### Video Transcript

πππ
ππ is a square based pyramid. The square base has sides of length 10 centimetres. π is the midpoint of ππ
. The apex π of the pyramid is vertically above π, not the centre of the base. ππ is equal to ππ
is equal to 13 centimetres. Part a) Show that ππ is equal to 12 centimetres.

Firstly, letβs know that weβre 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 thatβs ππ, ππ
, π
π, and ππ β are equal to 10 centimetres. Next, weβre told that π is the midpoint of ππ
. And so, this tells us that ππ is half the length of ππ
. And since we have that ππ
is equal to 10 centimetres, this means that ππ is equal to one-half of 10 centimetres or five centimetres.

Now, letβs consider the side weβre trying to find. So thatβs the side ππ. We know the length of side ππ since we just figured it out and also the length of side ππ. So letβs consider the triangle πππ. We know from the diagram that the angle πππ is 90 degrees or a right angle. And now, we can fill in the side lengths which we know.

We have that ππ is equal to five centimetres. And weβre given in the question that ππ 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 Pythagorasβs theorem.

Pythagorasβs 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 letβs label our triangle.

The hypotenuse is the longest length. So thatβs ππ. And we can label that with π. And now, we simply label the other two sides with π and π. It doesnβt matter which way around we do this.

Now, letβs substitute our values for π, π, and π into the equation from Pythagorasβs theorem. Our value of π is five centimetres. So we get five squared. Our value of π is the length ππ, giving us ππ 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 ππ squared is equal to 144. And our final step in finding ππ is simply taking the square root of both sides of the equation. This gives us that ππ 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 ππ is a length. And a length must have a positive value.

And so, now, weβve completed part a of the question by showing that ππ is equal to 12 centimetres.

Part b) Calculate the size of angle πππ.

Letβs start by drawing the triangle πππ on our diagram. Letβs also label angle πππ and call it π₯. Since π is vertically above π and ππ lies along the base of the pyramid, this tells us that angle πππ 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 ππ 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.

Letβs consider the triangle πππ. And we can draw this triangle out separately like this. Since angle πππ is the corner of a square, since itβs one of the corners of the square base, this means that it must be a right angle.

Since ππ is one of the sides of the square, its length must be 10 centimetres. And we found the length of ππ in part a has five centimetres. Here, we have a triangle where we know two of the side lengths. So we can use Pythagorasβs theorem to find the third side length.

The side ππ 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 ππ squared.

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

Now, we have found the length of ππ. And we are ready to consider the triangle πππ. Letβs draw out this triangle. Here, we have our triangle πππ. We have the angle πππ marked with π₯. And this is the angle weβre trying to find.

Now, we know that the length of ππ is 12 centimetres. And we have just calculated that the length ππ 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 weβre trying to find.

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

The hypotenuse is the longest side. So it must be ππ. The opposite is the side opposite the angle weβre trying to find. So thatβll be the side ππ. And then, the adjacent is the side next to the angle weβre trying to find. So that is side ππ.

So the two sides which weβre 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 weβll 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 πππ is equal to 47.0 degrees.