Question

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

• A. $S<T<C$
• B. $T<C<S$
• C. $T<S<C$
• D. $C<S<T$

Answer 1

The correct option is C $T<S<C$

Let the perimeter of each figure be $P$ cm.

Then, side of equilateral triangle $\frac{P}{3}$

Side of square $=\frac{P}{4}$

Radius of circle $=\frac{P}{2\pi }$

$\therefore T=\frac{\sqrt{3}}{4}\times {\left(\frac{P}{3}\right)}^2=\frac{\sqrt{3}{P}^2}{36}$

$S={\left(\frac{P}{4}\right)}^2=\frac{P^2}{16}$

$C=\pi \times {\left(\frac{P}{2\pi }\right)}^2=\frac{P^2}{4\pi }=\frac{7P^2}{88}=0.0795P^2$

$\Rightarrow T=\frac{\sqrt{3}{P}^2}{36}=\frac{1.732}{36}{P}^2=0.0481P^2$

$S=0.0625P^2$

$C=0.0795P^2$

$\therefore C>S>T$