Question:
The height of an equilateral triangle measures $9 \mathrm{~cm}$. Find its area, correct to 2 places of decimal. Take $\sqrt{3}=1.732$.
Solution:
Height of the equilateral triangle = 9 cm
Thus, we have:
Height $=\frac{\sqrt{3}}{2} \times$ Side
$\Rightarrow 9=\frac{\sqrt{3}}{2} \times$ Side
$\Rightarrow$ Side $=\frac{18}{\sqrt{3}}=\frac{18}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=6 \sqrt{3} \mathrm{~cm}$
Also,
Area of equilateral triangle $=\frac{\sqrt{3}}{4} \times(\text { Side })^{2}$
$=\frac{\sqrt{3}}{4} \times(6 \sqrt{3})^{2}$
$=\frac{108}{4} \sqrt{3}$
$=27 \sqrt{3}$
$=46.76 \mathrm{~cm}^{2}$