Question

Find the area of an isosceles triangle each of whose equal sides measure 13 cm and whose base measures 20 cm. Write your answer correct to 1 place of decimal.

Answer 1

Given: Equal sides of an isosceles triangle = a = 13 cm
Base = b = 20 cm
Area of an isosceles triangle = $\left(\frac{1}{4}b\sqrt{4a^2-b^2}\right)$
$$\begin{gathered} =\frac{1}{4}\times 20\sqrt{4\times \left({13}^2\right)-(20{)}^2} \\ =5\sqrt{676-400} \\ =5\sqrt{276} \\ =5\times (16.61) \\ =83.05 \\ =83.1{\mathrm{cm}}^2 \end{gathered}$$