Circular triangle

Geometry Level pending

$\Gamma$ is a semicircle with $A$ and $B$ as endpoints of the diameter and $O$ as the center. $C$ is a point on $\Gamma$ such that $\angle AOC = 162^\circ$ and $D$ is the midpoint of $\stackrel{\frown}{AC}$ . If the radius of $\Gamma$ is $10$ and the area of the region bounded by $\stackrel{\frown}{BC}$ , $CD$ and $DB$ can be expressed as $M \pi$ , what is the value of $M$ ?

Details and assumptions

All line segments are straight, unless otherwise denoted by the arc symbol e.g. $\stackrel{\frown} {AC}$ .

The answer is 5.

1 solution

Calvin Lin Staff
May 13, 2014

Since $D$ is the midpoint of $\stackrel{\frown}{AC}$ , thus $\angle AOD = \angle COD = \frac{\angle AOC}{2} = 81 ^\circ$ . We also have that $\angle OCB = \angle OBC = \frac{180^\circ - \angle COB}{2} = \frac{\angle AOC}{2} = 81^\circ$ . Thus $\angle COD = \angle OCB$ , which implies that $OD$ is parallel to $BC$ . This means that triangles $OCB$ and $DCB$ have equal areas as they have equal base and height.

Therefore the area of the region bounded by $\stackrel{\frown}{BC}$ , $CD$ and $DB$ is equal to the area of sector $COB$ , which is $\pi\cdot 10^2 \cdot \left(\frac{18^\circ}{360^\circ}\right) = 5 \pi$ . Hence $M = 5$ .

Circular triangle

The answer is 5.

1 solution

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