Difference between revisions of "2010 AMC 8 Problems/Problem 19"
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<math> \textbf{(A)}\ 36 \pi \qquad\textbf{(B)}\ 49 \pi\qquad\textbf{(C)}\ 64 \pi\qquad\textbf{(D)}\ 81 \pi\qquad\textbf{(E)}\ 100 \pi </math> | <math> \textbf{(A)}\ 36 \pi \qquad\textbf{(B)}\ 49 \pi\qquad\textbf{(C)}\ 64 \pi\qquad\textbf{(D)}\ 81 \pi\qquad\textbf{(E)}\ 100 \pi </math> | ||
− | == Solution == | + | == Solution == |
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> |
Revision as of 20:36, 12 January 2020
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
Note: The length is necessary information, as this tells us the radius of the larger circle. The area of the annulus 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.