A copper wire when bent in the form of a square encloses an area of 484 cm^{2}. The same wire is now bent in the form of a circle. Find the area enclosed by the circle.

In this question the wire is first bent in the shape of square and then same wire is bent to form a circle. The point to be noticed is that the same wire is used both the times which implies that the perimeter of square and that of circle will be equal.

Let the square be of side ‘a’ cm and radius of the circle be ‘r’

Given the area enclosed by the square = 484 cm^{2}

Also, we know that Area of square = Side × Side

Area of the square = a^{2}

⇒ a^{2} = 484

⟹ a = √484

⇒ a = 22 cm

Therefore, side of square, ‘a’ is 22 cm.

Also, circumference of the circle = Perimeter of square → eqn1

Perimeter of square = 4 × side

Perimeter of square = 4×22

⇒ Perimeter of square = 88 cm → eqn2

Also, we know, Circumference of circle = 2πr → eqn3

Put values in equation 1 from equation 2 & 3, we get

2πr = 88

On rearranging,

⇒ r = 14 cm

So, the radius ‘r’ of the circle is 14 cm.

Area of circle = πr^{2}

Where r = radius of the circle

= π(14^{2})

= 4312/7

= 616 cm^{2}

Area of the circle is 616 cm^{2}.

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