An isosceles triangle has base 10 cm, and its area is 60 cm². What are the lengths of the equal sides?

$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}$

$$60 = \frac{1}{2} \times 10 \times h$$

$$60 = 5h$$

$$h = 12 \text{ cm}$$

$$\text{Equal side} = \sqrt{\left(\frac{\text{Base}}{2}\right)^2 + h^2}$$

$$= \sqrt{5^2 + 12^2}$$

$$= \sqrt{25 + 144}$$

$$= \sqrt{169}$$

$$= 13 \text{ cm}$$