SSC CGL 11 Jun 2019 Shift 1 Paper

Question : 67 of 100

Marks: +1, -0

Chords AB and CD of a circle, when produced, meet at a point P outside the circle. If AB = 6 cm, CD = 3 cm and PD = 5 cm, then PB is equal to :

Solution:

P B \×P A=P D \×P C

\⇒\ x \×(6+x)=5 \×8

\⇒6 x+x^{2}=40

\⇒x^{2}+6 x-40=0

\⇒x^{2}+10 x-4 x-40=0

\⇒x=4 \ {\cm}

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