The hypotenuse of an isosceles right triangle is 10. What are its other two sidelengths? [Hint: Find the area of the square composed of two such right triangles.]

Question

Homi · Accepted Answer

We can understand the relationship between the sides of an isosceles right triangle by imagining it as half of a square.

Step 1 — Visualize the square

An isosceles right triangle has two equal sides. Let's call the length of these equal sides 'a'. The hypotenuse is the longest side. It is opposite the right angle. The problem tells us the hypotenuse is 10 units long. The hint asks us to think about a square made of two such triangles. Imagine a square. If you draw a line from one corner to the opposite corner (this line is called a diagonal), it cuts the square into two identical isosceles right triangles. This means that for our triangle, its two equal sides ('a') are the sides of this square. Its hypotenuse (which is 10 units) is the diagonal of this square. So, we have a square whose diagonal is 10 units.

Diagram 1

Step 2 — Calculate the area of the square

We know the diagonal of the square is 10 units. There is a special way to find the area of a square if you know its diagonal. The area of a square is half the square of its diagonal. Let's use this formula to find the area of our square.

$\text{Area of the square} = \frac{1}{2} \times (\text{diagonal})^2$

$= \frac{1}{2} \times (10)^2$

$= \frac{1}{2} \times 100$

$\boxed{50 \text{ square units}}$

Step 3 — Find the side length of the square

We found that the area of the square is 50 square units. We also know that the area of a square is found by multiplying its side length by itself. Let the side length of the square be 'a'. So, $a^2$ must be equal to the area we just calculated.

$a^2 = 50$

To find 'a', we need to calculate the square root of 50. We can simplify $\sqrt{50}$ by looking for factors that are perfect squares. We know that $50 = 25 \times 2$ . And 25 is a perfect square ( $5 \times 5 = 25$ ).

$a = \sqrt{50}$

$a = \sqrt{25 \times 2}$

$a = \sqrt{25} \times \sqrt{2}$

$\boxed{a = 5\sqrt{2} \text{ units}}$

Step 4 — State the sidelengths of the triangle

From Step 1, we understood that the two equal sides of our isosceles right triangle are the same as the side length of this square. We found the side length of the square to be $5\sqrt{2}$ units. So, the other two sidelengths of the isosceles right triangle are both $5\sqrt{2}$ units.

Answer

(i) The first equal sidelength is $5\sqrt{2}$ units. (ii) The second equal sidelength is $5\sqrt{2}$ units.

Question 3

Answer

More questions in FIO