The base of an isosceles triangle is 18 cm and its area is 108 cm². Find its perimeter.

Given,

Base (b) = 18 cm

Area = 108 cm².

Let the length of equal sides of the isosceles triangle be a cm.

By formula,

$\Rightarrow \text{Area of an isosceles triangle} = \dfrac{1}{4}b\sqrt{4a^2 - b^2} \\[1em] \Rightarrow 108 = \dfrac{1}{4} \times 18 \times \sqrt{4a^2 - 18^2} \\[1em] \Rightarrow 432 = 18 \times \sqrt{4a^2 - 324} \\[1em] \Rightarrow \sqrt{4a^2 - 324} = \dfrac{432}{18} \\[1em] \Rightarrow \sqrt{4a^2 - 324} = 24 \\[1em] \Rightarrow 4a^2 - 324 = 24^2 \\[1em] \Rightarrow 4a^2 - 324 = 576 \\[1em] \Rightarrow 4a^2 = 576 + 324 \\[1em] \Rightarrow 4a^2 = 900 \\[1em] \Rightarrow a^2 = \dfrac{900}{4} \\[1em] \Rightarrow a^2 = 225 \\[1em] \Rightarrow a = \sqrt{225} \\[1em] \Rightarrow a = 15 \text{ cm}.$

Perimeter = Sum of all sides of triangle

= 15 + 15 + 18

= 48 cm.

Hence, perimeter of the isosceles triangle = 48 cm.