In the given figure, if AC = 3 cm and CB = 6 cm, find the length of CR.

In △ABQ and △ACK,

CK || BQ

∴ ∠ACK = ∠ABQ [Corresponding angles are equal]

∠KAC = ∠QAB (Common)

∴ △ACK ~ △ABQ

From figure,

AB = AC + CB = 3 + 6 = 9 cm.

In similar triangles,

The ratios between the lengths of corresponding sides are equal.

$\therefore \dfrac{AC}{AB} = \dfrac{AK}{AQ} = \dfrac{CK}{BQ} \\[1em] \Rightarrow \dfrac{3}{9} = \dfrac{AK}{AQ} = \dfrac{CK}{7.5} \\[1em] \Rightarrow \dfrac{AK}{AQ} = \dfrac{3}{9} \\[1em] \Rightarrow \dfrac{AK}{AQ} = \dfrac{1}{3} \space ……….(1)$

Also,

$\Rightarrow \dfrac{CK}{7.5} = \dfrac{1}{3} \\[1em] \Rightarrow CK = \dfrac{7.5}{3} = 2.5$

In △KRQ and △APQ,

AP || KR

∴ ∠KRQ = ∠APQ [Corresponding angles are equal]

∠KQR = ∠AQP (Common)

∴ △KRQ ~ △APQ

In similar triangles,

The ratios between the lengths of corresponding sides are equal.

$\therefore \dfrac{KR}{AP} = \dfrac{KQ}{AQ} \\[1em] \Rightarrow \dfrac{KR}{4.5} = \dfrac{AQ - AK}{AQ} \\[1em] \Rightarrow \dfrac{KR}{4.5} = \dfrac{AQ}{AQ} - \dfrac{AK}{AQ} \\[1em] \Rightarrow \dfrac{KR}{4.5} = 1 - \dfrac{1}{3} \text{ (From 1)} \\[1em] \Rightarrow \dfrac{KR}{4.5} = \dfrac{2}{3} \\[1em] \Rightarrow KR = 4.5 \times \dfrac{2}{3} \\[1em] \Rightarrow KR = 3 \text{ cm}.$

From figure,

CR = CK + KR = 2.5 + 3 = 5.5 cm

Hence, CR = 5.5 cm