SOLUTION: The diameter AD of a circle is perpendicular to side BC of the equilateral triangle ABC with D lying on BC. If the length of BC is 4, find the area of the shaded part of the circle

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Question 1150723: The diameter AD of a circle is perpendicular to side BC of the equilateral triangle ABC with D lying on BC. If the length of BC is 4, find the area of the shaded part of the circle that is outside the triangle. (Sorry if you see double of this question! I forgot to include a diagram)
https://imgur.com/2pXi73q
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


The two shaded areas are congruent, so find the area of one and double it to get the answer.

Call the center of the circle O; let E be the point where AB intersects the circle.

Draw radius OE.

Draw altitude OF of triangle AOE.

The area of the shaded piece on the left is the area of a sector of a circle, minus the area of triangle AOE.

Angle BAO is 30 degrees, so arc DE is 60 degrees; then arc EA is 120 degrees. So the area of the sector of the circle cut off by radii OA and OE is 1/3 of the area of the circle.

With BC=4, BD=2; then AD=2%2Asqrt%283%29, so the radius of the circle is sqrt%283%29.

The area of the circle is then %28pi%29r%5E2+=+3%28pi%29; so the area of the sector of the circle cut off by radii OA and OE is %28pi%29.

With the radius of the circle sqrt%283%29, the altitude OF of triangle AOE is sqrt%283%29%2F2, and AF = OF%2Asqrt%283%29+=+3%2F2. The area of triangle AOE is then (half base) times (altitude) = %283%2F2%29%2A%28sqrt%283%29%2F2%29=3%2Asqrt%283%29%2F4.

Then the area of each of the two shaded regions is

pi-3%2Asqrt%283%29%2F4

And so to combined area of the two shaded regions is

2%28pi%29-3%2Asqrt%283%29%2F2