Difference between revisions of "2010 AMC 8 Problems/Problem 19"
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dot((0,0),ds); label("$A$", (-0.19,-0.23),NE*lsf); dot((2,0),ds); label("$B$", (1.97,-0.31),NE*lsf); dot((2,1),ds); label("$C$", (1.96,1.09),NE*lsf); dot((4,0),ds); label("$D$", (4.07,-0.24),NE*lsf); clip((-3.1,-7.72)--(-3.1,4.77)--(11.74,4.77)--(11.74,-7.72)--cycle); | dot((0,0),ds); label("$A$", (-0.19,-0.23),NE*lsf); dot((2,0),ds); label("$B$", (1.97,-0.31),NE*lsf); dot((2,1),ds); label("$C$", (1.96,1.09),NE*lsf); dot((4,0),ds); label("$D$", (4.07,-0.24),NE*lsf); clip((-3.1,-7.72)--(-3.1,4.77)--(11.74,4.77)--(11.74,-7.72)--cycle); | ||
</asy> | </asy> | ||
| + | |||
<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> | ||
| + | |||
| + | Note: The length <math>AC</math> is necessary information, as this tells us the radius of the larger circle. The area of the annulus is <math>\pi(AC^2-BC^2)=\pi AB^2=64\pi</math>. | ||
| + | |||
| + | |||
| + | == Video Solution == | ||
| + | https://youtu.be/Q6rnoQChiyU. Soo, DRMS, NM | ||
| + | |||
| + | ==Video Solution by OmegaLearn== | ||
| + | https://youtu.be/51K3uCzntWs?t=3206 | ||
| + | |||
| + | ~ pi_is_3.14 | ||
| + | |||
| + | |||
| + | ==Video by MathTalks== | ||
| + | |||
| + | https://www.youtube.com/watch?v=KSYVsSJDX-0&feature=youtu.be | ||
| + | |||
| + | ==Video Solution by WhyMath== | ||
| + | https://youtu.be/yjhitUYSAI0 | ||
| + | |||
| + | ~savannahsolver | ||
| + | |||
| + | ==See Also== | ||
| + | {{AMC8 box|year=2010|num-b=18|num-a=20}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 08:04, 10 March 2023
Contents
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?
![[asy] unitsize(45); import graph; size(300); real lsf = 0.5; pen dp = linewidth(0.7) + fontsize(10); defaultpen(dp); pen ds = black; pen xdxdff = rgb(0.49,0.49,1); draw((2,0.15)--(1.85,0.15)--(1.85,0)--(2,0)--cycle); draw(circle((2,1),2.24)); draw(circle((2,1),1)); draw((0,0)--(4,0)); draw((0,0)--(2,1)); draw((2,1)--(2,0)); draw((2,1)--(4,0)); dot((0,0),ds); label("$A$", (-0.19,-0.23),NE*lsf); dot((2,0),ds); label("$B$", (1.97,-0.31),NE*lsf); dot((2,1),ds); label("$C$", (1.96,1.09),NE*lsf); dot((4,0),ds); label("$D$", (4.07,-0.24),NE*lsf); clip((-3.1,-7.72)--(-3.1,4.77)--(11.74,4.77)--(11.74,-7.72)--cycle); [/asy]](http://latex.artofproblemsolving.com/2/c/f/2cfa82378d1bafff2108201dd94835ed7a9aab8d.png)
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 
.
Video Solution
https://youtu.be/Q6rnoQChiyU. Soo, DRMS, NM
Video Solution by OmegaLearn
https://youtu.be/51K3uCzntWs?t=3206
~ pi_is_3.14
Video by MathTalks
https://www.youtube.com/watch?v=KSYVsSJDX-0&feature=youtu.be
Video Solution by WhyMath
~savannahsolver
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. 
