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 cm
Length 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