<s>[INST] Given the length of two altitudes and one side , find the area of triangle. Segments $BE$ and $CF$ are the altitudes in $\triangle ABC$.
 $E$ is on line $AC$ and $F$ is on line $AB$.
$BC = 65$, $BE = 60$ and $CF = 56$.
Find $A(\triangle ABC)/100$.  
 
By the Pythagorean theorem , $CE=25$ , and $BF= 33$.
If the length of altitude from $A$ to B$C$ can be calculated then the area of $\triangle ABC$ can be calculated since the length of $BC$ is known.
But I'm stuck here , so any hints are apreciated .
 [/INST] We have that $\cot C=25/60$ and $\cot B=33/56$.
Now use the area formula:
$$\text{Area}=\frac{a^2}{2(\cot B + \cot C)}$$
Thus:
$$\frac{(ABC)}{100}=\frac{65^2}{200(\frac{25}{60} + \frac{33}{56})}=\frac{13^2}{8(\frac{25}{60} + \frac{33}{56})}=\frac{13^2}{\frac{10}{3} + \frac{33}{7}}=\frac{13^2}{\frac{70+99}{21}}=21$$
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