Solution:-

From the question it is give that,

Consider the ∆RLQ and ∆PLN,

∠RLQ = ∠NLP [vertically opposite angles are equal]

∠RQL = ∠LNP [alternate angle are equal]

Therefore, ∆RLQ ~ ∆PLN

So, QR/PN = RL/LP = 2/3

QR/PN = 2/3

10/PN = 2/3

PN = (10 × 3)/2

PN = 30/2

PN = 15 cm

Therefore, PN = 15 cm

(ii) Name a triangle similar to triangle RLM. Evaluate RM.

Solution:-

From the figure,

Consider ∆RLM and ∆QLP

Then, ∠RLM = ∠QLP [vertically opposite angles are equal]

∠LRM = ∠LPQ [alternate angles are equal]

Therefore, ∆RLM ~ ∆QLP

Then, RM/PQ = RL/LP = 2/3

So, RM/16 = 2/3

RM = (16 × 2)/3

RM = 32/3

RM =