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A wire when bent in the form of an equilateral triangle encloses an area of $121\sqrt{3}\; cm^2$ . If the wire is bent in the form of a circle, find the area enclosed by the circle. $\left ( \text{Use }\pi=\frac{22}{7} \right )$ .

Answers (1)

$\text{Area of equilateral }\bigtriangleup =\frac{\sqrt{3}}{4}a^2$

where a is a side of equilateral $\bigtriangleup$ .

Given :

$\text{Area of equilateral }\bigtriangleup =121\sqrt{3}\; cm^2$

$\Rightarrow \frac{\sqrt{3}}{4}a^2 =121\sqrt{3}$

$\Rightarrow a^2 =121\times 4$

$\Rightarrow a=\sqrt{121\times 4}$

$\Rightarrow a=11\times 2$

$\Rightarrow a=22\; cm$

$\text{Perimeter of circle }=\text{Perimeter of }\bigtriangleup$

$2\pi r=3a$

$2\times \frac{22}{7}\times r=3\times 22$

$r=\frac{3\times 22\times 7}{2\times 22}=\frac{21}{2}=10.5\; cm$

$r=10.5\; cm$

$\text{Area enclosed by circle }=\pi r^2$

$=\frac{22}{7}\times 10.5\times 10.5$

$=346.5\, cm^2$

Posted by

Safeer PP

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