Question

The closed curve in the figure is made up of $9$ congruent circular arcs each of length $\dfrac{2\pi}{3},$ where each of the centers of the corresponding circles is among the vertices of a regular hexagon of side $2.$ What is the area enclosed by the curve?

$2\pi + 6$

$2\pi + 4\sqrt{3}$

$3\pi + 4$

$2\pi + 3\sqrt{3} + 2$

$\pi + 6\sqrt{3}$

Solution:

Each arc has length $\dfrac{2\pi}{3}$ on a unit circle, so it is a $120^\circ$ sector. The nine equal sectors can be reassembled so that the enclosed region equals the regular hexagon of side $2$ plus one full circle of radius $1.$

A regular hexagon of side $2$ splits into $6$ equilateral triangles of side $2,$ so its area is $6 \cdot \dfrac{\sqrt3}{4} \cdot 2^2 = 6\sqrt3.$

Adding the unit circle's area $\pi$ gives $\pi + 6\sqrt3.$

Thus, the correct answer is E.

2012 AMC 12A Problem 14

Solution:

Problem 14 in Other Years