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 $72.7 \mathrm{~cm}$.

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Updated on: 10-Oct-2022

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