Question:
The height of an equilateral triangle is 6 cm. Its area is
(a) $12 \sqrt{3} \mathrm{~cm}^{2}$
(b) $6 \sqrt{3} \mathrm{~cm}^{2}$
(c) $12 \sqrt{2} \mathrm{~cm}^{2}$
(d) $18 \mathrm{~cm}^{2}$
Solution:
(a) $12 \sqrt{3} \mathrm{~cm}^{2}$
Height of equilateral triangle $=\frac{\sqrt{3}}{2} \times$ Side
$\Rightarrow 6=\frac{\sqrt{3}}{2} \times$ Side
$\Rightarrow$ Side $=\frac{12}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=\frac{12}{3} \times \sqrt{3}=4 \sqrt{3} \mathrm{~cm}$
Now,
Area of equilateral triangle $=\frac{\sqrt{3}}{4} \times(\text { Side })^{2}$
$=\frac{\sqrt{3}}{4} \times(4 \sqrt{3})^{2}$
$=\frac{\sqrt{3}}{4} \times 48$
$=12 \sqrt{3} \mathrm{~cm}^{2}$