Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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
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.

share|cite|improve this answer

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


share|cite|improve this answer
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 '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 (Triangle Calculator) Example:

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.