Question

A copper wire when bent in the form of a square, enclosed an area of 121 m^2. If the wire is bent to form a circle, the area enclosed by it would be-

Answer 1

A of square = s^2
121=s^2

$\sqrt{121 } = s$
s = 11
P of square = 4s
P = 4 × 11
Length of wire is equal to perimeter
therefore, length of wire = 44m
perimeter of square = perimeter of circle
circumference of circle =
$2\pi \: r$
$44= 2 \times \frac{22}{7} \times r$
$44 \times 7 = 2 \times 22 \times r$
$44 \times 7 = 44 \times r$
$7 = r$
area of circle =
$\pi {r}^{2}$
$= \frac{22}{7} \times 7 \times 7$
$22 \times 7$
$= 154m$
therefore area of circle = 154 m^2

Answer 2

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

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

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

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

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

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

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

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

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

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

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

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

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