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So I know the sine and cosine theorem and I tried using them but I got nowhere. (I got to an equation which I can't solve and I know there must be an easier method since we have not studied how to solve such equations.) I would really appreciate it if you can answer soon as I need to know how to do this for my exam today.

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2 Answers 2

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$$ 4\Delta = \sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}$$ by Heron's formula, hence by dividing both sides by $4\Delta^2$ we get (since $\frac{a}{2\Delta}=\frac{1}{h_a}$): $$ \frac{1}{\Delta} = \sqrt{\left(\frac{1}{h_a}+\frac{1}{h_b}+\frac{1}{h_c}\right)\left(-\frac{1}{h_a}+\frac{1}{h_b}+\frac{1}{h_c}\right)\left(\frac{1}{h_a}-\frac{1}{h_b}+\frac{1}{h_c}\right)\left(\frac{1}{h_a}+\frac{1}{h_b}-\frac{1}{h_c}\right)}$$ and we may compute the area of a triangle given its heights lenghts.

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    $\begingroup$ With a simplification in this case since two of the heights are equal. $\endgroup$ May 26, 2016 at 4:12
  • $\begingroup$ It might enhance understanding if you mentioned that $\Delta = \frac12 a h_a$. $\endgroup$ May 26, 2016 at 4:14
  • $\begingroup$ I see. I removed my comments as I saw my mistake. $\endgroup$
    – indjev99
    May 26, 2016 at 4:23
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If you don't know Herron formula, you can find this directly. Let the triangle be $ABC$. With $|AB|=|AC|=b$ and $|BC|=a$. Let $AH$ and $BH'$ be the two heights that your are given $|AH|=h_a$ and $|BH'|=h_b$. Note that $H$ is exactly in the middle of $BC$. Look at the triangle $AHC$, for which you have Pythagorean theorem: $$ \left(\frac{a}{2}\right)^2+h_a^2 = b^2 (*) $$ On the other hand the area of this triangle is $$ S=\frac{h_a a}{2}=\frac{h_b b}{2}\Longrightarrow b=\frac{h_a}{h_b}a $$ Put this in (*) to get $$ \frac{1}{4}a^2+h_a^2 = \frac{h_a^2}{h_b^2}a^2\Longrightarrow a= \frac{2h_ah_b}{\sqrt{4h_a^2-h_b^2}} $$ which means $$S=\frac{h_a^2h_b}{\sqrt{4h_a^2-h_b^2}}$$

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