In △ABC, let M be the midpoint of BC. If AM = MB = 6.5 and AB = 5, determine the area of triangle ABC.
+128460 If MB = 6.5 then BC = 2(6.5) = 13
Using the Law of Cosines we have that
AB^2 = AM^2 + MB^2 - 2(AM*MB) cos (AMB)
5^2 = (6.5)^2 + (6.5)^2 - 2(6.5)^2 * cos (AMB)
[ 25 - 6.5^2 - 6.5^2] / [ - 2* 6.5^2 ] = cos (AMB)
-59.5 / -84.5 = 119/169
Then the sin of AMB = sqrt [ 169^2 - 119^2] / 169 = 120/169
Using the Law of Sines
sin(AMB) /AB = sin (ABM) / 6.5
(120/169) /5 = sin (AMB) /6.5
6.5 ( 120/169) / 5 = sin AMB = 12/13
Then the area of ABC =
(1/2) AB * BC sin (AMB) =
(1/2) (5)(13) * (12/13) =
(1/2) 60 =
30 units^2
