2014 AMC 10A Problems/Problem 12

Problem

A regular hexagon has side length 6. Congruent arcs with radius 3 are drawn with the center at each of the vertices, creating circular sectors as shown. The region inside the hexagon but outside the sectors is shaded as shown What is the area of the shaded region?

$[asy] size(125); defaultpen(linewidth(0.8)); path hexagon=(2*dir(0))--(2*dir(60))--(2*dir(120))--(2*dir(180))--(2*dir(240))--(2*dir(300))--cycle; fill(hexagon,lightgrey); for(int i=0;i<=5;i=i+1) { path arc=2*dir(60*i)--arc(2*dir(60*i),1,120+60*i,240+60*i)--cycle; unfill(arc); draw(arc); } draw(hexagon,linewidth(1.8));[/asy]$

$\textbf{(A)}\ 27\sqrt{3}-9\pi\qquad\textbf{(B)}\ 27\sqrt{3}-6\pi\qquad\textbf{(C)}\ 54\sqrt{3}-18\pi\qquad\textbf{(D)}\ 54\sqrt{3}-12\pi\qquad$

$\textbf{(E)}\ 108\sqrt{3}-9\pi$

Solution

The area of the hexagon is just simply $\dfrac{3(6)^2\sqrt{3}}{2}=54\sqrt{3}$ .

We note that each interior angle of the regular hexagon is $120^\circ$ which means that each sector is $\dfrac{1}{3}$ of the circle it belongs to. The area of each sector is $\dfrac{9\pi}{3}=3\pi$ . The area of all six is $6\times 3\pi=18\pi$ . Therefore, our answer is $\boxed{\textbf{(C)}\ 54\sqrt{3}-18\pi}$

2014 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 11	Followed by Problem 13
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

Page

Toolbox

Search

2014 AMC 10A Problems/Problem 12

Problem

Solution

See Also