Answer :

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 cm2


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 = πr2


Where r = radius of the circle


Area of circle = π(10.52)





= 346.5 cm2


Area of the circle is 346.5 cm2.

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