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.
A circle is centered at O and has an area of 48pi. 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.
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OQ = OR = r = 4√3
∠QOR = 30º ∠QOP = 15º ∠QPR = 60º
QR = 2 (OQ * sin∠QOP) ==> QR = 6*sqrt(3)
Height of ΔPQR h = sqrt[PQ2 - (QR/2)2] = 12*sqrt(3)
[PQR] = h * QR / 2 = 8*sqrt(3)
+122 Using this diagram, you should be able to find the area of the equilateral triangle QRP.
Area of a triangle QRP = \(\sqrt[3]3\)
Credit to Dragan; https://web2.0calc.com/members/dragan/
+1490 Hi, cryptoaops!
That diagram is ok, but my answer "Area of a triangle QRP = 3√3" is NOT correct!!!
Link: https://web2.0calc.com/questions/geometry_49612
The correct answer is [QRP] = 5.569219381 square units!!! 
Link: https://web2.0calc.com/questions/quick-hard-question-help-asap
(I don't know how to convert that decimal into radical.)
