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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

3 Answers 3

Using Heron's Formula: $s=\dfrac{a+b+c}2=\dfrac{14+15+13}2=21$. $$ \begin{align} A &=\sqrt{s(s-a)(s-b)(s-c)}\\ &=\sqrt{21(7)(6)(8)}\\ &=\sqrt{7056\vphantom{()}}\\ &=84 \end{align} $$

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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

Take a $9-12-15$ right triangle and a $5-12-13$ right triangle and glue the $12$ sides together. This gives us the $13-14-15$ triangle. So the area is $\dfrac{1}{2} \cdot 9 \cdot 12 + \dfrac{1}{2} \cdot 5 \cdot 12 = 84$.

EDIT: Does this count as analytic geometry? If not:

Let $A = (9,0)$, $B = (-5,0)$, $C = (0,12)$. Using the distance formula, we see that $AB = 14$, $BC = 13$, $CA = 15$, so these coordinates are correct.

Then, using the Shoelace Theorem, we see that the area of the triangle is $\dfrac{1}{2}\left|(9 \cdot 12 + 0 \cdot 0 + -5 \cdot 0) - (9 \cdot 0 + -5 \cdot 12 + 0 \cdot 9)\right| = 84$

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