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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
    

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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:

$$\Delta=\frac{4}{\sqrt{\left(\frac{1}{h_A}+\frac{1}{h_B}+\frac{1}{h_C}\right)\left(-\frac{1}{h_A}+\frac{1}{h_B}+\frac{1}{h_C}\right)\left(\frac{1}{h_A}-\frac{1}{h_B}+\frac{1}{h_C}\right)\left(\frac{1}{h_A}+\frac{1}{h_B}-\frac{1}{h_C}\right)}}.$$

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There is mistake somewhere. I can't get right values out of it. –  nazar554 Oct 30 '12 at 19:00

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