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Help with tricky geometry please.

 

Triangle PQR is equilateral.  A semicircle with center O is drawn with its diameter on PR so that one end is at P and the curved edge touches QR at X.  The radius fo the semicircle is sqrt(3).

 

What is the length of QX?

 

 Dec 22, 2023
 #1
avatar+26 
+1

This problem is very similar: 

https://www.quora.com/A-point-P-is-given-on-the-circumference-of-a-circle-of-radius-R-The-chord-QR-is-parallel-to-the-tangent-at-P-What-is-the-maximum-possible-area-of-a-triangle-PQR

 

See if that helps you

 Dec 22, 2023
 #2
avatar+126665 
+1

Draw OX = sqrt (3)  = OP 

 

Since QR  is a tangent  meeting the circle at X , then angle OXR = 90

And since the triangle is equilateral, then angle PRQ  = 60

So triangle ORX   is a 30 -60 -90 right triangle

 

So XR  =  OX /sqrt 3  =   sqrt (3) / (sqrt (3)    = 1

OR = 2

 

PR  = OR + OP  =   2 + sqrt (3)  =  QR

 

QR  -  XR  = QX  

 

2 + sqrt 3  -  1  =  QX  =      1  + sqrt 3

 

cool cool cool

 Dec 23, 2023
edited by CPhill  Dec 23, 2023

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