Question

A wire is bent in the form of a square encloses an area of 121 sq.cm. if the same wire is bent in the form of a circle, then the area it enclosed is equal to

Answer 1

Step-by-step explanation:

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.

Answer 2

__________

$\large \boxed{ \textsf{given:-}}$

$\texttt {\: enclosed area of steel wire when bent to form square = 121 \: sq.cm }$

$\large \boxed{ \textsf{to find out:-}}$

$\textsf{the area of circle}=??$

$\large \boxed{ \rm \: solution:-}$

$\rm \: Side \: a \: square \: = \sqrt{121}cm = 11cm$

$\rm \: perimeter \: of \: square = (4 \times 11)cm = 44cm$

$\rm \therefore \: length \: of \: the \: wire \: = 44cm$

$\rm \therefore \: circumference \: of \: the \: circle \: = length \: of \: wire = 44cm$

$\textsf{let the radius of the circle be }r \rm \: cm$

$\rm \: then \: 2 \pi \: r = 44 \implies \: 2 \times \large \frac{22}{7} \small r = 44 \implies \: r = 7$

$\rm \therefore \: area \: of \: the \: circle = \pi \: r {}^{2}$

$\large \rm \: = \huge( \small \frac{22}{7} \times 7 \times 7 \huge) \small \:cm {}^{2} = 154 \: cm {}^{2}$