Question

A copper wire when bent in the form of a square, encloses an area of \[484c{m^2}\]. If the same wire is bent in the form of a circle, find the area enclosed by it.

Answer 1

Hint:
The hack in this question is to understand that when the same wire is bent to circle from square shape then perimeter won’t change. We’ll use the formula of area of square to find it’s side. Then with the side we can find the perimeter of the circle then area.
Perimeter of square = perimeter of circle.

Complete step by step solution:
We have to find the area enclosed by the circle. For that dimension of circle like radius or diameter is needed.
 A copper wire is bent in the form of a square and the same copper wire is bent to form the circle also.
So the perimeter in each case should be the same.
Perimeter of square = perimeter of circle.
But we do not have a side of square or any dimension of circle also.
Let’s find it.
Given a square encloses an area of \[484c{m^2}\].
Area of square \[ = side \times side\]
\[ \Rightarrow 484 = side \times side\]
Taking square root on both sides,
\[ \Rightarrow 22 = side\]
This side of the square is 22cm.
Now,
Perimeter of square = perimeter of circle.
\[
   \Rightarrow 4 \times side = 2\pi r \\
   \Rightarrow 4 \times 22 = 2 \times \dfrac{{22}}{7} \times r \\
   \Rightarrow r = 14cm \\
\]
We found the radius of circle r=14cm.
Now find the area enclosed,
Area enclosed by copper wire bent in circle,
\[
   \Rightarrow \pi {r^2} \\
   \Rightarrow \dfrac{{22}}{7} \times 14 \times 14 \\
   \Rightarrow 22 \times 28 \\
   \Rightarrow 616c{m^2} \\
\]

So, the area enclosed is \[616c{m^2}\].

Note:
1) Areas of the shapes cannot be equated directly because they are not the same.
2) Perimeters are equated because the same wire is used in both cases.