Question

The height of an equilateral triangle measures $9cm$. Find its area, correct upto $2$ places of decimal. Take $\sqrt{3}=1.732$

Answer 1

Height $=\frac{\sqrt{3}}{2}\left(side\right)=9$

$\Rightarrow$ side $=\frac{18}{\sqrt{3}}$

Now,

Area $=\frac{\sqrt{3}}{4}(side{)}^2$

$=\frac{\sqrt{3}}{4}\times \frac{18\times 18}{\sqrt{3}\times \sqrt{3}}$

$=3\sqrt{3}\times 9$

$=3\times 1.732\times 9$

$=5.196\times 9$

$=46.764c{m}^2$