Question

In the given triangle PQR, LM is parallel to QR and PM : MR = 3 : 4.

Calculate the value of ratio:

1. `(PL)/(PQ)` and then `(LM)/(QR)`
2. `"Area of ΔLMN"/"Area of ΔMNR"`
3. `"Area of ΔLQM"/"Area of ΔLQN"`

Answer 1

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)`