SSC CGL Tier II 25 Oct 2015 Paper 1

Question : 37 of 100

Marks: +1, -0

ABCD is a cyclic quadrilateral. AB and DC when produced meet at P, if PA = 8 cm, PB = 6 cm, PC = 4 cm, then the length (in cm) of PD is

Solution:

(4)

Clearly,
AP × BP = PD × PC
⇒ 8 × 6 = PD × 4
⇒ PD =

{8×6}/4=12\cm

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