Difference between revisions of "2010 AMC 8 Problems/Problem 19"

Revision as of 19:29, 2 January 2023

Problem

The two circles pictured have the same center $C$ . Chord $\overline{AD}$ is tangent to the inner circle at $B$ , $AC$ is $10$ , and chord $\overline{AD}$ has length $16$ . 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$", (4.07,-0.24),NE*lsf); clip((-3.1,-7.72)--(-3.1,4.77)--(11.74,4.77)--(11.74,-7.72)--cycle); [/asy]$

$\textbf{(A)}\ 36 \pi \qquad\textbf{(B)}\ 49 \pi\qquad\textbf{(C)}\ 64 \pi\qquad\textbf{(D)}\ 81 \pi\qquad\textbf{(E)}\ 100 \pi$

Solution

Since $\triangle ACD$ is isosceles, $CB$ bisects $AD$ . Thus $AB=BD=8$ . From the Pythagorean Theorem, $CB=6$ . Thus the area between the two circles is $100\pi - 36\pi=64\pi$ $\boxed{\textbf{(C)}\ 64\pi}$

Note: The length $AC$ is necessary information, as this tells us the radius of the larger circle. The area of the annulus is $\pi(AC^2-BC^2)=\pi AB^2=64\pi$ .

Video Solution

https://youtu.be/Q6rnoQChiyU. Soo, DRMS, NM

Video Solution by OmegaLearn

https://youtu.be/51K3uCzntWs?t=3206

~ pi_is_3.14

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.

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