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Given that AB=12, CD=1, angle BAC = 90 degrees, and the semicircle is tangent to BC, find the radius of the semicircle.

 

 Jan 27, 2021
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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

 

 

cool cool cool

 Jan 27, 2021

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