2011 AMC 10A Problems/Problem 18: Difference between revisions
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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 shades 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{2 \ \mathbf{(C)}}</math>. | 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 shades 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{2 \ \mathbf{(C)}}</math>. | ||
== See Also == | |||
{{AMC10 box|year=2011|ab=A|num-b=17|num-a=19}} | |||
Revision as of 09:47, 8 May 2011
Problem 18
Circles 
and 
each have radius 1. Circles 
and 
share one point of tangency. Circle has a point of tangency with the midpoint of 
. What is the area inside Circle but outside circle and circle ?
Solution
Draw a rectangle with vertices at the centers of and and the intersection of 
and 
. Then, we can compute the shades area as the area of half of plus the area of the rectangle minus the area of the two sectors created by and . This is 
.
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 | ||
