Answer:
100π - (96 + (13/2)√231 ) m^2
Explanation:
We will solve this by finding the area of the triangles (the quadrilateral is cut by the diameter) and subtract it from the area of the circle.
There is a circle theorem which states that if a triangle is in the semicircle and the hypothenuse extends to the length of the diameter, the angle at B here is 90 degrees.
Since angle B is 90 degrees we can use the Pythagorean theorem (a^2 + b^2 = c^2) to find Line AC.
so 144 + 256 = 400
√400 = 20
So diameter is 20
Finding the area of the top triangle:
1/2(12x16) = 96cm^2
For the bottom triangle, we need the side DC to find the area. To find it we will apply the theorem once again.
400 = 169 + DC^2
DC^2 = 231
DC = √231
so the area of the bottom triangle is 1/2(13√231) = (13/2)√231
Now we add the area of the triangles for the quadrilateral and subtract from the circle area.
The circle area is 100π (πr^2)
Hence, the area of the shaded region is:
100π - (96 + (13/2)√231 ) m^2