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:

Note that the region enclosed by the curve but outside the hexagon consists of $3$ sectors with angle $240^{\circ}.$

This means that together they form $2$ whole circles with radius $1.$ Now to find the area of the region inside both the hexagon and the curve.

This area can be found by finding the area of the hexagon and subtracting out the areas of the sectors outside the curve.

There are three $\dfrac{1}{3}$ circles that together form a whole circle. The area of the hexagon can be given by $\dfrac{3}{2} \cdot 2^2 \cdot \sqrt{3} = 6\sqrt{3}.$

The desired area is then $2\pi + (6\sqrt{3} - \pi) = \pi + 6\sqrt{3}.$

Thus, E is the correct answer.

2012 AMC 10A Problem 18

Solution:

Problem 18 in Other Years