Question

If the area of an isosceles right triangle is $8c{m}^2$, what is the perimeter of the triangle?

• A. $8+4\sqrt{2}c{m}^2$
• B. $8+\sqrt{2}c{m}^2$
• C. $4+8\sqrt{2}c{m}^2$
• D. $12\sqrt{2}c{m}^2$

Answer 1

The correct option is A $8+4\sqrt{2}c{m}^2$

Let $ABC$ be an isosceles right triangle such that its height $=$ base $=x$



Area of triangle $=8c{m}^2$ (given)

$\therefore$ Area $=\frac{1}{2}\times$ base $\times$ height

$8=\frac{1}{2}\times x\times x$

$x^2=16$

$\Rightarrow x=\sqrt{16}=4cm$

By Pythagoras theorem,

$A{C}^2=A{B}^2+B{C}^2$
$A{C}^2=(4{)}^2+(4{)}^2$
$A{C}^2=16+16$
$A{C}^2=32$
$AC=\sqrt{32}$
$=4\sqrt{2}cm$

$\therefore$ Perimeter of an isosceles triangle
$=AB+BC+CA$
$=4+4+4\sqrt{2}=8+4\sqrt{2}cm$

Hence, $\left(b\right)$ is the correct answer.