Difference between revisions of "2010 AMC 8 Problems/Problem 19"
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Since <math>\triangle ACD</math> is isosceles, <math>CB</math> bisects <math>AD</math>. Thus <math>AB=BD=8</math>. From the Pythagorean Theorem, <math>CB=6</math>. Thus the area between the two circles is | Since <math>\triangle ACD</math> is isosceles, <math>CB</math> bisects <math>AD</math>. Thus <math>AB=BD=8</math>. From the Pythagorean Theorem, <math>CB=6</math>. Thus the area between the two circles is | ||
<math>100\pi - 36\pi=64\pi</math> <math>\boxed{\textbf{(C)}\ 64\pi}</math> | <math>100\pi - 36\pi=64\pi</math> <math>\boxed{\textbf{(C)}\ 64\pi}</math> | ||
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+ | ==See Also== | ||
+ | {{AMC8 box|year=2010|num-b=18|num-a=20}} |
Revision as of 16:15, 5 November 2012
Problem
The two circles pictured have the same center . Chord is tangent to the inner circle at , is , and chord has length . What is the area between the two circles?
Solution
Since is isosceles, bisects . Thus . From the Pythagorean Theorem, . Thus the area between the two circles is
See Also
2010 AMC 8 (Problems • Answer Key • Resources) | ||
Preceded by Problem 18 |
Followed by Problem 20 | |
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 AJHSME/AMC 8 Problems and Solutions |