### Video Transcript

Consider the isosceles right triangle π shown in the figure. Squares have been added to each side of π. Let the area of π be one area unit. The areas of the squares on the legs are π΄ sub one and π΄ sub two, and the area of the square on the hypotenuse is π΄ sub three. Find a relationship between the area of the square on the hypotenuse and the area of the squares on the legs.

If we have a look at the diagram, we can see that there are two squares π΄ sub one and π΄ sub two on the legs of the triangle π and a third square on the hypotenuse of triangle π. Weβre asked to find a relationship between these three squares. As weβre told that the area of triangle π is one area unit, letβs see if we can find any actual values for the area of these three squares. We should remember that to find the area of a triangle, we multiply a half by the base by the perpendicular height.

Here, weβre told that triangle π is an isosceles triangle, so that means that both the legs on the triangle will be the same length. That means that if we were working out the area and we have the base and the height, because these are the same length, we could replace our value β with the value of π for the base. So, the area would be equal to a half times the base times the base, or indeed a half multiplied by the base squared. Weβre told that the area of π is one area unit. So, we rearrange one equals a half times π squared by multiplying both sides of this equation by two. Therefore, two is equal to π squared. And taking the square root of both sides would give us that π is equal to the square root of two, and the units would be length units.

So, we have now calculated that both of the legs on this triangle π have a length of root two length units. So, that means that the squares π΄ sub one and π΄ sub two will each have side length of root two length units. We can remember that to find the area of a square, we multiply the length by the length, π squared. So, the area of π΄ sub one is equal to the square root of two squared, which gives us two area units. The two squares π΄ sub one and π΄ sub two have the same areas as both of these squares have the same side length of the square root of two length units.

We now need to think about how we would find the area of the third square, π΄ sub three. If we need the length of the hypotenuse on triangle π, this would enable us to find the length of the side length of π΄ sub three. Observing that triangle π is a right triangle, this means we could apply the Pythagorean theorem. The Pythagorean theorem tells us that the square on the hypotenuse is equal to the sum of the squares on the other two sides.

Applying the Pythagorean theorem, we have π squared, thatβs the hypotenuse that we wish to find out, is equal to the square root of two squared plus the square root of two squared. As the square root of two squared is two, then we have that π squared is equal to four. In order to find the hypotenuse π, we take the square root of both sides, which gives us that π is equal to two. And because itβs a length, then these will be in length units.

Now weβve established that the length of the hypotenuse is two length units, this means that the square π΄ sub three has a side length of two length units. To find the area of π΄ sub three, once again, we apply the formula for the area of a square. The length squared is two squared, and two squared is four. So, π΄ sub three has an area of four area units.

Letβs now look at the areas of our three squares. The two squares π΄ sub one and π΄ sub two on the legs of our triangle π have equal areas of two area units. The area of our third square π΄ sub three has an area of four area units. Perhaps, the most obvious equation or relationship we can write that relates our three squares to each other is that π΄ sub three is equal to π΄ sub one plus π΄ sub two. And this would be our answer for a relationship between the areas. However, any rearrangement of this equation would also be a perfectly valid answer.