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

  • 0
    Use Heron's formula2012-10-30
  • 0
    @i.m.soloveichik: Heron's formula is for three side lengths, not for three heights. You need the [area theorem](http://en.wikipedia.org/wiki/Altitude_(triangle)#Area_theorem).2012-10-30
  • 0
    @i.m.soloveichik 3 heights, not 3 sides2012-10-30
  • 0
    http://mathforum.org/library/drmath/view/61893.html2013-10-14

3 Answers 3

-1

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.

1

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

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

  • 0
    There is mistake somewhere. I can't get right values out of it.2012-10-30
  • 0
    @nazar554: it looks perfectly fine to me. What's wrong with that? It comes straight from the Heron's formula (http://en.wikipedia.org/wiki/Heron%27s_formula) and the substitution $a=\frac{2\Delta}{h_A}$.2015-06-01
0

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