Question

The perimeter of an isosceles triangle is 32 cm. The ratio of one of the equal sides to its base is 3 : 2. Find the area of the triangle.

Answer 1

Let the equal sides of the isosceles triangle (a) be 3x cm and base of the triangle (b) be 2x cm.

(i) Perimeter = 32 cm
or, 3x + 3x + 2x = 32
or, 8x = 32
or, x = 4

∴ Equal sides of the triangle = 3x = 3(4) = 12 cm
Base of the triangle = 2x = 2(4) = 8 cm
Now,
$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{isosceles}\mathrm{triangle}=\frac{b}{4}\sqrt{4a^2-b^2} \\ \mathrm{Here}, \\ a=12\mathrm{cm}\mathrm{and}b=8\mathrm{cm} \\ \mathrm{Thus},\mathrm{we}\mathrm{have}: \\ =\frac{8}{4}\times \sqrt{4(12{)}^2-8^2} \\ =2\times \sqrt{576-64} \\ =2\times \sqrt{512} \\ =2\times \sqrt{256\times 2} \\ =2\times 16\sqrt{2} \\ =32\sqrt{2}{\mathrm{cm}}^2 \end{gathered}$$