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) $8 \sqrt{3} \mathrm{~cm}^{2}$
Solution:
(b) $8 \sqrt{5} \mathrm{~cm}^{2}$
Area of isosceles triangle $=\frac{b}{4} \sqrt{4 a^{2}-b^{2}}$
Here,
$a=6 \mathrm{~cm}$ and $b=8 \mathrm{~cm}$
Thus, we have:
$\frac{8}{4} \times \sqrt{4(6)^{2}-8^{2}}$
$=\frac{8}{4} \times \sqrt{144-64}$
$=\frac{8}{4} \times \sqrt{80}$
$=\frac{8}{4} \times 4 \sqrt{5}$
$=8 \sqrt{5} \mathrm{~cm}^{2}$