2008 AMC 12B Problem 4

Question

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?

\dfrac{2}{9}

\dfrac{1}{4}

\dfrac{5}{18}

\dfrac{7}{24}

\dfrac{3}{10}

\angle AOC,

\angle COD,

and

\angle DOB

together make the straight angle

\overline{AB}

A sector’s fraction of the circle is its central angle divided by

360^\circ

Since

\angle AOC,

\angle COD,

and

\angle DOB

fill the straight angle over diameter

\overline{AB},

\begin{aligned} \angle COD &= 180^\circ - 30^\circ - 45^\circ \\ &= 105^\circ. \end{aligned}

The sector’s share of the circle is

\dfrac{105^\circ}{360^\circ} = \dfrac{7}{24}.

Thus, the correct answer is D.

Problem 4 in other years