# Video: AQA GCSE Mathematics Higher Tier Pack 1 β’ Paper 3 β’ Question 27

AQA GCSE Mathematics Higher Tier Pack 1 β’ Paper 3 β’ Question 27

07:21

### Video Transcript

πππππ is a square based pyramid. The square base has sides of length 10 centimetres. π is the midpoint of ππ. The apex π of the pyramid is vertically above π, not the centre of the base. ππ is equal to ππ is equal to 13 centimetres. Part a) Show that ππ is equal to 12 centimetres.

Firstly, letβs know that weβre told that the square base has sides of length 10 centimetres. This tells us to all the side lengths on the base of the pyramid β so thatβs ππ, ππ, ππ, and ππ β are equal to 10 centimetres. Next, weβre told that π is the midpoint of ππ. And so, this tells us that ππ is half the length of ππ. And since we have that ππ is equal to 10 centimetres, this means that ππ is equal to one-half of 10 centimetres or five centimetres.

Now, letβs consider the side weβre trying to find. So thatβs the side ππ. We know the length of side ππ since we just figured it out and also the length of side ππ. So letβs consider the triangle πππ. We know from the diagram that the angle πππ is 90 degrees or a right angle. And now, we can fill in the side lengths which we know.

We have that ππ is equal to five centimetres. And weβre given in the question that ππ is equal to 13 centimetres. And so now, we can see that we have a right-angled triangle, where we know two of the side lengths, which means that we can use Pythagorasβs theorem.

Pythagorasβs theorem tells us that for any right-angled triangle, π squared plus π squared is equal to π squared, where π is the hypotenuse of the triangle which is the longest length. And π and π are the two shorter lengths in the triangle. So letβs label our triangle.

The hypotenuse is the longest length. So thatβs ππ. And we can label that with π. And now, we simply label the other two sides with π and π. It doesnβt matter which way around we do this.

Now, letβs substitute our values for π, π, and π into the equation from Pythagorasβs theorem. Our value of π is five centimetres. So we get five squared. Our value of π is the length ππ, giving us ππ squared. And then, this must be equal to π squared. And our π is 13, giving us 13 squared.

Now, five squared is equal to 25 and 13 squared is equal to 169. Next, we simply subtract 25 from both sides of the equation to obtain that ππ squared is equal to 144. And our final step in finding ππ is simply taking the square root of both sides of the equation. This gives us that ππ is equal to 12 centimetres.

Now, although we have taken a square root here which should give us both a positive and negative solution, we can ignore the negative solution since ππ is a length. And a length must have a positive value.

And so, now, weβve completed part a of the question by showing that ππ is equal to 12 centimetres.

Part b) Calculate the size of angle πππ.

Letβs start by drawing the triangle πππ on our diagram. Letβs also label angle πππ and call it π₯. Since π is vertically above π and ππ lies along the base of the pyramid, this tells us that angle πππ is a right angle. And so, we can label it on our diagram.

So we can see that π₯ is an angle within a right-angled triangle. And we know the length ππ since we found it in part a. So in order to use right angle trigonometry on this triangle, we simply need to find the length of one of the other sides.

Letβs consider the triangle πππ. And we can draw this triangle out separately like this. Since angle πππ is the corner of a square, since itβs one of the corners of the square base, this means that it must be a right angle.

Since ππ is one of the sides of the square, its length must be 10 centimetres. And we found the length of ππ in part a has five centimetres. Here, we have a triangle where we know two of the side lengths. So we can use Pythagorasβs theorem to find the third side length.

The side ππ is the hypotenuse. So we can label it π. And we labelled the other two sides π and π. Substituting these values into the formula, we get five squared plus 10 squared is equal to ππ squared.

Then five squared is 25 and 10 squared is 100. And we obtained that ππ squared is equal to 125. And now we take the square root of both sides which gives us that ππ is equal to five root five centimetres. And again, since ππ is a length, we ignore the negative solution here.

Now, we have found the length of ππ. And we are ready to consider the triangle πππ. Letβs draw out this triangle. Here, we have our triangle πππ. We have the angle πππ marked with π₯. And this is the angle weβre trying to find.

Now, we know that the length of ππ is 12 centimetres. And we have just calculated that the length ππ is five root five centimetres. We can see we have a right-angled triangle where we know two of the side lengths and an angle which weβre trying to find.

In order to find this angle, we will be using SOHCAHTOA. Letβs label the opposite, adjacent, and hypotenuse on our triangle.

The hypotenuse is the longest side. So it must be ππ. The opposite is the side opposite the angle weβre trying to find. So thatβll be the side ππ. And then, the adjacent is the side next to the angle weβre trying to find. So that is side ππ.

So the two sides which weβre going to use in our calculation will be the opposite and the adjacent since these are the two sides which we know the length of. And we can see that TOA contains both the opposite and adjacent. So weβll be using TOA in order to find π₯. What TOA tells us is that tan of π₯ is equal to the opposite over the adjacent.

And since our opposite side is 12 centimetres and the adjacent side is five root five centimetres, this tells us that tan of π₯ is equal to 12 over five root five. Finally, we can find π₯ by using the tan inverse operation to get that π₯ is equal to tan inverse of 12 over five root five.

By rounding our answer to three significant figures, we find that angle πππ is equal to 47.0 degrees.