What area do these intersecting circles share?

What area do these intersecting circles share?

Question

Two equal circles, each of radius $10\,\text{cm}$, have centres at $A$ and $B$ respectively and the length of $AB$ is $17.4\,\text{cm}$. The circles intersect at $C$ and $D$. Find

1. the angle $CBA$ and express it in radians correct to three significant figures,

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1. the area of the minor sector $BCD$,

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1. the area common to both circles.

Solution

The line segment $CD$ is the perpendicular bisector of $AB$, and so we have the following right-angled triangle.

Thus, $\cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{8.7\,\text{cm}}{10\,\text{cm}} = 0.87$ and so $\theta = \arccos(0.87) = \text{}0.51559\!\ldots\text{ radians}$ which, to three significant figures, is $0.516$ radians.

By symmetry, $\theta = \angle CBA = \angle DBA$, and so $\angle DBC = \angle DBA + \angle CBA = 2\theta = 2 \arccos(0.87).$

The area of a sector of radius $r$ and enclosed angle $\theta$ (in radians) is $\dfrac{1}{2}r^2\theta$, so the area of the sector $CDB$ is $\dfrac{1}{2}10^2(2 \arccos(0.87)) = 51.559...,$

which to $3$sf is $51.6\,\text{cm}^{2}.$

By symmetry, the area common to both circles is twice the area of this shaded region, and the area of this shaded region is equal to $\text{area of the minor sector }BCD\text{} - \text{area of }\triangle BCD\text{}.$

The area of a triangle is $\dfrac{1}{2}ab \sin C$, so the triangle $BCD$ has area $\dfrac{1}{2}100 \sin (2\arccos(0.87))=42.895...$

Thus the area of the shaded region is $51.559... - 42.895... = 8.664...,$ and so the area shared between the circles is ($3$sf) $17.3\,\text{cm}^{2}.$