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.
 A: Using Stewart's theorem, you can find that $BM$ and $CM$ are both $\sqrt{37}$ and not $4\sqrt2$, which means our third side ($BC$) is of length $2\sqrt{37}$ (= $\sqrt{148}$) 
Then, using the less messier form of the Heron's formula given by: 
$$16 \;|\triangle ABC|^2 = 2a^2 b^2+2b^2c^2+2c^2a^2-a^4-b^4-c^4$$
you can calculate the area to be equal to exactly $42$!
A: You can avoid Heron's formula by also computing the measure of $∠AMB$. Let $\theta=m∠AMB$, $x=BM$. Then (as in the derivation of Stewart's/Apollonius's theorem) we have
$$
49=100+x^2-20x \cos \theta\\
225=100+x^2+20x\cos \theta
$$
and so $20x \cos \theta=\frac{225-49}{2}=88$, and $x^2=\frac{49+225}{2}-100=37$.
Then the area of the triangle can be computed as
$$
10x\sin \theta = 10\sqrt{x^2-x^2\cos^2 \theta}
$$
A: 
Let $|AB|=7=c$,
$|AC|=15=b$,
$|AM|=10=m_a$,
$|BM|=|MC|=\tfrac{a}2=x$.
Then by  Stewart’s theorem for $\triangle ABC$
\begin{align} 
b^2x+c^2x&=2x(m_a^2+x^2)
,\\
x&=\sqrt{\tfrac12(b^2+c^2)-m_a^2}
=\sqrt{\tfrac12(49+225)-100}
=\sqrt{37}
,\\
a&=2x=2\sqrt{37}
.
\end{align}  
And the area
\begin{align} 
S&=\tfrac14\sqrt{4a^2b^2-(a^2+b^2-c^2)^2}
\\
&=\tfrac14\sqrt{4\cdot4\cdot37 \cdot 225
-(4\cdot37+225-49)^2}
.
\end{align}  
So, the answer is$\dots$ still 42. 
A: 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 $S = 4\sqrt{89}.$
