Subject: General Aptitude · Chapter: Area · Exam: · Marks: · Difficulty:
Two identical circles intersect so that their centres, and the points at which they intersect, form a square of side 1 cm. The area (in sq. cm) of the portion that is common to the circles is
(a)$\frac{\pi}{4}$
(b)$\frac{\pi}{2}-1$
(c)$\frac{\pi}{5}$
(d)$\sqrt{2}-1$
Answer
Answer (as printed): B
Explanation
Clearly, radius of each circle = 1 cm. Area of sector $O A C B O=$ Area of sector $O^{\prime} A D B O^{\prime}$ $=\left(\frac{1}{4} \times \pi \times 1^{2}\right) \mathrm{cm}^{2}=\left(\frac{\pi}{4}\right) \mathrm{cm}^{2}$. Area of square $O A O^{\prime} B=(1 \times 1) \mathrm{cm}^{2}=1 \mathrm{cm}^{2}$. ∴ Required area $=($ Area of sector $O A C B O+$ Area of sector $O^{\prime} A D B O^{\prime}$ - Area of square $O A O^{\prime} B$ ) $=\left(\frac{\pi}{4}+\frac{\pi}{4}-1\right) \mathrm{cm}^{2}=\left(\frac{\pi}{2}-1\right) \mathrm{cm}^{2}$.