Question

The base of an isosceles triangle is 8 cm long and each of its equal sides measures 6 cm. The area of the triangle is
(a) $16\sqrt{5}{\mathrm{cm}}^2$
(b) $8\sqrt{5}{\mathrm{cm}}^2$
(c) $16\sqrt{3}{\mathrm{cm}}^2$
(d) $8\sqrt{3}{\mathrm{cm}}^2$

Answer 1

(b) $8\sqrt{5}{\mathrm{cm}}^2$
$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{isosceles}\mathrm{triangle}=\frac{b}{4}\sqrt{4a^2-b^2} \\ \mathrm{Here}, \\ a=6\mathrm{cm}\mathrm{and}b=8\mathrm{cm} \\ \mathrm{Thus},\mathrm{we}\mathrm{have}: \\ \frac{8}{4}\times \sqrt{4(6{)}^2-8^2} \\ =\frac{8}{4}\times \sqrt{144-64} \\ =\frac{8}{4}\times \sqrt{80} \\ =\frac{8}{4}\times 4\sqrt{5} \\ =8\sqrt{5}{\mathrm{cm}}^2 \end{gathered}$$