Question

if a wire is bent in the form of square it encloses a area of 121 metre square. if same wire is bent in form of circle , find the area it encloses?

Answer 1

area of square =121 m
one side of square = root  121  i.e 11m
hence circumfrence of circle =length of wire =4*11
                                                              = 44 m
hence 2 * pie * r= 44
or
2* (22/7)* r=44
r (44*7)/(2*22)=7 m

:. area of circle = pie*r*r =(22/7)*7*7=154m2

Answer 2

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

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

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

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

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

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

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

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

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

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

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

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

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