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Difference between revisions of "2011 AMC 10A Problems/Problem 18"

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Not specific: Draw a rectangle with vertices at the centers of <math>A</math> and <math>B</math> and the intersection of <math>A, C</math> and <math>B, C</math>. Then, we can compute the shaded area as the area of half of <math>C</math> plus the area of the rectangle minus the area of the two sectors created by <math>A</math> and <math>B</math>. This is <math>\frac{\pi (1)^2}{2}+(2)(1)-2 \cdot \frac{\pi (1)^2}{4}=\boxed{ \mathbf{(C)} 2}</math>.
 
Not specific: Draw a rectangle with vertices at the centers of <math>A</math> and <math>B</math> and the intersection of <math>A, C</math> and <math>B, C</math>. Then, we can compute the shaded area as the area of half of <math>C</math> plus the area of the rectangle minus the area of the two sectors created by <math>A</math> and <math>B</math>. This is <math>\frac{\pi (1)^2}{2}+(2)(1)-2 \cdot \frac{\pi (1)^2}{4}=\boxed{ \mathbf{(C)} 2}</math>.
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Sidenote: Upon reading this solution I believe the writer meant triangle not rectangle.
 
  
 
== See Also ==
 
== See Also ==

Revision as of 13:07, 8 January 2017

Problem 18

Circles $A, B,$ and $C$ each have radius 1. Circles $A$ and $B$ share one point of tangency. Circle $C$ has a point of tangency with the midpoint of $\overline{AB}$. What is the area inside Circle $C$ but outside circle $A$ and circle $B$ ? [asy] pathpen = linewidth(.7); pointpen = black; pair A=(-1,0), B=-A, C=(0,1); fill(arc(C,1,0,180)--arc(A,1,90,0)--arc(B,1,180,90)--cycle, gray(0.5)); D(CR(D("A",A,SW),1)); D(CR(D("B",B,SE),1)); D(CR(D("C",C,N),1)); [/asy]

$\textbf{(A)}\ 3 - \frac{\pi}{2} \qquad\textbf{(B)}\ \frac{\pi}{2} \qquad\textbf{(C)}\ 2 \qquad\textbf{(D)}\ \frac{3\pi}{4} \qquad\textbf{(B)}\ 1 + \frac{\pi}{2}$

Solution

Not specific: Draw a rectangle with vertices at the centers of $A$ and $B$ and the intersection of $A, C$ and $B, C$. Then, we can compute the shaded area as the area of half of $C$ plus the area of the rectangle minus the area of the two sectors created by $A$ and $B$. This is $\frac{\pi (1)^2}{2}+(2)(1)-2 \cdot \frac{\pi (1)^2}{4}=\boxed{ \mathbf{(C)} 2}$.

See Also

2011 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
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

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