SOLUTION: You will have to go here for the diagram: http://img840.imageshack.us/img840/8545/prob2o.jpg In the figure, the circle PQR with radius r is enclosed in the sector OAB with centr

Algebra ->  Circles -> SOLUTION: You will have to go here for the diagram: http://img840.imageshack.us/img840/8545/prob2o.jpg In the figure, the circle PQR with radius r is enclosed in the sector OAB with centr      Log On


   

Question 466516: You will have to go here for the diagram:
http://img840.imageshack.us/img840/8545/prob2o.jpg
In the figure, the circle PQR with radius r is enclosed in the sector OAB with centre at O and radius 48 cm. Given that angle AOB = 60 degrees, find:
a) the value of r.
b) the area of the shaded region, giving your answer correct to 1 decimal place.
*Please answer as soon as possible bro :) =)
Answer by ccs2011(207) About Me  (Show Source):
You can put this solution on YOUR website!
radius = 16
area = 175.3
Sorry about the no-explanation.
Label the center of inscribed circle point C.
Line OP is tangent to circle, thus OP is perpendicular to CP
Form special right triangle OCP.
OC bisects 60 degree angle, thus mPOC = 30 degrees and mOCP = 60 degrees.
length of CP is r.
sin 30 = 1/2 = r/OC
=> OC = 2r
Length of CR is r, thus length of OR is 3r
OR = 48
=> 3r = 48
=> r = 16
Area of sector PCQ = %281%2F3%29pi%2816%5E2%29 = %28256%2F3%29pi
Draw line from P to Q forming triangle POQ
OP = OQ, thus it forms equilateral triangle
tan 30 = 1/sqrt3 = r/OP
=> OP = sqrt3 * r = 16sqrt3
Area of triangle POQ = %28sqrt%283%29%2F4%29%2816sqrt%283%29%29%5E2 = 192sqrt%283%29
Triangle PCQ forms an isosceles triangle
height = r/2 = 8
Area of triangle PCQ = %281%2F2%2916sqrt%283%29%2A8 = 64sqrt%283%29
Now subtract this area from area of sector to get left side of circle
=%28256%2F3%29pi+-+64sqrt%283%29 = 157.2313
Now subtract this from Area of triangle POQ
=192sqrt%283%29+-+157.2313 = 175.32
Area = 175.3