In △ABC, AB = AC = x, BC = 10 cm and the area of △ABC is 60 cm². Find x.

Let AD be the altitude. In isosceles triangle, the altitude to base bisects the base.

∴ BD = CD = 5 cm.

From figure,

In right triangle ABD,

AB² = AD² + BD²

x² = AD² + 5²

AD² = x² - 25.

AD = $\sqrt{x^2 - 25}$

$\text{Area of △ABC } = \dfrac{1}{2} \times \text{AD} \times \text{BC} \\[1em] \Rightarrow 60 = \dfrac{1}{2} \times \sqrt{x^2 - 25} \times 10 \\[1em] \Rightarrow 120 = 10\sqrt{x^2 - 25} \\[1em] \Rightarrow \sqrt{x^2 - 25} = 12 \\[1em] \Rightarrow x^2 - 25 = 144 \\[1em] \Rightarrow x^2 = 169 \\[1em] \Rightarrow x = \sqrt{169} = 13.$

Hence, x = 13 cm.