Let $ABC$ be a triangle of area $\Delta$ and $A'B'C'$ be the triangle formed by the altitudes $h_a, h_b, h_c$ as its sides with area ${\Delta}'$ and $A''B''C''$ be the triangle formed by the altitudes of $A'B'C'$ as its area ${\Delta}''$. If ${\Delta}'=30, {\Delta}''=20$. Find $\dfrac{\Delta}{9}$?
I am getting $\dfrac{5}{4}$ while the answer is given as $5$.
My process: $\Delta=\dfrac{1}{2}ah_a \implies h_a=\dfrac{2\Delta}{a}$.
Similarly, $h_b=\dfrac{2\Delta}{b}, h_c=\dfrac{2\Delta}{c}$.
$\Delta'=\dfrac{1}{2}h_ah_a' \implies h_a'=\dfrac{2\Delta'}{h_a}$
Similarly, $h_b'=\dfrac{2\Delta'}{h_b}, h_c'=\dfrac{2\Delta'}{h_c}$
$\Delta''=\sqrt{(h_a+h_b+h_c)(h_a+h_b-h_c)(h_a-h_b+h_c)(-h_a+h_b+h_c)}=\dfrac{\Delta'^2}{4\Delta}$
If $a,b,c$ are the lengths of the edges of $ABC$ then $A'B'C'$ has edges $\frac{2\Delta}{a}, \frac{2\Delta}{b},\frac{2\Delta}{c}$.
Therefore $A''B''C''$ has edges $\frac{\Delta'a}{\Delta}, \frac{\Delta'a}{\Delta},\frac{\Delta'a}{\Delta}$ and so $$\Delta''=\left(\frac{\Delta'}{\Delta}\right)^2\times \Delta$$ and so $$\frac{\Delta}{9}=5.$$