A wire when bent in the form of an equilateral triangle endoses an area of 121√3 cm^{2}. The same wire is bent to form a circle. Find the area enclosed by the circle.

In this question the wire is first bent in the shape of equilateral triangle 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 equilateral triangle and that of circle will be equal.

Let the equilateral triangle be of side ‘a’ cm and radius of the circle be ‘r’.

Given: Area enclosed by equilateral triangle = 123√3 cm^{2}

Also, we know that Area of equilateral triangle

Where ‘a’ = side of equilateral triangle

⟹ a = √484

⇒ a = 22 cm

Therefore, side of equilateral triangle, ‘a’ is 22 cm.

Also, circumference of the circle = Perimeter of equilateral triangle → eqn1

Perimeter of equilateral triangle = 3×side

= 3×22

= 66 cm → eqn2

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

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

2πr = 66

(put π = 22/7)

On rearranging,

⇒ r = 10.5 cm

So, the radius ‘r’ of the circle is 10.5 cm.

Area of circle = πr^{2}

Where r = radius of the circle

⇒ Area of circle = π(10.5^{2})

= 346.5 cm^{2}

Area of the circle is 346.5 cm^{2}.

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