2019 Mock AMC 10B Problems/Problem 12

Observe that the triangle is a 30-60-90 triangle because of the 10-10sqrt(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 = 5sqrt(3). Because ABC = 30 degrees, then BIO = 30 degrees (radii are the two legs of the triangle -> isosceles triangle). That means that COI is 60 degrees and IOB is 120 degrees. We know that the area of the 60-degree sector is 1/6*(5sqrt(3)^2)pi = 25pi/2. 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 15/2. Therefore, the area of that triangular portion = 5*sqrt(3)*15pi/4 = 75sqrt(3)/4. Therefore, the total area = 75sqrt(3)/4 + 25pi/2 = (75*sqrt(3)+50pi)/4. Our answer = 75 + 3+ 50 + 4 = 132.