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Difference between revisions of "2015 AMC 12B Problems/Problem 14"

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==Solution==
 
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The area of the circle is <math>\pi \cdot 2^2 = 4\pi</math>, and the area of the triangle is <math>\frac{4^2 \cdot\sqrt{3}}{4} = 4\sqrt{3}</math>. The difference between the area of the region that lies inside the circle but outside the triangle and the area of the region that lies inside the triangle but outside the circle is the same as the difference between the area of the circle and the area of the triangle (because from both pieces we are subtracting the area of the two shapes' intersection), so the answer is <math>4\pi-4\sqrt{3}  = \boxed{\textbf{(D)}\;  4(\pi-\sqrt{3})}</math>.
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2015|ab=B|num-a=15|num-b=13}}
 
{{AMC12 box|year=2015|ab=B|num-a=15|num-b=13}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 14:02, 4 March 2015

Problem

A circle of radius 2 is centered at $A$. An equilateral triangle with side 4 has a vertex at $A$. What is the difference between the area of the region that lies inside the circle but outside the triangle and the area of the region that lies inside the triangle but outside the circle?

$\textbf{(A)}\; 8-\pi \qquad\textbf{(B)}\; \pi+2 \qquad\textbf{(C)}\; 2\pi-\dfrac{\sqrt{2}}{2} \qquad\textbf{(D)}\; 4(\pi-\sqrt{3}) \qquad\textbf{(E)}\; 2\pi-\dfrac{\sqrt{3}}{2}$

Solution

The area of the circle is $\pi \cdot 2^2 = 4\pi$, and the area of the triangle is $\frac{4^2 \cdot\sqrt{3}}{4} = 4\sqrt{3}$. The difference between the area of the region that lies inside the circle but outside the triangle and the area of the region that lies inside the triangle but outside the circle is the same as the difference between the area of the circle and the area of the triangle (because from both pieces we are subtracting the area of the two shapes' intersection), so the answer is $4\pi-4\sqrt{3} = \boxed{\textbf{(D)}\; 4(\pi-\sqrt{3})}$.

See Also

2015 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 13 		Followed by
Problem 15
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 12 Problems and Solutions

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