Question

Two circles, each of unit radius, are so drawn that the centre of each lies on the circumference of the other. The area of the region, common to both the circles, is
(a). \[\dfrac{4\pi -3\sqrt{3}}{12}\]
(b). \[\dfrac{4\pi -6\sqrt{3}}{12}\]
(c). \[\dfrac{4\pi -3\sqrt{3}}{6}\]
(d). \[\dfrac{4\pi -6\sqrt{3}}{6}\]

Answer 1

Hint: For finding the common area between the two circles, we can calculate the area of the segments and then we can add the areas of the two segments to get the complete area of the common region.
An important formula that is used in this question is as follows
\[area=2\left( area\ of\ \sec tor\ -area\ of\ \Delta ABC \right)\]
And also, the area of the sector can be calculated using the following formula
\[Area=\dfrac{1}{2}\times \angle ABC\times {{r}^{2}}\]

Complete step-by-step solution -
As mentioned in the question, we have to find the area that is common between the two circles.
Now, consider the \[\Delta ABD\],
Here, we can say that the length of DB=0.5 units
(Because of the symmetry in the figure)
Also, the length of AB=1 unit
(Because it is the radius of the circle and the circles are unit circles)
Now, for calculating the length of AD, we can use Pythagoras theorem as follows
\[\begin{align}
  & AD=\sqrt{{{1}^{2}}-{{\left( 0.5 \right)}^{2}}} \\
 & AD=\sqrt{\dfrac{3}{4}}=\dfrac{\sqrt{3}}{2} \\
\end{align}\]
Now, the total area of \[\Delta ABD\] and \[\Delta ABD\]
\[Area=\dfrac{1}{2}\times 1\times \dfrac{\sqrt{3}}{2}=\dfrac{\sqrt{3}}{4}\]
Now, in \[\Delta ABD\],
\[\angle ABD={{\cos }^{-1}}\left( \dfrac{0.5}{1} \right)={{\cos }^{-1}}\left( \dfrac{1}{2} \right)=\dfrac{\pi }{3}\]
So, the we can say by symmetry that
\[\angle ABC=2\times \dfrac{\pi }{3}=\dfrac{2\pi }{3}\]
​​Now, using the formula mentioned in the hint, we can calculate the area of the sector as follows
\[\begin{align}
  & Area=\dfrac{1}{2}\times \angle ABC\times {{r}^{2}} \\
 & Area=\dfrac{1}{2}\times \dfrac{2\pi }{3}=\dfrac{\pi }{3} \\
\end{align}\]
Now, area of the common region can be calculated as follows
\[\begin{align}
  & area=2\left( area\ of\ \sec tor\ -area\ of\ \Delta ABC \right) \\
 & area=2\left( \dfrac{\pi }{3}\ -\dfrac{\sqrt{3}}{4} \right)=\dfrac{4\pi -3\sqrt{3}}{6}c{{m}^{2}} \\
\end{align}\]

Note: -The students can make an error while solving this question if they don’t know any of the mentioned properties that are given in the hint.
Also, one should also know what Pythagoras theorem is and its applications for solving this question.