Question

A steel wire which bent in the form of square enclosed an area of 121 cm2 if the same wire is bent in circle​

Answer 1

Answer:154cm2

Step-by-step explanation:

Area of square= a^2

a=√121

a=11cm

Now, perimeter of square=4a

=4×11

=44 .

Now, perimeter of square=perimeter of circle .

So, 44=2Πr

Πr=22

r =22×7/22

r=7

Hence area of circle=Πr^2

=22/7×7^2

=22×7 = 154cm2

Answer 2

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

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

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

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

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

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

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

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

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

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

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

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

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