In the given diagram ∆ADB and ∆ACB are two right angled triangles with ∠ADB = ∠BCA = 90°. If AB = 10 cm, AD = 6 cm, BC = 2.4 cm and DP = 4.5 cm

(a) In ∆APD and ∆BPC,

⇒ ∠APD = ∠BPC (Vertically opposite angles are equal)

⇒ ∠ADP = ∠BCP (Both equal to 90°)

Hence, proved that ∆APD ∼ ∆BPC.

(b) In ∆ADB,

By pythagoras theorem,

⇒ AB² = AD² + BD²

⇒ 10² = 6² + BD²

⇒ BD² = 100 - 36

⇒ BD² = 64

⇒ BD = $\sqrt{64}$ = 8 cm.

⇒ PB = BD - PD = 8 - 4.5 = 3.5 cm

Hence, BD = 8 cm and PB = 3.5 cm.

(c) In ∆APD,

By pythagoras theorem,

⇒ AP² = AD² + DP²

⇒ AP² = 6² + (4.5)²

⇒ AP² = 36 + 20.25

⇒ AP² = 56.25

⇒ AP = $\sqrt{56.25}$ = 7.5 cm

Hence, length of AP = 7.5 cm.

(d) We know that,

Ratio of area of similar triangles is equal to the square of the corresponding sides.

$$\begin{aligned}\therefore \frac{\text{Area of △APD}}{\text{Area of △BPC}} = \frac{AD^2}{BC^2} \\[1em] = \frac{6^2}{(2.4)^2} \\[1em] = \frac{6 \times 6}{2.4 \times 2.4} \\[1em] = \frac{1 \times 1}{0.4 \times 0.4} \\[1em] = \frac{10 \times 10}{4 \times 4} \\[1em] = \frac{100}{16} \\[1em] = \frac{25}{4} \\[1em] = 25 : 4.\end{aligned}$$

Hence, area ∆APD : area ∆BPC = 25 : 4.