+1439 In triangle $ABC$, $AB = 7$, $AC = 17$, and the length of median $AM$ is $12$. Find the area of triangle $ABC$.
+129895 A
7 17
B M C
We can show that such a triangle cannot exist
angles AMB and AMC are supplemental....so cos (AMB) = -cos (AMC)
Since AM is a median, then BM = CM = x
Using the Law of Cosines
AC^2 = CM^2 + AM^2 - 2 (AM * MC)cos (AMC)
AB^2 = BM^2 + AM^2 - 2 (AM * BM) (-cos (AMC)
17^2 = x^2 + 12^2 - 2 (AM * x) cos (AMC)
7^2 = x^2 + 12^2 + 2(AM * x) cos (AMC) add these
338 = 2x^2 + 288
50 = 2x^2
25 = x^2
5 = x
Then BM + CM = 5 + 5 = 10 = BC
But, by the Triangle Inequality, AB + BC must be > AC = 17
However AB + BC = 7 + 10 = 17
Which means that no such triangle can exist
