$\begingroup$Given the lengths of 3 heights in a triangle, I need to find its area.
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$\begingroup$Since $h_A=\frac{2\Delta}{a}$, by Heron's formula we have:
$$\small\frac{1}{\Delta}=\sqrt{\left(\frac{1}{h_A}+\frac{1}{h_B}+\frac{1}{h_C}\right)\left(-\frac{1}{h_A}+\frac{1}{h_B}+\frac{1}{h_C}\right)\left(\frac{1}{h_A}-\frac{1}{h_B}+\frac{1}{h_C}\right)\left(\frac{1}{h_A}+\frac{1}{h_B}-\frac{1}{h_C}\right)}.$$
$\endgroup$2$\begingroup$You know that base times height gives you area. Let the triangle have sides $a,\ b$ and $c$ with corresponding altitudes $h_a,\ h_b,\ h_c$. Then $$ah_a = bh_b = ch_c = 2A$$ where $A$ is the area of the triangle. Substitute these relations into Heron's formula and solve for $A$.
Edit: I didn't know the resulting formula had a name, but apparently as joriki mentions, it is the area theorem.
$\endgroup$$\begingroup$$$ t=area, x=ha, y=hb, z=hc $$ $$ t=\frac{x^2*y^2*z^2}{\sqrt{(xy+yz+zx)(-xy+yz+zx)(xy-yz+zx)(xy+yz-zx)}} $$ or
$$ t=\frac{1}{\sqrt{\frac{2}{x^2*y^2}+\frac{2}{y^2*z^2}+\frac{2}{z^2*x^2}-\frac{1}{x^4}-\frac{1}{y^4}-\frac{1}{z^4}}} $$
Use TrianCal.esy.es (Triangle Calculator) Example:
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