Question

Find the area of the shaded portion in the circle.

click pic below

Answer 1

its 18pi + ?

Answer 2

i think 24 pi but not sure dont enter that its just a suggestion
but its 18pi + something

Answer 3

Shaded area is the area of a half circle plus the area of the indicated segment.\[\frac{\pi r^2}{2}+\frac{1}{2}r{}^{\wedge}2(\theta -\text{Sin}[\theta ])\text{ where } \theta =\pi -2\text{ArcSin}\left[\frac{6}{12}\right]\text{and } r=6 \]\[18 \left(-\frac{\sqrt{3}}{2}+\frac{2 \pi }{3}\right)+18 \pi =78.6593 \]

Answer 4

The area of the shaded part is the are of the circle minus the area of the bright part; so it's better to find the area of the bright part instead.
Connect the 3 lines as shown in the attached photo.
You will have 2 equal triangles and 2 equal pieces of circle.
The radius is 12/2 = 6.
Using the arc functions, find the angle "a".
Now You can easily find the area of the bright part and hence, also the shaded part.
Area of the triangle is 3*6/2 = 9.
Area of the small piece of circle is: \[\Pi(6^{2})\div \tan^{-1} (3/6) =4.25\]
Add them and multiply by two to find the total area of the bright part and we have: 26.51.
So the area of the shaded part is: π(36)- 26.51 = 86.53

Answer 5

@Parsa
A detailed solution with comments using the Calculus and your approach is attached.

You might recheck the dimensions on your triangle. The one dimensioned 6 on one of sides is actually 3*radical 3.

The solution first posted by me as 78.6593 above is correct and comes from a general formula for a circle's segment area. Both of the attached solutions support 78.6593 as the numerical answer and the following symbolic answer:\[30 \pi -9 \sqrt{3} \]