Question

a steel wire is bent in the form of a square encloses an area of 121cm2. if the the same wire is went is bent in the form of a circle ,then the circumference of the circle is :

Answer 1

we know that even if the shape of the wire changes and area it encloses changes, the perimeter will remain same......
so side of square=
$\sqrt{121}$
=11 cm
hence perimeter of square =4*side=4*11=44
circumference=44cm
2*22/7*r=44
r=7cm

hope it helps.....

Answer 2

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

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

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

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

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

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

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

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

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

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

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

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

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