Question taken from RD sharma

Class 10th

Chapter no. 4

Chapter name:- Triangles

Exercise :- 4.2

This is very basic and important questions.

In this question we have In a ΔABC, D and E are points on AB and AC respectively,

Also it is given that DE ∥ BC.

AD = 2.4 cm, AE = 3.2 cm, DE = 2 cm and BC = 5 cm.

Now we have to Find BD and CE.

Understanding and learning CBSE maths

RD sharma, DHANPAT RAI publication

Given:In Δ ABC,

Length of side AD = 2.4 cm, AE = 3.2 cm, DE = 2 cm and BE = 5 cm

Also, DE ∥ BC

To find: Length of side BD and CE.As DE ∥ BC, AB is transversal,

∠APQ = ∠ABC (corresponding angles) – equation 1

As DE ∥ BC, AC is transversal,

∠AED = ∠ACB (corresponding angles) – equation 2

In Δ ADE and Δ ABC,

Now from equation 1 and 2 we get,

∠ADE = ∠ABC

∠AED = ∠ACB

∴ ΔADE = ΔABC (By AA similarity criteria)

Now, we know that

Corresponding parts of similar triangles are proportional.

Therefore,

⇒ AD/AB = AE/AC = DE/BC

AD/AB = DE/BC

2.4/ (2.4 + DB) = 2/5 [Since, length of side AB = AD + DB]

2.4 + DB = 6

DB = 6 – 2.4

DB = 3.6 cmLength of side DB is 3.6 cm

In the same way, we get

⇒ AE/AC = DE/BC

3.2/ (3.2 + EC) = 2/5 [Since AC = AE + EC]

3.2 + EC = 8

EC = 8 – 3.2

EC = 4.8 cm∴ BD = 3.6 cm and CE = 4.8 cm.

Length of side BD is 3.6 cm and CE is 4.8 cm