An equilateral triangle,a square,and a circle | vedclass

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(C) Let the perimeter be $P$. Since the perimeters are equal,we have $3a = 4b = 2\pi r = P$,where $a$ is the side of the triangle,$b$ is the side of t

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$T < S < C$

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(C) Let the perimeter be $P$. Since the perimeters are equal,we have $3a = 4b = 2\pi r = P$,where $a$ is the side of the triangle,$b$ is the side of the square,and $r$ is the radius of the circle.$1$. Area of the equilateral triangle: $T = \frac{\sqrt{3}}{4} a^2 = \frac{\sqrt{3}}{4} \left(\frac{P}{3}\right)^2 = \frac{\sqrt{3}}{36} P^2 \approx 0.0481 P^2$.$2$. Area of the square: $S = b^2 = \left(\frac{P}{4}\right)^2 = \frac{1}{16} P^2 = 0.0625 P^2$.$3$. Area of the circle: $C = \pi r^2 = \pi \left(\frac{P}{2\pi}\right)^2 = \frac{P^2}{4\pi} \approx 0.0796 P^2$.Comparing the coefficients of $P^2$: $0.0481 Thus,$T < S < C$.

An equilateral triangle,a square,and a circle have equal perimeters. If $T$ denotes the area of the triangle,$S$ the area of the square,and $C$ the area of the circle,then

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