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.
+128448 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
