The base of an isosceles triangle is 12cm  and its perimeter is 32cm. Then Find its area.

Area of triangle $=\frac{1}{2}\times \text{ base }\times \text{ height }$
Let the congruent sides of the isosceles triangle =a and the base =b.

Given, $b=12 \mathrm{cm}$
Perimeter of the triangle = Sum of all sides =a+a+b=32
$\Rightarrow 2a+12=32$
On solving, $a=10 \mathrm{cm}$
Since the height of the triangle, divides it into two right angled triangles,
${\mathrm{a}}^2={\mathrm{h}}^2+{\left(\frac{\mathrm{b}}{2}\right)}^2\Rightarrow {10}^2={\mathrm{h}}^2+6^2$
$h^2=100-36=64$
$\mathrm{h}=\sqrt{64}=8 \mathrm{cm}$
$\text{ Area of triangle }=\frac{1}{2}\times \text{ base }\times \text{ height }=\frac{1}{2}\times 12\times 8=48\mathrm{sqcm}$