Question

A metal wire, when bent in the form of an equilateral triangles of largest area, encloses an area of $484\sqrt{3}c{m}^2$. If the same wire is bent into the form of a circle of largest area, find the area of this circle.

• A. 4772 $c{m}^2$
• B. 2597 $c{m}^2$
• C. 1386 $c{m}^2$
• D. 1126 $c{m}^2$

Answer 1

The correct option is C 1386 $c{m}^2$

Given the area of wire band in equilateral triangle is $484\sqrt{3}c{m}^2$

We know that area of equilateral triangle=$\frac{\sqrt{3}}{4}{a}^2$ where a is the side of equilateral triangle.

$\therefore \frac{\sqrt{3}}{4}{a}^2=484\sqrt{3}$

$\Rightarrow a^2=1936$

$\Rightarrow a=44$

So side of equilateral triangle=44 cm

Then perimeter of equilateral triangle.=$3\times side=3\times 44=132cm$

If equilateral triangle band in circle the his perimeter =132 cm

We know that perimeter of circle =$2\pi r$ where r is radius of circle

$\therefore 2\pi r=132\Rightarrow r=21$

Then area of circle =$\pi r^2=\frac{22}{7}\times (21{)}^2=1386c{m}^2$