Let the radius of circle is 'r' and a side of a square is 'a', then given condition
\( \Large 2\pi r \)=4a => a=\( \Large \frac{\pi r}{2} \)
Area of square
=\( \Large \left(\frac{\pi r}{2}\right)^{2}=\frac{\pi^{2}r^{2}}{4}=\frac{9.86 r^{2}}{4}=2.46 r^{2} \)
and area of circle = \( \Large \pi r^{2}=3.14 r^{2} \)
and let the side of equilateral triangle is x.
Then, given condition,
3x=\( \Large 2\pi r \)=> x=\( \Large \frac{2\pi r}{3} \)
Area of equilateral triangle = \( \Large \frac{\sqrt{3}}{4}x^{2} \)
=\( \Large \frac{\sqrt{3}}{4}\)\( \Large \times \)\( \Large \frac{4\pi^{2}r^{2}}{9} \)
=\( \Large \frac{\pi^{2}}{3\sqrt{3}}r^{2}=1.89 r^{2} \)
Hence, Area of circle > Area of square > Area of equilateral triangle.