Question

The base of an isosceles triangle measure $24cm$ and the area in $60c{m}^2$ find its perimeter.

Answer 1

Let the the triangle be $ABC$ in which $D$ is mid-point of $AC$

$AC=24$ , therefore $AD=12cm$

Area $=60c{m}^2$

and we know that area of triangle $=\frac{1}{2}\times AC\times BD$

then, $60=\frac{1}{2}\times 24\times BD$

$BD=5cm$

Now, by Pythagoras theorem

$A{B}^2=A{D}^2+B{D}^2$

$AB=13$

Now perimeter of $\Delta ABC=AB+BC+AC=2AB+AC$ (two sides of isosceles triangle are equal)

Perimeter $=26+24=50cm$