10th Class
Mathematics
Areas Related to Circles
Question Bank
Areas Related to Circles
question_answer
AB is a line segment of length 4 cm. P is the mid-point of AB. Circles are drawn with A, P and B as centres and radii AP = PB (see figure). The area of the shaded portion (in\[c{{m}^{2}}\]) is _____.
A) \[6\sqrt{3}\]
B) \[2\pi -6\sqrt{3}\]
C) \[2\pi -3\sqrt{3}\]
D) \[3\sqrt{3}\]
Correct Answer:
C
Solution :
Join DP and CP. Now, 3 sectors are formed each subtending an angle of \[{{60}^{o}}\] at P. Area of each sector \[=\frac{\theta }{{{360}^{o}}}\times \pi {{r}^{2}}\] \[=\frac{{{60}^{o}}}{{{360}^{o}}}\times \pi \times {{(2)}^{2}}=\frac{4\pi }{6}=\frac{2\pi }{3}c{{m}^{2}}\] And area of \[\Delta \,DCP=\frac{\sqrt{3}}{4}\,{{(PC)}^{2}}\] \[=\frac{\sqrt{3}}{4}\times 4=\sqrt{3}\,c{{m}^{2}}\] Therefore, area of shaded portion \[=\left( \frac{2\pi }{3}-\sqrt{3} \right)\times 3=2\pi -3\sqrt{3}\]