Question

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

Answer 1

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

Step 1: Find the length of the base and perpendicular

We know that area of an isosceles triangle $=\frac{1}{2}\times$base $\times$ 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, ${\left(\mathrm{hypotenuse}\right)}^2={\left(\mathrm{base}\right)}^2+{\left(\mathrm{perpendicular}\right)}^2$

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

$=\sqrt{32}$

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