SSC CHSL Model Paper 17

Question : 52 of 100

Marks: +1, -0

Two chords AB and CD of a circle when produced, meet at a point P outside the circle. If AB = 6 cm, PB = 5 cm, PD = 4 cm, then CD is equal to :

Solution:

AB = 6 cm, PB = 5 cm, PD = 4 cm
AP × PB = CP × PD
⇒ (6 + 5) × 5 = CP × 4

⇒\CP={11×5}/4=13.75

CD = CP - PD = 13.75 - 4 = 9.75 cm

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