Question

Find the area of an isosceles triangle each of whose equal sides is 13 cm and whose base is 24 cm.

Answer 1

Area of an isosceles triangle:
$$\begin{gathered} =\frac{1}{4}b\sqrt{4a^2-b^2}(\mathrm{Where}a\mathrm{is}\mathrm{the}\mathrm{length}\mathrm{of}\mathrm{the}\mathrm{equal}\mathrm{sides}\mathrm{and}b\mathrm{is}\mathrm{the}\mathrm{base}) \\ =\frac{1}{4}\times 24\sqrt{4{\left(13\right)}^2-{24}^2} \\ =6\sqrt{4\times 169-576} \\ =6\sqrt{676-576} \\ =6\sqrt{100} \\ =6\times 10 \\ =60{\mathrm{cm}}^2 \end{gathered}$$