Difference between revisions of "2014 AMC 10A Problems/Problem 12"

Revision as of 15:18, 30 January 2016

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?

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

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

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

Solution

The area of the hexagon is equal to $\dfrac{3(6)^2\sqrt{3}}{2}=54\sqrt{3}$ by the formula for the area of a hexagon.

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 $6\times 3\pi=18\pi$ .

The shaded area is equal to the area of the hexagon minus the sum of the area of all the sectors, which is equal to $\boxed{\textbf{(C)}\ 54\sqrt{3}-18\pi}$

2014 AMC 10A (Problems • Answer Key • Resources)
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Difference between revisions of "2014 AMC 10A Problems/Problem 12"

Revision as of 15:18, 30 January 2016

Problem

Solution

See Also

@@ Line 7: / Line 7: @@
 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);
+fill(hexagon,grey);
 for(int i=0;i<=5;i=i+1)
 {