+359 In triangle $ABC$, $AB = 7$, $AC = 15$, and the length of median $AM$ is $7$. Find the area of triangle $ABC$.
+34 Using Stewart's theorem, you can find that BM and CM are both √37 and not 4√2, which means our third side (BC) is of length 2√37 (= √148)
Then, using Heron's formula given by:
\(16| \triangle ABC|^2 = 2a^2b^2+2b^2c^2+2c^2a^2-a^4-b^4-c^4\)
you can calculate the area to be equal to exactly 42
