2020 AMC 10B Problem 14

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

Answer: D

Concepts:regular polygon sector area decomposition

Difficulty rating: 1530

Small hint:

Use the symmetry of the regular hexagon to split the shaded region into six equal pieces

Big hint:

Each piece is two equilateral triangles minus a

60^\circ

sector of radius

1

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.