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

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



cool cool cool

CPhill  Jun 30, 2018

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