geometryyyyy 2

Let r be the radius of the circle.

Obviously, OC = r.

Let M be the mid-point of CD.

Since $\angle ABD = 90^\circ$, ABMO is a rectangle. Therefore, OM = 12.

Since M is the midpoint of CD, CM = 18/2 = 9.

Then, by Pythagorean theorem, 

$OC^2 = OM^2 + MC^2\\ r^2 = 12^2 + 9^2 = 225\\ r = 15$

Circle O is tangent to AB at A, and angle ABD = 90 degrees. If AB = 12 and CD = 18, find the radius of the circle.

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r = 1/2{sqrt[(2*AB)² + CD²]}

r = 15