+130511 Using the Law of Cosines, we can find angle BAC
6^2 = 5^2 + 5^2 - 2(5)(5)cosBAC
[36 - 50] / [-2(25)] = cosBAC
[-14/ -50] = cosBAC
[7/25] = cosBAC
And arccos(7/25) = BAC
And angle OAC will be 1/2 of this = (1/2)arccos(7/25)
But, since O is the circumcenter of triangle ABC, OC = OA
Then, angle OAC = angle OCA
And angle AOC = [pi - arcos (7/25)]
So.....using the Law of Sines, we can find OC, thusly
OC/sinOAC = AC /sin AOC
OC / sin [(1/2)arccos(7/25)] = AC / sin [ pi - arccos(7/25)]
OC / sin [(1/2)arccos(7/25)] = 5 / sin [ pi - arccos(7/25)]
OC = 5 * sin [(1/2)arccos(7/25)] / sin [ pi - arccos(7/25)] = 3.125 = 25/8
And, by symmetry OC = BO
So....using Heron's formula....the area of triangle BOC can be found thusly :
Semi-perimeter = [ 2*25/8 + 6] / 2 = 49/8 = s
And the area =
√[ s * (s - a) * (s - b) * (s - c)] where a = 6 and b and c = 25/8 .... so we have
√[ (49/8) * ( 49/8 - 6) * (49/8 - 25/8)^2 ] =
√[ (49/8) * (49/8 - 48/8) * (24/8)^2 ] =
√[ (49/8) * (1/8) * ( 3)^2 ] =
√[ (49/8) * (9/8) ] =
√49 * √9 * 1 / √64 =
[7 * 3 ] / 8 =
(21/8) units^2
