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Given the lengths of 3 heights in a triangle, I need to find its area.

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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

3 Answers 3

up vote -1 down vote accepted

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.

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


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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 ( and the substitution $a=\frac{2\Delta}{h_A}$. –  Jack D'Aurizio Jun 1 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 (Triangle Calculator) Example:

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