Question

As shown in the figure below, six semicircles lie in the interior of a regular hexagon with side length 2 so that the diameters of the semicircles coincide with the sides of the hexagon. What is the area of the shaded region — inside the hexagon but outside all of the semicircles?

$6\sqrt3-3\pi$

$\dfrac{9\sqrt3}{2}-2\pi$

$\dfrac{3\sqrt3}{2}-\dfrac{\pi}{3}$

$3\sqrt3-\pi$

$\dfrac{9\sqrt3}{2}-\pi$

Solution:

By symmetry, the shaded region is made of six congruent pieces. One such piece is the union of two equilateral triangles with side length $1$ , minus a $60^\circ$ sector of a circle of radius $1$ .

The two equilateral triangles have total area $2\cdot\frac{\sqrt3}{4}=\frac{\sqrt3}{2}.$ The sector has area $\frac{60^\circ}{360^\circ}\cdot\pi(1)^2=\frac{\pi}{6}.$ Thus one shaded piece has area $\frac{\sqrt3}{2}-\frac{\pi}{6}$ , and the total shaded area is $6\left(\frac{\sqrt3}{2}-\frac{\pi}{6}\right)=3\sqrt3-\pi.$

Thus, D is the correct answer.

2020 AMC 10B Problem 14

Solution:

Problem 14 in Other Years