39;re talking about the bags having a volume of two-thirds of a cubic foot everything&amp;#39;s got to be in terms of feet right now everyone should know at this point that 12 inches is one foot all right so at least they did that at least they didn&amp;#39;t say 14 inches or 18 inches or something like that but right keep in mind that still means that the radius R is 0.5 feet and man this is critical right if you use six as the radius you&amp;#39;re going to get this problem completely wrong now you won&amp;#39;t lose all the points on it but come on you know what I mean you just you just can&amp;#39;t you can&amp;#39;t do that right make it all the same unit and the unit you really want it to all be in is feet anyway let&amp;#39;s figure out now the volume of just one of the single ones of these pilings okay pi times the radius squared times H all right let&amp;#39;s do it let&amp;#39;s see what pi times 0.5 squared times H is here we go my Pi there it is times 0.5 squared got to make sure to square it times 4 enter and as many people will recognize this is actually just equal to PI right like and again we could figure out why that was but it&amp;#39;s not worth our time and we&amp;#39;re actually getting a little bit short on time we&amp;#39;ve now been going for two hours and 25 minutes and I&amp;#39;m not going past three hours because that&amp;#39;s all the time we get for this exam anyway so what I know now is that a single one of these is 3.1415 ETC now of course we need to make 10 footings right so I need 10 times V which is going to be 31.415 Etc cubic feet all right and of course on the calculator we can always do that by just multiplying by 10. so 34 31.145926 Etc now this is another tricky part we&amp;#39;re actually trying to find the minimum number of concrete bags that we need to buy and we were told that each one holds 2 2 3 of a cubic foot now if they said each one of them held 1.5 cubic feet I bet you&amp;#39;d take this number and divide by 1.5 if they said certainly if they said each one of them contained five cubic feet by the way that would be a heck of a lot of concrete right if they said each one of them contained you know one point you know like two cubic feet you divide all this by two we need to take this answer and divide by two-thirds and that&amp;#39;s really pretty easy on most modern calculators we can literally say divide by two-thirds right and that&amp;#39;s how many bags we need right we need 47.123 Etc now let me just do this really quick all right 10v divided by two thirds let me just skip over here really quick is 47.123 ETC now last part of this problem that&amp;#39;s a bit tricky right what is the minimum number of bags of concrete needed to make all 10 footings it would be very tempting right now to say 47 bags because definitely if they just said round to the nearest bag or round to the nearest whole number then definitely the answer is 47. the problem is 47 bags wouldn&amp;#39;t be enough right it wouldn&amp;#39;t be enough because we need 47.123 bags which means we need 48 bags of concrete all right and that&amp;#39;s that and that was three part four questions which means with just a little over a half an hour left we are on our last question so let&amp;#39;s just take a moment all right a little bit more lemonade and then we&amp;#39;ll do our last problem [Applause] okay so part four questions on the geometry exam one part for a question it&amp;#39;s worth six points sometimes unfortunately it can be a euclidean geometry proof all right but we already did that one sometimes it can be a really involved volume density problem but we actually had a volume density problem that was actually a multiple choice problem all right and it&amp;#39;s not one of those what it is is what&amp;#39;s called a coordinate geometry proof problem and just like some of the rest of these things I am going to go back over to this presentation software and do it here let&amp;#39;s take a look at our last problem the coordinates of the vertices of triangle a b c are a negative two comma four B negative seven comma negative 1 and C negative three comma negative 3. prove that triangle a b c is isosceles parentheses the use of the set of axes on the next page is optional all right so I you know I kept you know repeating this set of axes so I wasn&amp;#39;t like flipping back and forth and stuff like that your set is on the next sheet and that&amp;#39;s fine anyway so we don&amp;#39;t actually have to plot this triangle but I would always suggest doing so okay because if you do it it will give you a sense of what&amp;#39;s going on now for me I&amp;#39;m just going to make it sort of magically appear okay so there&amp;#39;s my triangle ABC and it asks me to prove that ABC is isosceles so the first thing that you have to know is what an isosceles triangle is and if you don&amp;#39;t you&amp;#39;re going to be sunk in the water all right that&amp;#39;s just the plain fact if they say prove it&amp;#39;s a parallelogram and you don&amp;#39;t know what a parallelogram is it&amp;#39;s going to be game over same thing with a rectangle a rhombus a trapezoid a square anything it&amp;#39;s just going to be a problem all right but what we should know because we&amp;#39;ve discussed it repeatedly tonight is that an isosceles triangle is any triangle with at least two sides that are equal in length now one of the great things about plotting the thing is it makes it fairly easy to see that the two equal lengths if they are equal are side a b and AC and if I want to prove that they have the same length I&amp;#39;m going to use the distance formula all right so I want to find the length of a b and I want to find the length of AC now remember the distance formula basically says to find the distance between any two coordinate points we do we subtract our x coordinates Square subtract their Y coordinates and square add those two together and take the square root now all this is is the Pythagorean theorem okay but in the coordinate plane so I want to find the length of a b and I want to find the length of AC so a b equals all right so if I look at A and B right I&amp;#39;m going to have negative 7 minus negative 2 which will be negative seven plus two plus 4 minus negative 1 which will be four plus one all right so a b will be the square root of negative 5. squared plus 5 squared that will be the square root of 25 plus 25 and that will be the square root of 50. now the square root of 50 most certainly simplifies it simplifies as 5 times the square root of 2. but there&amp;#39;s absolutely no need for you to simplify that we just want to prove that the length of a b and the length of AC are the same and it doesn&amp;#39;t matter what form we have it in all right I would try to avoid putting things down like equals and then putting down some kind of rounded decimal that could be problematic because you&amp;#39;re going to be stating that the length of a b is something rounded and that&amp;#39;s a little bit questionable anyway let&amp;#39;s take a look at AC all right now AC right that&amp;#39;s a little bit harder but I&amp;#39;m going to like I&amp;#39;ll do this kind of underline those two that will be the square root now of negative 2 minus negative three lots of negatives in this problem so squared and then 4 minus negative 3 which will be 4 plus 3 squared so AC will be the square root of 1 squared plus 7 squared which will be the square root of 1 plus 49 which will be the square root of 50. all right now I need to have a little conclusion here I can now say therefore three dots I love using the three dots for therefore triangle ABC is isosceles because it has two sides that are the same length there we go now you might say but what about BC don&amp;#39;t you have to calculate the length of that and the answer is absolutely not an isosceles triangle is a triangle with at least two sides that are the same length I have shown that two sides are the same length and therefore this triangle is isosceles done all right let&amp;#39;s take a look at the second part of this problem which is not a coordinate geometry proof and should be very easy for you let&amp;#39;s take a look State the coordinates of triangle a prime B Prime C Prime the image of triangle ABC after a translation five units to the right and five units down all right I think they threw this one in there to kind of make it a six point problem we just want to take these three points right and we want to move them five units to the right and five units down now keep in mind that the way translations work is if you&amp;#39;re moving to the right that means you&amp;#39;re going to add 5 to each x coordinate and if you&amp;#39;re moving down by five you&amp;#39;re going to subtract 5 from each y coordinate so if you will that is our algebraic rule for this particular translation so this should be really easy a at negative two comma four will get mapped to a prime negative 2 plus 5 is 3. and four minus 5 is negative one so I&amp;#39;m just applying this rule to each one of those coordinate points B which is at negative seven negative one will get mapped to B Prime so now I want to add 5 to the x coordinate that&amp;#39;ll be negative 2. I&amp;#39;m going to subtract 5 from the y coordinate and that&amp;#39;ll be negative 6. and finally C which is at negative 3 comma negative 3. we&amp;#39;ll go to C Prime and again adding 5 to the x coordinate we&amp;#39;ll go to 2 and subtracting 5 from the y coordinate will go to negative 8. and that&amp;#39;s it now don&amp;#39;t get me wrong in a six point problem my bet is that this is worth two points this is worth one point and what&amp;#39;s still yet to come is worth the other three points all right now what&amp;#39;s to come is on the last sheet of the exam but I&amp;#39;ve just kind of put it all on here so the final part of this problem and the final part of this test prove that quadrilateral a a Prime C Prime C is a rhombus the use of the set of axes below is optional now one of the things that&amp;#39;s a little bit unfortunate is that they only give you one grid for the whole thing right and so that can be a little bit tricky you do have extra graph paper on the test so you could go to that all right but again really really worthwhile to take a look at it now if I plot points a a Prime C Prime and C I&amp;#39;m going to get something that looks like this okay now normally have we not done the first part of this problem normally I would say okay everybody I want you to think about what a rhombus is and what a rhombus is is it&amp;#39;s any quadrilateral that has four sides that are the same length four sides the same length so then I would say hey everybody let&amp;#39;s calculate the length of a a prime a Prime C Prime C Prime C and AC and let&amp;#39;s just show that they&amp;#39;re all equal and that is a great way of doing this there is nothing wrong with doing that again using the distance formula to calculate 1 2 3 4 distances actually you&amp;#39;d only have to do three because you&amp;#39;ve already figured out the length of AC earlier in this problem but then we would just be doing a lot more distance formulas so I&amp;#39;d like to look with the limited time that we have left and a different way of doing this problem but again if your initial reaction is hey man I&amp;#39;m just going to do the distance formula three times yes yes yes yes yes okay but again let&amp;#39;s look at a different way all right there&amp;#39;s almost always more than one way to prove these particular figures so let&amp;#39;s say I drew in the diagonals of this rhombus now a rhombus is a parallelogram okay and the diagonals of all parallelograms have this cool property is in that they bisect each other meaning that they have the same midpoint so one way of proving that a four-sided figure is a parallelogram forget about a rhombus one way of proving it&amp;#39;s a parallelogram is showing that the two diagonals a c Prime and a Prime C have the same midpoint let me do that really quick so I&amp;#39;m going to take a look at a c Prime all right I&amp;#39;m going to find their midpoint okay and the midpoint remember we looked at this earlier is the average of the two x&amp;#39;s and the average of the two y&amp;#39;s all right negative two plus two is zero and zero divided by two is zero four plus negative 8 is negative 4 and negative 4 divided by 2 is negative 2. so the midpoint of a c Prime is 0 negative two which you can kind of see here on the graph right that&amp;#39;s zero negative two and now let&amp;#39;s take a look at the midpoint of a Prime C foreign C that&amp;#39;s going to be 3 plus negative 3 divided by 2 and negative 1 plus negative 3 divided by 2. 3 plus negative three is zero zero divided by two is zero and negative 1 plus negative three is negative four negative four divided by two is negative two all right so what has this done for us you know finding the two midpoints of the two diagonals and seeing they&amp;#39;re the same well what that tells me is that a a Prime C Prime C is a parallelogram because it&amp;#39;s diagonals bisect each other or have the same midpoint either way yeah you&amp;#39;re probably saying to yourself but wait wait a second I need to prove it&amp;#39;s a rhombus well the way you can prove something is a rhombus is by first proving it&amp;#39;s a parallelogram and then showing that its diagonals are perpendicular to each other all right so all parallelograms have diagonals that bisect each other but we needed to do this just to prove that it was in fact a parallelogram now let&amp;#39;s prove that the two diagonals are perpendicular to each other and we&amp;#39;re going to do that using slopes so now I&amp;#39;m just kind of pan down a little bit here so now I&amp;#39;m going to go a c Prime and I&amp;#39;m going to calculate its slope all right so I&amp;#39;m going to do negative 8 minus 4 Change in y all divided by 2 minus negative 2 change in X that&amp;#39;s going to be negative 12 divided by 4. and that&amp;#39;s going to be a slope of negative 3 . now let me figure out the slope of a Prime C I&amp;#39;m trying to do a colon there let&amp;#39;s see m is negative 3 minus 1 change in X or sorry change in y divided by the change in x negative 3 minus 3 and that&amp;#39;s going to be negative four nope that&amp;#39;s something I did something wrong there um hold on negative three minus oh there it is that&amp;#39;s negative 3 plus 1. that&amp;#39;s not working negative 2 over negative 6 and that we&amp;#39;ll simplify to one-third right so I&amp;#39;ve got one slope that&amp;#39;s negative three one slope that&amp;#39;s positive one-third so I can now say that a Prime C A C Prime is perpendicular to ah come on this is making it very hard to finish this exam a c Prime is perpendicular to a Prime C because slopes are negative reciprocals and finally I can say therefore oh here we go a a Prime C Prime C is a rhombus let me see if I can spell it right a rhombus because it is a parallelogram with with perpendicular diagonals all right now again don&amp;#39;t get me wrong if your initial sort of urge here is to do four distance calculations or three distance calculations because you&amp;#39;ve already done the one for AC you&amp;#39;d find that all four lengths were the square root of 50 just like in the first part of the problem all right and you&amp;#39;d conclude it was a rhombus because it was a quadrilateral with four sides that were the same length or four congruent sides okay and that is fine it&amp;#39;s good there&amp;#39;s nothing wrong with it awesome 100 good job all right I just wanted to do these two because they could easily come up in a problem a couple days from now all right in other words if you&amp;#39;re asked to prove that a four-sided figure is a parallelogram one great way to do that is to find the midpoint of both diagonals and if the midpoint is the same You can conclude it&amp;#39;s a DOT it&amp;#39;s a parallelogram because the diagonals bisect each other then if you already know it&amp;#39;s a parallelogram this part right whoops almost fell then you can calculate the slopes of the two diagonals and if those slopes are negative reciprocals then you know the two diagonals are perpendicular to one another and you can say it&amp;#39;s a rhombus because it&amp;#39;s a parallelogram with perpendicular diagonals all right let&amp;#39;s get out of here and go back to the test itself because that is the last problem right that&amp;#39;s it let me go all the way back up to the beginning whoo that took us two hours and 45 minutes to go through now you probably aren&amp;#39;t going to take that long on the exam because you&amp;#39;re not going to be explaining it to somebody as you work through it then again you will have also not worked through it two or three different times like I had before even coming in here tonight all right so you know it&amp;#39;s taking it right the plain fact is if you&amp;#39;ve really been present in your geometry class all throughout the year right there&amp;#39;s a lot of stuff you should have learned right you know and It&amp;#39;s tricky because unlike algebra in Geometry there is a lot of memorization granted it&amp;#39;s it&amp;#39;s kind of tactile right it&amp;#39;s Figures it&amp;#39;s parallelism it&amp;#39;s perpendicularity but there&amp;#39;s a lot of stuff you have to know you have to know the definitions of a parallelogram a rectangle a rhombus a square a trapezoid an isosceles triangle a right triangle you need to know the three trig ratios apparently right especially on this exam they came up I think four times which seems like quite a bit to me you need to know the Pythagorean theorem and when to apply it you have to feel comfortable with volume formulas for pyramids cones cylinders spheres hemispheres right there&amp;#39;s a lot of content but you&amp;#39;ve got three hours on this exam right it won&amp;#39;t be nearly as calculated as intensive as either the algebra 1 exam or eventually the algebra 2 exam will be for you but it will be a lot of thinking about the problem understanding what information they give you and what it tells you and what it doesn&amp;#39;t tell you right think about those problems where they they tried to get you to say that something was true even when they didn&amp;#39;t give you the information to really know that it was true all right so read the questions carefully take your time on the exam and you are going to do great on it okay and you&amp;#39;re going to be all set up to take Algebra 2 next year all right well I think I&amp;#39;ve talked your ear off tonight so I&amp;#39;m going to just say good night and good luck a couple days from now my name is Kirk Weiler this has been the emath instruction live geometry regents review good luck in a couple days keep thinking and keep solving problems

Transcript for:Geometry Regents Review (June 2, 2022) - Live on YouTube

Transcript for:
Geometry Regents Review (June 2, 2022) - Live on YouTube