In the adjoining figure, PQ || CA and all lengths are given in centimeters. The length of BC is

1. 6.4 cm

2. 7.4 cm

3. 8 cm

4. 9 cm

Given PQ || CA.

Considering △ABC and △PBQ,

∠ B = ∠ B (Common angles)
∠ CAB = ∠ QPB (Corresponding angles are equal)

Hence, by AA axiom △ABC ~ △PBQ.

Let length of QC be x cm.

Since triangles are similar,

$\therefore \dfrac{BQ}{BC} = \dfrac{BP}{BA} \\[1em] \Rightarrow \dfrac{BQ}{BQ + QC} = \dfrac{BP}{BP + PA} \\[1em] \Rightarrow \dfrac{5}{5 + x} = \dfrac{4}{4 + 2.4} \\[1em] \Rightarrow \dfrac{5 \times 6.4}{4} = x + 5 \\[1em] \Rightarrow 8 = x + 5 \\[1em] \Rightarrow x = 3.$

BC = BQ + QC = 5 + x = 5 + 3 = 8 cm.

Hence, Option 3 is the correct option.