+250 In triangle $ABC$, $AB = 7$, $AC = 17$, and the length of median $AM$ is $12$. Find the area of triangle $ABC$.
+128578 A
7 12 17
B M C
Law of Cosines
12^2 = 7^2 + (BC/2)^2 - 2 (7 * BC /2) cos ( ABC)
17^2 = 7^2 + BC^2 - 2 (7 * BC) cos (ABC)
Equate cosines and simplify
[144 - 49 - BC^2/4] / [ 7 BC ] = [289 - 49 - BC^2 ] / [ 14 BC]
[ 95 - BC^2 / 4 ] / 7 = [ 240 - BC^2 ] / 14
[95 - BC^2/4 ] = [ 240 - BC^2 ] / 2
95 -BC^2 / 4 = 120 - BC^2/2
120 - 95 = BC^2 / 4
25 = BC^2 /4
BC^2 = 25 * 4
BC = 5 * 2 = 10
Impossible because of the trianle inequality, AB + BC > AC
But
AB + BC =
7 + 10 not greater than 17
