Question

In fig
seg BE ⊥ seg AB and
seg BA ⊥ seg AD.

If BE = 6 and AD =9 find

$\frac{A(∆ABE)}{A(∆BAD)}$

Answer 1

We have:
BE = 6 and AD = 9
Also,
Height of ΔABE = AB = Height of ΔBAD
We know that the ratio of the areas of two triangles with the same height is equal to the ratio of their corresponding bases.
Thus, we get:

$\frac{\mathrm{A}(∆\mathrm{ABE})}{\mathrm{A}(∆\mathrm{BAD})}=\frac{\mathrm{BE}}{\mathrm{AD}}=\frac{6}{9}=\frac{2}{3}$