<s>[INST] Area of $\triangle ABC$, given $AB=7$, $AC=15$, and median $AM=10$ I've been working on this interesting problem for a while already, and here it is: 

In $\triangle ABC$, $AB = 7$, $AC = 15$, and median $AM = 10$. Find the area of $\triangle ABC$.

I have figured out that $BM$ and $CM$ are both $4\sqrt2$ using Stewart's Theorem. Now, I tried to use Heron's Formula to calculate the area, which was a mess.
Any help is appreciated. Thanks.
 [/INST] Are you sure? $p = \dfrac{7+15+8\sqrt{2}}{2} = 11+4\sqrt{2}\implies S^2 = p(p-15)(p-7)(p-8\sqrt{2}) = (11+4\sqrt{2})(11-4\sqrt{2})(-4+4\sqrt{2})(4+4\sqrt{2}) = (121-32)(32-16) = 89\cdot 16.$ 
So $S = 4\sqrt{89}.$
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