Question

The unequal side of an isosceles triangle is $48cm$, and its perimeter is $108cm$. Find the area of the triangle.

Answer 1

Let $AB=AC=0$

$\Rightarrow 2a+48=108$

$\Rightarrow 2a=60+a=30$

Perpendicular AD to BC also bisect BC(Isosceles triangle prop)

$\Rightarrow BD=24cm$ By Pythagoras theorem,

$A{D}^2=a^2-{24}^2=900-576=324$

$\Rightarrow AD=18cm$

area of triangle $=\frac{1}{2}\times Base\times Height=\frac{1}{2}\times 48\times 18=432c{m}^2$.