In the circle with center $Q$, radii $AQ$ and $BQ$ form a right angle. The two smaller regions are tangent semicircles, as shown. The radius of the circle with center $Q$ is 14 inches. What is the radius of the smaller semicircle? Express your answer as a common fraction.

Guest Jun 30, 2018

#1**+1 **

Let M be the midpoint of AQ and N be the midpoint of BQ

Thus....M and N will be the centers of both of the smaller semi circles

Connect M and N

By Euclid, a segment connecting the centers of two tangent circles or semi-circles will pass through the point of tangency

So MQ will be the radius of the semicircle with the diameter AQ = 14

So MQ = 7

And call the radius of the smallest semi-circle, r

So....MQN will form right triangle with leg MQ = 7, leg QN = 14 - r and

hypotenuse MN = ( 7 + r )

So we have that, by the Pythagorean Theorem

7^2 + (14 - r)^2 = (7 + r)^2

49 + r^2 - 28r + 196 = r^2 + 14r + 49

196 = 42r

196/42 = r

14/3 = r = the radius of the smaller semi-circle

CPhill
Jun 30, 2018