The base of an isosceles triangle is 24 cm and its area is 192 sq. cm. Find its perimeter.

Given:

Base (b) = 24 cm

Area = 192 cm²

Let s be the length of the equal sides of the triangle.

Area of an isosceles triangle = $\dfrac{1}{4} \times b \times \sqrt{4s^2 - b^2}$

$⇒ 192 = \dfrac{1}{4} \times 24 \times \sqrt{4s^2 - 24^2}\\[1em] ⇒ 192 = 6 \times \sqrt{4s^2 - 576}\\[1em] ⇒ \sqrt{4s^2 - 576} = \dfrac{192}{6}\\[1em] ⇒ \sqrt{4s^2 - 576} = 32\\[1em] ⇒ 4s^2 - 576 = 1024\\[1em] ⇒ 4s^2 = 1024 + 576\\[1em] ⇒ 4s^2 = 1600\\[1em] ⇒ s^2 = \dfrac{1600}{4}\\[1em] ⇒ s^2 = 400\\[1em] ⇒ s = \sqrt{400}\\[1em] ⇒ s = 20$

Perimeter = sum of all sides of triangle

= 20 + 20 + 24 cm

= 64 cm

Hence, the perimeter is 64 cm.