Question

A circle of radius $2$ is centered at $O.$ Square $OABC$ has side length $1.$ Sides $\overline{AB}$ and $\overline{CB}$ are extended past $B$ to meet the circle at $D$ and $E,$ respectively. What is the area of the shaded region in the figure, which is bounded by $\overline{BD},$ $\overline{BE},$ and the minor arc connecting $D$ and $E?$

$\dfrac{\pi}{3}+1-\sqrt{3}$

$\dfrac{\pi}{2}(2-\sqrt{3})$

$\pi(2-\sqrt{3})$

$\dfrac{\pi}{6}+\dfrac{\sqrt{3}-1}{2}$

$\dfrac{\pi}{3}-1+\sqrt{3}$

Solution:

Since $OA=1$ and $OD=2$ with $D$ on the line $x=1,$ we get $\angle AOD=60^\circ,$ and likewise $\angle COE=60^\circ,$ so $\angle DOE=30^\circ.$

The sector $DOE$ has area $\tfrac{30}{360}\pi(2^2)=\tfrac{\pi}{3}.$

The region is this sector minus triangles $OBD$ and $OBE.$ With $BD=BE=\sqrt3-1,$ each triangle has area $\tfrac12(\sqrt3-1)(1),$ totaling $\sqrt3-1.$

So the shaded area is $\tfrac{\pi}{3}-(\sqrt3-1)=\tfrac{\pi}{3}+1-\sqrt3.$

Thus, the correct answer is A.

2006 AMC 10B Problem 19

Solution:

Problem 19 in Other Years