# How to find area of isosceles triangle when given two heights? [closed]

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.

$$4\Delta = \sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}$$ 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.
• It might enhance understanding if you mentioned that $\Delta = \frac12 a h_a$. May 26, 2016 at 4:14
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 $$S=\frac{h_a^2h_b}{\sqrt{4h_a^2-h_b^2}}$$