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

Calculate the value of ratio:

(i) $\dfrac{PL}{PQ} \text{ and then } \dfrac{LM}{QR}$

(ii) $\dfrac{\text{ Area of Δ LMN }}{\text{ Area of Δ MNR }}$

(iii) $\dfrac{\text{ Area of Δ LQM }}{\text{ Area of Δ LQN }}$

(i) Given,

$\dfrac{PM}{MR} = \dfrac{3}{4}$

Let PM = 3x and MR = 4x.

From figure,

PR = PM + MR = 3x + 4x = 7x.

$\dfrac{PM}{PR} = \dfrac{3x}{7x} = \dfrac{3}{7}$.

In ΔPLM and ΔPQR,

As LM || QR, corresponding angles are equal.

∠PLM = ∠PQR

∠PML = ∠PRQ

∴ ∆PLM ~ ∆PQR [By AA]

Since, corresponding sides of similar triangles are proportional to each other we have :

$\therefore \dfrac{PL}{PQ} = \dfrac{PM}{PR} = \dfrac{LM}{QR} \\[1em] \therefore \dfrac{PL}{PQ} = \dfrac{LM}{QR} = \dfrac{3}{7}.$

Hence, PL : PQ = 3 : 7 and LM : QR = 3 : 7.

(ii) As ΔLMN and ΔMNR have common vertex at M and their bases LN and NR are along the same straight line.

$\therefore \dfrac{\text{Area of ΔLMN}}{\text{Area of ΔMNR}} = \dfrac{\text{LN}}{\text{NR}}$ …..(1)

Now, in ΔLMN and ΔRNQ we have,

⇒ ∠NLM = ∠NRQ [Alternate angles are equal]

⇒ ∠LMN = ∠NQR [Alternate angles are equal]

∴ ∆LNM ~ ∆RNQ [By AA]

Since corresponding sides of similar triangle are proportional to each other, we have :

$\dfrac{MN}{QN} = \dfrac{LN}{NR} = \dfrac{LM}{QR} = \dfrac{3}{7}$

Substituting value in (1) we get :

$\dfrac{\text{Area of ΔLMN}}{\text{Area of ΔMNR}} = \dfrac{3}{7}$.

(iii) From part (ii) we get :

$\dfrac{MN}{QN} = \dfrac{3}{7}$

Let MN = 3a and QN = 7a

From figure,

MQ = MN + QN = 3a + 7a = 10a.

As ΔLQM and ΔLQN have common vertex at L and their bases QM and QN are along the same straight line.

$\dfrac{\text{ Area of ΔLQM }}{\text{ Area of ΔLQN }} = \dfrac{QM}{QN} = \dfrac{10a}{7a} = \dfrac{10}{7}$.

Hence, $\dfrac{\text{ Area of ΔLQM }}{\text{ Area of ΔLQN }} =\dfrac{10}{7}$