Question

in ∆ ABC PQ || AB , (AP) / PC =2/5 then find BQ / QC =? ​

Answer 1

SOLUTION :
Given :  AP = 2.4 cm, AQ = 2 cm, QC = 3 cm, and BC = 6 cm.
By using Basic proportionality theorem,
AP/PB = AQ/QC
2.4/ PB = 2/3
2PB = 2.4 x 3  
PB = (2.4×3) /2  
PB = 1.2 × 3  
PB = 3.6 cm
Now, AB = AP + PB
AB = 2.4 + 3.6
AB = 6 cm
In Δ APQ and Δ ABC,
∠APQ =∠ABC  (corresponding angles)
[PQ || BC, AB is transversal]
∠AQP =∠ACB  (corresponding angles)
[PQ || BC, AC is transversal]
Δ APQ  ~ Δ ABC (AA similarity)
Therefore,  AP/AB = PQ/BC = AQ/AC
[In similar triangles corresponding sides are proportional]
AP/ AB = PQ/BC
2.4/ 6 = PQ/6
PQ = 2.4 cm

Hence, AB = 6 cm  and PQ = 2.4 cm.

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