Given that AB=12, CD=1, angle BAC = 90 degrees, and the semicircle is tangent to BC, find the radius of the semicircle.
+128460 Using the Pythagorean Theorem
BC^2 = 12^2 + ( 2r + 1)^2 where r is the radius of the semi-circle
BC^2 =144 + 4r^2 + 4r + 1
BC^2 = 4r^2 + 4r + 145
BC = sqrt (4 r^2 + 4r + 145)
Call the center of the semi-circle, O
From O draw radius OP perpendicular to BC
And triangles CAB and CPO are similar
So
AB/ CB = OP / CO
12/ sqrt ( 4r^2 + 4r + 145) = r / (r + 1)
12 (r + 1) = r sqrt (4r^2 + 4r + 145) square both sides
144 ( r^2 + 2r + 1) = r^2 ( 4r^2 + 4r + 145)
4r^4 + 4r^3 + 145r^2 = 144r^2 + 288r + 144
4r^4 + 4r^3 + r^2 - 288r - 144 = 0
Solving this for the positive value of r gives that r = 4
