Question

The area of isosceles right angled triangle is $16c{m}^2$ then the length of the hypotenuse is

• A. $8cm$
• B. $9cm$
• C. $10cm$
• D. $11cm$

Answer 1

The correct option is A

$8cm$

Step 1: Find the equal sides of the triangle

Given

The area of isosceles right angled triangle is $16c{m}^2$

We know that area of isosceles right angle triangle with equal sides of length $a$is given by the formula

$\begin{array}{rcl}A& =& \frac{1}{2}\times a\times a\\ & =& \frac{a^2}{2}\end{array}$

Now,

$$\begin{gathered} \Rightarrow 16=\frac{a^2}{2} \\ \Rightarrow 32=a^2 \\ \Rightarrow a=\sqrt{32} \\ \Rightarrow a=4\sqrt{2}cm \end{gathered}$$

Step 2: Find the hypotenuse of the triangle

We know that from Pythagoras' theorem

${\left(\mathrm{hypotenuse}\right)}^2={\left(\mathrm{perpendicular}\right)}^2+{\left(\mathrm{base}\right)}^2$

For an isosceles right-angle triangle with side length$a$, using Pythagoras theorem

Here $\mathrm{base}=\mathrm{perpendicular}=\mathrm{a}$

So,

$\begin{array}{rcl}{\left(hypotenuse\right)}^2& =& a^2+a^2\\ & =& 2a^2\end{array}$

$\begin{array}{rcl}& \Rightarrow & hypotenuse=\sqrt{2a^2}\\ & =& a\sqrt{2}\end{array}$

Thus, the hypotenuse has length $a\sqrt{2}$

We have, $a=4\sqrt{2}$

So,

$\begin{array}{rcl}a\sqrt{2}& =& 4\sqrt{2}\times \sqrt{2}\\ & =& 4\times 2\\ & =& 8cm\end{array}$

Hence, the length of the hypotenuse is $8cm$

Therefore, the correct option is (A).