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terricherri
Find the area of the shaded portion in the circle. click pic below
i think 24 pi but not sure dont enter that its just a suggestion but its 18pi + something
Shaded area is the area of a half circle plus the area of the indicated segment.\[\frac{\pi r^2}{2}+\frac{1}{2}r{}^{\wedge}2(\theta -\text{Sin}[\theta ])\text{ where } \theta =\pi -2\text{ArcSin}\left[\frac{6}{12}\right]\text{and } r=6 \]\[18 \left(-\frac{\sqrt{3}}{2}+\frac{2 \pi }{3}\right)+18 \pi =78.6593 \]
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
@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:\[30 \pi -9 \sqrt{3} \]