Question

In the given figure, DE║BC and DE: BC = 3:5. Calculate the ratio of the areas of ΔADE and the trapezium BCED. 

 

Answer 1

It is given that DE || BC.
∴ ∠𝐴𝐷𝐸= ∠𝐴𝐵𝐶 (𝐶𝑜𝑟𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑖𝑛𝑔 𝑎𝑛𝑔𝑙𝑒𝑠)
∠𝐴𝐸𝐷= ∠𝐴𝐶𝐵 (𝐶𝑜𝑟𝑟𝑒𝑠𝑝𝑜𝑛𝑑𝑖𝑛𝑔 𝑎𝑛𝑔𝑙𝑒𝑠)
Applying AA similarity theorem, we can conclude that Δ ADE ~ ΔABC.  

∴`( ar(ΔABC))/(ar(ΔADE))=(BC)^2/(DE)^2` 

Subtracting 1 from both sides, we get: 

`(ar(ΔABC))/(ar(ΔADE))-1=5^2/3^2-1` 

⇒`( ar(ΔABC)-ar(ΔADE))/(ar(ΔADE))=(25-9)/9` 

⇒ `(ar(BCED))/(ar(ΔADE))=16/9`  

Or, `(ar(ΔADE))/(ar(BCED))=9/16`