An isosceles triangle has perimeter 50 cm and each of its two equal sides is 13 cm long. Find its area in ${\text{cm}}^2$.

Base = $50-(13+13)=24$ cm. The height drawn to the base bisects it, so height = √$(13^2-12^2)=$ √(169 − 144) = √25 = 5 cm, and the area = ½ × $24\times 5=60{\text{cm}}^2$. (Using the 13 cm slanting side as the height gives $156{\text{cm}}^2$; using half the base, 12 cm, as the height gives $144{\text{cm}}^2$; $120{\text{cm}}^2$ forgets the ½.)