Find the area of an isosceles triangle, each of whose equal sides is 13 cm and base 24 cm.

Each of equal sides (a) = 13 cm and base (b) = 24 cm

By formula,

$\Rightarrow \text{Area of an isosceles triangle} = \dfrac{1}{4}b\sqrt{4a^2 - b^2} \\[1em] \Rightarrow \dfrac{1}{4} \times 24 \times \sqrt{4 × 13^2 - 24^2} \\[1em] \Rightarrow 6 \times \sqrt{676 - 576} \\[1em] \Rightarrow 6 \times \sqrt{100} \\[1em] \Rightarrow 6 \times 10 \\[1em] \Rightarrow 60 \text{ cm}^2.$

Hence, area of the triangle = 60 cm².