In the given figure, P is a point on AB such that PB : AP = 3 : 4 and PQ || AC.

(i) Calculate PQ : AC.

(ii) If AR ⊥ CP, QS ⊥ CB and QS = 6 cm, calculate the length of AR.

(i) Given, AP : PB = 4 : 3.

Since, PQ || AC. By using Basic Proportionality Theorem,

$\Rightarrow \dfrac{AP}{PB} = \dfrac{CQ}{QB} \\[1em] \Rightarrow \dfrac{CQ}{QB} = \dfrac{4}{3} \\[1em] \Rightarrow \dfrac{BC - BQ}{BQ} = \dfrac{4}{3} \\[1em] \Rightarrow 3(BC - BQ) = 4BQ \\[1em] \Rightarrow 3BC - 3BQ = 4BQ \\[1em] \Rightarrow 3BC = 7BQ \\[1em] \Rightarrow \dfrac{BQ}{BC} = \dfrac{3}{7} \qquad \text{[….Eq 1]}$

Considering △PBQ and △ABC,

∠QPB = ∠CAB (Corresponding angles are equal)

∠PQB = ∠ACB (Corresponding angles are equal)

Hence by AA axiom △PBQ ~ △ABC. Since triangles are similar so the ratio of the corresponding sides are equal,

$\Rightarrow \dfrac{PQ}{AC} = \dfrac{BQ}{BC} \\[1em] \Rightarrow \dfrac{PQ}{AC} = \dfrac{3}{7} \qquad \text{[From Eq 1]}.$

Hence, PQ : AC = 3 : 7.

(ii) Considering △ARC and △QSP,

∠ARC = ∠QSP (Both are equal to 90°)

∠ACR = ∠SPQ (Alternate angles are equal)

Hence by AA axiom △ARC ~ △QSP. Since triangles are similar so the ratio of the corresponding sides are equal,

$\Rightarrow \dfrac{AR}{QS} = \dfrac{AC}{PQ} \\[1em] \Rightarrow AR = \dfrac{AC}{PQ} \times QS \\[1em]$

We calculated PQ : AC = 3 : 7 above.

$\therefore \dfrac{AC}{PQ} = \dfrac{7}{3}$

Putting this value of $\dfrac{AC}{PQ}$ we get,

$\Rightarrow AR = \dfrac{7}{3} \times 6 \\[1em] \Rightarrow AR = 7 \times 2 \\[1em] \Rightarrow AR = 14.$

Hence, length of AR = 14 cm.