2020 AMC 10B Problem 14

Attempt Problem 14 of the 2020 AMC 10B below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2020 AMC 10B solutions, or check the answer key.

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14.

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?

633π6\sqrt3-3\pi

9322π\dfrac{9\sqrt3}{2}-2\pi

332π3\dfrac{3\sqrt3}{2}-\dfrac{\pi}{3}

33π3\sqrt3-\pi

932π\dfrac{9\sqrt3}{2}-\pi

Answer: D
Concepts:regular polygonsectorarea decomposition
Difficulty rating: 1530
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 1,1, minus a 6060^\circ sector of a circle of radius 1.1.

The two equilateral triangles have total area 234=32.2\cdot\frac{\sqrt3}{4}=\frac{\sqrt3}{2}. The sector has area 60360π(1)2=π6.\frac{60^\circ}{360^\circ}\cdot\pi(1)^2=\frac{\pi}{6}. Thus one shaded piece has area 32π6,\frac{\sqrt3}{2}-\frac{\pi}{6}, and the total shaded area is 6(32π6)=33π.6\left(\frac{\sqrt3}{2}-\frac{\pi}{6}\right)=3\sqrt3-\pi.

Thus, D is the correct answer.

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