Plz solve q no. 6

Plz solve q no. 6 nvno

Dear Student,
Here we have triangle as shown:

(i) Now we know that line parallel to base divides the other two sides in the same ratio. So PM:MR=PL:LQ=3:4

So, $\frac{P L}{P Q} = \frac{P L}{P L + L Q} = \frac{3}{3 + 4} = \frac{3}{7}$

Now triangles PLM and PQR are similar, so we get:

$\frac{P L}{P Q} = \frac{L M}{Q R} \Rightarrow \frac{L M}{Q R} = \frac{3}{7}$
(ii) Now,

LM is parallel to QR

So ∠PLM = ∠PQR and ∠PML = ∠PRQ ( corresponding angles)

Also ∠LPM is common so ΔPLM ∼ ΔPQR

we know that ratio of areas of aimilar triangles is equal to ratio of squares of their sides

⇒ $\frac{a r e a P L M}{a r e a P Q R} = \frac{{P M}^{2}}{{M R}^{2}} = \frac{3^{2}}{4^{2}} = \frac{9}{16}$

Regards.