The perimeter of an isosceles triangle is $32 \, cm$. The ratio of the equal side to its base is $3:2$. Find the area of the triangle.

(D) Let the equal sides of the isosceles triangle be $3x$ and $3x$,and the base be $2x$.
The perimeter of the triangle is the sum of its sides: $3x + 3x + 2x = 8x$.
Given that the perimeter is $32 \, cm$,we have $8x = 32$,which gives $x = 4$.
Thus,the sides of the triangle are $12 \, cm, 12 \, cm$,and $8 \, cm$.
The semi-perimeter $s$ is given by $s = \frac{12 + 12 + 8}{2} = \frac{32}{2} = 16 \, cm$.
Using Heron's formula,Area $= \sqrt{s(s-a)(s-b)(s-c)}$.
Area $= \sqrt{16(16-12)(16-12)(16-8)} = \sqrt{16 \times 4 \times 4 \times 8}$.
Area $= \sqrt{16 \times 16 \times 8} = 16 \sqrt{8} = 16 \times 2\sqrt{2} = 32\sqrt{2} \, cm^2$.