ABC26GN4623 · Area
Subject: General Aptitude · Chapter: Area · Exam: 2004 · Marks: · Difficulty:
The area of the greatest circle which can be inscribed in a square whose perimeter is 120 cm, is
(a)$\frac{22}{7} \times\left(\frac{7}{2}\right)^{2} \mathrm{cm}^{2}$
(b)$\frac{22}{7} \times\left(\frac{9}{2}\right)^{2} \mathrm{cm}^{2}$
(c)$\frac{22}{7} \times\left(\frac{15}{2}\right)^{2} \mathrm{cm}^{2}$
(d)$\frac{22}{7} \times(15)^{2} \mathrm{cm}^{2}$
Answer
Explanation
Side of the square $=\frac{120}{4} \mathrm{cm}=30 \mathrm{cm}$. $$\begin{aligned} & \text { Radius of the required circle }=\left(\frac{1}{2} \times 30\right) \mathrm{cm}=15 \mathrm{cm} . \\ & =\pi \times 2 \\ & =\left[\frac{22}{7} \times(15)^{2}\right] \mathrm{cm}^{2} . \end{aligned}$$
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