PROBLEM #4 SOLUTION
A number of approaches to this problem are possible.
Firstly, we draw a line from C to the center of the circle O.
Then draw radii OD and OE to the points of contact of tangents CB and CA, respectively. The radii meet
the tangents at right angles.
Using Angle Bisector Theorem
By symmetry, CD = CE. Clearly OD = OE = radius of semicircle.
Hence
ODC is similar to
OEC, and so
OCE =
OCD.
Applying the Angle Bisector Theorem to
ABC, we have AC/BC = AO/BO.
Hence 11/13 = AO/(20 − AO), and so 13AO = 11(20 − AO) = 220 − 11AO. Therefore AO = 55/6.
Applying the law of cosines (also known as the cosine rule) to
ABC, we have
2
2
2
13 = 11 + 20 − 2·11·20·cos A. Hence cos A = 4/5, and so, by Pythagoras' Theorem, sin A = 3/5.
In
EAO, sin A = OE/AO.
Hence OE = (3/5) · (55/6) = 11/2.
Therefore, the diameter of the semicircle is 11 units.
Using Area of Triangles
Applying Heron's Formula to
ABC, area of ABC = [22 · (22 − 11) · (22 − 13) · (22 − 20)] = 66.
Let the radius of the semicircle be r, so that OD = OE = r.
Then area of
AOC = ½ · AC · r = 11r/2.
Similarly, area of
BOC = ½ · BC · r = 13r/2.
Since area of
ABC = area of
AOC + area of
BOC, we have 66 = 12r, and hence r = 11/2.
Therefore, the diameter of the semicircle is 11 units.