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
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 PQ
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 cm
Now, considering ΔAPQ and ΔABC
∠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.
⇒ 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