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The area of circle inscribed in an equilateral triangle is 154c{{m}^{2}}. Find the perimeter of the triangle.
Areas Related To Circles | CBSE | Class 10 | Exercise 15.1 | Maths | RD Sharma
The area of circle inscribed in an equilateral triangle is 154c{{m}^{2}}. Find the perimeter of the triangle.

Assume the circle inscribed in the equilateral triangle be with a center O and radius r.

We know that, formula of area of a Circle =\pi {{r}^{2}}

Now, the given that area is 154c{{m}^{2}}.

\left( 22/7 \right)\times {{r}^{2}}=154

{{r}^{2}}=\left( 154\times 7 \right)/22=7\times 7=49

r=7cm

From the given figure we can conquer that,

At point M, BC side is tangent and also at point M, BM is perpendicular to OM.

We all know that,

In an equilateral triangle, the perpendicular from vertex divides the side into two halves.

BM=\frac{1}{2}\times BC

Assume the side of the equilateral triangle be x cm.

BM=\frac{1}{2}x=\frac{x}{2}

O{{B}^{2}}=B{{M}^{2}}+M{{O}^{2}}

OB=\sqrt{{{r}^{2}}+\frac{{{x}^{2}}}{4}}=\sqrt{49+\frac{{{x}^{2}}}{4}}

BD=\frac{\sqrt{3}}{2}(\operatorname{si}de)=\frac{\sqrt{3}}{2}x=OB+OD

\frac{\sqrt{3}}{2}x-r=\sqrt{49+\frac{{{x}^{2}}}{4}},r=7

We get the answer as

x=14\sqrt{3}cm

Perimeter =3x=3\times 14\sqrt{3}=42\sqrt{3}cm

Hence, the perimeter of the triangle 42\sqrt{3}cm=42\left( 1.73 \right)=72.7cm

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