How to find the area of a triangle with lengths of heights? Given the lengths of 3 heights in a triangle, I need to find its area. 
 A: Since $h_A=\frac{2\Delta}{a}$, by Heron's formula we have:
$$\small\frac{1}{\Delta}=\sqrt{\left(\frac{1}{h_A}+\frac{1}{h_B}+\frac{1}{h_C}\right)\left(-\frac{1}{h_A}+\frac{1}{h_B}+\frac{1}{h_C}\right)\left(\frac{1}{h_A}-\frac{1}{h_B}+\frac{1}{h_C}\right)\left(\frac{1}{h_A}+\frac{1}{h_B}-\frac{1}{h_C}\right)}.$$
A: 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.
A: $$
t=area, x=ha, y=hb, z=hc
$$
$$
t=\frac{x^2*y^2*z^2}{\sqrt{(xy+yz+zx)(-xy+yz+zx)(xy-yz+zx)(xy+yz-zx)}}
$$
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: http://triancal.esy.es/?lang=en&tip=2&x=45&y=60&z=36
