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Question
In a ΔABC, P and Q are points on sides AB and AC respectively, such that PQ || BC. If AP = 2.4 cm, AQ = 2 cm, QC = 3 cm and BC = 6 cm, find AB and PQ.
Solution
We have || BC
Therefore, by BPT
We have,
`"AP"/"PB"="AQ"/"QC"`
`2.4/"PB"=2/3`
`rArr"PB"=(3xx2.4)/2=(3xx24)/2=(3xx6)/5=18/5`
⇒ PB = 3.6 cm
Now, AB = AP + PB
= 2.4 + 3.6 = 6cm
Now, In ΔAPQ and ΔABC
∠A = ∠A [common]
∠APQ = ∠ABC [∵ PQ || BC ⇒ Corresponding angles are equal]
⇒ ΔAPQ ~ ΔABC [By AA criteria]
`rArr"AB"/"AP"="BC"/"PQ"` [corresponding sides of similar triangles are proportional]
`rArr"PQ"=(6xx2.4)/6`
⇒ PQ = 2.4 cm
Hence, AB = 6 cm and PO = 2.4 cm
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