2019 Mock AMC 10B Problems/Problem 12

We need to find the area of the intersection between the triangle and the semicircle. Observe that the triangle is a $30$ - $60$ - $90$ triangle because of the $10$ to $10\sqrt{3}$ ratio. Let the center of the circle be $O$ , the upper left point of the triangle be $A$ , the right angle be $C$ , the point on the right be $B$ , and the intersection between the hypotenuse of the triangle and the curve of the semicircle be $I$ . Then $OI$ and $OB$ are the radii of the semicircle $= 5\sqrt{3}$ . Because $\angle ABC = 30$ degrees, then $\angle BIO = 30$ degrees (radii are the two legs of the triangle $\rightarrow$ isosceles triangle). That means that $\angle COI$ is $60$ degrees and $\angle IOB$ is $120$ degrees. To find the area of the overlappingregion, we can find the area of the 60-degree sector and the area of the remaining region, which is a triangle. We know that the area of the 60-degree sector is $\frac{1}{6} \cdot (5\sqrt{3})^2 \pi = \frac{25}{2} \pi$ . We also can draw the altitude of the remaining area we need to find, which forms a $30$ - $60$ - $90$ triangle, giving us the height of the triangle is $\frac{15}{2}$ . Therefore, the area of that triangular portion = $5\sqrt{3} \cdot \frac{15}{2} \pi \cdot \frac{1}{2} = \frac{75\sqrt{3}}{4}$ . Therefore, the total area $= \frac{75\sqrt{3}}{4} + \frac{25}{2} \pi = \frac{75\sqrt{3}+50\pi}{4}$ . Our answer $= 75 + 3+ 50 + 4 = \boxed{132}$ .

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2019 Mock AMC 10B Problems/Problem 12