Question

Find the area of an isosceles triangle each of whose equal sides measures 13 cm and whose base measures 20 cm.

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}: \\ a=13\mathrm{cm}\mathrm{and}b=20\mathrm{cm} \\ \therefore \mathrm{Area}\mathrm{of}\mathrm{isosceles}\mathrm{triangle}=\frac{b}{4}\sqrt{4a^2-b^2} \\ =\frac{20}{4}\times \sqrt{4(13{)}^2-{20}^2} \\ =5\times \sqrt{676-400} \\ =5\times \sqrt{276} \\ =5\times 16.6 \\ =83.06{\mathrm{cm}}^2 \end{gathered}$$