Question

A copper wire, bent in the form of a square, encloses an area of 484cm^2 .
if the same wire is bent in the form of a circle, find the area enclosed by it.

help

Answer 1

Given that Area of the square formed by bending a copper wire = 484cm^2.

We know that side of the square = s^2

                                 s^2 = 484

                                 s = 22cm.


We know that perimeter of the square = 4 * s

                                                                 = 4 * 22

                                                                 = 88cm.


Therefore the length of the wire = 88cm.

Now,

Given that the same wire is bent into a form of circle.

Circumference of circle = Length of wire

2pir = 88

2 * 22/7 * r = 88

44/7 * r = 88

44 * r = 88 * 7

44 * r = 616

r = 616/44

r = 14.


We know that Area of circle = pir^2

                                                = 22/7 * (14)^2

                                                = 22 * 14 * 2

                                                = 616cm^2.


Therefore the area enclosed by the circle = 616cm^2.



Hope this helps! ---------------- Gud Luck

Answer 2

$area \: of \: square = side \times side$
$area = {side}^{2}$
$484 = {side}^{2}$
$side = \sqrt{484}$
=22cm

to find the area enclosed by it, we have to first find the perimeter of square.
why??
because it is given that the same wire is bent in a square.
so,
perimeter of square=circumference of the circle

perimeter of square=4×side
=22×4
=88cm

now,

let the radius be 'r'

$circumference of \: circle \\ = 2\pi \: r$
$88 = 2 \times \frac{22}{7} \times r$
$88 \times \frac{1}{2 } \times \frac{7}{22}$=r
radius=14cm

$area \: of \: circle = \pi {r}^{2}$
$= \frac{22}{7} \times 14 \times 14$
=
$616 {cm}^{2}$
is the area