Video: Pack 3 • Paper 2 • Question 7

Pack 3 • Paper 2 • Question 7


Video Transcript

Here is a right-angled triangle. Four such triangles are placed as shown in the diagram to form the square 𝐴𝐵𝐶𝐷. Find the area of the square 𝐴𝐵𝐶𝐷 in terms of 𝑥 and 𝑦.

There are several elements to this question. Remember, the area of a rectangle is found by multiplying its width by its height. In the case of a square, since its width and its height are equal, it’s the same as finding its width and multiplying it by itself, or squaring it. So how do we find the width of the square?

Well, the trick is to remember that these are all right-angled triangles. Each side of the square is made up of the longest side, the hypotenuse, of a triangle. We’ve been given the lengths of the other two sides of this triangle. So we can use Pythagoras’s theorem to help us find an expression for the length of the hypotenuse.

Remember, 𝑐 is always the hypotenuse. In this case, that’s what we’re trying to find. It’s the length 𝐴𝐵, which is the width of the square. That means we can rewrite Pythagoras’s theorem as 𝑥 squared plus 𝑦 squared is equal to 𝐴𝐵 squared.

Usually, we would then find the square root of both sides of this equation to find the length of the hypotenuse. But remember, we said that, to find the area of the square, we can take the width and square it. However, the width is actually 𝐴𝐵. And when we square that, we get 𝐴𝐵 squared. So the area of the square is given by 𝐴𝐵 squared. But we already said that that was the same as 𝑥 squared plus 𝑦 squared. The expression for the area of the square is, therefore, 𝑥 squared plus 𝑦 squared.

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