Okay, so starting this off we're having a look at working out the length AB. I've got two different triangles here, two different questions, so let's identify where AB is on the first one. So A to B is this length here.
When it comes to right angled triangles, if we've got no angles involved we can use Pythagoras to work out the missing lengths. So Pythagoras'theorem is a squared plus b squared equals c squared. Now that c represents this longest length here which we call the hypotenuse.
So this is the hypotenuse which I'll label as h. HYP, so the hypotenuse, and we've got the shorter sides, A and B, so 4 and 7. So to work out this length of this hypotenuse, we're going to do 4 squared, add 7 squared, and work that out on the calculator. So 4 squared plus 7 squared is 65. Now that is obviously the length of C squared, so to find out what C is, we just need to square root our answer. So to finish that off, square root of 65, which gives us a decimal for this particular answer.
and that gives us a length of 8.06225, and some more decimals. And again, if it asked us to round it to one or two decimal places, we'd do that in the question. Now moving on to the next one, we've actually got the length of the hypotenuse this time, so we've got 13, and we're trying to find out that this length this time, so a to b this time is one of the shorter sides.
So rather than add them together like we did down here, we're just gonna subtract them. So we're gonna do 13 squared, take away the shorter side. side which is 12 squared and again we're going to square use our answer.
This is quite a good one because it's actually one that could be done without a calculator. We get 169 for 13 squared, take away 144. And 169 take away 144 is 25. OK, so as you can see, that's given us a square number. So when we finish this question off, to do the square root of 25, we get the answer 5. This is all in centimetres, so that would be 5 centimetres. So over here, our answer would be 5 centimetres.
OK, so Pythagoras, square the sides, add them together before square rooting for the longest side, and subtract them for one of the shorter sides before square rooting. OK, so working out the size of angle X again. I've got two here and we're looking at parallel lines as you can see by these arrows here these lines are parallel.
So there are three rules that we're going to really look at when we're looking at parallel lines as well as all our normal angle rules. We've got alternate angles, corresponding angles and co-interior angles. So if we look at this one on the left here, 110 there's lots of angles we can find now.
So all around this point here we can find all of these angles. Vertically opposite the 110 is also 110 and then next to the 110 on the straight line there, remembering angles in a straight line add up to 180. these would all be 70. There we go. So it's up to you really how you actually find this final one. I'm going to have a look at finding this Z shape here, which is an alternate angle, and alternate angles are equal.
So this 70 here is the same as this 70 here. So X would equal 70 degrees in this one. And we've normally been asked to give our reasons there.
So my reason may have been angles on a straight line equal 180, and that's how I got the 70. And then I would have said alternate angles are equal, and that's really important the way you say that. So alternate angles are equal. Okay, there are also other ways that I could have done that if I split to another colour.
We could have said okay well if this is 110, well then this here is also 110 because those angles are corresponding and again I'll try and draw this on a different colour but they make this F shape okay so that 110 and the other 110 are corresponding angles so we could have done it that way and then did angles on a straight line let's have a look at this other one so 105 degrees down in the bottom here so straight away just above that we can find that this angle is 75 because angles on a straight line equal 180. Now if you have a look we've got a triangle in this question. Now there's these two symbols here which tell us that this is an isosceles triangle and base angles in an isosceles are equal and it's underneath these two little what I call little eyebrows. Okay so underneath this will also be 75 because base angles in an isosceles are equal. We can then find this missing angle here because angles in a triangle equal 180 so those two 75s equal 150 and if we take that away from 180 It leaves us with the other angle in that triangle being 30. There we go.
And there's all the angles that we found just using basic angle facts. Now to get that angle X, there's a couple of ways we could do it. We could find this angle here and then do angles on a straight line.
Or we could again identify that actually we've got one of these alternate angles. going up making this shape again. So that's an alternate angle with the 30. So that's 30 degrees as well. So for this one we'd say x equals 30. And again we'd probably give this same reason over here that alternate angles are equal.
Unless of course we took a different approach which we could have done. Again, lots of different ways that you can tackle these questions, but a few there that you need to make sure that you try and remember. So working out the size of the angle between this regular octagon and regular hexagon.
So there's different ways that you can approach this, but we need to know what the angles inside a hexagon all add up to and what the angles... inside an octagon add up to. There are different ways of doing this, you can do the 180 list rule so you can keep adding 180, so 3 sides is 180, 4 sides is 360, 5 sides is 540 and you can keep doing it this way or you can apply a bit of a formula.
So I'm going to go for the formula even though I've started writing this down but that is an option there that you can use. So for 6 sides in a hexagon we take away 2, so 6 take away 2 which is 4 and then 4 times 180. It tells us the total amount of... oh, that should be a time sign.
Let's change that. 4 times 180 gives us the total amount of angles in a hexagon, which is 720. So all the angles inside a hexagon equal 720, and there are 6 equal angles in there. So if we divide that by 6, 720 divided by 6 gives us...
120 degrees. So all of these angles in the hexagon are 120. Obviously we're concerned about this little area here in this question because we're trying to find the angle between them but we've got the angle in the hexagon and that's really important. When you see these words regular hexagon and regular octagon always just work out the interval. interior angle and label it all over the diagram.
So moving on to the octagon, same process, eight sides, take away two is six, and then times that by 180. There you go, and six times 180, which is okay if you have a calculator, if not you just have to work that out, but six times 180 is 1080. And again it's got eight equal angles, so if we divide that by eight, 1080, divided it by 180 gives us 135 degrees, and again we can just label that all over the diagram. 135, 135, and lots of these, but I'm concerned again with this little area here. So, angles around a point equal 360, and at the moment we've got two of these angles, kind of like a jigsaw. So if we add these two together, 120 plus 135, at the moment we've got 255 degrees, and we need it to add up to 360. So to finish that off, if we do 360 and take away that 255 that we've just worked out there, we get a missing angle of 105 degrees, and that's it.
That's our final answer there, 105 degrees, which we could always label on the diagram as well, 105. And there we go. So whenever you see these questions where it says regular octagon, regular hexagon, just spot these bits of language here, just work out their interior angles and label it all over the diagram. So when we're looking at circles, there's two formulas that we need to know. Circumference equals pi times diameter, and area equals pi r squared.
Remember the units for area have squared in, don't they, when we used to... centimeters we have centimeter squared so I always like to remember the little formula for area there is the pi r squared the one with the squared in so we're going to work out the area and the circumference of this circle so to work out the area we're going to do pi r squared if we start with that so the area equals pi r squared and we just need to know what the radius is so over here the diameter is eight so we're going to first half that for the radius so I'll just write r equals four and we'll just got to type that in so r pi times four squared four squared is 16 so it'd be times 16, and you can just write that as 16 pi. So if it asks you to give your answer in terms of pi, you can just do that, but we want to type that into our calculator if not. So 16 times pi will give us a decimal, which is 50 point, I'm going to round this, but obviously only round it if it says two in the question, it'd be 50.27 if I round it centimetres squared, and that'd be an area there to two decimal places.
Let's have a look at working out the circumference. So circumference up here is pi times diameter. Now we've been given that our diameter is 8, so to get the circumference we just do pi times 8 and we could just say 8pi.
So we could just say 8pi which just means pi times 8 but again if it asks us to work it out we're going to do pi times 8 on the calculator. So 8pi, press equals and we get... And if I round this to one decimal place this time, I get 25.1. And circumference is a distance.
It's around the outsides, which is just centimetres, 25.1 centimetres. But again, rounding it to how the question asks. And remember, you can write it in terms of pi, like down here.
16 pi and up here 8 pi as well. Okay, so working out the area of a circle sector. Now we're gonna do this in a very similar way to what we did with the circle. We're still gonna do area equals pi r squared, but because it's a fraction of a normal circle, we're gonna times it by whatever fraction of the circle that is. So it says over here, It's 102 degrees, so it's 102 degrees out of a normal 360 degree circle.
So we're going to plug the numbers into there and we'll get our area. So it says down here the radius is 8, which is quite nice, so it's pi times 8 squared. multiplied by this fraction of the circle 102 over 360. You can type that into the calculator so pi times 8 squared times 102 over 360 and that gives us an area here of 50 56.96754679. And I'll tell you what, let's round it to two decimal places. So that would be 56.97.
And again, units centimetre squared. And there's the area of the sector. Again, you could actually work out this here as well if you're asked to. It's called the arc length. So we would do the same thing, but we'd use our circumference formula.
So we'd do the circumference equals pi times diameter, multiplied by the fraction. there which would be 102 over 360. So we could work out both just using the appropriate formula in the same way. Right working out the area of a trapezium there's a formula for the area of a trapezium it's a plus b over 2 times by the height or it can be written in a slightly different way you can say half a plus b again times in by the height. Now a and b are the parallel sides which in this case is the 5 and the 9 so we add those together to start with so 5 plus 9 we halve that so 5 plus 9 is 14 half of it is 7 and then we times that by the height and they've just got to be careful here because you are given these two lengths on the side and they're really there just to throw you they're not the height they're diagonal lengths this 4 in the middle is what we want the perpendicular height from the base indicated by that red arrow there so 4 is our height so we're going to do 7 times 4 which gives us 28 and it is an area it's in centimeters again so centimeter squared so just be careful with the trapezium there that you do identify the perpendicular height from the base. So we've got to work out the surface area of this cuboid.
Now there's only three surfaces we can see. We've got this one on the front but there's also one on the back that's exactly the same, back here that we can't see. We've got the one on the top, number two, which is the same as the one on the bottom, and we've got this one on the side here which is the same as this one that we can't see on the side.
So if we work out the area of these three faces we can just double our answer to get the ones on the back as well. So I'm going to do this in steps. I'm going to do the face number one which over here is 8 times 12. So 8 times 12. times 12 is 96 centimeters squared.
Face number two, the one on the top, we've got 14 along there and it's not labeled but this is 12 as well, the same as the bit below just down there. So that's 14 times 12. So 14 times 12, we've just got to work that out, and that's 168. So that's 168 centimetres squared. And then the last one, face number 3 on the side, it's not labelled, but we can label that the height's 8, and the length going backwards is 14, the same as the length on the top there.
So we've got 8 times 14. So 8 times 14 for our last one. 8 times 14 is 112. So that's 112 centimetres squared. And if we add all of these up, 112 plus 168 plus 96, we get the answer 376 centimetres squared.
Now you've just got to be careful at this point. Obviously, don't forget there are two of each face. So we just need to double our answer or times that by 2. So times that by 2 gives us a total surface area here of 752 centimetres squared.
And there's our final answer for the surface area. Working out the volume. of a cuboid is quite nice and simple. I do like to do it in a particular way though just so it's the same for all of these shapes where we have a constant cross-section. So this cross-section I'm gonna have a look at this front face just here.
So I'm first going to just work out the area of that which is 8 times 12. So 8 times 12 gives me 96. That is the area of the cross-section, 96 centimetres squared. And to get the volume we just times it by however far back the shape it goes which is the same as that 14 right there. So times that by 14, 96. times 14 will give us our volume.
So 96 times 14 is 1344. And it's a volume, so we put centimetres cubed. And that's how you work out the volume there. That works for any shape. We've got a constant cross-section.
If you imagine, sometimes we have these triangular prisms, and sometimes we could even have a trapezium face prism there. Okay, anything with that constant cross-section, you always work out the area of this front face to start with, the cross-section, and times it by the distance it goes. back.
That same logic can be applied to a cylinder, so our cross-section this times a circle. So all we've got to do is work out the area of the circle, remembering area equals pi r squared. So working out the area of that circle, it's pi times 4 squared, which again is 16 pi.
We can write that as a decimal. Now what I'm going to do is I'm just going to leave it at 16 pi for the moment. No any rounding errors here, but you can write it as an estimate at this point, but I'm just going to leave it as 16 pi. Now once you've got the area of the cross section, you just times it by how far it goes through the shape, which in this case is this 15. So all I've got to do is times that by 15. So 16 pi times 15. which comes out as 240 pi, which is also a decimal, so it just depends on how the question wants the answer here.
So it's 240 pi, or if I write that as a decimal, remember that S to D button or F to D button on your calculator, depending on which... which one you have. You get 753. I'm going to round it to two decimal places again.
0.98 centimetres cubed for volume. And there's the volume of my cylinder. So surface area of a cylinder is a little bit more complicated because we've got these two circles to work out. And then we've got this shape wrapped around the outside of the cylinder.
So the area of the circle we've actually already worked out. Pi times r squared is 16 pi. And that equals the circle. Okay.
And not forgetting, we've got two of those. So let's times that by 2, which gives us 32 pi, which again is a number, but I'm just going to leave it till the end for the area of my two circles. Now, the next thing that I have is this shape wrapped around the outside.
And if we were to unravel that, we'd have a nice big rectangle. And the height of that rectangle there is 15. The length of the rectangle is actually the circumference of the circle. So we just looked at circumference of a circle. The circumference is pi times diameter.
So I'll do 8 times pi or 8 pi. pi of the diameter there being 8, it's given us a radius of 4, so we could put the diameter equals 8, plus 8 pi for the length. And we work out the area of a rectangle just by doing length times width.
So to work out the area of this, we'll do 15 times the 8 pi, and that gives us 120 pi for the area of that rectangular part wrapped around the outside. Now what we've got to do to finish this off is add these two together, so we have 32 pi. Add 120 pi, and in total that gives us 152 pi.
Again, we could just write this as a decimal, but 152 pi comes out as 477.5 to one decimal place centimetres squared for surface area. And there's our final answer. So you've just got to remember you've obviously got the circles, top and bottom, and then to work out the area of the curved part, you need to think about that circumference there being the length of the rectangle. So we've got that these two shapes are mathematically similar, work out the length of A and B. Now that there, mathematically similar, means one is an enlargement of the other, which means there's a scale factor between them.
So if we can identify a similar side on both of them, which we've got here for and 10, we can find the scale factor. So we always just do the big one divided by the little one, so 10 divided by 4, and 10 divided by 4, 4 fits in twice and a half, 4, 8, and then an extra 2, so it's 2.5. You might just get a whole number there, but there's my scale factor, and I always label that, I just say sf, so the scale factor is 2.5.
So to get from a small one, let's have a look at a to start with, we've got to get from 7.5 back down to a, so that's going from the big shape down to the little one. one. So to get from the big one down to the little one, we divide by the scale factor, so divide by 2.5, and 7.5 divided by 2.5 will be 3 centimetres.
And that's actually going to show the same for this other one here, look, because this one's 3 as well. But to get from 3 to b, we actually just times by that scale factor, so times by 2.5. And 3 times 2.5, which we already have there, is 7.5 centimetres. Okay, so you just got to remember, divide from the big to the small. one and then multiply by that scale factor to go from the small to the big one.
It's all about identifying that scale factor that we did up here, that 2.5. Here we go, working out some bearings. So write down the bearing of B from A.
So if I'm at B and I want to go from A, then I'm standing over here at point A. Now there's three rules for bearings. They're always measured in a clockwise direction. There's always three digits.
I always measure it from north. So they are my three bearing rules. So look in here, if I'm standing facing north, which is shown by the north line, I turn clockwise, I'd have to turn this full 60 degrees.
So that thing there, 60 degrees, but it doesn't have three digits, so I stick a zero in front of the 60. So it'd be 060 degrees or 060 degrees. Now the next part of the question says write down the bearing to A from B. So that means now I'm standing at B.
So let's just get rid of these little bits. So this time, A from B means I'm standing at B. Obviously following these rules, if I face north, I've got to turn all the way around this way to get back to the direction of going back to A. So let's have a look at how we can do this. Now what I quite like to do with these sorts of questions is to think about...
If I can extend the line slightly. So if I extend the line, because these two north lines are parallel, we've actually got a corresponding angle there. So this bit here is actually 60 degrees.
So we can think about other angle rules while we're doing this. So from that bit just down to my little dotted line I've put in is 60 degrees. Now because this is a straight line, this extra bit here is an extra 180 degrees.
So actually what I've got there is 60 to get me down to here, and then an extra 180. And if we add those together, 60 plus 180, we get a total of 240 degrees. So my answer for that would be 240 degrees, which does have three... digits so I can just leave it like that.
Of course there is another way that you can look at this as well. We could think about these two angles here as being co-interior that add up to 180, these two add up to 180. So this one here must have been 120. So actually I could take away 120 from 360 because that full turn there is 360 and we've got this little bit here as being 120. So I could actually do it like that as well. So there's two ways that we could approach it. Okay, so translating a shape by a vector. Now that top number in the vector means left and right, and the bottom number means up and down.
Okay, thinking about positive numbers moving towards these positive numbers on the axes, and negative numbers moving towards the negative numbers as well. So 4 moves it right towards the negative numbers by 4, so all I ever do is pick a point, you can pick any, and I move it right by 4, so 1, 2, 3, 4, it's going to end up there. The next thing I do is look at the other number, which is minus 2. 2 moves down towards the negative numbers.
So 1, 2 down, and that means that new point is going to end up just there. So all I have to do is draw this triangle in exactly as it is from that top point. So it goes down 3 and across 2, and there we go.
We can just draw it in like that. Not forgetting as well, you might actually be asked to describe a translation, in which case you'd say it just like it says above you. You'd say, translate the shape by a vector, and you'd say what the vector is. Okay so reflect the shape in the line x. x equals 1. Now this is your x-axis and this is your y-axis, so x equals 1 is right there.
And the only line that you can do from that point, other than going across the x-axis, and it doesn't say the x-axis, is to go up and down. So x equals 1 is this line here. If you think about any coordinates on that line, the x coordinate's always 1. So that's our reflection line there. Now we're going to reflect it in that line, which is quite nice and easy to do. Pick a point, and we go 1, 2 to the line, so 1, 2 away.
And just follow that process for all the the points so picking this point that is four away so another four away gets me to there and the same one at the top that's four away and four away gets me to there and then just joining it all in obviously using a pencil for this one and that would be my triangle I haven't drawn that in particularly neatly there we go okay not forgetting as well though you could have a line it could say the line I don't know y equals three which would be across at y equals three which would be across here there's another couple of lines that we could have as well. We could have this one, y equals x. The line y equals x is the diagonal line here where all the x and y coordinates are equal.
So we could have that one as well where we have a diagonal. There's one more that we could have as well. Let's pick a different colour for this.
We could have y equals negative x, which is very similar to the one above, but it's pointing in the opposite direction. If I do that in blue as well, that's going down this way. So we could also have to reflect in a diagonal line. But again, you just follow the same process, counting diagonally.
Okay so rotate the shape 90 degrees clockwise about the point minus one zero. So first thing to do is obviously to locate that point which is minus one across, nut up and down so minus one zero is just there. The next bit this is a lot easier if we have some tracing paper.
So you stick your tracing paper nice and flat over your shape, trace it in, so trace over your shape, stick your pen on that rotation point and just rotate the tracing paper not moving it away from the pen. So we would rotate it 90 degrees. So it would be facing this way and just making sure it stays nice and flat. And it would rotate the shape.
Let's have a look. It would go to here on your tracing paper. And then you would just lift your tracing paper up nice and carefully and draw in at that spot.
And again, the same process if you have to describe one of these, you stick your tracing paper over the top and just move your pen into different points. Pick in places like here and here and moving it around until you get the one that matches when you swizzle your paper around 90 degrees. And remember, you want to be careful. remembering obviously in your description you need to put the amount of degrees, the direction that it's gone in and that rotation point along with the words rotation. Okay so enlarge the shape by a scale factor of 2 from the point.
So let's identify that point first and then we're just going to enlarge it by a scale factor of 2. So what I do is I pick a point on the shape, normally the closest one, let's say this one And I just think, right, how do I get there from the point? So we go two across and one up. And I'm just going to do exactly the same again. So two across and one up.
And that there is a scale factor of two. The first movement to the shape is my first scale factor. The second is... my second scale factor there. So if I had a scale factor of 3, I'd do the same again.
I'd go 2 across, 1 up, and that would be a scale factor of 3. But this one has just set a scale factor of 2, so I'm going to leave it there. Now if I have a look at the shape that's drawn originally, it's 2 along and it's 3 up. and this is going to enlarge it by a scale factor of 2 which doubles those numbers so it's no longer going to be 3, it's going to be 6 and it's no longer going to be 2, it's going to be 4 so from this point here that we've got to, all I need to do is redraw that shape in remembering it was the bottom left corner, this one here so we just need to go 4 across, 1, 2, 3, 4 and 6 up, 1, 2, 3, 4, 5, 6 and then join it all up, nice and neat with a pencil and a ruler and there it is enlarged by a scale factor of 2 just remember if you have to describe one of these, it's a really nice way of doing it it as well, you just pick the two corners, get your ruler and a pencil, join them up really nice and neat, pick another one, match it with the appropriate one, and then again join that up really nice and neatly, and it will show you down here, look, where that enlargement point came from. And you can find the scale factor by just looking at the sides, so the side length is 6 here, and the side length is 3 there, and 6 divided by 3 gives us a scale factor of 2. So if it was already drawn in, we found our enlargement point, we just say it's an enlargement, a scale factor of 2, and we can just go back and scale factor 2 and from this point down here just like it did in the question. Enlargement, scale factor 2 and from this particular point.
OK, so moving on to some geometry. So we're going to start off with this volume of a frustum. Now you would be given the volume of a cone formula and that is volume equals one third pi r squared h where r is the radius and h is the height. So if we have a look at this...
frustum here, the only thing it doesn't give us is this length across the top, but if we have a look, the height of the little cone on top is 20 and the height of the big cone if it was still there is 40. So it's half the size, so the diameter there is 15. Now in order to work out the volume of a frustum, you've got to work out the volume of the the volume of the big cone if it was there and we're going to take away the little cone that's been chopped off the top. So I'm going to start off with the big one so I'm just going to label this the big cone and I'm just going to plug the numbers into my calculator so it's one third times by pi times by the radius squared now careful because it's not 30 it's 15 half of that so 15 squared times by the height which is 40. So typing that one into the calculator and I can leave that in terms of pi or I can write out the full decimal and it comes out as nine. four two four point seven seven seven nine six one centimeters cubed now for the small cone on top there we go we can work out the volume of that as well so it's one third times pi times the radius squared careful it's half of 15 so 7.5 squared and times that by the height of that cone which is 20 and that gives us a volume and again if we just type that in let's have a look so one third multiplied by pi multiplied by 7.5 squared times 20 and that gives us a volume again I could leave it in terms of pi but 1178.097245 and again that's centimetre cubed.
So in order to work out the volume of what's left when the little cone's been cut off the top we just need to do the big cone subtract the volume of the little cone there so I'm just going to subtract this away from that so subtract that. And again I'm just going to do that on the calculator. So 9424.777961. Take away that answer there. Leaves us with a total volume of 8246. going to round it to two decimal places although you would be asked how to round it in a question but I'm going to go for 0.68 centimetre cubed.
Just be careful what the question asks for obviously if it said one decimal place it'd be 0.7 it might even just say four or three significant figures in which case you just got to make sure you just round it how the question questions asked. But there you go, that's using the volume of a comb formula on a frustum now, which is a little bit more interesting than just doing a normal comb. Next one here we have a hemisphere, and it says work out the volume of the hemisphere.
Now again we'd be given the formula to work out the volume of the hemisphere, and the volume of the hemisphere is 4 thirds pi r cubed, and that is our final volume there. Oh, just make sure that says, that's a rubbish three, let's change that. 4 thirds pi r cubed.
So all you've got to do is stick the numbers in there. in the formula again. So if I want to work out the volume, all I have to do is in the calculator I do four thirds multiplied by pi multiplied by eight cubed, and we'll get our answer here straight away. So fraction button, four thirds multiplied by pi multiplied by eight to the power of three.
And we get a volume here of two. 1.46 again if I round it to two decimal places, centimeters cubed. Just remember obviously to write all the numbers down and only round it once you've been asked to.
Okay I could be asked to work out the surface area though, so I've not included it in here but again you'd be given the formula for this. If I was to work out the surface area, the surface area of a sphere Okay, so once you've got your total volume of your sphere there, just remember this is a hemisphere, so it's half the size of a sphere, exactly half, so I just need to divide my answer by two here, and once we divide that by two, we get a final answer of, let's have a look. 1072, so 1072.33.
Again, I've rounded it there, centimetres cubed. And that would be your final answer here for the volume of a hemisphere. Not forgetting as well, you could be asked to work out the surfaces.
area and if you're asked to work out the surface area again it gives you the formula of a sphere and the surface area of a sphere is 4 pi r squared. So again if I was to work out the surface area of this one it's a little bit more complicated but only in the sense that we work out the full surface area so we do 4 times pi times 8 squared. If I just go for this let's work it out 4 times pi times 8 squared that gives us a surface area of a sphere being 8 and it's 0.25 as well.
Again we'd have to halve it because it's half. Okay so half of that would give us a surface area of 402, 0.12 and that is obviously a surface area so it's centimeter squared but if you're working out surface area you've got to make sure you don't forget as well the circle sitting on top there because it's the total surface and the area of a circle is pi r squared so we can work out the area of the circle as well. well. If we just do that, pi times 8 squared, we get an area of the circle there of 201.06, and you would just add these two numbers together to get the total surface area. So if you add 201.06, we get a total surface area of 102.12 added to that.
It gives us 603.18 centimetres squared. Again, so I know we were looking at volume there, but just thinking about if you had to do the surface area as well, how you'd approach that when you've got a hemisphere and you've got to make sure you halve them and add on the extra circle. Okay, so we've got some similar shapes. Two cones are mathematically similar.
The height of cone A is 4 and the height of cone B is 10. Now you can always imagine what this is going to look like. You can always draw yourself a picture if you haven't been given one. And we've got a bigger cone, so we've got a height of the smaller one being 4 and a height of the larger one being 10. Now straight away you can work out a scale factor between these two. If they are mathematically similar, it just means one's an enlargement of the other. So I can do the bigger length, 10 divided by 4, and it gives us a scale factor of 2.5.
Now when it comes to scale factors, that is our length scale factor, I always label it length scale factor. Now there's two more scale factors we could want to look at, and that's the area scale factor, or the volume scale factor. And to get from the length scale factor to the area scale factor, you square it.
Okay, so we square the scale factor. To get from the length scale factor down to the volume scale factor, you have to cube it. So if we just have a look at this question here, it says the volume of Kona is... 40 work out the volume of cone B. So that's cone A is the smaller one in this case I should probably label that A and B and it says the volume of this one is 40. So in order to find my volume scale factor I need to do 2.5 and cube it to get a volume scale factor.
So we're going to do that on the calculator 2.5 cubed gives me 15.625 and that's my scale factor in terms of the volume. So the volume of cone B is going to be 15.625 times bigger. So in order to get that that volume there.
We're just going to multiply 40 by 15.65 So times that by 40 and we get a volume here of 625 centimetres cubed. There you go. Just remembering as well, if we were going from the bigger down to the smaller, we'd have divided by the scale factor.
But as we're going smaller to bigger, we're going to multiply. So let's have a look at a different one. So a similar question, two cylinders are mathematically similar.
We've got cylinder A and cylinder B. Now, so cylinder A is 160 and cylinder B is 20. So A is my larger one. So if I just draw a little basic diagram of this, we've got A.
And we've got B. There we go. And this time it says, it actually gives me the volumes. So it says the volume of A is 160 and the volume of B is 20. So again, if I find a scale for that, I can do the bigger one divided by the smaller one.
160 divided by 20 gives me a scale factor of 8. Now, if you're thinking about this area scale factor, length scale factor and volume scale factor, There we go. I've got this time, I've got the volume scale factor and that is 8. But I can't get from volume straight to area. Now if you remember to get from length to volume we had to cube it.
Okay so to get back from vol- to length we have to cube root it. So if I do the cube root of 8 that tells me the length scale factor is 2. Now the question here is saying it then starts it asks me to work out the surface area so I need the area scale factor and if you remember just from the last one to get from area, it's length down to area, we square it, so that'll be 2 squared, which would be 4, so my area scale factor is 4, okay, so whenever you're given an area or a volume, you have to go back to length first, to then get back to an area or a volume, so my scale factor is 4 for the area, and it says the surface area of cylinder A is 40, so that's the bigger one, I'll just write area rather than surface area, so to get from the bigger one down to the smaller one, we're going to have to divide by that scale factor, which is 4. for our area and 40 divided by 4 gives us 10 so we get a surface area of 10 centimeters squared. Right there we go there's thinking about some similar shapes.
Okay so we've got a transformation here it says enlarge shape P with a scale factor of negative 1 half with the center of enlargement 0 0 so always mark out your center of enlargement first and an enlargement with a negative scale factor and a fraction here means that we've just got to do this very very carefully. Now the first things first you need to pick a point on the shape and just figure out how you get from the centre of enlargement to that point there. And if I just count that, that's 1, 2 to the right, and 1, 2, 3, up.
Now all I need to do to do this, and it's going to be quite small in this diagram, is I just count in the opposite direction by whatever the scale factor is, that's what the negative is. So rather than going from 0, 0, rather than going right and up, I'm going to go left and down. But it's a scale factor of a half, so I need to also halve those distances, just as if it was negative 2, I would double that. those distances but for negative a half so I'm going to go left one rather than two and I'm going to go down rather than going up three I need to go down 1.5 or one and a half so one and a half gets me to there there we go so that's one and a half I'm going to do the next one in a different colour so I'm going to pick this point here and let's just think how we get to that let's get rid of some of these markings here there we go so to get to that we go one two three four across and one, two, three, up. So I'm just going to halve that again.
I'm going to go two left rather than four right. So one, two, and then that one and a half down, which again gets me down to there. There we go.
And that is that point. Then again, I just need to repeat the process for the last one. You might be able to start to do these in your head once you've had some practice, but if we do the last point here, being the top one, I should just see how we get there. So it's one, two, cross just like it was to that red one and then we've got to go up one two three four five six seven so half of that's going to be 3.5 so I'm going to go one to the left okay so we need to go down 3.5 one two three and a half just there there you go obviously you can do all this in pencil to keep this all nice and tidy but when you've got a negative scale factor you're just going to go backwards in the opposite direction by whatever that scale factor is obviously join it all up nice and neat and there we go and that is that one there just always go back and double check so it was seven up so we went 3.5 down that's absolutely fine the other one was three up and we went 1.5 down half of that and the other one there perfect okay so you can see with these it does actually actually get bigger and bigger, it gets bigger or smaller, and it rotates 180 degrees, and that's what a negative enlargement does to a shape. Okay, so we're going to have a look at another one here.
Okay, so it says describe the single transformation that maps shape B onto shape A. This is quite nice, hopefully you can tell that it's an enlargement, okay, because obviously one's got bigger and smaller, and we're going from B to A. Now what you can do is you can get your ruler, and I've got to think, I'm going to have to try and do this quite carefully, but you get your ruler, and you join up the similar points.
Now the similar points are here. and here. Okay, obviously it's been rotated 180 degrees, and if you get your ruler and join that up, you end up with a little line like that. And then you just do that for all the other points.
So I can probably get away with just doing two, we'll have to wait and see. So if I join up these two similar points here, there we go, they join up just like that, and what you'll find is that there's this little crossover point, and that tells you where the centre of enlargement is. So I know that it's an enlargement, we've already got that, so I'm going to have to state it's an enlargement.
Okay, so it's an enlargement. I've got to get what the scale factor is, I haven't got that quite yet but I'll have a look. We know it's negative as it's getting this 180 flip, so we know it's negative, we'll come up with the number in a sec. And I know where the centre of enlargement is, so centre of enlargement. I'm going to write down what that coordinate is and it's 2, 2. So centre enlargement, 2, 2. Now to count that scale factor, we're just going to count this side here maybe and this side here and just see how much bigger it's got.
So that's gone from a length of 1 to a length of 3. so that is a negative 3 scale factor so it's an enlargement scale factor negative 3 center of enlargement 2 2 and you can use that approach for any of these enlargements whether it's negative or not also you might just want to note here that if it had been from from A to B, it would have been the other way around. It would have gone from 3 down to 1 and my scale factor instead, if I was going from A to B, it would be exactly the same description there but my scale factor would be negative 1 third because it's getting 1 third of the size going the other way. Okay, so two different ones but the description would be exactly the same there.
Okay so we're gonna look at some circle theorems. For the purpose of keeping this video as short as we can I'm only going to go over a couple of them and just how to approach the question. If you want to have a look at every single circle theorem and how to tackle every type of question you'll have to look at my circle theorems video. Now some information to go along with this I'm going to say that A and B are tangents.
There we go. So when it comes to tangents they meet at equal length that's one of our circle theorems so these two two lines here are equal length, meaning that AB, that triangle there, is an isosceles, meaning that the base angles, these two, are the same. So there we go, we can start to figure out some of the angles here, because if we do 180, the angles in the triangle, take away 86, it leaves us with 94. That we can split in half to share it between our two triangles there, so 94 divided by 2 is 47. So both the angles at the bottom of the triangle there are 47. There we go, and as with all these questions, it always says...
to state your reasons. So you would say tangents meet at equal length, therefore this triangle is an isosceles, okay, and writing that down. Okay, on to the next bit.
We've also got, and it doesn't say it here, but O is the centre of the circle. Now O to B there is a radius, looking at this line just here, and a radius meets a tangent at 90 degrees, meaning, and always draw this on the diagram when you see a tangent meeting a radius, we've got a right angle just here. So we can work out the value of x because if the full angle's 90. We're going to do 90, take away the 47 and that leaves us with 43 degrees. The important part with this question is to make sure you write down all those reasons.
So we would write that tangents meet at equal length, isosceles base angles are the same, and then our final reason for this bit of the working out, the tangent meets the radius at 90 degrees, and therefore we could do 90 take away 47. OK, a different circle theorem question with some different circle theorems within this one. So if you have a look, we've got these points A, D, C, B around the circle, and that forms a quadrilateral. And these are called acyclic quadrilaterals.
And the. The rule here is that opposite angles add up to 180, so we've got the 70 over there, so opposite that is the y. And opposite angles have to add up to 180, so we could do 180 take away 70, and it gives us our answer of 110 degrees.
And our reason for that, which we would have to write down again, is that opposite angles in a cyclic quadrilateral add up to 180. The next one here is to find this angle x, and that's going to involve one of our other circle theorems. And when you've got these points made at the centre, and I'm just having a look at points d and y. and B, I always do this with a little highlighter, but if I make this angle 70 here I can also from the same two points make this angle here at the center and that's one of our other circle theorems.
Angles at the center are double angles made at the circumference from the same two points. So to work out angle X and this first one we did was Y, to work out angle X we would just do 70 times 2, double it and that would give us 140 degrees and again that that would be accompanied by the reason angles at the centre are double angles at the circumference when they're made from the same arc. So our two answers here are 110 degrees and 140. Again, just a few little bits of circle theorems there to be having a thinking about and making sure that you are writing down all those reasons. That's absolutely crucial for these.
Okay, so we're on to some congruent triangles. It says prove angle Ab is congruent to CB. Now, if you have a look, Ab is the triangle on the left and CB is the triangle on the right.
right. Now in order to prove triangles are congruent we need to look for similar sides and similar angles. Now the first one that pops into my head is the line B to D, this line in the middle.
Okay I'll highlight that. This line B to D in the middle is the same for both triangles. So I could prove a side straight away.
So I can say a side, I've got B is the same in both. So both triangles already share a similar side or exact same side. B is the same in both. And remember, we're trying to prove that they're congruent, which means that they're exactly the same. So let's get rid of that.
We can have a look at another one now. Can we see any other sides that are the same? Well, we've got these little symbols. and I've already got an arrow pointing to two of them really, we've got these little symbols on all of them.
Which means that all of those lengths are the same length. So we could state any of these are the same now. We've got AB, this one here, number one, is the same as B. number two so I can do another side so I can say that AB equals B and actually that's given to us in the question so I would just have to say given in question There we go. And actually using that same logic there, we could actually prove that these two down the bottom, which are labeled 3 and 4, are the same as well because they're all the same length.
So I could prove another side. So I could say that A equals C, and again that's given in the question. There we go. So what I have to do is state why they're congruent there.
And I can just say, and I'm going to write this down in a box, and I can say, therefore, they're congruent. OK, because the SSS rule here, so side, side, side, congruent, and we're using the SSS rule. OK, so there's other things that you can look for as well. You can have a look for angles.
Obviously, there's a few of these that you can use. This is just an idea of how to approach it. how to lay it out, just labeling what you found.
But because these are obviously isosceles triangles, so this one on the left here, we've got two sides at the same length, so we know that the base angles are the same here. Therefore, as they're all the same length, those base angles must also be the same as the base angles. the base angles on the other triangle.
So we could use that as well. And if those angles are the same, then this angle at A has to be the same as the angle at C, so actually we could prove all of them. So there's a lot of different types of questions here, a lot of different proofs, but there's just an idea of one of them and how to lay it all out. Okay, so we're going to have to work out some missing lengths in this triangle. Now it's not a right-angled triangle, so to find missing lengths and angles we can use either the sine rule or the cosine rule.
And these are rules that you're going to have to remember. But we're going to have a look at one to start with and how we know when to use them. use it. Now the first thing you do is you try and identify in a triangle for first of all what we're looking for which is AB.
So let's label that X. Now what I look for straight away is do I have pairs of opposites. So I have this pair of opposites here and I've got both of those and then I've also got X opposite to this angle. So I don't have X but it's in one of my pairs of opposites. So when we've got this scenario where there's two pairs of opposites we can use the sign rule and we only ever need part of the sign rule so I'm just going to use A over sine A and equals b over sine b.
Also equals c over sine c, but we only ever need to use two of them here. So let's have a look. We need to label this up.
Now I'm going to completely ignore the letters that are on the actual triangle itself. I'm going to label this angle a, which it already is, and this side little a opposite that, and then this angle b, and the one opposite little b. And all I need to do now is stick all the numbers into the formula. So let's have a look.
a is 12, so it's 12 over sine the one opposite that, 15. So let's have a is equal to b which is our x over sine b. Obviously you should already know there's two variations of this formula we could have it flipped over so we could have sine a over a equals sine b over b but this is our one for side lengths and we know that because our unknown piece is on the top so we're able to isolate this now quite easily. Sorry I've written b there it should be 20, there we go, sorry there we are 20. So what we need to do is times both sides now by sine 20. What you could do is work this out on your calculator and times your own.
by sine 20 but I'm going to multiply it straight over so I can type it all in in one go so times by sine 20 and if we do that we get 12 sine 20 it goes on to the top there over sine 55 there we go and all we have to do is type that into your calculator obviously just being careful that you put these angles in brackets some calculators are going to need you to do that so if we type that all into the calculator again not forgetting you could just work out 12 over sine 55 first and then times it by sine 20 but I'm just going to go for it like this. So 12 sine 20 on the top, closing your bracket, over sine 55. And on the calculator, just writing down what you got, you got 5.010353998. Now a question would normally say how to round this.
So if we imagine it was two decimal places for this one, it would be 5.01. And it's a length, so centimetres. Okay. So just obviously, just be careful of the question.
and says let's have a look at an idea where we've got to find the angle. Okay, so in this question, let's have a look. Work out the size of angle BA. So let's identify that.
That is here, BA. Okay, so we're going to use the formula the other way around this time. So sine A over A equals sine B.
That does not say sine B. So I'm B over B. Okay, so plugging in our numbers, let's just label it up.
So let's call this one A as the A's next to it, that's fine. And again, I'm just going to write over this one, I'm just going to put B and B. Okay, just because of the way I've written my formula.
So then sticking in all the numbers. what have we got? Sine a is sine x, so we have sine x over 20 equaling sine 43 over 14. Okay, so exactly the same approach as we did before.
we can isolate the sine x by times in both sides by 20. And again you could work that right hand side out and times it by 20 but I'm just going to go stick it up the top there. So we end up with sine x equals 20 sine 43 over 14. If we type that into the calculator now what do we get? 20 sine 43 over 14 and we get an answer here.
Let's write it over here. So we get sine sine x equals 0.97428 and a few more decimals. And obviously, just like normal trigonometry when you're doing Sohcahtoa, to get the actual x here, we have to do the inverse of sine.
So we're leaving that number on your calculator. you do shift sign which gets you sine minus one, type in that answer or just press your answer button. So shift sign answer press equals and I get an answer here of 76.97. seven, seven, nine.
And again, a question would ask us to round it here. Just depends, so let's just go to the nearest degree. We'd go for 77 degrees.
Obviously just making sure what the question says here, but let's just round it to the nearest degree there, so 77. Okay, so that's how you use the sine rule. Right, let's see how this question's different then. So work out the length of AB.
So this one over here. Now straight away, looking at this, look, we've got a pair of opposites there, but I don't have any other pairs of opposites. So I've not got any other pairs of opposites.
got anything opposite my 15, I've not got anything opposite the 12. So I can't actually use the sine rule here and that is your clue, that is your hint here that the sine rule is not going to work and we're going to have to use the cosine rule. So another rule that you need to know, so the cosine rule is a squared equals... b squared plus c squared minus 2bc cos a. OK, so a being the side we're looking for, so we'll label this little a and this one big A, ignoring the letters on the triangle, and then labelling the other two sides, and they can be b and c, however you like.
And from there, all you've got to do is stick these numbers in. It's actually quite simple to use when you know it. So a squared equals b squared, which is 15 squared. plus c squared, which is 12 squared, minus, and I'm going to stick this bit in brackets, 2 times 15 times 12 cos a, which is down here, which is 20. There we go.
Right, so sticking that all in the calculator, let's have a look what we get. So 15 squared plus 12 squared minus 2 times 15 times 12 cos 20. Okay. Press equals, and I get a squared equals 30.7106, and a few more decimals.
Now, obviously, that is a squared, and we don't want a squared, we want to know what a is, so we just need to square root both sides now. So square root, leaving the answer on the calculator, square root answer, and we get a equals 5.541719. And again, obviously, you'd be asked to round this in a particular way in the question. Let's go for two decimal places, so a equals 5.54 centimetres.
Right, there we go, and there's using a bit of the cosine rule. OK, so working out the size of angle BA, which again is this one at the top, and again just having a look, there are definitely no opposites because we've got no other angles. But the angle that we're looking for is going to be our a, and it is opposite 9. There we go, so the others can be B and C again. Now obviously here we're looking for an angle, so it's up to you if you choose to learn the formula.
So let's go ahead and learn the formula. I tend to find that I just learn this formula, a squared equals b squared plus c squared minus 2bc cos a. And I'm quite happy just rearranging that to get cos a on its own. So in order to do that, I'm going to get this whole minus 2bc cos a.
I'm going to add it to the other side. So we get a squared plus 2bc. PC Coz A equals b squared plus c squared. Now I can get rid of that a squared so we can minus a squared from both sides so minus a squared and you get 2bc cos a equals b squared plus c squared minus a squared now and I can finish off this rearrangement you can divide by 2bc just to leave you with cos a so cos a equals b squared plus c squared minus a squared over 2bc. It's up to you.
You can choose to just learn that formula if you want, but that's the formula we're going to use to find an angle. So plugging in all these numbers then just into my formula there, we'll get cos a equals b squared plus c squared, so 10 squared plus 5 squared. Minus a squared, so minus 9 squared, all over 2 times b times c, so 2 times 10 times 5. Nice and easy typing that into the calculator.
So fraction button, 10 squared plus 5 squared, minus 9 squared. all over 2 times 10 times 5 and that equals, let's have a look, so cos A equals 0.44 and then same process again we need to do the inverse of cos, so cos minus 1 of your answer and you get, let's have a look, shift cos, answer and I get 63.896 There we go, degrees. And again we could round that so we could just say 63 point, and let's just go to one decimal place, 63.9 degrees. Again, just reading the question. And that's how to use the cosine rule for finding angles.
Moving on to the area of a triangle. triangle formula obviously not half base times height because we don't have the height here so for any triangle the area is half a b sine c so half a b sine c another formula there that you need to know and let's just have a look at how we apply that so sine c this time the angle that we're going to use is going to be our c, so we'll call that c, this would then be little c, and the others will be a and b, and they can be a and b in either order. And then it's as simple as just sticking those numbers into a formula.
So it's half times 5 times 8 times sine 41. There you go. Type that into the calculator, so half times 5 times 8 times sine 41. And we get the answer 13. 13.12 if I round it to two decimal places. It's area, so that's metres squared, just being careful of the units. So 13.12 metres squared. There we go, there's using the area of a triangle formula.
A question like this though, you might be given the area of the triangle. So it says the area is 80, so we're working backwards a little bit here. So we're going to use half a, b, sine c again. So half a, b, sine c, but this time it gives us the answer.
So the answer to that half a, b, sine c has to equal 80. So let's just plug in. plug in all the numbers here, again, c being my angle, and then a and b being my other lengths. So if I was to type this into the formula, we'd have a half times 11 times 16 times sine c, and it would equal 80. Now I can't actually work that out, but what I can do is I can work this bit out. So if I do a half times 11 times 16, we get the answer 88. So we have 88 times sine c.
equals 80. So we can rearrange this look, we can get both sides and divide them by 88 and that will just give us sine C on its own. So sine C equals and 80 divided by 88 is Let's leave it as a fraction. Actually, it is a decimal there, but it comes out as 0.90, and that is actually 9-0 recurring. There we go.
I'm just going to leave it on the calculator screen. And then obviously to finish that off, you do the inverse sign again, just like we have on these other questions. So I'm going to do that.
over here to the left, so sine minus 1 of this number here, which is actually also a fraction, it's 10 over 11, sine minus 1 of 10 over 11, all that 0.90 recurring, and we get the answer. of 65.38 degrees. There we go, so we can find an angle as well, just working backwards if we're given the answer. And you can apply that logic there to lots of different questions where it gives you the answer and you can use your formula backwards.
Just don't do it. dividing everything over to the other side. Okay, so we've got a vectors question.
It says B is the midpoint of A. So that's this point here is the midpoint. So straight away that tells us look as you can see some of the vectors on here that this must also be A and again there are so many different variations of vectors here.
This is just one idea of how to approach it. There's so many different types of questions you can get. We could spend an hour to two hours just looking at vectors.
M is the midpoint of PB. So we've got a midpoint here, so that's halfway along. And then show that NMC is a straight line.
So straight away I'd like to draw NMC in and just think about that. So that it's saying is show that that's a straight line. And we're going to have a look at some of the vectors on this.
line and see if we can get a common multiple or a common bracket however you've looked at this before. So all I want to have a look to start with is that line N to C I've got three vectors I can look at I've got the full line from N to C we've got from N just to M so I can go from N to M or I could go from M to C and they're all little parts of this line okay the full line there let's just highlight this the full line the first one that I've written is N to C all the way the second one I've drawn is N to M from there to there and the third one I can have a look at is from M to C from there to there now it's up to you which one you choose sometimes some of them are easier than others but the first one there so this N to C is quite simple for us to do. Now to get from N to C, I've got to move backwards up the line this way and then I can go forwards along to C there. So that first part of the journey, going from, let's get rid of that, going from N up to A, and let's draw this in, to go from N to A, so I'm just going to write this down, is minus 2B. You have to go backwards through that 2B.
So from N to C, first of all, it's minus 2B. Then I can go from A over to C, doing two As, and that's positive. So plus 2a. Now, here I've got minus 2b plus 2a.
Now, whenever it comes to these vectors here, you always just want to look to factorise them. So I'm going to move this out of the way. I'm just going to see what does that factorise to. Now, I can factorise it by 2. And I would get 2 lots of minus b plus a.
And obviously, we could write that in a slightly different way. I could swap the letters around if I want. I could write 2 lots of a minus b.
Now, if I can prove that one of these other vectors... vectors here that I'm looking at below has this bracket a minus b then they must be on the same straight line because therefore they're going in the same direction. So let's have a look at one of them.
So I can either look at n to m or m to c. It doesn't really matter I'm just going to go with m to c. Either one here are going to be just as difficult for us to do but that first one there is nice and easy so always make sure you have a go at finding some of the lines.
Now that m letter here is halfway along pb so I'm going to have to find the a vector for p to b and that's going to be my absolutely key one here p to b because if I can't find the full length of the line I'm definitely not going to find halfway along it. So if we look at how do we get from p to b well that's actually quite simple you can go from p up to a and that's minus three b's so let's write that down you've got you've got the minus b to get to n and another two to get up to a so minus three b's and then to get from a to b you do a plus a moving along that vector there so plus a. So that is my vector P to B.
That's a very good one to always make sure you get that missing line where you've got a halfway point. Now in order to move halfway along that line, I'd have to do half of this vector. So to get from P to M, or from M to B, which is half of that, let's go for M to B, as that's the quickest way for us to move there, M to B. We'd have to do half of this vector, so to times that by a half, I'm just going to write it as a half lots of minus 3B plus A. So half of that.
Now if I expand that... I can actually add stuff to this as well, because that only gets us from M to B. If I draw that on, that gets us from here to here, doing half of that line. So you expand that out. Minus three times a half is minus three halves.
So minus three over two, B, plus half of A, so half A. And now we can finish that off, because now we're at B, we can do our final little bit of the line, which is to go over here to C, and that's adding an additional A. in there. Okay so that's our last little bit plus a and if we add this all together and let's see what we get. So we have minus 3 over 2b.
I'm going to write this over here. That's not changing. So we've got minus 3 over 2b and then we've got half a add an a and that gives us 1.5a or another 3 over 2a. There we go 3 over 2a.
So let's write that down over here. here. We have minus 3 over 2b. plus 3 over 2 a.
So that is our vector there to get from m to c. Now what I need to do is just factorise this one. If you have a look they both divide by 3 over 2. So I can take this 3 over 2 out the bracket and if we divide them both by 3 over 2 we get minus b plus a.
There we go and I rewrote that a different way as well. I could write that as 3 over 2 brackets a minus b the other way around and there we go. We've got that matching bracket there in both of them. OK, so both share that common vector there, that common movement between the points on the diagram N, M and C.
Again, you could have found N to M as well. We could have done the same thing. And actually, because we found this half, this one just here, we can actually look, just think, if I was moving from n to m, so n to m, we would do plus b and then half of that vector just there. And if we did do that, let's just do it over here to the side.
So this is from n to m, not that we didn't need it this one, we only ever need two. So from n to m I could do plus b, or just b, and then add to that this vector down here that we did in the green down there. So add to that minus 3 over 2b. plus half a.
And if we add that all together there, minus 1.5 plus 1 gives us minus a half, so we'd have minus a half b plus a half a. And again, we could factorise that by a half, so we'd get a half. minus b plus a and again we get that minus b plus a there which we can write in a different way so we could have done any three of them all three of them eventually give us this same bracket here okay so we have the same multiple it means it's on the same line and there's a bit of vectors for you.
And the last question here is just involving some circle sectors but also involving a little bit of trigonometry as well. It says that this is an arc of a circle sent to zero, it wants us to work out this shaded region. So if we can work out the full sector we can take away this unshaded triangle.
So to work out the area of a sector we would have to do pi r squared, so not forgetting the area of a circle formula, pi r squared and multiply that by whatever fraction of the circle this is. Now it says 35 degrees over there. 35 over 360. So if we plug those numbers in it will give the area of the sector.
So we've got pi times 80 squared, that's in metres, multiplied by 35 over 360. If we type that in on the calculator, let's see what we get. Pi times 80 squared times 35 over 360 and we get the answer. 1954.777.
Obviously just be careful what the question says again, and that's metre squared. Then we can work out the area of the triangle, and obviously not forgetting just because it's not just a triangle on its own, we can still use our half a b sine c. So we can use half a b sine c to get this one. And if we plug the numbers in, 35 there is our c. And because it's a circle sector, a and b here are the same because it's the radius of the circle.
So we can do half times 80 times 80 times sine. 35. There we go, and if we type all that in, 0.5 times 80 times 80 times sine 35, we get the area of that triangle there, which is 1835.44. And then to finish that off we can do the bigger shape of the sector, subtract the area of the triangle, so 1954.77 take away this 1835. 0.44 and that leaves us with an overall area here. Let's work it out. 1954.77 takeaway answer and we get 119.325 meters squared.
and there we go and obviously just rounding that however you'd be asked in the question so there's quite a large selection of geometry questions there but we're done on the geometry