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

Area of an isosceles triangle:

$=\frac{1}{4} b \sqrt{4 a^{2}-b^{2}}$ (Where $a$ is the length of the equal sides and $b$ is the base)

$=\frac{1}{4} \times 24 \sqrt{4(13)^{2}-24^{2}}$

$=6 \sqrt{4 \times 169-576}$

$=6 \sqrt{676-576}$

$=6 \sqrt{100}$

$=6 \times 10$

$=60 \mathrm{~cm}^{2}$