# The base of an isosceles triangle is 8 cm long and each of its equal sides measures 6 cm.

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:

$\frac{8}{4} \times \sqrt{4(6)^{2}-8^{2}}$

$=\frac{8}{4} \times \sqrt{144-64}$

$=\frac{8}{4} \times \sqrt{80}$

$=\frac{8}{4} \times 4 \sqrt{5}$

$=8 \sqrt{5} \mathrm{~cm}^{2}$