AB and CD are two chords of a circle intersecting at point P outside the circle. If

(i) PA = 8 cm, PC = 5 cm and PD = 4 cm, determine AB.

(ii) PC = 30 cm, CD = 14 cm and PA = 24 cm, determine AB.

(i) We know that,

If two chords of a circle intersect externally, then the product of the lengths of the segments are equal.

⇒ AP × PB = CP × PD

⇒ 8 × PB = 5 × 4

⇒ PB = $\dfrac{20}{8}$ = 2.5

From figure,

AB = PA - PB = 8 - 2.5 = 5.5 cm.

Hence, AB = 5.5 cm.

(ii) From figure,

⇒ PC = CD + PD

⇒ 30 = 14 + PD

⇒ PD = 16 cm.

We know that,

If two chords of a circle intersect externally, then the product of the lengths of the segments are equal.

⇒ AP × PB = PC × PD

⇒ 24 × PB = 30 × 16

⇒ PB = $\dfrac{30 \times 16}{24}$ = 20

From figure,

AB = PA - PB = 24 - 20 = 4 cm.

Hence, AB = 4 cm.