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(∆DEF)/ar(∆DEB)
Answers
Step-by-step explanation:
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
.....[Usin
Answer:
here it is... plz anybody tell it's correct answer
