Question drawn from very renowned book RD sharma of class 10th, Chapter no. 4

Chapter name:- Triangles

Exercise :- 4.2

This is very important question of triangle.

In this question we have been given that In a ΔABC, P and Q are the points on sides AB and AC,

Also it is given PQ ∥ BC.

And if AP = 2.4 cm, AQ = 2 cm, QC = 3 cm and BC = 6 cm.

Now we have to Find AB and PQ.

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RD sharma, DHANPAT RAI publication

Class 10th, triangle

Given:In ΔABC,

Length of side AP = 2.4 cm, AQ = 2 cm, QC = 3 cm, and BC = 6 cm.

Also, PQ ∥ BC.

To find: Length of side AB and PQNow,

Since it’s given that PQ ∥ BC

So by using Thales Theorem, we get

AP/PB = AQ/ QC

2.4/PB = 2/3

2 x PB = 2.4 × 3

PB = (2.4 × 3)/2 cm

⇒ PB = 3.6 cm

Therefore, Length of PB is 3.6 cm

Now finding, AB = AP + PB

AB = 2.4 + 3.6

⇒ AB = 6 cm

Therefore, Length of AB is 6 cmNow, considering ΔAPQ and ΔABC

We have,

∠A = ∠A (Common angle)

∠APQ = ∠ABC (Corresponding angles are equal and PQ||BC and AB being a transversal)

Thus, ΔAPQ and ΔABC are similar to each other by AA criteria.

Now, we know that corresponding parts of similar triangles are propositional.

Therefore,

⇒ AP/AB = PQ/ BC

⇒ PQ = (AP/AB) x BC

= (2.4/6) x 6 = 2.4

∴ PQ = 2.4 cm.Therefore, Length of PQ is 2.4 cm and AB is 6cm