AB is the diameter of the semicircle. C is a point on the circumference.O is the center if the semicircle.

If the radius of the semicircle is 5, ∠COA=60∘. Find the sum of the area of the region R and S. note region R and S is not the semicircle with the diameter AC and BC

Guest Jun 12, 2020

#1**+1 **

**AB is the diameter of the semicircle. C is a point on the circumference. O is the center if the semicircle.**

**If the radius of the semicircle is 5, \(\angle COA=60^\circ\). Find the sum of the area of the region R and S. **

**Note region R and S is not the semicircle with the diameter AC and BC**

\(\text{Let segment of the circle $AC = A_1$} \\ \text{Let segment of the circle $CB = A_2$}\)

\(\begin{array}{|rcll|} \hline \mathbf{A_1 + A_2} &=& \mathbf{\text{semicircle AB} - \triangle ABC} \\\\ R+S &=& (\text{semicircle AC}-A_1) + (\text{semicircle CB}-A_2) \\ R+S &=& \text{semicircle AC}+\text{semicircle CB}-(A_1+A_2) \\ R+S &=& \text{semicircle AC}+\text{semicircle CB}-(\text{semicircle AB} - \triangle ABC) \\ R+S &=& \text{semicircle AC}+\text{semicircle CB}-\text{semicircle AB} + \triangle ABC \\ R+S &=& \dfrac{\pi \left(\dfrac{5}{2}\right)^2 }{2} +\dfrac{\pi \left(\dfrac{5\sqrt{3}}{2}\right)^2 }{2}-\dfrac{\pi 5^2 }{2} + \triangle ABC \\\\ R+S &=& \dfrac{5^2}{8}\pi +\dfrac{3*5^2}{8}\pi -\dfrac{5^2}{2}\pi + \triangle ABC \\\\ R+S &=& \dfrac{5^2}{8}\pi +\dfrac{3*5^2}{8}\pi -\dfrac{4*5^2}{8}\pi + \triangle ABC \\\\ R+S &=& \dfrac{4*5^2}{8}\pi-\dfrac{4*5^2}{8}\pi + \triangle ABC \\\\ \mathbf{R+S} &=& \mathbf{\triangle ABC} \\ \hline R+S &=& \dfrac{5*5\sqrt{3}}{2} \\\\ \mathbf{ R+S } &=& \mathbf{\dfrac{25\sqrt{3}}{2}} \\ \hline \end{array}\)

heureka Jun 12, 2020