In Figure, C is the center of the semicircle, and the area of the semicircle is `8pi`. What is the area of triangle ABC in terms of `theta` ?

Options

A. `2 sin theta tan theta`
B. `4 sin theta`
C. `8 cos theta`
D. `8 sin theta`

Answer

D

Explanation

The area of any triangle is `(1)/(2)xx b xx h`. Call the base of triangle ABC side AC and say that the height drawn from point B to base AC meets base AC at point D.

Base AC of triangle ABC is a radius of the circle. The area of the semicircle is `8pi`. So if the radius of the semicircile is r (in particular, r = AC = BC), then
`(1)/(2)pi r^(2)=8pi, pi r^(2)=16 pi, r^(2)=16`, and r = 4. So AC = 4 and BC = 4.
You can find the height BD of the triangle in terms of `theta` by saying that `(BD)/(BC)=sin theta`, so `BD = BC sin theta = 4 sin theta`. Thus the base AC is 4, and the height BD is 2 sin `theta`. The area of the triangle is `(1)/(2)xx AC xx BD=(1)/(2)xx 4xx(4 sin theta)=8 sin theta`.