Question

15. D and Eare points on the sides AB and
AC respectively of AABC such that DE
is parallel to BC and AD : DB = 4: 5,
CD and BE intersect each other at F.
Then, the ratio of the areas of ADEF
and ACBF is
(a) 16 : 25
(b) 16 : 81
(C) 81 : 16
(d) 4:9​

Answer 1

Given, D and Eare points on the sides AB and AC respectively of ∆ABC (as shown in figure). CD and BE intersect each other at point F.

Also, DE || BC and AD:DB = 4:5

We have to find the ratio of the areas of ∆DEF and ∆CBF.

In ∆DEF and ∆CBF

∠DFE = ∠BFC (vertically opposite angles)

∠FED = ∠FBC (alternate interior angles)

By AA property

∆DEF ~ ∆CBF

The ratios of area of two similar triangles is equal to square of their corresponding sides.

So, (Ar. ∆DEF)²/(Ar. CBF) = (DE)²/(BC)²

Now, In ∆ABC and ∆ADE

∠ABC = ∠ADE (corresponding angles)

∠ACB = ∠AED (corresponding angles)

By AA property

∆ABC ~ ∆ADE

Corresponding sides of two similar triangles are in proportional.

So, (Ar. ∆ADE)/(Ar. ∆ABC) = AD/AB = DE/BC

Also, AD/DB = 4/5

→ DB/AD + 1 = 5/4 + 1

→ (DB + AD)/AD = 9/4

→ AB/AD = 9/4

→ AD/AB = 4/9

Now,

→ 4/9 = DE/BC

Also, (Ar. ∆DEF)²/(Ar. CBF) = (DE)²/(BC)²

→ (Ar. ∆DEF)²/(Ar. CBF) = (4)²/(9)²

→ (Ar. ∆DEF)²/(Ar. CBF) = 16/81

Option b) 16:81

Answer 2

$\huge\mathfrak\green{Answer:-}$

Given:

It is mentioned in the question that D and E are points on sides AB and AC respectively of such that DE || BC and the ratio of AD : DB is 4:5.

CD and BE intersect each other at F.

To Find:

Ratio of areas of ADEF and ACBF.

Solution:

In ∆DEF and ∆CBF

∠FED = ∠FBC [Alternate interior angles]

Also,

∠DFE = ∠BFC [Vertically opposite angles]

Therefore, ∆DEF~∆CBF by AA Property.

We knoe that the ratio of areas of two similar triangles is equal to the square of their corresponding sides. Hence,

(Area of ∆DEF²)/(Area of ∆CBF)

= (DE)²(BC)²

In ∆ABC and ∆ADE,

∠ABC = ∠ADE [Corresponding angles]

Also,

∠ACB = ∠AED [Corresponding angles]

Therefore, ∆ABC~∆ADE

So,

(Area of ∆ADE) / (Area of ∆ABC)

= AD/AB = DE/BC

AD/DB = 4/5 [Given]

DB/AD +1 = 5/4 + 1

(DB + AD)/AD = 9/4

AB/AD = 9/4

AD/AB = 5/4

we know,

4/9 = DE/BC

Also,

(Area of ∆DEF²)/(Area of ∆CBF)

= (DE)² / (BC)²

(Area of ∆DEF²)/(Area of ∆CBF)

= (4/9)² = 16/81

Therefore,the ratio of the areas of ADEF

and ACBF is 16 : 81