Difference between revisions of "2015 AMC 12B Problems/Problem 14"
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==Problem== | ==Problem== | ||
| + | A circle of radius 2 is centered at <math>A</math>. An equilateral triangle with side 4 has a vertex at <math>A</math>. 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? | ||
| − | + | <math>\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}</math> | |
==Solution== | ==Solution== | ||
| − | + | 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 15:02, 4 March 2015
Problem
A circle of radius 2 is centered at 
. An equilateral triangle with side 4 has a vertex at . 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?
Solution
The area of the circle is 
, and the area of the triangle 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 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 
.
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 | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
