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The area of a circle inscribed in an equilateral triangle is $154\ cm^2$. Find the perimeter of the triangle. [Use $\pi = \frac{22}{7}$ and $\sqrt3= 1.73$]
Given:
The area of a circle inscribed in an equilateral triangle is $154\ cm^2$.
To do:
We have to find the perimeter of the triangle.
Solution:
Let a circle of radius $r$ be inscribed in an equilateral triangle $ABC$.
This implies,
$OE=r$
$\Rightarrow \pi r^{2}=154$
$\Rightarrow \frac{22}{7} r^{2}=154$
$\Rightarrow r^{2}=\frac{154 \times 7}{22}$
$\Rightarrow r^{2}=49$
$\Rightarrow r^{2}=(7)^{2}$
$\Rightarrow r=7 \mathrm{~cm}$
Therefore,
$\mathrm{OE}=7 \mathrm{~cm}$
In the equilateral triangle $\mathrm{ABC}$, $AD\ \perp\ BC$ which bisects $BC$ at $D$.
$AD=3OD$
$=3 \times 7$
$=21 \mathrm{~cm}$ Let $a$ be the side of the triangle $ABC$.
$\Rightarrow \frac{\sqrt{3}}{2} a=21$
$\Rightarrow a=\frac{21 \times 2}{\sqrt{3}}$
$\Rightarrow a=\frac{21 \times 2 \times \sqrt{3}}{(\sqrt{3})^2}$
$\Rightarrow a=\frac{21 \times 2 \sqrt{3}}{3}$
$\Rightarrow a=7 \times 2 \sqrt{3}$
$\Rightarrow a=7 \times 2(1.73)$
$\Rightarrow a=24.22 \mathrm{~cm}$
Perimeter of the triangle $=3a$
$=3 \times 24.22$
$=72.66 \mathrm{~cm}$
$=72.7 \mathrm{~cm}$
The perimeter of the triangle is $72.7 \mathrm{~cm}$.