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A semicircle is inscribed in triangle $XYZ$ so that its diameter lies on $\overline{YZ}$, and is tangent to the other two sides.   If $XY = 10,$ $XZ = 10,$ and $YZ = 10 \sqrt{2},$ then find the area of the semicircle.

 

 Jan 7, 2024
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avatar+129388 
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Since YZ  =10 sqrt 2 and XY = XZ  = 10....then XYZ is a right triangle with the right angle at  X

 

Call the center of the  semi-circle O  

 

Let the tangent  XZ  intersect the circle at P 

 

Draw OX.....this bisects angle YXZ

 

So  we have right triangle POX

 

Angle  OXP   =   45

Angle OPX  = 90

So angle XOP = 45

 

So triangle OPX  is a 45 -45- 90 right triangle  with OX =  (1/2)YZ = (1/2) 10sqrt (2)  = 5sqrt2

 

And OP  = XP  =  5sqrt 2 / sqrt 2 =  5 

 

So   OP is the radius of the semi-circle  

 

Area  of semi-circle  =   (1/2) pi  * (5)^2  =  12.5 pi  ≈  39.27

 

cool cool cool

 Jan 7, 2024
edited by CPhill  Jan 8, 2024

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