ABC26GN4658 · Area

Subject: General Aptitude · Chapter: Area · Exam: 2010 · Marks: · Difficulty:

Three equal circles are described with vertices of the triangles as centres. If the radius of each circle is $r$, the sum of areas of the portions of the circles intercepted in a triangle is

(a)$2 \pi r^{2}$

(b)$\frac{3}{2} \pi r^{2}$

(c)$\pi r^{2}$

(d)$\frac{1}{2} \pi r^{2}$

Answer

Answer (as printed): D

Explanation

We have: $$\begin{aligned} \text { Required area } & =\frac{\pi r^{2} \theta_{1}}{360}+\frac{\pi r^{2} \theta_{2}}{360}+\frac{\pi r^{2} \theta_{3}}{360} \\ & =\frac{\pi r^{2}}{360}\left(\theta_{1}+\theta_{2}+\theta_{3}\right) \\ & =\frac{\pi r^{2} \times 180}{360}=\frac{\pi r^{2}}{2} \cdot\left[\because \theta_{1}+\theta_{2}+\theta_{3}=180^{\circ}\right] \end{aligned}$$

Explanation as extracted from the printed page; notation may be imperfect.

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