Question

A steel wire is bent in the form of a square that encloses an area of 121 sq.cm
The same wire is bent in the form of a circle. Find the area of the circle.

Answer 1

Area of the Square = 121 cm²

Side of Square = Square root of Area

= √(121) cm

= 11 cm

∴ Perimeter of Square = 4 × Side

= (4 × 11) cm

= 44 cm

According to Question,

Perimeter of Square = Circumfernce of Circle

∴ Circumference of Circle = 44 cm

Let's assume the radius as x cm

According to Question,

Circumference of Circle = 44 cm

2Πr = 44 cm 【Π = 22 / 7】

⇒ 2 × 22 / 7 × x = 44

⇒ 44 / 7 × x = 44

⇒ x = 44 × 7 / 44

⇒ x = 7

As, the radius of circle is x cm

Hence, radius of circle is 7 cm

∴ Area of Circle = Πr²

= [ 22 / 7 × 7 × 7 ] cm²

= [ 22 × 7 ] cm²

= 154 cm²

Hence, the area of circle is 154 cm²

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Answer 2

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

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

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

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

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

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

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

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

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

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

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

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

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