Question

In the adjoining figure, DE II BC and AD : DB = 5 : 4. find: (i) DE : BC (ii) DO : DC (iii) area of △DOE : area of △DCE (iv) area of △DOE : area of △COB​

Answer 1

Answer:

ANSWER

In △ABC, we have

DE||BC

⇒ ∠ADE=∠ABC and ∠AED=∠ACB [Corresponding angles]

Thus, in triangles ADE and ABC, we have

∠A=∠A [Common]

∠ADE=∠ABC

and, ∠AED=∠ACB

∴ △AED∼△ABC [By AAA similarity]

we have

AD/DB=5/4

⇒ DB/AD= 4/5

⇒DB/AD+1=4/5+1

⇒ DB+AD/AD = 9/5

⇒ AB/AD= 9/5⇒AD/AB=5/9

∴DE/BC=5/9

In △DFE and △CFB, we have

∠1=∠3 [Alternate interior angles]

∠2=∠4 [Vertically opposite angles]

Therefore, by AA-similarity criterion, we have

△DFE∼△CFB

⇒ Area(△DFE)/Area(△CFB)=DE ^2/BC^2

⇒ Area(△DFE)/Area(△CFB)=(5/9)^2=25/81....[Using (i)]