Question

An isosceles right triangle has an area of $8{\mathrm{cm}}^2$. The length of its hypotenuse is

• A. $\sqrt{32}\mathrm{cm}$
• B. $\sqrt{16}\mathrm{cm}$
• C. $\sqrt{48}\mathrm{cm}$
• D. $\sqrt{24}\mathrm{cm}$

Answer 1

The correct option is A

$\sqrt{32}\mathrm{cm}$

Explanation of the correct option

Step 1: Find the length of the base:

Given that the area of an isosceles right triangle is $8{\mathrm{cm}}^2$.

We know that area of an isosceles triangle $=\frac{1}{2}\times \mathrm{base}\times \mathrm{perpendicular}$

Since in an isosceles right triangle, base and perpendicular are of equal length. Let us say that they both measure $\text{'}\mathrm{x}\text{'}$ then,

Area $=\frac{1}{2}\times \mathrm{x}\times \mathrm{x}$

$$\begin{gathered} \Rightarrow \frac{1}{2}\times {\mathrm{x}}^2=8 \\ \Rightarrow {\mathrm{x}}^2=16 \\ \Rightarrow \mathrm{x}=4 \end{gathered}$$

Thus base $=4\mathrm{cm}$ and perpendicular $=4\mathrm{cm}$

Step 2: Find the length of the hypotenuse

In a right-angled triangle, ${\left(\mathrm{Hypotenuse}\right)}^2={\left(\mathrm{Base}\right)}^2+{\left(\mathrm{Perpendicular}\right)}^2$

$$\begin{gathered} ={\left(4\right)}^2+{\left(4\right)}^2 \\ =16+16 \\ =32 \end{gathered}$$

$\Rightarrow \mathrm{Hypotenuse}=\sqrt{32}$

Therefore the length of the hypotenuse is $\sqrt{32}\mathrm{cm}$.