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The sides of triangle XYZ are XY = XZ = 25 and YZ = 40. A semicircle is inscribed in triangle XYZ so that its diameter lies on line YZ, and is tangent to the

other two sides. Find the area of the semicircle.

I'm not sure what to do. I dropped an altitude to split it in half but it didn't really help. Could someone please help? Thank you, and I hope you have a nice day and stay safe!

 Apr 2, 2020
 #1
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The radius of the semicircle is 10, so its area is 1/2*pi*10^2 = 50*pi.

 Apr 2, 2020
 #2
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Draw  altitude  XD   . . .(as  you did)

 

We  can find  this altitude  as

 

XD  =  sqrt  [ 25^2 - 20^2 ]=   sqrt [ 625 - 400 ]  =  sqrt (225)  = 15

 

So...

 

sin   XZY  =  15 / 25 =    3/5

 

From D, draw  DE  such that  E  is  the  point  of  tangency between the circle and the side  XZ  of the triangle

 

Then   DE  = the radius of the semi-circle

 

And  angle  DEZ  is right

 

So...by  the Law of Sines

 

DE / sin XYZ =  DZ  /sin 90

 

DE  / (3/5)  = 20

 

DE  =  20 (3/5)  =  12

 

This is the radius of the semi-circle.....can you take it from there  ???

 

 

cool cool cool

 Apr 2, 2020
edited by CPhill  Apr 2, 2020

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