We tackle geometric problems in this video. Standard geometric problems often have diagrams provided involving triangles, quadrilaterals and other polygons. It is important to know the properties of shapes as multiple questions involve knowing angle and length properties. The standard problem-solving framework should be used to approach geometry problems.

In math geometry covers a wide array of problems and we tend to split these into two different fields. One of these we call trigonometry, which generally deals with the actual numbers and equations around shapes. Well, shapes, which refers to the actual visual process of taking different objects and trying to give dimensions to them. We're going to begin with trigonometry because that's by far and away the more common of the two that come up in exam scenarios. So first of all, when we talk about Pythagoras'theorem, more generally, the simplest shape that we really consider, the triangle. So I'm going to move this one over here, which I made earlier. So in a triangle, it's got three shapes, three sides. That's not terribly interesting. However, we're going to look at the right angles. Triangle a right angle is this thing here, and it occurs whenever we have two lines that are perpendicular to one another. Perpendicular just means one of them is going vertically, the other is going horizontally. And where they meet, they make a right angle. Now, you may notice this is exactly one quarter of a way around. If we were to do another one, two, three.


So one right angle is a quarter of the circle. And we say that a circle is 360 degrees. That's just a number. That's fine. So this is one quarter of 360. This angle here is 90 degrees.


Okay.


Now, Pythagoras was the first to notice an interesting property with our angle triangles, which is where and this is always true if we have these three sides, ABC, we have it true that A squared and A is just measured in centimetres, same as B, and C plus B squared equals C squared.


Okay?


The most common example of this is where we have A equals three, B equals four. And that means that we're going to have three squared plus four squared that equals nine plus 16, which is 25, that equals C squared. So C is the square root of 25, which is five. Okay, this is called a Pythagorean triangle, and it's very common. It means if we have any two sides of one of these shapes, we can figure out the third. It turns out, actually, that with some more advanced trigonometry, we can figure out even more than that. If I take another triangle that I prepared earlier, we're going to have one that's got an angle here, x. Okay, I'm just going to redraw this angle here, x. And X is just any number. In this case, it clearly be between zero and 90 degrees. And sometimes X is used for these angles. Other times it's this Greek layer here, theta. Either way, what is interesting, this angle and right angle triangles of this interesting property, where if we change the sides by the same amount, the angle never changes. It's always substantially the same amount. It's always got the same thing.


What this means is because an angle is just a way of describing how much something increases. Say for example, how much is vertical side increases compared to the horizontal side. It means that we can make ratios out of these sides that are always true and it's going to be related to the angle in some way. So we're going to label these sides. We call the longest one the hypotenuse. Often that's just shortened to hip. Then we've got the length which connects the angle we're interested in the right angle which we call the adjacent which is adj and then the last remaining one is the opposite. And the ratios that we're interested in are well we have to pick some of them and the usual first one that we pick is whatever this y sign is the opposite divided by the longest edge. The opposite divided by the hypotenuse is always going to be a constant value for a given angle and we call that equal to sine of x. Sin is just a function. If you do sin of any angle in your calculator, it's always true. Say for example, you picked sine of 30 in your calculator.


Sine of 30 equals one half and what that means is that for a right angle triangle of 30 degrees, however long this vertical edges here the opposite, the hypotenuse is always twice as long as that's true for any right angle triangle with 30 degrees. So we can define some other trigonometric identities here, just fill out the high volumes. The next one that we pick is equal to the adjacent on the x side divided by the hypothesese. It's called the calls of x cosine of x. Again, very similar results. And lastly tan of x which is and there's not really any good reason that it's this way around and that's equal to the opposite divided by the adjacent aka how big is the vertical side compared to the horizontal. And what this means is if we know the angle and our calculators know a lot of different ways of doing sin of an angle and one other side, we can figure out all the information about the rest of the triangle. Very powerful tools. Okay, the last thing to talk about here in the sort of equation space is area, surface and volume.


Okay?


So when we have a shape, it can have a dimension. I know it sounds like complicated but it's not that bad. A line is one dimensional and it's going to have a length in this case say x. And then if we want to make it two dimensional 2D, we have to add another line here and we'll also call this x in this case and that means that suddenly we got a 2D shape and this thing's gone area. So whereas before we had a length, now we have an area which is this thing here. And if we introduce a third line, the third dimension and we can also make this a different number in this case, we're going to make a volume. The simplest case of this is where we have a rectangular prism here's what I made earlier with lengths A, B and C. Okay? And we've got a length A, which we can calculate here. And then each of the faces is going to have a different area. So in the case of, say, this one here, it's going to shade in this one here. This is the top surface that's going to have an area of A times B.


So to get an area, you have to multiply two numbers. That's a fact, no matter what kind of shape you're dealing with. If you want 2D objects, you need to have some kind of two numbers being multiplied together. And similarly, the whole volume of this be A times B times C. That's true of all shapes. It's the most simple in this case for something that's rectangular. But sometimes you might have different numbers in front of the values you're putting in. For example, for a sphere, the volume is four divided by three times Pi times by R cubed. So where R is the radius. So it's not quite the same as A times B times C. We've got a cube in there, but there is this number in the front. But that's the only real difference. Other than that, that ties up a lot of stuff we have to do here in Trigonometry, going over to shapes, there isn't too much that we actually do here at GCSE. There are these things here called circle theorems. We're not going to go into them too much here, but these are a variety of rules that happen to be true for circles.


You can prove yourself using triangles. The other thing we're interested in is transformations, which is if you have a shape that is on a graph of some kind, what can you do in order to, well, change the shape? Can you move it?


Can you squish it?


Can you rotate it? And in general, we see there are four different ways of transforming a shape. Okay, I'm just going to draw a little square here, and one way we could transform this thing is translating it, which means just moving it from side to side or top to down. We have rotating it, which is where in this case, about the centre, we move about access. We could enlarge it if we wanted to observe means to change its size. And then lastly, we could be reflecting it, which just means we draw a line somewhere and then we're going to get a similar looking shape on the other side. And depending on where we draw our mirror, we're going to get a sign that looks a bit different. And as I've just shown there, we can combine multiple transformations. Cloner having to draw this little triangle here sorry, little square, and then moving it randomly as a diamond. So we're going to go over some examples of these, in particular, these trigonometric problems, because these are the ones that are more likely to come up. So this is a fairly typical Pythagoras exam question and it is a little bit tricky, but you'll get the logic behind what we're doing in a minute.


So the rectangle and part A is the basis box, the cuboids fit in a cuboid box with a straight rod of length. One metre fit in the box. So the idea is the longest edge in this box is going to be this one here. That's where you could potentially try and put a rod. And we're trying to show woods on this, 100 centimetres fit. So to do that, we figure out that length there. So to do that well, where do we start? Well, what we can notice is that we're going to need probably one of this dash line here down at the bottom. And in this case, we know that this is 30 centimetres. That's 35. We can read it also to calculate that length there called it A. That means that we can use Pythagoras. So we say that A squared is going to be equal to 30 squared plus 35 squared.


Okay.


Now, if we want to, we can calculate it right now by doing A equals the square root of those. I'm actually going to leave it in this form for a minute. You'll see why in a second. Then you can use this trick as you do any more your maths. Okay, so the next bit we'll need to figure out this length here is we can notice that this year is 87 centimetres, same as the other one. And then we're interested in finding out this longest edge here. This one going up here, we'll call that B in this case. So to figure out B squared is going to be equal to well, it's going to be 87 squared plus A squared. So that's why I kept it as A squared. It turns out that Pythagoras just works in 3D without any additional adjustments. So B squared is going to equal 87 squared plus 30 squared plus 35 squared. So B is going to equal the square root of this entire thing here. Now, you can put this into a calculator. You would not be expected to figure this one out by yourself. You get a number that's about 98 centimetres.


Okay.


Now, that's actually less than one metre, which means the box could not contain a straight rod of one metre, just about.


Okay.


And that shows you could do this. And transforming Pythagoras into 3D is actually pretty straightforward. If we go to the next problem we've got here, we've got volumes and areas, which is a bit of a trigonometric problem. So the volume of this pyramid here is a third times the area of the base, times the height, perpendicular height, in this case, being this length here. So this is a rectangular based pyramid with volume three three six to calculate the angle between VB and the base. This is actually quite complicated question. So first of all, to figure out the angle, we need this length here and we need that there. So to do that, we're given the volume. So three three six equals one third multiplied by the area of the base, which in this case is eight times 14 times the height. And that height is what we need there. So this thing here, we'll call that height h. Okay? So rearranging h equals three times three three six divided by eight times by 14.


Okay.


And you can then put that into the calculator. Again, you would not be expected to figure that one out just by yourself. And if you do so, you find that H is equal to nine centimetres.


Okay?


So from there, we know that this is nine. And then the last part is that, well, we need one of the other sides of this triangle here and in this case we can use Pythagoras. Again, this will keep coming up. Where this length here. Aren't you going to change colour? This length here is what we need. And we know that this here is four centimetres. This is seven centimetres, which means that to calculate that edge there, it's going to be seven squared plus four squared all square rooted. Now, that's going to be 49 plus 16 square rooted. That's the square root of 65. That's not a nice number and that's okay, it doesn't have to be a nice number. So we can just keep that the square root of 65. So then to figure out, lastly this angle here, we take a look at what we've got. The angle there, we'll call it x. We've got the opposite, which is because this is the right angle. So we have the opposite, which is nine, the adjacent, which is square, is 65. So that means that we are opposite and adjacent. We want tan. So tan of x is equal to the opposite over the adjacent.


So on your calculator, you can actually put ten to the minus one of nine over route 65 equals x, which in turn gives an answer, which is about 48 degrees.


Okay.


And that's the answer. That's quite complicated, that one there, but still we could do it. The last thing to leave to you guys to do is this is a little proof of Pythagoras. So given this rather weird looking shape here, you're trying to prove A squared plus B squared equals C squared. To do that, try and find two ways of expressing the area of this thing here, and in doing so, you'll end up getting a good answer if you also include these triangles. I encourage you to leave your answers in the comments below. And if you have any questions, please also leave a comment. Thank you very much and I hope you will do well in your geometry questions in your exams.


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