Question:
The height of an equilateral triangle is 6 cm. Find its area. $[$ Take $\sqrt{3}=1.73]$
Solution:
Let the side of the equilateral triangle be x cm.
As, the area of an equilateral triangle $=\frac{\sqrt{3}}{4}(\text { side })^{2}=\frac{x^{2} \sqrt{3}}{4}$
Also, the area of the triangle $=\frac{1}{2} \times$ Base $\times$ Height $=\frac{1}{2} \times x \times 6=3 x$
So, $\frac{x^{2} \sqrt{3}}{4}=3 x$
$\Rightarrow \frac{x \sqrt{3}}{4}=3$
$\Rightarrow x=\frac{12}{\sqrt{3}}$
$\Rightarrow x=\frac{12}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}$
$\Rightarrow x=\frac{12 \sqrt{3}}{3}$
$\Rightarrow x=4 \sqrt{3} \mathrm{~cm}$
Now, area of the equilateral triangle $=3 x$
$=3 \times 4 \sqrt{3}$
$=12 \sqrt{3}$
$=12 \times 1.73$
$=20.76 \mathrm{~cm}^{2}$