Subject: General Aptitude · Chapter: Area · Exam: · Marks: · Difficulty:
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
(a)$\mathrm{S}<\mathrm{T}<\mathrm{C}$
(b)$\mathrm{T}<\mathrm{C}<\mathrm{S}$
(c)$\mathrm{T}<\mathrm{S}<\mathrm{C}$
(d)C < S < T
Answer
Answer (as printed): C
Explanation
Let the perimeter of each be $a$. Then, side of the equilateral triangle $=\frac{a}{3}$; side of the square $=\frac{a}{4}$; radius of the circle $=\frac{a}{2 \pi}$. $$\begin{aligned} T & =\frac{\sqrt{3}}{4} \times\left(\frac{a}{3}\right)^{2}=\frac{\sqrt{3} a^{2}}{36} ; S=\left(\frac{a}{4}\right)^{2}=\frac{a^{2}}{16} ; C \\ \therefore \quad & \pi \times\left(\frac{a}{2 \pi}\right)^{2}=\frac{a^{2}}{4 \pi}=\frac{7 a^{2}}{88} . \end{aligned}$$ So, $C>S>T$.
Explanation as extracted from the printed page; notation may be imperfect.