In the given figure, PQ || AB; CQ = 4.8 cm, QB = 3.6 cm and AB = 6.3 cm. Find :

(i) $\dfrac{CP}{PA}$

(ii) PQ

(iii) If AP = x, then the value of AC in terms of x.

(i) Given PQ || AB,

By basic proportionality theorem :

$\therefore \dfrac{CP}{PA} = \dfrac{CQ}{BQ} \\[1em] \Rightarrow \dfrac{CP}{PA} = \dfrac{4.8}{3.6} \\[1em] \Rightarrow \dfrac{CP}{PA} = \dfrac{4}{3}.$

Hence, ratio = 4 : 3.

(ii) In ∆CPQ and ∆CAB,

∠CPQ = ∠CAB [As PQ || AB, corresponding angles are equal.]

∠PCQ = ∠ACB [Common angle]

∴ ∆CPQ ~ ∆CAB [By AA].

From figure,

CB = CQ + QB = 4.8 + 3.6 = 8.4

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

$\therefore \dfrac{PQ}{AB} = \dfrac{CQ}{CB} \\[1em] \Rightarrow \dfrac{PQ}{6.3} = \dfrac{4.8}{8.4} \\[1em] \Rightarrow PQ = 6.3 \times \dfrac{4}{7} = 3.6 \text{ cm}.$

Hence, PQ = 3.6 cm

(iii) As, ∆CPQ ~ ∆CAB.

We have,

$\dfrac{CP}{AC} = \dfrac{CQ}{CB} \\[1em] \dfrac{CP}{AC} = \dfrac{4.8}{8.4} \\[1em] \dfrac{CP}{AC} = \dfrac{4}{7}.$

So, if AC is 7 parts and CP is 4 parts, then PA is 3 parts.

Given, AP = x

or, 3 parts = x

⇒ 1 part = $\dfrac{x}{3}$

⇒ 7 parts = $\dfrac{7x}{3}$.

Hence, AC = $\dfrac{7x}{3}$.