AB and CD are two chords of a circle which intersect each other externally at P

(i) If AB = 4 cm BP = 5 cm and PD = 3 cm, then find CD.

(ii) If BP = 3 cm, CP = 6 cm and CD = 2 cm, then find AB.

If two chords of a circle intersect each other internally or externally, then area of rectangle contained by the segment of one chord is equal to the area of rectangle contained by the segment of other chord.

Using above theorem,

PA × PB = PC × PD …(1)

(i) Given: AB = 4 cm

BP = 5 cm

PD = 3 cm

∵ AP = AB + BP

⇒ AP = 4 cm + 5 cm

⇒ AP = 9 cm

Putting the values in (1),

9 cm × 5 cm = PC × 3 cm

⇒ PC = 15 cm

∵ PC = CD + DP

⇒ 15 cm = CD + 3 cm

⇒ CD = 15 cm – 3 cm

⇒ CD = 12 cm

Hence, CD = 12 cm

(ii) Given: BP = 3 cm

CP = 6 cm

CD = 2 cm

∵ CP = CD + DP

⇒ DP = CP – CD

⇒ DP = 6 cm – 2 cm

⇒ DP = 4 cm

Putting the values in (1),

PA × 3 cm = PC × PD

⇒ PA × 3 cm = 6 cm × 4 cm

⇒ PA = 8 cm

∵ AP = AB + BP

⇒ 8 cm = AB + 3 cm

⇒ AB = 8 cm – 3 cm

⇒ AB = 5 cm

Hence, AB = 5 cm