Question

If the area of an isosceles right triangle is 8cm, what is the perimeter of the triangle.

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

Answer 1

The correct option is B

$8+4\sqrt{2}c{m}^2$

Let base = x
ABC an isosceles right triangle, which has 2 sides same
$\Rightarrow Height=x$

Area = 8 cm (Given)
$\therefore Area=\frac{1}{2}\times Base\times Height$
$8cm=\frac{1}{2}\times x\times x$
$x^2=16$
$\Rightarrow x=\sqrt{16}=4cm$
(Where x = base and height)
Now, by Pythagoras Theorem
$A{C}^2=A{B}^2+B{C}^2$
=$4^2+4^2$
= 16+16= 32 cm
$AC=\sqrt{32}$
$\Rightarrow AC=4\sqrt{2}cm$
Perimeter of an isosceles triangle
= AB+BC+CA
= $4+4+4\sqrt{2}cm$
= $8+4\sqrt{2}cm$