+195 1)An equilateral triangle has an area of 64sqrt3 cm^2 . If each side of the triangle is decreased by 4 cm, by how many square centimeters is the area decreased?
2) In triangle ABC , AB= AC=5 and BC=6. Let O be the circumcenter of triangle ABC. Find the area of triangle OBC .
1) The side length of the original equiateral triangle is 12, so the area is decreased by 20*sqrt(3) cm^2.
2) The circumradius is 4, so the area of OBC is 3*sqrt(7).
+129928 2. Using the Law of Cosines
6^2 = 5^2 + 5^2 - 2 (5)(5) cos BAC
36 = 50 - 50 cos BAC
[36 - 50]
_______ = cos BAC
-50
14
___ = cos BAC
50
7
__ = cos BAC
25
So sin BAC = 24 / 25
And angle BOC = twice angle BAC
So
cos (2 * BAC) = cos (BOC) = 1 - 2sin^2 (BAC)
cos (BOC) = 1 - 2 (24/25)^2 = 1 - 2 ( 576/625) = [ 625 - 2(576) ] / 625 = -527/625
And using the Law of Cosines again
6^2 = r^2 + r^2 - 2(r)(r) cos (BOC)
36 = 2r^2 - 2r^2 [ -527 / 625 ]
36 = 2r^2 [ 1 + 527/625 ]
18 = r^2 [ 1 + 527/625 ]
18 = r^2 [1152 /625 ]
r^2 = [625 * 18 ] / 1152
r^2 = 11250 / 1152
r^2 = 625 / 64
r = 25/8 = the circumradius
So the semiperimeter of triangle BOC is [ 25/8+ 25/8 + 6 ]/2 = 49/8
So...using Heron's Formula....the area of BOC is
sqrt [ (49/8) ( 49/8 - 25/8) (49/8 - 25/8) (49/8 - 6) ] =
sqrt [ (49/8) ( 24/8) (24/8) ( 1/8) ] =
(7 * 24 * 1) 168 21
__________ = ___ = ___ = 2.625 units^2
64 64 8
