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.


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. 
