Question

A wire when bent in the form of an equilateral triangle encloses an area of $121\sqrt{3}c{m}^2$ The same wire is bent form a circle Find the area enclosed by the circle.

• A. $564.5c{m}^2$
• B. $346.5c{m}^2$
• C. $446.5c{m}^2$
• D. $146.5c{m}^2$

Answer 1

Answer: (B)

$346.5c{m}^2$

Area of an equilateral triangle$=\frac{\sqrt{3}{a}^2}{4}$

where a=side of the triangle

Now, given that

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

$a^2=121\times 4$

$a^2=484$

$a=\sqrt{484}$

$a=22cm$

Therefore,

Length of wire$=a+a+a=66cm$

Now, this wire is bent into a circle, so circumference of circle=length of the wire

So,

$2\pi r=66$

$2\times \frac{22}{7}\times r=66$

$r=\frac{66\times 7}{2\times 22}$

$r=\frac{21}{2}$

Now,

Area of circle$=\pi r^2$

$=\frac{22}{7}\times {\left(\frac{21}{2}\right)}^2$

$=\frac{693}{2}$

$=346.5c{m}^2$