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Question
In the given triangle PQR, LM is parallel to QR and PM : MR = 3 : 4.
Calculate the value of ratio:
- `(PL)/(PQ)` and then `(LM)/(QR)`
- `"Area of ΔLMN"/"Area of ΔMNR"`
- `"Area of ΔLQM"/"Area of ΔLQN"`
Solution
i. In ∆PLM and ∆PQR,
As LM || QR, Corresponding angles are equal
∠PLM = ∠PQR
∠PML = ∠PRQ
∆PLM ~ ∆PQR
`=> 3/7 = (LM)/(QR)` ...`(∵ (PM)/(MR) = 3/4 => (PM)/(PR) = 3/7)`
Also, by using basic proportionality theorem, we have:
`(PL)/(LQ) = (PM)/(MR) = 3/4`
`=> (LQ)/(PL) = 4/3`
`=> 1 + (LQ)/(PL) = 1 + 4/3`
`=> (PL + LQ)/(PL) = (3 + 4)/3`
`=> (PQ)/(PL) = 7/3`
`=> (PL)/(PQ) = 3/7`
ii. Since ∆LMN and ∆MNR have common vertex at M and their bases LN and NR are along the same straight line
∴ `"Area of ΔLMN"/"Area of ΔMNR" = (LN)/(NR)`
Now, in ∆LNM and ∆RNQ
∠NLM = ∠NRQ ...(Alternate angles)
∠LMN = ∠NQR ...(Alternate angles)
∆LMN ~ ∆RNQ ...(AA Similarity)
∴ `(MN)/(QN) = (LN)/(NR) = (LM)/(QR) = 3/7`
∴ `"Area of ΔLMN"/"Area of ΔMNR" = (LN)/(NR) = 3/7`
iii. Since ΔLQM and ΔLQN have common vertex at L and their bases QM and QN are along the same straight line
`"Area of ΔLQM"/ "Area of ΔLQN" = (QM)/(QN) = 10/7`
`(∴ (NM)/(QN) = 3/7 => (QM)/(QN) = 10/7)`
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