Question

A copper wire when bent in the form of an equilateral triangle encloses an area of $121\sqrt{3}$ $c{m}^2$. If the same wire is bent in the form of a circle then the area of the circle is:

• A. $121c{m}^2$
• B. $342c{m}^2$
• C. $346.5c{m}^2$
• D. $154.8c{m}^2$

Answer 1

Answer: (C)

$346.5c{m}^2$

Given that, a copper wire when bent in the form of an equilateral triangle encloses an area of $121\sqrt{3}c{m}^2$

To find out: Area of a circle that the same wire can enclose.

We know that, area of an equilateral triangle having side $a$ is $\frac{\sqrt{3}{a}^2}{4}$

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

$\Rightarrow a^2=484$

$\Rightarrow a=22cm$

Side of the triangle is $22cm$.
We know that, perimeter of an equilateral triangle is $3a$

Hence, the perimeter of the triangle is $66cm$.

The same wire is enclosing the circle.

Hence, perimeter of the triangle $=$ circumference of the circle

$\therefore 66cm=$ Circumference of the circle.

We know that, circumference of a circle is $2\pi r$

$\therefore 2\pi r=66$

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

$\therefore r=10.5cm$

We know that, area of a circle is $\pi r^2$
$\therefore$ Area $=\pi (10.5{)}^2$$=346.5{cm}^2$

Hence, the area of the circle is $346.5c{m}^2$.