Question

In ∆ABC , DE is drawn parallel to BC such that AD = 3.5 cm, DB = 6.5 cm and DC intersects BE at F . Find

(a) ar(∆DEF)/ar(∆BCF)

(b) ar(∆DEF)/ar(∆ECF)

(c) ar(∆DEF)/ar(∆DEB)
​

Answer 1

Answer:

n. 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]

⇒

AB

AD

=

BC

DE

We have,

DB

AD

=

4

5

⇒

AD

DB

=

5

4

⇒

AD

DB

+1=

5

4

+1

⇒

AD

DB+AD

=

5

9

⇒

AD

AB

=

5

9

⇒

AB

AD

=

9

5

∴

BC

DE

=

9

5

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(△CFB)

Area(△DFE)

=

BC

2

DE

2

⇒

Area(△CFB)

Area(△DFE)

=(

9

5

)

2

=

81

25

Answer 2

Step-by-step explanation:

$\small{\mathcal{\purple{A}{\green{N}{\pink{S}{\blue{W}{\purple{E}{\green{R}{\pink{!}}}}}}}}}$