In the above image, what will be the ratio of areas of triangle $A$ and $B$?
From Googling, I've found that:
$\operatorname{Ar}(A) = \dfrac{a^2h}{2(a+b)}$
and
$\operatorname{Ar}(B) = \dfrac{b^2h}{2(a+b)}$
but how do I get these formulas from the classic formula of $\dfrac{\rm base \times height}2$?!
Basically, how were the formulas in this image:
figured out?
Well, what is the height of the triangle $\Delta a$? It is similar to the triangle $\Delta b$ (why?), so and the heights $h_a$ of $\Delta a$ and $h_b$ of $\Delta b$ are related in the following way:
First, we have $h = h_a + h_b$. This is obvious if you draw them on your figure. Secondly, from similarity, we have that $$ \frac{h_b}{h_a} = \frac{b}{a}. $$ By manipulating the last equation, and with a little help from the first to rid ourselves of $h_b$, we get $$ \frac{h-h_a}{h_a} = \frac{b}{a} \\\\ \frac{h}{h_a} - 1 = \frac{b}{a} \\\\ \frac{h}{h_a} = 1 + \frac{b}{a} \\\\ h_a = \frac{h}{1 + \frac{b}{a}} = \frac{ha}{a + b} $$ This means that the area of $\Delta a$ is equal to $$ \frac{1}{2}\cdot a \cdot h_a = \frac{1}{2}\cdot a \cdot \frac{ha}{a + b} = \frac{a^2h}{2(a+b)} $$