Question

in the given fig ,DE parallel BC. if DE:BC=3:5,Find ar(∆ADE)/ar(Trap.BCED)

Answer 1

it is given that DE || BC
so angle ADE = angle ABC [ corresponding angles]
angle AED = angle ACB [ corresponding angles]

see the rest in inserted image

Answer 2

$\dfrac{ΔADE}{ΔBCED}=[tex]\dfrac{9x}{16x}$

Step-by-step explanation:

The missing figure is:

In ΔADE ≅ ΔADE

∴ $\dfrac{ΔADE}{ΔADE}=\dfrac{DE^2}{BC^2}$

$=(\dfrac{DE}{BC})^2=(\dfrac{3}{5})^2=\dfrac{9}{25}$

Let ar(ΔADE) = 9x and

ar(ΔADE) = 25x  

∴ ar(Trap.BCED) = ar(ΔADE) - ar(ΔADE)

= 25x - 9x = 16x

∴ \dfrac{ΔADE}{ΔBCED}=$\dfrac{9x}{16x}$

Hence,  $\dfrac{ΔADE}{ΔBCED}=[tex]\dfrac{9x}{16x}$