Question

In the given figure, DE || BC and DE : BC = 3 : 5. Calculate the ratio of the areas of $\Delta ADE$ and the trapezium BCED.

Answer 1

Let DE = 3x and BC = 5x

In ΔADE and ΔABC, we have

∠ADE = ∠ABC (corresponding angles)

∠AED = ∠ACB (corresponding angles)

ΔADE ~ ΔABC (by AA similarity)

So, $\frac{ar\left(\Delta ADE\right)}{ar\left(\Delta ABC\right)}=\frac{D{E}^2}{B{C}^2}={\left(\frac{3x}{5x}\right)}^2=\frac{9}{25}$

Let, ar(ΔADE) = $9x^2$ units

Then, ar(ΔABC) = $25x^2$ units

Therefore, ar(trapezium BCED)=ar(ΔABC)−ar(ΔADE)= $25x^2-9x^2=16x^2$ units

Therefore, $\frac{ar\left(\Delta ADE\right)}{ar(trap.BCED)}=\frac{9x^2}{16x^2}=\frac{9}{16}$