The base of an isosceles is 14 cm and its perimeter is 36 cm. Find its area.

Answer: Option A

Let each equal side of isosceles triangle be x cm
Perimeter of an isosceles triangle = 36 cm
$$\begin{array}{rl}& \therefore x+x+14=36\\ & \Rightarrow 2x=36-14\\ & \Rightarrow x=\frac{22}{2}\\ & \Rightarrow x=11cm\end{array}$$

BD = DC = 7cm
From ΔABD
By using Pythagoras theorem :
$$\begin{array}{rl}& AD=\sqrt{A{B}^2-B{D}^2}\\ & =\sqrt{{11}^2-7^2}\\ & =\sqrt{121-49}\\ & =\sqrt{72}\\ & =3\times 2\sqrt{2}\\ & =6\sqrt{2}cm\end{array}$$
∴ Area of ΔABC
$$\begin{array}{rl}& =\frac{1}{2}\times BC\times AD\\ & =\frac{1}{2}\times 14\times 6\sqrt{2}\\ & =42\sqrt{2}\text{ sq}\text{. cm}\end{array}$$