# 2010 AMC 8 Problems/Problem 19

## 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

~ pi_is_3.14

~savannahsolver