If the area of an isosceles triangle is 60 cm² and the length of each of its equal sides is 13 cm, find its base.

Let the length of equal sides be a cm and length of base be b cm.

Area of isosceles △ABC = $\dfrac{1}{4}b\sqrt{4a^2 - b^2}$, where a is length of equal sides and b is the length of other side.

Substituting values in above equation we get,

$\Rightarrow 60 = \dfrac{1}{4}b\sqrt{4(13)^2 - b^2} \\[1em] \Rightarrow 240 = b\sqrt{4 \times 169 - b^2} \\[1em] \Rightarrow 240 = b\sqrt{676 - b^2}$

Squaring both sides we get,

$\Rightarrow 57600 = b^2(676 - b^2) \\[1em] \Rightarrow 676b^2 - b^4 - 57600 = 0 \\[1em] \Rightarrow b^4 - 676b^2 + 57600 = 0 \\[1em] \Rightarrow b^4 - 576b^2 - 100b^2 + 57600 = 0 \\[1em] \Rightarrow b^2(b^2 - 576) - 100(b^2 - 576) = 0 \\[1em] \Rightarrow (b^2 - 100)(b^2 - 576) = 0 \\[1em] \Rightarrow (b^2 - 100) = 0 \text{ or } (b^2 - 576) = 0 \\[1em] \Rightarrow b^2 = 100 \text{ or } b^2 = 576 \\[1em] \Rightarrow b = 10 \text{ or } b = 24.$

Hence, the length of base = 10 cm or 24 cm.