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Explanation:

Let radius of the circle be r, a side of the equilateral triangle be a, and a side of the square be x.

The circumference/perimeter of the circle, triangle and square are equal. Hence,

2πr = 3a = 4x = k

r=k2π,a=k3, and x=k4

The areas of the circle, triangle and square are c, t, s respectively. Hence,

c=πr2=πk24π2=k24π,t=34a2=34×k29=k2123s=x2=k2161π>14>133c>s>t

Hence, option (c).

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