In a ∆ABC, 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 the AB and PQ.
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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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=> PQ = 1.2 cm
=> AP/AB = PQ/BC
=> 2.4/AB = 1.2/6
=> AB = 2×6 = 12 cm
____________[ANSWER]
____◦•●◉✿[ सधन्यवाद]✿◉●•◦____
=> AP/AB = PQ/BC
=> 2.4/AB = 1.2/6
=> AB = 2×6 = 12 cm
____________[ANSWER]
____◦•●◉✿[ सधन्यवाद]✿◉●•◦____
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