Question

A wire, when bent in a form of a square, encloses a region of area 121 cm2. If the same wire is bent to form a circle the area of the circle will be?

Answer 1

Let the side of the square formed = x

∴ Area = x²

∴ x² = 121 cm²

x = 11 cm

Length of wire = Perimeter of Square

∴ Length of wire = 4 × 11 cm (4 × side)

Length of wire = 44 cm

Since same wire is bent to form a circle

∴ Perimeter of circle (circumference) = Length of wire

∴ 2 π r = 44 cm (circumference of circle = 2πr)

2 × (22/7) × r = 44 cm

r = 7 cm

Area of circle = π r²

∴ Area of the circle so formed = (22/7) × 7 cm × 7 cm

                                                 = 154 cm²

Answer 2

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

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

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

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

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

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

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

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

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

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

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

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

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