2008 AMC 12B Problems/Problem 4

Problem

On circle $O$ , points $C$ and $D$ are on the same side of diameter $\overline{AB}$ , $\angle AOC = 30^\circ$ , and $\angle DOB = 45^\circ$ . What is the ratio of the area of the smaller sector $COD$ to the area of the circle?

$[asy] unitsize(6mm); defaultpen(linewidth(0.7)+fontsize(8pt)); pair C = 3*dir (30); pair D = 3*dir (135); pair A = 3*dir (0); pair B = 3*dir(180); pair O = (0,0); draw (Circle ((0, 0), 3)); label ("$C$", C, NE); label ("$D$", D, NW); label ("$B$", B, W); label ("$A$", A, E); label ("$O$", O, S); label ("$45^\circ$", (-0.3,0.1), WNW); label ("$30^\circ$", (0.5,0.1), ENE); draw (A--B); draw (O--D); draw (O--C); [/asy]$

$\textbf{(A)}\ \frac {2}{9} \qquad \textbf{(B)}\ \frac {1}{4} \qquad \textbf{(C)}\ \frac {5}{18} \qquad \textbf{(D)}\ \frac {7}{24} \qquad \textbf{(E)}\ \frac {3}{10}$

Solution

$\angle COD = \angle AOB - \angle AOC - \angle BOD = 180^\circ - 30^\circ - 45^\circ = 105^\circ$ .

Since a circle has $360^\circ$ , the desired ratio is $\frac{105^\circ}{360^\circ}=\frac{7}{24} \Rightarrow D$ .

2008 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 3	Followed by Problem 5
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All AMC 12 Problems and Solutions

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2008 AMC 12B Problems/Problem 4

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