​If perimeters of a square, circle and equilateral triangle are same, then minimum area will be, of:​

The perimeters of a square, circle and equilateral triangle are same
Let:

​$$P = 44, \quad \pi = \frac{22}{7} \\[8pt]$$​​

Square:
​$$P = 4a \implies a = \frac{44}{4} = 11 \\[4pt]A{_s} = a^2 = 11^2 = 121 \\[12pt]$$​​
Circle:
​$$P = 2\pi r \implies r = \frac{44}{2 \times \frac{22}{7}} = \frac{44 \times 7}{44} = 7 \\[4pt]A_c = \pi r^2 = \frac{22}{7} \times 7^2 = \frac{22}{7} \times 49 = 154 \\[12pt]$$​​
Equilateral Triangle:
​$$P = 3a \implies a = \frac{44}{3} \approx 14.67 \\[4pt]A_t = \frac{\sqrt{3}}{4}a^2 = \frac{\sqrt{3}}{4}\left(\frac{44}{3}\right)^2= \frac{\sqrt{3}}{4} \times \frac{1936}{9} \approx 93.3 \\[12pt]$$​​
Comparison
​$$A_t \approx 93.3 \; (\text{minimum}), \quad \\A_s = 121, \quad \\A_c = 154 \; (\text{maximum}) \\[8pt]$$​​
Correct answer is (C) – Equilateral triangle
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