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

𝑃𝑄𝑅𝑆𝑇 is a square based pyramid. The square base has sides of length 10 centimetres. 𝑀 is the midpoint of 𝑄𝑅. The apex 𝑃 of the pyramid is vertically above 𝑀, not the centre of the base. 𝑃𝑄 is equal to 𝑃𝑅 is equal to 13 centimetres. Part a) Show that 𝑃𝑀 is equal to 12 centimetres.

Firstly, let’s know that we’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 that’s 𝑄𝑇, 𝑄𝑅, 𝑅𝑆, and 𝑇𝑆 β€” are equal to 10 centimetres. Next, we’re told that 𝑀 is the midpoint of 𝑄𝑅. And so, this tells us that 𝑄𝑀 is half the length of 𝑄𝑅. And since we have that 𝑄𝑅 is equal to 10 centimetres, this means that 𝑄𝑀 is equal to one-half of 10 centimetres or five centimetres.

Now, let’s consider the side we’re trying to find. So that’s the side 𝑃𝑀. We know the length of side 𝑄𝑀 since we just figured it out and also the length of side 𝑃𝑄. So let’s consider the triangle 𝑃𝑀𝑄. We know from the diagram that the angle 𝑃𝑀𝑄 is 90 degrees or a right angle. And now, we can fill in the side lengths which we know.

We have that 𝑄𝑀 is equal to five centimetres. And we’re given in the question that 𝑃𝑄 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 Pythagoras’s theorem.

Pythagoras’s theorem tells us that for any right-angled triangle, π‘Ž squared plus 𝑏 squared is equal to 𝑐 squared, where 𝑐 is the hypotenuse of the triangle which is the longest length. And π‘Ž and 𝑏 are the two shorter lengths in the triangle. So let’s label our triangle.

The hypotenuse is the longest length. So that’s 𝑄𝑃. And we can label that with 𝑐. And now, we simply label the other two sides with π‘Ž and 𝑏. It doesn’t matter which way around we do this.

Now, let’s substitute our values for π‘Ž, 𝑏, and 𝑐 into the equation from Pythagoras’s theorem. Our value of π‘Ž is five centimetres. So we get five squared. Our value of 𝑏 is the length 𝑃𝑀, giving us 𝑃𝑀 squared. And then, this must be equal to 𝑐 squared. And our 𝑐 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 𝑃𝑀 squared is equal to 144. And our final step in finding 𝑃𝑀 is simply taking the square root of both sides of the equation. This gives us that 𝑃𝑀 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 𝑃𝑀 is a length. And a length must have a positive value.

And so, now, we’ve completed part a of the question by showing that 𝑃𝑀 is equal to 12 centimetres.

Part b) Calculate the size of angle 𝑃𝑇𝑀.

Let’s start by drawing the triangle 𝑃𝑇𝑀 on our diagram. Let’s also label angle 𝑃𝑇𝑀 and call it π‘₯. Since 𝑃 is vertically above 𝑀 and 𝑀𝑇 lies along the base of the pyramid, this tells us that angle 𝑃𝑀𝑇 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 𝑃𝑀 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.

Let’s consider the triangle 𝑄𝑇𝑀. And we can draw this triangle out separately like this. Since angle 𝑀𝑄𝑇 is the corner of a square, since it’s one of the corners of the square base, this means that it must be a right angle.

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

The side 𝑀𝑇 is the hypotenuse. So we can label it 𝑐. And we labelled the other two sides π‘Ž and 𝑏. Substituting these values into the formula, we get five squared plus 10 squared is equal to 𝑀𝑇 squared.

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

Now, we have found the length of 𝑀𝑇. And we are ready to consider the triangle 𝑃𝑇𝑀. Let’s draw out this triangle. Here, we have our triangle 𝑃𝑇𝑀. We have the angle 𝑃𝑇𝑀 marked with π‘₯. And this is the angle we’re trying to find.

Now, we know that the length of 𝑃𝑀 is 12 centimetres. And we have just calculated that the length 𝑀𝑇 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 we’re trying to find.

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

The hypotenuse is the longest side. So it must be 𝑃𝑇. The opposite is the side opposite the angle we’re trying to find. So that’ll be the side 𝑃𝑀. And then, the adjacent is the side next to the angle we’re trying to find. So that is side 𝑀𝑇.

So the two sides which we’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 we’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 𝑃𝑇𝑀 is equal to 47.0 degrees.

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