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We have a $\triangle ABC$ with:

  • Base $AB$ with length 14

  • $AC$ with length 15

  • $BC$ with length 13

Find the area of the triangle using analytic geometry.

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Is Heron's formulat analytical? –  Jean-Sébastien Nov 10 '12 at 16:36
en.wikipedia.org/wiki/Heron's_formula –  user31280 Nov 10 '12 at 16:37
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1 Answer

Drop a height from vertex A to side BC. Let the height intersect BC at point X and have size h. Let XB have size x and XC have size y. By Pythagoras theorem, we know that $h^2+x^2=14^2$ and $h^2+y^2=15^2$. Now subtract both equations to get $y^2-x^2=15^2-14^2$, therefore $y-x=(15^2-14^2)/(x+y)=(15^2-14^2)/BC=(15^2-14^2)/13$, since $x+y=13$, therefore we can solve for $x,y$ from the last 2 equations. After getting $x,y$ we can get $h$ from $h^2+x^2=14^2$. Finally we find that the area is $(1/2)(BC)h$

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I would not be surprised if, in the general case, this proved equivalent to proving Hero's formula for the area given the sides. –  marty cohen Mar 25 '13 at 3:46
It is equivalent, but the "factoring" of the general expression we get is non-obvious. –  André Nicolas Jul 31 '13 at 0:49
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