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) Prove that △APD ∼ △BPC.
(b) Find the lengths of BD and PB.
(c) Hence, find the length of PA.
(d) Find area of △APD : area of △BPC.
(a) In △APD and △BPC:
∠APD = ∠BPC because they are vertically opposite angles.
∠ADP = ∠BCP = 90°.
Therefore, △APD ∼ △BPC by the AA criterion.
(b) In right-angled △ADB, by the Pythagoras theorem:
AB² = AD² + BD²
⇒ 10² = 6² + BD²
⇒ 100 = 36 + BD²
⇒ BD² = 64
⇒ BD = √64
⇒ BD = 8 cm
Since BD = BP + PD:
PB = BD − PD
⇒ PB = 8 − 4.5
⇒ PB = 3.5 cm
(c) In right-angled △APD:
AP² = AD² + DP²
⇒ AP² = 6² + 4.5²
⇒ AP² = 36 + 20.25
⇒ AP² = 56.25
⇒ AP = √56.25
⇒ AP = 7.5 cm
(d) The ratio of the areas of similar triangles equals the square of the ratio of corresponding sides.
\frac{\text{Area of }\triangle APD}{\text{Area of }\triangle BPC}=\frac{AD^2}{BC^2}⇒ Area of △APD : Area of △BPC = 6² : 2.4²
⇒ 36 : 5.76
Multiplying both terms by 100:
⇒ 3600 : 576
Dividing by 144:
⇒ 25 : 4
Therefore, area of △APD : area of △BPC = 25 : 4.
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