The altitude of an isosceles triangle is $8cm$ and the perimeter is $32cm.$ Calculate the area of the triangle.

The correct option is A: $48c{m}^2$

Let $ABC$ be the isosceles triangle, the $AD$ be the altitude

Let $AB=AC=x$ then $BC=32-2x$ [Given, Perimeter $=32cm,$ and Altitude $=8cm$]

Since, in an isosceles triangle the altitude bisects the base

So, $BD=DC=16-x$

In $\Delta ADC,$ $(AC{)}^2$ = $(AD{)}^2$ + $(DC{)}^2$ [Using Pythagoras theorem]

$\Rightarrow$ $x^2=(8{)}^2+(16-x{)}^2$

$\Rightarrow$ $x^2=64+256+x^2-32x$

$\Rightarrow 32x=320$

$\Rightarrow x=10$

$BC=32-2x=32-20=12cm$

Hence, required area of triangle $ABC=\frac{1}{2}\times$ base $\times$ height

$=\frac{1}{2}$ $\times BC$$\times AD$

$=\frac{1}{2}$ $\times 12$$\times 8=48c{m}^2$