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A circle is centered at O and has an area of 48 pi.  Let  Q and R be points on the circle, and let P be the circumcenter of triangle QRO. If P is contained in triangle QRO and triangle PQR is equilateral, then find the area of triangle PQR.

 Aug 21, 2022
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The radius^2 of the  circle =  48 =  OQ^2

 

Note that since P  is the  circumcenter of triangle  QOR......then  OP   =  QP  =  RP

Call this distance , x

 

Since triangle  PQR is  equilateral, then angle QPR = 60

Let the altitude  be  PS

So angle QPS =  30

 

 

So....the angle  formed by OPQ  = 180 - angle QPS  = 180 - 30 =  150

 

Using the Law of Cosines

 

OQ^2  =  OP^2 + QP^2 - 2 ( OP * QP) *cos (150)           {cos 150  = -sqrt (3) / 2 }  

 

48 = x^2 + x^2  + 2 (x^2) (sqrt(3) / 2)

 

48 = 2x^2  + x^2 sqrt (3)

 

48 = x^2 ( 2 + sqrt (3))

 

48  / ( 2 + sqrt 3)  =   x^2

 

Area of triangle  PQR  =   (1/2) (QP * RP) sin (60)  =

 

(1/2) x^2  (sqrt 3)  / 2  =

 

(1/4) 48 sqrt (3)  / (2 + sqrt 3)  =

 

[12sqrt (3)] / [ 2 + sqrt (3) ] =

 

(12 sqrt (3) ) ( 2 - sqrt (3) )

_____________________  =

         4  -  3

 

24sqrt (3) - 36

 

 

cool cool cool

 Aug 21, 2022

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