AB= 12m, BC= 16 cm and AD= 13 m.
Find the area of the shaded region.

100π - (96 + (13/2)√231 ) m^2

Step-by-step 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

hi there!

how can i assist you?

step-by-step explanation:

just plug in each number for x and see which ones work and don't