Transcript for:
Geometry Exam Tips and Standards

should I do the intro again sure [Music] hello everyone and welcome to the e-math instruction geometry regions review my name is Kirk Wiler and tonight I'll be going through the June 21st 2024 geometry regions exam i highly suggest that you have a copy of this exam in front of you because it's going to make it much easier to follow along as we go through this now I know that we're starting a little bit later than we thought i apologize hopefully you clicked on the Take Two uh um streaming channel we had a little bit of technical difficulty to begin uh but now we're going to kind of get up and get running i'd be a little bit remiss if I didn't mention the fact that this is a brand new geometry regions exam that's going to be given on Wednesday of this week and it's brand new because this year the next generation learning math standards for New York State are going into effect now those are very similar to the older common core state standards that New York State has used for the last decade or so but there are some key differences so after going through this exam I've got a few additional supplementary problems to take a look at some of the new additions in these standards another big difference between the nextG test which you're going to take on Wednesday and the common core exams that were given in the past is the formula sheet that they give you in the past the formula sheet was always uniform for algebra 1 geometry and algebra 2 but this year the formula sheet is going to be very very specific to geometry alone so before we even jump into this exam I want to take a look at the new formula sheet so here we go this is what that reference sheet is going to look to you now it's very very bare bones and I know this is small on the screen i'm going to zoom in just in a second in fact let me do that right now all right so that you can see it a little bit better all they give you on the formula sheet all you're going to have there are going to be volume formulas for a cylinder a general prism that's basically a box a sphere a cone and a pyramid not only that but they give you these generic formulas like the volume formula for a cylinder is the B * H where B is the area of the base so you need to know ah cylinder's got a circular base and I know the area formula for a circle is p r 2 and so the volume of a cylinder is p r 2 * h all right so this is very very bare bones in terms of formulas in fact the best formula that they give you is sort of the volume formula of a sphere which we're going to be using almost immediately in this exam speaking of which let's launch right into it and get into the June 2024 geometry regions exam let's begin all right here we go so problem number one in the diagram below triangle B R I is the image of triangle J O E after a translation triangle C A T is the image of triangle B R I after a line reflection which statement is always true all right so let's get into it now first let's just kind of visualize what's going on here right we start with triangle J O E and we translate it up to triangle B R I and then we reflect that over a line to get triangle C A T and we're trying to figure out which of these statements is true so the first statement says that angle R is congruent to angle T well angle R is right there and when I reflect that triangle across that line it's going to land on angle A so angle R and angle T are not always congruent they could be but in this case R is most certainly just congruent to angle A so second statement angle J is congruent to angle A well angle J when it gets translated it becomes congruent to angle B and when angle B is reflected it is then congruent to angle C so angle J is not necessarily congruent to angle A let's take a look at this segment J E is congruent to segment R I well here's J E i translate it it's congruent to BI i reflect it it's congruent to CT so nope the final one OE so OE is right here when we translate it it's congruent to R I and when we reflect it it's congruent to a so after all of that what we find is that the correct choice is choice number four all right let's take a look at problem number two right away here we go number two a right cylinder is cut parallel to its base the shape of this cross-section is a what all right so a right cylinder now really I wish that they would be a little bit more mathematically correct and call it a right circular cylinder but we're just going to have to assume that's what they're talking about so what we've got is we've got our cylinder here okay now if I if I take my poorly drawn cylinder and I slice it parallel to the base I will get a cross-section that is congruent to its base and also to the top which is another base of it so that is then simply a circle all right let's take a look at problem number three what is the minimum number of degrees that a regular hexagon must rotate about its center to carry it onto itself carry it onto itself all right well the first thing that you have to know in this problem is what a regular hexagon is and that is a regular sixsided figure so you're expected to know that a pentagon has five sides a hexagon has six sides a septagon 7 an octagon 8 a nonagon 9 a decagon 10 right you're supposed to know those things so what does a regular six-sided figure look like well it kind of looks something like this something like that with a center like that now the idea is if I took this thing and I rotated it what would be the minimum angle that I would rotate so that it would fall back onto itself carry it onto itself well it's pretty easy to visualize right if I kind of draw these I'm going to call them radi they're not really radi but they are all the same length right we all know that if we rotate all the way around in a complete circle we have 360° so if I just want to sort of rotate it the minimum amount that this vertex will then get mapped onto this one well I just have to divide 360° by 6 and that will give me the minimum angle of 60° now there's a lot of variations on these types of problems that show up on the regions exam often times they'll say things like which of the following is an angle of rotation now any multiple of 60 would be a correct angle so 60 120 180 240 etc but the way that we can always always find that minimum angle that sort of base angle if you will is by taking 360 and divide by the number of sides that the regular polygon contains all right let's keep going ah here we go number four in the diagram below a sphere is inscribed inside a cube the cube has edge lengths of 18 what is the volume of the sphere in terms of pi all right so here's our first volume question this one though is fairly straightforward because it's actually on this sheet right the volume of a sphere is 4/3 * * r3 and I know that that's pretty small but of course volume has to have cubic units so it couldn't be r^2 it would have to be at least r cubed right for a sphere all right nope not that one here all right so the first thing I'm going to do is just write down my volume formula v equ= 4/3 pi r cubed now of course the issue is I need to know what the radius is well the key here is that this sphere is inside of a cubed right inscribed inside a cube whose edge length is 18 now watch out it would be tempting to think that r therefore is 18 but 18 is in fact the diameter of this sphere not the radius right so in other words here all the way to here would be 18 but here to here gives me a radius that's just equal to 9 so my volume is going to be 4/3 * * 9 3r now notice all of these answers are in terms of pi okay if these answers were in terms of some kind of decimal or even rounded let's say then I might type this into my calculator kind of all just like that but because it's in terms of pi what I'm going to do is I'm now going to bring out my TI inspire calculator and maybe I'll move it so that you can hopefully I'll move it there it is so that you can see that and now what I'm going to do is I'm just going to go in and put in 4/3 but I'm going to leave the pi out all right and I'm just going to put 9 to the 3r 4/3 * 9 cubed and I know that that's awfully small well I apologize for that let me let me go side by side here okay 4/3 * 9 cubed i hit enter and I have 972 so 972 is what gets to multiply the pi all right and in fact then we can see that that is choice three 972 pi all right let's keep moving on number five in the diagram below EM intersects H A at J ea is perpendicular to HA and EM is perpendicular to HM if EAL= 7.2 EJ= 9 AJ= 5.4 and HM= 3.29 29 what is the length of MJ to the nearest 100th all right well there is a ton of information given here let me get some of it marked down and I'm going to begin by marking down these two perpendicularity facts those are pretty obvious but obviously what that means is that I've got a right angle here i've got a right angle here now this little piece that they tell us the reason they tell us that is because that allows us to then know that these two angles are vertical angles all right in other words we know that EM is a segment and we know H A is a segment and they intersect at J giving us two equal vertical angles there all right let's fill in some of those side lengths ea is 7.2 2 EJ is 9 AJ is 5.4 and HM is 3.29 and what I want is the length of MJ so I'm going to call that X to the nearest 100th so what's going on in this problem what is the big picture well the big picture is not the Pythagorean theorem although that would be tempting to think because we've got two right triangles here the big picture is that these two right triangles are similar due to the fact that we've got the angle angle similarity going on right that angle is equal to that angle and those two vertical angles are equal now what does that allow me to do it allows me to set up a proportion and solve for X and there's a lot of different ways you can set up this proportion for instance I could say that X / 5.4 is equal to 3.29 / 7.2 2 or I could say x / 3.29 is equal to 5.4 / 7.2 lots of different ways to do it if I just use the fact that the ratio of corresponding sides are proportional then I would say something like this x / 5.4 is equal to 3.29 / 7.2 now if you feel more comfortable cross multiplying you could do that at this point but what I'm going to do is just multiply by 5.4 on both sides right i don't really need to cross multiply and that's going to give me x= 3.29 / 7.2 * 5.4 let's bring out our calculator and see what that gives us there we go let me just see if I can just slide it over here all right there's our calculation here we go 3.29 divided by 7.2* 5.4 and that's [Music] 2.4675 let me minimize that again and to the nearest hundredth that is 2.47 all right so our first similarity problem definitely not our last let's keep going okay number six which equation represents the line that passes through the point 2 -7 and is perpendicular to the line whose equation is y = 3/4 * x + 4 all right great now one thing to note when I look at all four choices these are all in what are known as the point slope version of a line and the point slope version of a line says if you know the slope of a line oh that's almost m and you know one point that lies on the line then the equation of the line can be expressed as y - y1 = m * x - x1 well the point that they're telling us lies on the line is 2 -7 so we're going to use 2 for x1 we're going to use -7 for y1 that's the easy part well actually I think both parts are fairly easy so what's the slope well we're told that the line not only contains that point but is perpendicular to this line now one thing that you need to know for the geometry exam and it's going to come up again later on is the relationship between the slopes of perpendicular lines slopes of perpendicular lines will always be negative reciprocals of one another assuming the lines aren't a horizontal and vertical line so what does that mean well this thing has a slope of 3/4s which means the m for my line is going to be -4/3 and now it's just sort of a plugand check right so we have y - -7 = -4/3 * x - 2 and if I just make that double negative into a positive I have y + 7 is = -4/3 * x - 2 let me find that y here it is choice three awesome let's take a look at problem number seven stole a little bit of its room but that's okay here we go in triangle RHM below the measure of angle R is 110° and the measure of angle M is 40° if triangle RHM is reflected over side HM to form quadrilateral RH RP prime M which statement is always true awesome well this problem is all about really understanding rigid motions right rigid motions are transformations that don't change the size or the shape of an object and there are three primary rigid motions and we saw two of them even in problem one translations reflections and rotations in this particular case we're going to be taking triangle RHM and just reflecting it across line segment MH or HM now when we do that what happens is we get something that looks like this right where this becomes r prime but keep in mind that when we do reflections all angles and segment lengths are preserved so if that's a 40° angle that has to be a 40° angle now one thing I maybe should have filled in right from the beginning but didn't feel the need to until right now is the fact that this angle would have to be 30° how do I know that well I know that because the three angles of any triangle always add up to 180° now of course if that's a 30° angle this must be a 30° angle okay so let's now take a look at what has to be true well number one quadrilateral RH prime M is a parallelogram it's tempting to think it is a parallelogram this is driving me nuts right now it's It's tempting to think it is a parallelogram except it can't be right and it can't be a parallelogram because opposite angles in a parallelogram are always equal and even though the 110 and the 110 must be equal this entire angle right is 80° this entire angle is 60° so it can't be a parallelogram it can't be right because if it were then th that 80° angle would have to be an 80° angle over here all right so let's take a look at number two the measure of angle MH um RP prime is 40 m H R prime is 40 now that angle is 30° right let's take a look at this one how about the measure of H MRP prime h M R prime is 40° yes that is true all right one last one to just take a look at mr is congruent to HR no not at all right mr has to be congruent to MR prime but MR would not have to be congruent to HR prime not unless this was a parallelogram and again it can't be because opposite angles in a parallelogram must be congruent all right that is the last problem on that page okay let's keep going our second volume problem number eight the funnel shown below can be used to decorate cookies with melted chocolate m chocolate the funnel can be modeled by a cone whose radius is 6 cm and height is 13 cm the baker uses 2 cm of chocolate to decorate each cookie when the funnel is completely filled what is the maximum number of cookies that can be decorated with the melted chocolate all right so this is a classic volume question right i need to figure out how much volume is in this cone in terms of cubic cm right and then I need to take note that the baker uses 2 cubic cm of chocolate to decorate each one of the cookies and I got to watch out for that but first let's talk about the volume of a cone now if I go over to my less than helpful formula sheet right what I end up seeing is that the volume formula for a cone is 1/3 B * H where B is the area of the base but the base is a circle right the base is a circle so this really means that the volume is 1/3 pi r well man that I don't know where my my pen has gone there it is little lag little lag let's try that again 1/3 pi r 2 that's that area of the base times the height and my red pen decided to to make an appearance awesome you never know when the red pen's going to come out could be anytime all right so let's go back over here i'm going to try to write that formula down one more time so the volume of the cone is 1/3 pi r 2 * h so let's figure out the volume of the cone it's pretty straightforward 1/3 pi my r is 6 so nothing tricky there in terms of a diameter or anything like that my height is 13 and now I just need to figure out what that is in terms of my calculator so let's go over here and let's get ourselves [Applause] a 1/3 this time I actually want to put the pi in * 6^ 2 * h and so the volume right in my cone is about 490 cubic cm about 490.1 but I'm just going to say 490 cm now one thing that is really key on any kind of test but especially on math tests is watch out for getting an answer looking at your choices and going "There it is choice 3 490 let's move on to problem 9." Right you've done the hard part this cone contains 490 cubic centimeters of melted chocolate but we need two cubic centimeters of chocolate per cookie so I need to divide that by by two so uh I guess cookies if you will will be volume divided by two so that's 490 / 2 and that is 245 cookies and that's a shame too because I guarantee there was a sizable number of students that on this problem did everything right to get that 490 and then chose number three which would make total sense and got the wrong answer i've been known to do that as well all right let's take a look at problem number nine in circle O below chords CT and BN intersect at point A chords CB and NT are drawn which statement is always true all right so first things first we have to understand what is going on with these two triangles now you did a lot of geometry with circles right you learned all these theorems about when this chord intersects this chord their products are equal and da da da da da right well let's start understanding these two triangles one thing I can note is that angle B right angle B intersects ark CN right so it must be half that measure angle T also intersects arc CN so it must be half the measure of arc CN which means that angle B and angle T are the same measure and you may have even been given some you know circle theorem sheet by your teacher that said you know two inscribed angles that intercept the same arc have equal measures right and that's true so this and this are equal now note right that we've got a pair of vertical angles here that must also be equal and finally if you want angle C and angle N both intersect arc BT so they must be equal and then what that tells us is that these two triangles are similar to one another actually by the time we had the two pairs of angles marked they were similar to each other but I wanted to throw the third one in there anyway all right so then what does that tell us now one thing I don't know is that these two triangles are congruent but I most certainly know that they are similar and this gets into these sort of proportions right and also into angles being equal so let's take a look at the first proportion that says that NT / TA is equal to CB / BA so kind of kind of let's take a look right and I'm going to really quickly let me like highlight NT in blue right nt and let me go to red divided by TA let me go back to blue is equal to CB divided by AB all right and that is absolutely true it's always going to be true and this gets into this principle that basically says that the ratios of similarly positioned sides are always proportional in two similar triangles so in other words if you want to think about like NT as being the side opposite of the double marked angle right and a being the side across from the triple marked angle right so I've got like double marked side divided by triple marked side is equal to double marked side divided by triple marked side right you could almost think about it as like these are two right triangles and we've got like the the trig ratios maybe this is the the the sign ratio or the tangent ratio let's not go there it's they're not right triangles anyway the point though is that the this proportion is true because this side divided by this side is similarly sort of positioned as this side divided by that side so for once in one of these like which of these four choices is always true it's number one which is awfully nice all right let's keep going number 10 i love problems like this let's take a look in the diagram below of triangle ABC CBF sorry ray CBF is drawn segment AB bisects angle FBD and segment BD is perpendicular to AC if the measure of angle C is 42° what is the measure of angle A i love these kind of problems they're what I call fill in the angle problem so let's let's play around a little bit let's fill in some angles number one we know that BD is perpendicular to AC so that's pretty easy that tells me that that angle is 90° and that angle is 90° okay now they also told me simply that measure of angle C is 42 so I'm going to write that in here now one thing that you should feel very comfortable with is the idea that the that the two acute angles in a right triangle are complmentary of one another they add up to 90° by the way that's going to come up again in a few problems from now so angle D B C right this particular angle has to be equal to 48° because well 48 and 42 is equal to 90 all right now I'm looking for the measure of angle A so somehow I have to get into this right triangle now it would be very tempting to say well if that's 48 then that's also 48 but I don't know that right one thing I do know is that this entire angle right here right angle F B D let's get that must be 180°US 42° which is 138° there's my red pen again 138° all right it's got to be that entire thing now way back in this problem it tells me that AB right bicts angle FB and that's awesome because it means that FBA this angle and ABD all but done this angle right must each be half of 138° so the one that I really want is the measure of angle A B D must be 12 of 138° so if I just really quickly do 138 / 2 right I get 69° and that's going to be right in here now finally what I can do something doesn't seem right about this what What did I do wrong um 170 give me a second folks 130 it's 132 wait why is it 48 oh that's why i'm like looking at this going where where is the 138 coming from should have been 180 minus 48 my apologies for that i would have gotten all the way to the end going "Well I'm just not finding the right answer there." All right let's try that again 132° still red love it 132° that's going to be 66° okay label it on my diagram now 66° hey we all make mistakes all right finally I can figure out what measure of angle A is right because again measure of angle A and measure of angle ABD must be complimentary they must add up to 100 they must add up to 90° now I'm flustered so the measure of angle A must be 90 - 66° and that's 24° wow that was fun all right here I'm like this is one of my favorite problems and I can't get it right all right here we go let's take a look at number 11 in circle O below O A is equal to 6 and the measure of angle CO A is equal to 100° what is the area of the shaded sector all right so you can be asked two types of questions about sectors of circles and sectors of circles are sort of like if you will either pizza slices of circles or Pac-Man right if you will right the sort of the the other portion of the the pizza all right now these problems right when you get sectors you're either going to be asked what the area of the sector is or what the arc length around the sector is okay now in either way you're going to do this by first thinking about what the total area or the total circumference of the circle is in this case the total area so let's figure that out to begin right we know that the area of a circle I hope because this isn't on the formula sheet is p r^ 2 so in this case that's * 6^ 2 or 36 pi and I'm going to label this a with a little subscript t for total total area so the entire area of this circle is 36 pi now the question is how much of the circle is shaded well you can always think about that in terms of degrees all right now I know that's 100° so this must be 260° right because 360° in a full rotation around a circle now what I'm going to do is I'm going to set up a proportion and I'm going to literally say the area of the sector divided by area total is equal to um 2° divided by 360° right so it's simply proportional to how much of the circle you're occupying and that's 2° out of 360° so now I can say the area of my sector is to 36 pi as 260 is to 360 let me move this up a bit i'm going to multiply by 36 pi on both sides all right cancel out now again notice all of my answers are in terms of pi so if I want to do this on my calculator if I don't want to worry about reducing fractions and all of that which I could I could um you know then what I'm going to do is I'm going to pull this thing out uh and I'm going to put in let me um let me just do this really quick handheld only so that we can kind of see it what I'm going to do is I'm going to put in my fraction 260 over 360 and then I'm just going to multiply it by 36 all right and let me now put this in like that so that you can kind of see it notice I'm leaving off the pi right i don't want the pi in there right now because it would give me just a messy decimal at the end and all of these answers are in terms of pi now when I actually do that notice it gives me 26 you can probably see that many of you right because 260 over 360 if you divided out a common factor of 10 you'd have 26 over 36 then the 36 down here would cancel the 36 there leaving you with a 26 and so my final answer is going to be 26 pi all right and that's that let's keep going here we go number 12 in rectangle ABC diagonal AC is drawn the measure of angle A CD is 37° and the length of BC is 7.6° 7.6 cm not 7.6° what is the length of AC to the nearest tenth of a centimeter all right great so let's draw a good picture that is absolutely key in a problem like this would be great if they had given us a little bit more room but they didn't so here we go we've got rectangle A B C D remember always kind of stay in the same direction with diagonal AC drawn all right we know that angle A C D is 37° all right and we know the length of BC is equal to 7.6 now this is a rectangle so opposite sides have the same length and of course interior angles are equal to 90° so that's where we're sitting right now now what they ask us for is the length of diagonal AC to the nearest tenth of a centimeter implying that something ugly is going to happen here right there's going to be some kind of math involved that's going to make this answer a little ugly but let me put an X here right just so that we have it now what we're really dealing with is the first example of right triangle trigonometry that we've seen come up on this exam specifically we're dealing with right triangle A D C right and we hone in on that 37° angle right we note that the 7.6 is opposite the 37 and that the x is equal to the hypotenuse all right so what involves opposite and hypotenuse that's going to be the sign ratio right soa s is opposite over hypotenuse cosine is adjacent over hypotenuse and tangent is opposite over adjacent all right so what do we know we know that the s of 37° is equal to the side opposite 7.6 / x right now if we cross multiply that will give me x * the s of 37° is equal to 7.6 and I'm going to go down here that means that x will be equal to 7.6 divided by the s of 37° all right let's go to our calculator now right now if I type in this expression if I type in 7.6 / the s of 37 let's take a look at what we get trig sin 37 all right I I I get an insane answer i get -1.809 i mean I can't have a negative side length plus it's not even one of the answers right so what's going on here well what's going on here is that currently my calculator is in what's known as radian mode all right now pretty much all of you will have your calculator memories reset or you'll be given a schoolisssued calculator when you walk into the exam in all likelihood your calculator will be in radian mode and depending on what kind of calculator it is whether it's a TI Inspire or a TI 84 Plus or a Casio or a Huelet Packard or whatever there must be a way for you to go from radian mode into degree mode in the Inspire it's very easy you just kind of take your mouse and you click on the radian and it becomes a degree and now if I go up and I grab the same calculation or if I try to grab the same calculation there it is and I hit enter now I have something that's perfectly reasonable 12.628 etc and that will be closest not to H that can't be right hold on i wrote it down wrong 12.62 little I I'm like I'm pretty sure I did everything right there 628 and that will be closest to 12.6 all right so moral of the story right when you're doing right triangle trig and we're going to see right triangle trig I think come up three times on this exam make sure your calculator is in degree mode or you're going to get answers that aren't even vaguely correct right and and you know obviously you don't want that to happen okay let's take a look at 13 a peanut butter manufacturer would like to use a cylindrical jar with a volume of 1,180 cm the jar has a height of 10 cm what is the diameter of the jar to the nearest tenth of a centimeter okay great so again it's another volume problem so the first thing that we always want to do whenever we're working with volume is make sure to really understand what is the three-dimensional object you're working with here we're working with a cylinder right so yet again I go over to my formula sheet and it's not very helpful right because if I look at a cylinder it's just volume equals base time height where B the base is the area of the base i shouldn't I shouldn't say base time height i should just say capital B * H where B is the area of the base so we need to know we need to know here that B the area of the base is p r 2 so the volume is pi r 2 * h man my pen my pen is not playing along all right so let's write that down right we know that pi r 2h must be equal to 1,180 all right we also know that the height of the jar is 10 cm so I can put the 10 in there right so I'll get p r^2 * 10 is equal to 1,180 all right now I need to know what the diameter of the jar is but hopefully I'm very aware that the diameter will be twice the radius so this formula right I can isolate the r 2 and specifically I can do that by dividing both sides by 10 * or p * 10 but I'm going to write it as 10 * pi because that makes me feel better all right and then I can simply take the square root of both sides and I'll find that the radius is the square of 1,180 / 10 pi and this is something that this is a good use of your calculator right the square of 1,180 / 10 pi so I'm going to close that window i don't even know why that was there all right let's take a look square of 1 180 divided by 10 pi all right now remember that is my radius right my radius is about 6.12 etc i'm not even writing it down i'm now going to do time 2 and there's my diameter right my diameter is 12.25 257 etc and where is it 12.3 all right there we go okay let's take a look at problem 14 triangle KLM is dilated by a scale factor of three to map onto triangle DRS which statement is not always true okay so let's kind of get a sense for this right dilations are when we take an object and we stretch it so that it becomes either larger or we compress it so that it becomes smaller but it stays the same shape imagine taking an image on your phone right or on your computer and just stretching it enlarging it to make it larger or shrinking it to make it smaller right it's going to be larger or smaller but the overall shape is going to stay the same okay so let let's kind of get a sense for this right we've got this like triangle I don't know uh triangle KLM and we dilate it right dilate it by a factor of three so it's going to get much larger and it's going to become uh let's see KLM and D R S now again it says which statement is not and this is key not always true we're looking for the one that is possibly incorrect the first one says angle K is congruent to angle D and that is absolutely always true right the whole point of a dilation is that angle sizes are preserved but lengths get scaled accordingly right next one km is 1/3 the length of DS so KM is 1/3 the length of DS that is absolutely positively true right ds is three times as long as KM but that's the same as saying that KM is 1/3 the length of DS now number three the area of DRS is 3 * the area of KLM well that's the first one that we wonder about right now clearly because we've you know dilated this by a factor of three this particular triangle right has a larger area than this one the question is is it three times as large i mean we've dilated it by a factor of three is it three times as large and the answer is no it's actually nine times as large all right in fact if you want to know how much the area is scaled by you always take the scaling factor and you square it so if I scaled it by a factor of two the area would go up by a factor of 2^2 or 4 if I scale it by a factor of three the area would go up by a scaled factor of 3^ 2 or 9 and it's really cool you can actually kind of see this if I took this little triangle right and I stretched it by a factor of three it would kind of look like this right and what would happen is I could literally put nine of these triangles in here right here's my original triangle right here and there are nine of them now 1 2 3 4 5 6 7 8 nine of them that fit inside the larger triangle right and you just think about it right this thing got three times as long this thing got three times as long right if you even thought about the area formula for a triangle you'd get two factors of three going in there for an overall scaling factor of 9 now I just want to kind of like really quickly that means that three is the one that is not always true and by the way three is never true it's not only not always true it's never true but then choice four the perimeter of DRS is three times the perimeter of KLM that's absolutely true so that's a linear that's just a distance measure measurement so if all distances go up by a factor of three then the perimeter which is just adding up all the distances also goes up by a factor of three all right let's keep going let's take a look at 15 a rectangle with dimensions of 4t x 7 ft is continuously rotated about one of its 4ft sides the resulting three-dimensional object is a what so one of the things that you were supposed to study in geometry was what happens when you take a solid a two-dimensional solid a triangle a rectangle you know whatever and you rotate it about some line all right so let's take a look at this particular thing we've got a rectangle with dimensions of 4tx 7 ft so maybe I'll I'll kind of draw it like that and it is continually see rotated about one of its 4ft sides so in other words if I took like a little rod like that you know and I attached it like that and I spun it around right so I got this and I spin it around right this side that that probably is not helping at all as a visual aid but if I did that then what would happen is I'd get this kind of mirror image over here i don't know what was just going on there and I'd get something that would look like this right so what is that well that is a cylinder right whose height is 4t and whose radius is 7 ft so a cylinder whose height is 4 feet and whose base radius is 7 feet all right let's keep going number 16 in right triangle ABC altitude CD is drawn to hypotenuse AB if A is 4 and CD is 8 the length of BD is what all right well we need a good picture here so let's try to draw a good picture now we've got a right triangle and we're told that the hypotenuse is AB now if you're going to draw this as a suggestion do not draw an isosles right triangle draw a right triangle where there is a shorter leg and a longer leg all right now it says that altitude CD this has got to be C is drawn to hypotenuse AB and altitude is a line segment or a line drawn from a vertex perpendicular to the opposite side now these problems probably look familiar to you this is a problem where you've taken a right triangle and you've divided it into two smaller right triangles both of which are similar to the larger right triangle so this is actually a similarity problem all right now what do we know we know that a d is equal to 4 we know that CD is equal to 8 and we want to know the length of BD all right now again the key is that this small triangle this mediumsiz triangle and this large right triangle are all similar to one anothers and notice each one of them each one of those three right triangles has a shorter leg all that are given and the one that I want to know they are the shorter leg of this right triangle the longer leg of this right triangle the shorter leg of this right triangle and the longer leg of this right triangle the sort of the big right triangle is not even kind of come into play here so what I can say is I can say well the short leg here to the longer leg here right so four is to 8 right so short is to long as short is to long so is equal to 8 overx okay I can now solve this by cross multiplying 4x is equal to 64 i divide both sides by 4 and x is equal to 16 now a lot of teachers when they go through this particular topic and we're going to see it come up one more time on this exam they'll have you memorize all sorts of different things like one thing they might have you memorize is that that altitude is always the geometric mean between the two segments that are partitioned by the altitude there's all sorts of things like that again I'm going to go back each and every time to the fact that the three right triangles that are created are all similar to one another all right let's keep going problem number 17 if ABC D is a parallelogram which additional information is sufficient to prove that ABC D is a rectangle all right so in other words right this thing called sufficiency right i start with a parallelogram let me kind of draw a very generic parallelogram and I'm going to call it AB C D right now what else do I need to know what is sufficient information to now conclude that this isn't only a parallelogram but it's also a rectangle well well if two consecutive sides of a parallelogram are the same length it tells you that that parallelogram is also a rhombus right not a rectangle okay so the first thing would tell us that this was a rhombus not a rectangle second thing AB is parallel to CD well that is always true in every parallelogram so this statement actually gives us no further information whatsoever about this picture we already knew that AB was parallel to BC let's take a look at AC is congruent to BD well those are the two diagonals ac is congruent to BD all right well in a generic parallelogram you always have a shorter diagonal and a longer diagonal in this case the way I've got it drawn AC is the shorter and BD is the longer but if you force those two diagonals to be the same length as in this particular picture then we have a rectangle all right so a rectangle is a parallelogram that has two diagonals that are the same length in fact that's one good way that people in the construction business check to see if a rectangle that they've created let's say for a wall or something like that is in fact a rectangle is in fact square they simply measure one diagonal another diagonal and if they're the same length then we know that we've got a rectangle assuming that we've got a parallelogram by the way the last condition AC is perpendicular to BD the two diagonals are perpendicular to each other that would be another one that would tell you that the parallelogram is a rhombus okay in a parallelogram if we force the diagonals to be perpendicular i.e intersecting at right angles then it must be a rhombus all right let's keep going number 18 line segment APB has end points at A5A 4 B 7 -4 what are the coordinates of P if A to PB is in the ratio of 1 to 3 all right great well just for a second let's kind of take a look at what's going on here we've got a line segment right that goes from A to B and it's got those coordinate points and somewhere in between there's point P and wherever P is located wherever it's located right then this is sort of one unit to three units all right one unit to three units which means right that P is 1/4th of the way along this particular line segment right not 1/3 1/4 right because it's 1 to three if they had said AP to AB was 1 to three then it would be 1/3 of the way but it is one quarter of the way so the first thing I'd like to know is just how much did x change when I go from a to b and how much did y change when I go from a to b so I'm going to actually calculate delta x when I go from a to b and that's going to be simply 7 - -5 and that's 12 units in other words like we go 12 units this way right and delta y is equal to -4 - 4 and that's8 which means we're going down 8 units right maybe I shouldn't put the negative there but you understand so how do I figure out the coordinates of P well if P is 1/4 of the way over then it's actually 1/4 of that way and it's one quarter of this way so watch how I can just figure out what this is i'm going to use a little bit of space here i'm going to say x of p equals I'm going to take my starting x value -5 and I'm going to add to it 1/4 of the change in x right so that's going to be -5 + 3 and that's going to be at -2 now y of p is going to be the starting ycoordinate which is 4 plus 1/4 in the change of y which is8 so that's going to be 4 + -2 and that's going to be 2 so the coordinates of P must be -2 comma 2 and there they are all right let's take a look at number 19 and we'll just you know we'll kind of ignore all of that writing number 19 in the diagram below A B and C D intersect at E and CA and DB are drawn if CA is parallel to BD which statement is always true all right well let's take a look at what we know we know that C D and AB intersect at point E which immediately tells me that these two angles are congruent now we know that CA is parallel to EB now or sorry is parallel to BD my apologies there that tells us that alternate interior angles are congruent and that tells me that angle C is congruent to angle D and angle A is congruent to angle B now that's all I know so let's start taking a look at these choices first choice AE is congruent to be E let me try to get this up so that we can you can see it a little bit my arm not isn't getting in the way now it's tempting i look at this picture and I think sure yeah Ae and BE are maybe they're congruent because these two triangles are congruent except there is absolutely nothing in this statement that says that E is the midpoint of AB nothing at all but they want you to think it in fact the second one CA is congruent to DB this is congruent to this well again there's there's nothing about this that tells us that choice three triangle AE C is similar to triangle B D that's our winner right absolutely now they've drawn the picture to make it look like these two triangles are not only the same shape but they're also the same size but we don't know that the only way we would know that is if we were told that point E was the midpoint of either AB or CD either one of those and I'm golden and I can say these two triangles are congruent and then that's true and that's true and that's true everything is true actually in here if I know that which is problematic all by itself all right but for all I know you know the situation I really have is something that looks more like this right where maybe this triangle is much smaller than that one i don't know i just know that E is located at the intersection of CD and AB if s of 3x + 9° is equal to cosine of 5x - 7° what is the value of x all right this is a tricky problem because you really have to recognize what factoid they're really getting at here and what they're getting at is that if I know that the sign of a is equal to the coine of b then a + b must be equal to 90° at least in this course all right now once you go to algebra 2 and above this is there no it could be any there could be a lot of different relationships actually that they have but in this course the s of a is equal to cosine of b then the two angles must add up to 90° and that's true because if I looked at right triangle trigonometry let's say let's say I had angle A here and angle B here and let's just call this C then obviously right with this being the hypotenuse right the S of A which is this side / the hypotenuse is equal to the cosine of B which is this side / the hypotenuse and those two acute angles add up to 90° once I know that then I can simply set up the following equation 3x + 9 + there we go 5x - 7 is equal to 90 now I can add like terms 3x and 5x is 8x 9 + -7 is pos2 that's all equal to 90 8x is equal to 88 and it just wouldn't be one of my videos if the red pen wasn't coming out a bunch and we get x is equal to 11 awesome let's take a look at problem 21 which set of integers could represent the lengths of the sides of an isosles triangle all right well first things first we have to know what an isosesles triangle is right and an isosesles triangle is any triangle where two sides have the same length but the thing is when you look at these four choices each one of them right 1a 1a 3 2a 2a 5 3a 3a 6 4a 4a 7 i mean like they all seem to fulfill the category of being an isosles triangle because there's two of the three lengths that are the same but here's the key right this problem is really about what's called the triangle inequality okay and the triangle inequality the triangle inequality states that when you add the lengths of any two sides of a triangle the sum must always be greater than the length of the third side so if I took any triangle forget about an isosesles triangle let me just do a random triangle ABC if I grabbed any two lengths let's say I grabbed the length of A and AC and I added them together it would have to be longer than the length of BC or if I took AC plus BC and I added them together it would have to be longer than AB the reason for that by the way is a very simple fact about physical space which is that the shortest distance between any two points is along a straight line so if a C plus BC was somehow shorter than AB well then I could get from A to B quicker by going or shorter by doing this than by doing that and that just doesn't make any sense right so when I look at be greater 3 + 3 is 6 well that's equal to but it's got to be greater 4 + 4 is 8 and that's greater than the 7 and therefore that could be the lengths of the sides of an isosesles triangle all right let's take a look at 22 oh another one of these in the diagram shown below altitude CD is drawn to the hypotenuse of right triangle ABC which equation can always be used to find the lengths to find the length of AC all right so yet again this is one of these similar right triangle things right so we've got three different Oh that is gorge gorgeous right angles there here let me do it all at once there we go right we've got three right triangles the big one ABC the medium-sized one A C and the small one B d all right and sometimes it's helpful in these problems if you don't have these theorems memorized or whatnot to kind of mark off the acute angles like if that's that and let me do that as that right then this thing has got to be the double and this has got to be the single all right and that kind of gives you a sense for the similar right triangles now which one of these proportions will always allow you to solve for AC okay and again let's kind of highlight AC maybe we can like just really get it there that we see it right ac okay back to blue excellent all right now we could go through each one of these in turn but I think what I'm going to do to save a little bit of time is I'm going to go right to the correct one and talk about why it's correct so the correct one is actually a divided by AC all right so let's think about this for a second ab / is the hypotenuse right of the large right triangle divided by the longer side of the large right triangle so I can I could say hypotenuse of the large divided by long side of the large equals and now let's take a look at this ac / a D ac is the hypotenuse of the medium if you will divided by a D which is the long side of the medium triangle right so I've got this like hypotenuse of the sorry the hypotenuse of the of the large triangle divided by the longer leg of the the larger right triangle is equal to the hypotenuse of the medium triangle divided by the long side of the medium right triangle and again there's a lot of different theorems like one theorem that you may have learned was that the legs will always be the legs of the large right triangle will always be the geometric means all right of the um of of the the hypotenuse and the segment right sort of like adjacent to that side that wasn't very well said but but hopefully hopefully it works and that's where we've got this AC being in that geometric mean position and then the AB and the AD being in the geometric extreme position i'm not sure how well that went but let's let's move on ah I dislike those all right rhombus well we did this exact same problem but with a rectangle not that long ago so let let's let's do it again let's draw a generic parallel the first statement MA is congruent to MK right m A is congruent to MK well that would be enough to know that this is a rhombus all right and it would be enough to know it's a rhombus because remember a rhombus is a parallelogram with all four sides being the same length but we already know in a parallelogram that opposite sides have the same length so and that side are equal because it's a parallelogram then I would be able to mark this up and that would be a rhombus ma is congruent to MK all right one more multiple choice problem then we're going to take a brief break number 24 a line whose equation is y = -2x + 3 is dilated by a scale factor of 4 centered at 0a 3 which equation represents the image of the line after the dilation all right well let me move up the board here and let's kind of get a sense for this right now I don't need to make a perfect sketch of this line but it actually has to be decent all right and I'll explain why in a moment this is a line whose y intercept is at three and has a negative slope of -2 all right so there's my line okay now whenever we dilate a line about a point that's not on the line then that line stretches out right it ends up being parallel to the original line so it has the same slope meaning that the answer's got to be one or two because they also have a slope of -2 all right but it kind of moves away from the original line but this is a particularly tricky problem because the center of dilation for this line is at 03 and that is a point on the line now that's a lot trickier because if we've got a point on the line then what happens is when we stretch this line out it simply lies on top of itself it's just the same line it literally every point just kind of moves out therefore the equation of the line is -2x + 3 most of the time right especially if you're dilating with respect to the origin right then you'd get some kind of line that would like kind of be out here and it would be a different line it would be parallel so it'd be have the same slope but it would be a different line but this one's a particularly tricky problem because the center of the dilation actually lies on the line itself all right let's go to part two and take a brief break brief Kirk break or what I like to call my lemonade break that's right this isn't just me doing this free e math instruction i'm also a shill for the National Lemonade Association pretty sure that organization exists but it might not be called that wouldn't that be great some trade organization out there it's got to exist right whose sole purpose is to simply make sure that people buy more and more lemonade right more and more lemonade um I bet exists you know there's probably only like four or five companies that actually still make lemonade i mean there's thousands and thousands of like elementary school peop children around the country that hawk lemonade out on the side of the road you know at any time of the year especially when it's hot outside um but but in terms of actual lemonade you know organizations I I bet there's one umbrella organization that that's like pretty happy with me or really upset i don't know which one anyway I do want to remind you that this review is being brought to you by EMath Instruction i would encourage you to go over and visit e-mathinstruction.com where you can find a ton of instructional videos that instruct in math from everything from grade six which hopefully if you're watching this video you don't need anymore all the way through algebra 2 all right i'd also like to take this moment to remind you um that we're going to be doing an algebra 1 review one week from today on June 16th and then an algebra 2 review one week from then on June 23rd each one of those reviews will be live from 6:00 to 9:00 p.m on Monday all right one more sip of lemonade uh and then we will return to our problems one of these days one of these days I'm going to take a big drink it's just going to go all over me all over me all right and then I'll unwisely whip my shirt off in the middle of a live video that's not going to happen all right so be that'd be terrible okay so I say we jump back in we're now going to be going into part two questions now part two questions are worth two points each just like the multiple choice questions are but the big difference is you can get partial credit on these problems and you can also lose complete and utter credit if you do things that are just crazy so let's take a look at problem number 25 in triangle ABC below the measure of angle C is equal to 90° ac is equal to 11 and AB is equal to 18 determine and state the measure of angle A to the nearest degree all right so notice we've got a right triangle here and we're trying to determine the measure of an angle to the nearest degree implying that we are going to do some rounding if you've got a right triangle and you're trying to figure out the measure of an angle and they tell you to round you are using right triangle trigonometry you are using soa now I'm just naturally in red okay here we go so right we've got to figure out which one of these ratios we're using and that's always about the angle the acute angle so in this case we try to figure out angle A we know side 11 and that's the adjacent side 18 is the hypotenuse adjacent and hypotenuse is coine right so what can I write down i can write down that the coine of angle A is equal to the side adjacent to it which is 11 / the hypotenuse 18 now that is what the coine of angle A is but what is angle A well angle A can now be found by simply doing what's known as the inverse cosine of the ratio 118 and for that we definitely need our calculator but it's simple enough i go into trig i go down to inverse cosine i hit enter i put in 118 remember very critical again that your calculator is in degree mode look the arrow is even pointing to it and I hit enter and my angle is 52.33 etc degrees so maybe I'll put down 52.33 dot dot dot degrees nearest degree nearest degree nearest degree watch your rounding 52° i cannot tell you how many times I have graded a geometry regions exam and I see kids do beautiful work and they are so geared into rounding to the nearest tenth that they write down 52.3° and they lose 50% credit right it's only a twopoint problem right so suddenly they're sitting at only one out of two points and that's a shame because they definitely know the geometry on this problem all right ah our first off the books problem let's take a look at actually we're going to go to sorry a supplementary problem here so that I actually have a compass and a uh ruler let's do it 26 use a compass and straight edge to construct an equilateral triangle inscribed in circle A below leave all construction marks so I'm always hesitant to do constructions during one of these reviews because whatever construction I'm going to do right now is not going to be on the test on Wednesday there's no way they're going to have the same construction two years in a row but let's do an equilateral triangle it's one of the easier constructions now you could be asked to construct an equilateral triangle a regular hexagon or a square inside of a circle here is how we're going to construct an equilateral triangle first thing I'm going to do is I'm going to take my straight edge and I am going to draw a line through the center of the circle in other words I'm going to construct a line or draw a line that has a diameter in it all right that's really simple enough right now we're just going to put our straight edge to the side i'm going to now bring in my compass and what I'm going to do is I'm going to set the length of my compass to No that almost worked to be the length of the radius all right and notice I'm actually going to put the pointy end i love referring to a compass that way the pointy end and the pencil end i'm going to put the pointy end actually on the edge of the circle and what I'm going to do is I'm going to draw an arc i'm going to try to draw an arc oh my goodness that was fun let's try that again oh come on come on now this is where I always claim that you guys sitting at home have an easier time than I do here there we go i'm going to draw an arc like that okay now believe it or not I basically have everything I need that's why this is such a simple construction or at least it should be these three points the two points where that arc intersects the circle plus this final point here are going to give me an equilateral triangle all right now I'm going to just like freehand in the side length of this triangle you should in no way shape or form do that on your test you should use the straight edge but this thing is so cumbersome when I was practicing that I don't want to waste five minutes of your time trying to get it lined up but that will end up being an equilateral triangle i know it's not the prettiest equilateral triangle especially on that end but that is an equilateral triangle by the way if you took this compass and you did the same thing from that end so that you had sort of an arc that looked like this all right and then you did this then you'd have that regular hexagon which almost looks better than the equilateral triangle that I drew all right so let's uh let's get out of this for a second and let's head back over here all right and let's take a look at 27 which ironically is going to be back over here so let's take a look at this one full screen all right 27th quadrilateral d a r and its image quadrilateral d e a d prime e prime a prime rp prime are graphed on the set of axes below let me move this out of the way describe a sequence of transformations that maps quadrilateral d a r onto quadrilateral d prime e- prime a prime rp prime all right let me get my pen set up for a slightly smaller width and let's talk about this now the first thing I want you to do if you ever see a problem like this is I want you to look at the letter order on this particular on the pre-image and I want you to like go around it sort of in a circular manner so if I went around D E A R like this notice I am going counterclockwise right i'm going opposite the direction that a clock would go in on the other hand now I don't like how wide that is let me go to this if I go to this one D prime E prime A prime RP prime right that is clockwise so notice in one case I'm going counterclockwise in the other case I'm going clockwise however you want to think about that if that direction changes then you must have a line reflection all right there's got to be a line reflection is it a vertical line reflection is a horizontal line reflection we're not going to worry about that right now but looking at probably a translation and or a rotation okay translations and rotations just shift directionction in the mirror the side of your head that the part is on switches right just like that the orientation of these letters will also switch so now let's talk about what the trans transformations are now there's actually an infinite number i'm going to just delete that so it's out of there and get rid of this so it's out of there um come on get get rid of that there it is great um so there's an infinite number of correct answers here but one thing that you might say right away is well let me let me take this thing and reflect it across the y ais right so if I did that let's take a look at what that would look like i'm going to here I go flip left right suddenly I'm looking at it like that and now you should be able to already see that it's in the right orientation so I'm going to say number one reflect across across the y axis okay and I'm right there great awesome now if I just translate this thing like that then I've mapped it onto itself so what is that translation well I want to take this thing and I go two units to the right and seven units down so I'm going to say translate two units right and seven units down and that's it now don't get me wrong all right you could translate it first all right I'm not going to kind of completely and utterly go through this let me just flip it back real quick right you could alternatively do something like this where you first translate it like that right and then you reflect it over the vertical line x= 2 all right but people tend to feel more comfortable reflecting across the y and the x-axis than they do across other sort of random vertical lines all right so this is probably the most natural one of these now little warning on a problem like this i saw one of these come up on an exam maybe like 2 3 years ago where they it was phrased exactly the same way it was like you know describe a sequence of transformations etc etc and then the weird thing was it was a single transformation it was simply a rotation that was one otherwise it's not a sequence it's just a single transformation but it was just a single transformation so just be careful on that danger danger all right let's go back here all right finally one that we stay here which means I got to change the line width again because it's slightly different here we go problem number 28 in circle P below tangent A L and secant A K E are drawn if A K is equal to 12 and K is equal to 36 state the length of A L all right well first let's just get these segment lengths in right we know that A K is equal to 12 and K E is equal to 36 all right well there were all sorts of theorems that you learned that related the lengths of tangents to secants and chords and all sorts of things like that and you probably you must have memorized them because they aren't on that formula sheet i'll tell you that right now which is weird you know of all the things you're like why wouldn't they put this one on the formula sheet that would have been nice anyway so what is the theorem well there's this general theorem where if you've got like two tangents or a tangent and a secant or two secants you can always say look I'm going to look at this thing and I'm going to take the product of the portion that's on the outside with the hole is equal to the product of the portion on the outside time whole is equal to outside times the hole now what does that mean in here well for the tangent that just means that a l * a l is equal to a k * a e all right so a l * a l the portion outside time the hole which is just itself is equal to a k * a e all right well that tells me that a l 2 is equal to a k which is 12 * a e this is the part that you have to be careful about which is 48 so it's not 12 * 36 it's 12 * 48 and therefore a l will be the square of 12 * 48 now you're more than welcome to figure out what 12 * 48 is and then find the square root or you can just do it all at once write the square of 12 * 48 and that's equal to 24 there we go all right so AL is equal to 24 excellent all right let's take a look at 29 the equation of a circle is x^2 + y^2 + 8 x - 6 y + 7 is equal to 0 determine and state the coordinates of the center and the length of the radius of the circle now sometimes on these problems they only will ask you to find the center of the circle or they'll ask you to find the radius this is quite a bit to ask for a twopoint problem but it is what it is and you probably remember doing a bunch of these and what we're going to do is we're going to complete the square twice but the first thing I'm going to do is I'm going to rearrange this a bit so that the x terms are grouped together x^2 + 8 x i'm going to leave a little space then I'm going to group the y terms together y^2 - 6 y and then I'm going to take that seven and I'm going to put it on the other side and make it a - 7 all right and that's absolutely critical right if I've got a constant hanging out on this side with the x squares and the y squars get it over to the other side but of course change the sign now we're going to complete the square recall completing the square I look at the coefficient on the x to the first term i find half of it that's equal to 4 i square it and I add it on and that's 16 now what I add to one side I need to add to the other side so 16 there and likewise over here I didn't give myself enough space so let me just erase that a little bit can I move these guys nope put it here all right here I'll take half of -6 which is -3 i'll square it and add on and that's 9 and now I'm sitting like that all right if I follow that faithfully take half of B square it and add it then what I get is perfect square tromials all right on this side I'm going to simplify this really quick -7 + 16 is 9 9 + 9 is 18 on this side x^2 + 8 x + 16 is x + 4^ 2ar and this is y - 3antity^ 2 i'm sure your teachers pointed this out to you plenty but you can always know what's inside of these parentheses because that will always be x plus 12 b y + 1/2 b remember 1/2 B was 4 pos4 1/2 B here was -3 -3 right there now we're ready right the center of this circle is going to be at -4 pos3 remember sort of if you will it's the opposite of the signs in here and now this is a little bit tricky r2 is equal to 18 which means R is the square of 18 so you can leave your answer as -4 3 and the of 18 you could simplify this as well if you wanted to this is equal to 3 * the 2 but for goodness sakes do not do not do not come up with a decimal for this and then write it down all right that will lose you one out of two points on this problem which is insane given how much work you've done to get all the way down here leave it as the of 18 leave it as 3 * the of two either way is fine don't give a decimal unless they say write the radius to the nearest 10th or 100th or what have you okay so let's keep going all right problem number 30 on the set of axes below triangle ABC is drawn with vertices that have coordinates A 2A3 B 4A 5 and C 5 sorry -5 comma 1 and what do they want they want us to determine and state the area of triangle ABC they just want to make sure that you don't just determine the area but never state it i'm not really sure how you would do that you know it'd be weird i can see how you might state the area but not determine it although that would be weird too all right now it would be very understandable to be like "Oh well I know the area formula for a triangle is 1/2 base time height so you know maybe I could use the distance formula to find the length of AC and maybe AB but those don't look like they're perpendicular to each other so what are you going to do?" All right if you have a figure in the coordinate plane like this right especially if it's triangle or some kind of parallelogram like fig figure the way that you find the area is what I call surround and subtract so let's take a look at how we surround and subtract what we're going to do is we are going to find a rectangle that inscribes or sorry circumscribes circumscribes this triangle that almost went through point A it didn't quite but There we go all right now I want to know this area and that's tough to do because that triangle is not oriented very nicely but what I can definitely do what I can definitely do is figure out the area of this rectangle and then I can figure out the area of those three unshaded triangles because they're right triangles i just want to be systematic about it so let's get some lengths all right this has got a length of nine this has got a length of 8 just wanted to make sure so one thing I can say right away is that the area of my rectangle which is base time height is 72 square units and I can just put it as 72 all right now let's be very very systematic i'm going to label these triangle one triangle 2 and triangle 3 okay and again the beauty of this method is that these triangles will all be right triangles so I can simply say well okay for triangle one you know I know that this is four i know that that's already nine so triangle 1 the area of 1 is going to be 12 of 4 * 9 and that's going to be 18 all right so I know that now I can go up i can look at triangle 2 you know triangle 2 has got this like little base of 2 and a height of 8 so the area of triangle 2 is going to be 12 of 2 * 8 and that's going to be 8 units all right so that's cool and now finally triangle 3 i know this is four and I know this is seven so I can say the area 3 is 12 of 4 * 7 which is 14 so these three triangles are all within that larger rectangle but outside of the area of triangle ABC so now if I add all of these together A1 + A2 + A3 I will get uh 40 units all right and therefore the area of triangle ABC must be 72 - 40 or 32 square units and that can be pretty much done with almost any figure that shows up in the coordinate plane you can surround it with a rectangle a bounding rectangle if you will whose area should be very easy to find just base times height and then all of the figures that are on the outside of that triangle typically end up being right triangles whose areas are very easy to find simply find the three areas or however many there are add them up and subtract them from the overall area to figure out the area of the figure that you want surround and subtract okay let's take a look problem number 31 remember these are still twopoint problems problem 31 in the diagram below AE is 15 EB is 27 AF is 20 and FC is 36 explain why EF is parallel to BC all right so for some reason right I have to explain why EF is parallel to BC and this really gets into what's known as the sides spplitter theorem or to be perfectly correct the converse of the sides spplitter theorem all right and the sides spplitter theorem basically said if you have some segment that is parallel to a side of a triangle then that segment will divide the two sides that it intersects proportionally what do I mean by that well let's take a look at the ratio of 15 to 27 right 15 / 27 if I divide both of those by 3 I get 5 9ths right 15 / 27 is 5 9ths now if I look at 20 / 36 right and let's say I divide both of those by four I also get five 9ths so EF has divided side AB and AC proportionally and there's this thing called the sides spplitter theorem and again more appropriately the converse of the sides spplitter theorem that says because BC divided No sorry not because BC EF Because EF divided AB and AC proportionally it must be parallel to EF all right so it's very critical right that you show those two ratios and that you make some kind of statement like this all right that gets you those full two points okay let's keep going ah part three all right well before we go into part three let me take just a lemonade break real quick right i got have a lemonade break between our sections not only is this dehydrating but it's also calorie like intensive i also played a lot of pickle ball earlier today and if you've ever played a lot of pickle ball which some of you may have right it can be pretty intense and exhausting so you know I might just give up at any moment i might just fall down if you recall a year from a year ago right about now I could barely stand because I had played way too much pickle ball and basically thrown out my back um that's not so much the case right now anyway let's get into our part threes now our part three problems are all fourpoint problems all right so they should be a little bit more involved there's a little bit more opportunity for partial credit all right there's also a lot of opportunity for going zero for four if you have no idea what's going on let's take a look at problem 32 a building is composed of a rectangular pyramid on top of a rectangular prism as shown in the diagram below the rectangular prism has a length of 38 ft a width of 15 ft and a height of 22 ft the rectangular pyramid sits directly on top of the rectangular prism and its height is 12 ft an air purification filter was installed that will clean all the air in the building at a rate of 2,400 cubic feet per minute determine and state how long it will take to the nearest tenth of a minute for the filter to clean the air contained in the building i wish it would have said for the filter to clean all the air contained in the building because quite frankly it's going to take almost no time at all for it to start cleaning air in the building but I quibble okay anyway now hopefully as soon as you see something like this cubic feet you know it's talking about volume and there's really two solids here there's the base right that rectangular prism and then there's the rectangular pyramid so let's kind of get these two volumes now the lower one I'm not going to even go to the formula sheet okay you've been finding the volume of boxes basically since you started finding volume all right and you should know that that's length time width time height so let's kind of let's let's go like volume one if you will right 38 * 15 * 22 and let me see if I've got that written down here somewhere i think I do i didn't okay so I'm just going to like like leave it like that i'm I'm not even going to put an equals there now how about that pyramid well if I go back over to my formula sheet again less helpful than you might think right the pyramid is 1/3 * the base area time the height all right so it's basically the same formula as the cone in fact it's exactly the same formula as a cone and it's really the same formula as the general prism but with a factor of 1/3 thrown in all right boy it'd be nice if it were 1/2 kind of like 1/2 base time height but it's not it's 1/3 okay so let's head back over here so well what do we have well the height is 12 and the base is 38x 15 so we'll get volume 2 is 1/3 of 38 * 15 that rectangular base times its height which is 12 ft all right now if I take my calculator and I'm not going to go to that because that's just kind of punching those buttons in and I type all of that in at once 38 * 15 * 22 + 1/3 * 38 * 15 + 12 or * 12 I get 14,820 cubic feet right so that is the volume that I have to move and I guarantee that if you get that far you're probably looking at three out of four points all right you're definitely getting at least two out of three points or two out of four points but probably three out of four but now I have to figure out how long it's going to take to move all of this air well I know it's being moved at a rate of 2400 cubic feet per minute right so it should be pretty easy now to know that the time is going to be the volume divided by the rate if you will so 14,820 cubic feet divided by 2400 cubic feet per minute right that's no different than if I took distance and I divided it by speed to get time now I'm doing volume divided by volumetric rate to get time that ends up being 6.175 minutes but it wants it to the nearest tenth of a minute watch your rounding so 6.2 minutes right and you really really want to watch that because again it's a shame to get this far you've calculated your volume correct you've calculated your time correct but maybe you just leave it as 6.175 and suddenly you have three out of four points so at least you know you got 75% of the credit but you know why not get 100% okay it's the proof all right so for the proof we're going to go back over to our our little supplement here and let me go view now you can you can expect a proof right to be either in the four-point section or the sixpoint section and you got to hope it's in the four-point section cuz if it's in the sixpoint section it's really long all right so let's take a look at what we're asked to do in proof on number 33 given triangle ABC triangle D E F AB perpendicular to BC DE perpendicular to EF AE congruent to DB and AC parallel to FD prove that triangle ABC is congruent to triangle D E F all right well at least it's not a CPCTC proof right so it's just a triangle congruence proof i need to prove that triangle ABC is congruent to triangle D E F let's start marking up my givens now first thing they just tell me is I've got triangle ABC and triangle DF and there ain't anything I can really do with that so now let's do AB is perpendicular to BC of course that's going to tell me that this is a right angle right de is perpendicular to EF that's going to tell me that's a right angle ae is congruent to DB i'm going to put that in kind of going like this and this that didn't work i mean nope still not wow that just that like there we go this uh and AC is parallel to FD all right so all of that now I need to use to somehow prove that ABC is congruent to DEF now the worst thing that you could do right now is to just start throwing things into the proof i always basically do the entire proof in my head if I can right before I start writing anything down all right so let's let's think about what's going on here and what more we can do now one thing I can see right away is that AB and D are going to be congruent to each other because both of them I can just simply add EB to right i can add the same length to and that's going to get me my AB congruent to my D all right so I can I can prove that AB is congruent to DE it's going to take a little work it's what's called a little mini partition proof in the middle of this but but I can do it right now how about that AC is perpendicular to F or sorry AC is parallel to FD well because those two are parallel right i could then say that angle A right here is congruent to angle D because those are alternate interior angles now watch out don't be telling me that angle F is congruent to angle C because these two lines are parallel those things are not in any way equal due to the fact that these two things are parallel all right they might be equal due to CPCTC but that's not going to come into this all right so let's take a look right angle A is congruent to angle D ab is going to be congruent to D because we can do that like kind of partition thing where we add you know EB to both AE and to DB and we also know that angle E and angle B are congruent because they're both right angles so I'm going to be able to prove now that A B C is congruent to D E F by using the angle side angle angle side angle congruence theorem all right and let's take a look at that okay now to save you uh I can't seem to move this up to save you my amazingly bad writing I'm just going to have these things mysteriously appear now the first thing I'm going to do is what I call a given dump so I'm going to just write down all of these different givens which I'm going to eventually use and then I'm going to say given all right now the first thing I'm going to do is I'm going to get those alternate interior angles congruent so in other words I am going to get the fact that C AE is congruent to FDB right and I'm going to say alternate interior angles formed by parallel lines are congruent okay great now the next thing I want to do me personally is I want to get AB congruent to DE and that's where I've got to put in this like little mini partition proof now I already know that Ae is congruent to DB because it's in my givens so what do I have to do now well I've got to really add EB to both of them all right so one thing I'm going to do is I'm just going to state EB is congruent to itself i wish you didn't have to do that but you do and that's known as the reflexive property all right i wish you could say that's the of course property but you can't all right now I need to add EB to AE and add EB to DB in other words I need to add this congruence statement to this congruent statement that's going to look a little bit like that and that is literally called the addition property right all I'm doing is adding this congruent statement to this congruent statement now I can now say that Ae plus EB is equal to AB and DB plus EB is equal to DE that is known as the partition property or some of you might call it the whole is the sum of its parts finally I can now state that AB is congruent to DE because I'm just replacing this with AB and I'm replacing this with DE all right and that is the substitution property okay and I had to do all of this to simply get the fact now that AB is congruent to DE it's quite a bit of work but you know it's done now now I need just one last piece which is I need to establish that those two right angles are congruent but before I can do that I have to establish the fact that I have two right angles so I'm going to say that angle CBA and FE are right angles right cba and FE are right angles and that is simply because perpendicular lines form right angles and now I can say CBA is congruent to feed because again very obvious all right angles are congruent and I know that's a little bit silly but it is what it is right i mean all 32° angles are congruent all 85° angles are congruent and certainly all right angles are congruent now finally I have it right because I've got an angle pair equal to an angle pair i've got a side pair equal to a side pair and I've got an angle pair equal to an angle pair so I can now say triangle ABC is congruent to triangle DF by the angle side angle congruence theorem all right and there's our proof okay which one's next let's go back here aha problem 34 let's take a look in the diagram below a boat at point A is traveling towards the most powerful waterfall in North America the Horseshoe Falls well you learn something new every day i I hope this is true the Horseshoe Falls has a vertical drop of 188 ft the angle of elevation from point A to the top of the waterfall is 15° after the boat travels toward the falls the angle of elevation at point B to the top of the waterfall is 23° determine and state to the nearest tenth of a foot the distance the boat traveled from point A to point B i I I want to be one of these problem makers who's like looking at this problem and saying "Hey I say we put the 15 in but let him just try to figure out where the 23 goes." You know what I mean you're like "Okay why?" Um so just to be very clear the 23° goes right there i again I just I just don't know i don't understand why i mean you either put them both or put put neither but anyway there's the 23 now they want us to figure out this length meanwhile I'd just be on the boat and wondering why I'm getting seasick but they want us to figure out that length now this is a classic double right triangle trig problem so this is a right triangle trig problem and there's two of them right so there's a smaller right triangle which is right here and there's a larger right triangle which is right here so let let me draw those two right triangles right one of them kind of looks like this all right with a 15° angle here right angled there and 188 ft there and then we've got sort of a smaller one which looks like this with a 23° angle and 188 ft there and let me make that actually look a little more like a three okay now I want to know the length of AB what I can certainly find using right triangle trig is I can find the length of this let me call this something different i always use different variable names for different quantities in the same problem so maybe I'll call that x maybe I'll call that y i can find both of those and then of course the actual distance right this is x this is y and then ab is going to simply be x minus y all right so let's do it now the beautiful thing is for both of these right what we have is we've got the side opposite of the angle and we want to know the side adjacent to the angle right in both situations and this is very often the case in these types of problems so because we're talking about opposite and adjacent we're talking about TOA [Music] or tangents right so I'm going to set up two different equations involving the tangent ratio in the first one I'm going to say the tangent of 15° is equal to 188 / x in this one I'm going to say the tangent of 23° is equal to 188 / y okay now in both situations I can cross multiply here i'll get x is equal to sorry put an intermediate step here x * the tangent of 15° is equal to 188 in which case x will be 188 divided by the tangent of 15° man I love it when they give me enough room to work here i'll get y * tangent of 23° is 188 so y will be 188 / tangent of 23° and remember right ultimately my final answer ab is going to be x minus y so if you really want to do this special if you want to do this so you don't have any rounding issues so that you get the final answer right away then you can literally say AB which is X - Y will be 188 / the tangent of 15US 188 / the tangent of 23 so let's do it all at once all right I'm going to have 188 divided by the tangent of 15 get out of there minus another fraction 188 divided by the tangent of 23 now don't get me wrong there's there's absolutely nothing wrong with figuring out the value of this figuring out the value of that both of them separate subtracting great there's nothing wrong with that i like it like this it's very clean we hit enter and there's our answer i mean we got to round it but there's our answer right 258.7 and what do they want me to round to the nearest foot so that's going to be 259 259 feet and again nothing wrong with and I know a lot of students would do it this way figuring out what 188 / tangent 15 is for the record let me see if I can actually find that i know I have it somewhere uh this one would be 701.6 this one is 442 nothing wrong with that at all okay let's go on i think we actually have just one more problem the part four problem on this particular exam then I'm going to do a few supplementary ones but let's take just a moment all right we've almost made it and we've only used up about 2 hours especially because we had a little bit of a technical glitch getting going for some reason we started our live stream it was going sort of well it was counting down and then it just stopped um and I wish I could tell you why but I don't know um so if you you know if you're anyway so part four problems on these exams these are six-point problems and there will be only one of them all right and that problem could be literally part A part B part C each one of them worth two points or as in this part there's going to in this question there's going to be two parts all right and they're going to roughly be broken down into three points each okay so let's jump right into it and see what they're asking us to do although yet again I'm going to go to my supplementary program cuz this will be definitely not what I wanted there we go that's better okay let's take a look at 35 triangle Joe triangle Joe i like it triangle J O E has vertices whose coordinates are J 4a 6 O -2A 4 and E 6A 0 prove that triangle J O E isosles parenthesis the use of the set of axes on the next page is optional now I put the axes right here so that we could see them anytime they say that the use of the axes is optional what they mean is we gave them to you for a reason but you don't have to use them but you'd be silly not to all right so let's do a real quick plot i'm not going to grab a ruler or anything like that i just want to throw this thing out there so 1 2 3 4 1 2 3 4 5 6 let's get a picture of Joe j O is -2 1 2 3 4 O and E 1 2 3 4 5 6 Oops come on nope it's just not going to go 1 2 3 4 5 6 fair enough whatever here we go all right so just freehanding and by the way it's completely okay to kind of freehand this because they're not going to grade anything over here you could put a giant smiley face you can doodle on this thing later on just don't do anything profane or that'll get you in trouble okay don't go there all right now we want to prove that J O E is isoclesles now an isosesles triangle is a triangle that has two sides that are of equal length and that's why I like to have the picture up here cuz I can save myself some time right they're not saying determine if JOE is isosles they're saying prove it is which means it is i just have to show that two sides are the same length and it's pretty obvious in this picture that the two sides that are going to be the same length are side J O and side J all right and I'm going to find their length using the distance formula right the distance formula being the square of x2 - x1^ 2ar + y2 - y1 squared all right so the first thing I'm going to do is I'm going to figure out the length of o which is kind of funny all by itself all right so I'm going to do the distance formula on these two points all right so I'm going to do -2 - 4 which is going to give me the sign distance between the x coordinates^ squar + 4 - 6^ squared so o j will be the square of -6^ 2ar + -2 2ar that's the of 36 + 4 and that's the square of 40 all right great so I know the length of OJ now if I'm looking at this picture correctly then I'm kind of hoping that the length of J is also the square of 40 don't feel like you have to simplify this don't make it into two of 10 certainly don't make it into a decimal just leave it as the square of 40 the only way I would simplify this is if it were a nice number like the square root of 25 I would make into five the square root of 81 I'd make into 9 etc anyway let's now figure out the length of J E so for that I want to go with the distance formula between these two and it doesn't matter which one you say is X1 and which one you say is X2 Y2 it just that's totally irrelevant all right so let's take a look at J i want to try to get it in here so you can still see it and maybe make I can't erase today all right so let's take a look at J all right the length of J I'm going to do 6 - 4^ 2 + 0 - 6^ 2 all right so J E is equal to the of 2^ 2 + -6^ 2AR which is the of 4 + 36 which is the of 40 so take a look we've got right two sides that are the same length so we should make some kind of conclusionary statement so I'm going to say triangle J O E isosles because it has two sides the same length all right great so that was sort of like the front side of the sheet and of course you know the the grid's on the back side so you might be doing this and this and this but the problem is not done right so question number 35 continued let's take a look point Y 2 2 is on OE prove that JY is the perpendicular bis sector of OE now on your copy of course you don't have these points sitting right here okay i just kind of copied them and put them here so that we would have the reference again so let's let's again draw the picture all right so 4 1 2 3 4 5 6 J -2 1 2 3 4 O and 1 2 3 4 5 6 E all right and then they say y which is at 2 comma 2 is on oe and it is and then they say prove that jy which is that segment is the perpendicular bis sector of oe now that's really two things right that means I've got to pro I have to prove that JY is perpendicular to OE and I have to prove that it bicts OE all right um I think we might be getting a little bit of lag on the video but hopefully we will get through this without too much of a problem i'm hoping that it's not a problem for you watching at home but so what we've got right is we've got to really prove two things we've got to prove that JY is perpendicular to OE and we've got to prove that JY bicts OE so let's first work on the perpendicular part and this really gets to slope right the slope right now what I'm going to do is I'm going to figure out the slope of OE first all right so OE let me figure out its slope remember that's the change in y / the change in x so for oe that's going to be 0 - 4 / 6 - -2 watch your signs here that's going to be -4 over pos8 and that's going to be a slope of - one2 all right now I also need to figure out the slope of JY okay let me make that look a little bit more like a J all right so JY let's take a look so here again change in Y over change in X all right i will have 6 - 2 all over 4 - 2 which is 4 halves which is two okay so we've got the slope of OE is - 1/2 we've got the slope of J Y which is pos2 that means that OE must be perpendicular to JY because their slopes are negative reciprocals let's get that down so I'm going to say OE is perpendicular to JY because slopes are come on negative reciprocals okay so I have one part of the perpendicular bis sector thing already done okay I've got the perpendicular part now how do I prove the bicting part well there's two ways to do it the harder way would be to use the distance formula to show that the length of O Y is the same as the length of Y all right but the easier way to do it is to simply show that Y is the midpoint of OE so let's find the midpoint of OE now remember to find midpoints all we have to do come on all we have to do is average the x and the y coordinates so I'm going to put -2 + 6 / 2 and pos4 + 0 / 2 that's going to be 4 / 2 4 / 2 which is 2 2 which is y right and so now I can say that JY is the perpendicular bis sector of OE because it is perpendicular and Y is the midpoint of O E i'd have to look at the grading guide i'm not entirely sure but it very well could be that the first part of this problem on the front side of the sheet was worth two points and this part of the problem was worth four points because this is far more involved right we had to do two slopes make a statement figure out the midpoint or two distances which would have been even longer and then make a final statement now all of that would then conclude this part of the this exam right this would have been the whole exam from last year but because this is based on a new set of standards there are some additional questions that we should look at all right and I promise they aren't going to take very long i have them on this sheet um you wouldn't have these problems because these problems basically were just published by the state so let's take a look at this particular problem in right triangle ABC below the measure of angle C is equal to 90° the measure of angle B is equal to 30° and CB is equal to 6 * the 3 the length of AB is what all right so one of the things that is new in the next generation standards compared to the common core standards is they want you to know two types of special right triangles one of them is what's known as the 4545 right triangle and the other one is the 3060 right triangle which is what we have right here all right 30 right triangle now one of the things that you should have either memorized or somehow kind of reasoned out is that in a 30 right triangle so a right triangle where you've got a 60° angle and a 30° angle you always have a relationship where if this is x if the hypotenuse is x then the shorter side is 1/2 x and the longer side is the 3 / 2 * x all right now another way of looking at that though is saying this 3 * x in other words the shorter side the hypotenuse is twice the length of the shorter side the hypotenuse is twice the length of the shorter side opposite the 60° angle right that means that the shorter side has to be just equal to six and the hypotenuse has to be 12 whoops make that look like a two they want to know the length a of a and so that's equal to 12 so watch for that not this exact problem but watch for problems that either involve a 4545 right triangle which has a different relationship and a 30 right triangle now it may not come up on the exam at all they added new things to the algebra 1 exam that haven't come up yet but they very well could be on Wednesday's exam let's take one right triangle ABC shown below ac is equal to 5 in bc is equal to 8 in and the measure of angle C is 57° what is the area of triangle ABC to the nearest tenth of a square in all right the fact that this is a non-right triangle is completely and utterly irrelevant i don't care about that fact at all what I do care about is that we've got a scenario that's what I would call a side angle side scenario in other words we have a side another side and we have the angle that's in between them and whenever you have that you can find the area of a triangle very simply by doing 1/2 of this side length times this side length times the sign of the angle in between them all right so your teacher might have presented this as something like area equals 12 A * the S of C but really it's just going to be the product of these two 12 the product of these two times the sign of the angle in between in other words that's going to be 12 * 5 * 8 time the s of 57° and the thing that's remarkable about this is that it works whether you've got an obtuse angle an acute angle a right angle it just doesn't matter if we have 12 * 5 * 8 it's got to be the sign function times the sign I think it was 57 yep times the sign of 57 there it is 16.773 let me just write that down 16.773 and that's equal to 16.8 all right one last thing that we're going to touch on i'm not going to go through the whole problem but I I do want to kind of take a look at it uh let's go view full screen move on to the last one let's take a look at this triangle XYZ is shown below using a compass and straight edge construct the circum center of triangle XYZ i'm not going to go through this whole construction because it's long and it's involved but I want to talk a little bit about this piece of terminology the circumcenter all right new to the next generation learning standards are the fact that you need to be able to construct what are known as the points of concurrency of triangles the points of concurrency now what is a point of concurrency a point of concurrency is a location where three different line segments or lines intersect that is a point of concurrency all right now what's remarkable for triangles is they've got four different types of concurrent points and you're supposed to not only know what these are but you're supposed to be able to construct them right now they asked us to construct the circumcenter the circumcenter is the intersection of the perpendicular bis sectors of the three sides of a triangle so it's remarkable and again I'm not going to I'm not going to go through this but if I take And I construct the perpendicular bis sector of this side let's say it's like that and the perpendicular bis sector of this side let's say it's that and the perpendicular bis sector of this side right they will all intersect at the same point now what's really cool is if you're actually doing this construction you would construct that perpendicular bis sector maybe that perpendicular bis sector maybe but you wouldn't have to construct the third one because it's guaranteed to go through that intersection point all right it's cool you can construct the third one but you wouldn't have to now the key is which one will show up on that test on Wednesday it might not be any of these it might be some other random construction it might be here's an angle construct the angle bis sector you know we don't know but these are sort of new they they were old then they went away and now they're coming back it's like what was old is new again kind of thing so there's also the in center and the in center is going to be the intersection of the angle bis sectors of the three angles of a triangle which is kind of cool so if you construct two of the three angle bis sectors you will find the in center as the intersection the centrid is probably the coolest one this is the intersection of the three medians of a triangle now remember a median is a line segment drawn from a vertex of a triangle to the midpoint of the opposite side and there's a lot with centrids that we didn't really see on this test you know the idea that actually if they intersect each other they partition each other into a 1:2 ratio all sorts of things like that but the centrid's really cool because if you construct the centrid and then like cut the triangle out it's the exact balancing point of the triangle really really kind of cool and finally there's something called the ortho center the orthocenter is the intersection of the three altitudes of a triangle now recall an altitude is a line segment that is drawn from a vertex of a triangle perpendicular to the opposite side and again it's remarkable the three altitudes of a triangle all intersect at a common point and again they could ask you to construct any one of these points all right and again what's kind of cool is because each one of them is the intersection of three of these things you just have to actually construct two of them now I'm not saying that they're easy right the first one's not bad because constructing a perpendicular bis sector is relatively easy constructing an angle bis sector is also relatively easy the median is a little bit harder because you'd actually have to do the perpendicular bis sector simply to find the midpoint to draw the median and the altitude is probably the hardest of them all so I certainly hope that they don't actually require you to construct the ortho center on Wednesday if they did I would almost think it would be like a six-point problem might be a four-pointer i don't know they'd probably make it two points though you know it's like so many different constructions anyway these are really really some very technical terms i think it's a little bit unfortunate that you have to memorize them and on top of that you have to memorize how to then do the constructions but they are in the standards you know it'd be great if they were on the formula sheet too but apparently the only formulas that you don't need to remember are the volume formulas for like various figures who knows anyway I didn't write the formula sheet if I had it would be way more extensive all right anyway let's head back just for a moment to the exam and come all the way back up here because we're basically done at this point um now you know you've got 3 hours on Wednesday at a bare minimum to get through this geometry exam and I know geometry can be hard it can seem like it is just a jumble of theorems and facts and techniques and all sorts of things like that and that is very true but it is also a wonderful wonderful topic that really helps to explain how two-dimensional and three-dimensional shape space sorry works you know the rules that it lives by all right you can get through this test you've got 3 hours work on it take your time make sure you've got your calculator handy make sure it's in degree mode instead of radian mode for those right triangle trig problems besides that mostly what you'll use your calculator for are routine calculations and those obnoxious volume calculations it's not nearly as much of a help for you on this exam as it is on the algebra 1 and the algebra 2 exam but you are still going to use it all right and I know that you can get through this test even though it's a brand new version of it it's probably going to look very similar to the one that we just did with a lot of similarity and congruence and parallel lines and perpendicular lines and all that stuff you can get through this especially if you worked hard all year and if you did and you're watching this review then you've done what you need to do so for those of you taking the geometry exam on Wednesday good luck i have full faith that you can make it through for right now I just want to thank you for joining me for the e-math instruction geometry regions review my name is Kirk Wiler and until I see you again keep thinking and keep solving problems