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

A. 42 $\sqrt{2}$ sq. cm

B. 42 sq. cm

C. 84 sq. cm

D. 48 sq. cm

Solution:

Let each equal side of isosceles triangle be x cm
Perimeter of an isosceles triangle = 36 cm

\begin{aligned} ∴ x + x + 14 = 36 \\ \Rightarrow 2 x = 36 - 14 \\ \Rightarrow x = \frac{22}{2} \\ \Rightarrow x = 11 c m \end{aligned}

BD = DC = 7cm
From ΔABD
By using Pythagoras theorem :

\begin{aligned} A D = \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} c m \end{aligned}

∴ Area of ΔABC

\begin{aligned} = \frac{1}{2} \times B C \times A D \\ = \frac{1}{2} \times 14 \times 6 \sqrt{2} \\ = 42 \sqrt{2} sq . cm \end{aligned}

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