# 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.

-
Use Heron's formula – i. m. soloveichik Oct 30 '12 at 14:38
@i.m.soloveichik: Heron's formula is for three side lengths, not for three heights. You need the area theorem. – joriki Oct 30 '12 at 14:39
@i.m.soloveichik 3 heights, not 3 sides – user31280 Oct 30 '12 at 14:39

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.

-

Since $h_A=\frac{2\Delta}{a}$, by Heron's formula we have:

$$\small\frac{1}{\Delta}=\sqrt{\left(\frac{1}{h_A}+\frac{1}{h_B}+\frac{1}{h_C}\right)\left(-\frac{1}{h_A}+\frac{1}{h_B}+\frac{1}{h_C}\right)\left(\frac{1}{h_A}-\frac{1}{h_B}+\frac{1}{h_C}\right)\left(\frac{1}{h_A}+\frac{1}{h_B}-\frac{1}{h_C}\right)}.$$

-
There is mistake somewhere. I can't get right values out of it. – nazar554 Oct 30 '12 at 19:00
@nazar554: it looks perfectly fine to me. What's wrong with that? It comes straight from the Heron's formula (en.wikipedia.org/wiki/Heron%27s_formula) and the substitution $a=\frac{2\Delta}{h_A}$. – Jack D'Aurizio Jun 1 '15 at 0:27

$$t=area, x=ha, y=hb, z=hc$$ $$t=\frac{x^2*y^2*z^2}{\sqrt{(xy+yz+zx)(-xy+yz+zx)(xy-yz+zx)(xy+yz-zx)}}$$ 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

-