Question

In the given figure, DE || BC and DE: BC = 3:5.
Calculate the ratio of the areas of AADE and
the trapezium BCED.​

Answer 1

Given :–

• DE || BC.
• DE : BC = 3 : 5.

To Find :–

• The ratio of the areas of ∆ADE and the trapezium BCED.

Solution :–

Given that,

DE || BC

Therefore,

→ $\angle ADE = \angle ABC$

It means,

→ $\angle AED = \angle ACB$

By AA similarity theorem,

• ∆ADE∼∆ABC

Therefore,

→ $\dfrac{ar( \triangle ABC)}{ar ( \triangle ADE)} = \dfrac{ {BC}^{2} }{ {DE}^{2} }$

Given,

• DE : BC = 3 : 5

Now, put the given values,

→ $\dfrac{ar( \triangle ABC)}{ar ( \triangle ADE)} = \dfrac{ {(5)}^{2} }{ {(3)}^{2} }$

Now, subtracting 1 both sides (L.H.S. and R.H.S.),

→ $\dfrac{ar( \triangle ABC)}{ar ( \triangle ADE)} - 1 = \dfrac{ {(5)}^{2} }{ {(3)}^{2} } - 1$

→ $\dfrac{ar( \triangle ABC)}{ar ( \triangle ADE)} - \dfrac{1}{1} = \dfrac{25}{9} - \dfrac{1}{1}$

→ $\dfrac{ar( \triangle ABC) -ar( \triangle ADE)}{ar ( \triangle ADE)} = \dfrac{25 - 9 }{ 9}$

→ $\dfrac{ar(BCED)}{ar ( \triangle ADE)} = \dfrac{16 }{ 9}$

OR,

→ $\dfrac{ar ( \triangle ADE)}{ar(BCED)} = \dfrac{9 }{16}$

Hence,

The ratio of the areas of ∆ADE and the trapezium BCED is 9 : 16.