Question

In a ∆ABC, point D is on side AB and point E is on side AC, such that BCED is a trapezium. If DE : BC = 3 : 5, then Area (∆ ADE) : Area (◻BCED) =

पर्याय

• 3 : 4

•  9 : 16

• 3 : 5

• 9 : 25

Answer 1

Given: In ΔABC, D is on side AB and point E is on side AC, such that BCED is a trapezium. DE: BC = 3:5.

To find: Calculate the ratio of the areas of ΔADE and the trapezium BCED.

In ΔADE and ΔABC,

\[\angle ADE = \angle B \left( \text{Corresponding angles} \right)\]
\[\angle A = \angle A \left( \text{Common} \right)\]
\[ \therefore ∆ ADE~ ∆ ABC \left( \text{AA Similarity} \right)\]

We know that

`(Ar(Δ ADE))/(Ar(ΔABC))=(DE)^2/(BC)^2`

`(Ar(Δ ADE))/(Ar(ΔABC))= 3^2/5^2`

`(Ar(Δ ADE))/(Ar(ΔABC))= 9/25`

Let Area of ΔADE = 9x sq. units and Area of ΔABC = 25x sq. units

`Ar [trap BCED]=Ar(Δ ABC) - Ar (Δ ADE)`

`= 25x -9x`

`16x`sq units

`Now ,`

`(Ar(Δ ADE))/(Ar(trapBCED))= (9x)/(16x)`

`(Ar(Δ ADE))/(Ar(trapBCED))= 9/16`

Hence the correct answer is `b`.