SSC CPO SI and ASI 2012 Paper

Question : 132 of 200

Marks: +1, -0

Chords AB and CD of a circle intersect externally at P. If AB = 6 cm, CD = 3 cm and PC = 5 cm, then the length of PA is

Solution:

(4)

Here, PA

×{PB}=\ {PC} ×{PD},

( {PB}- {AB}) ×{PB}=\ {PC} ×{PC}+ {CD}) ,

( {PB}-6) ×{PB}=5 \× 8

Therefore,

P B^{2}-6 P B-40=0 ,

(P B+4)(P B-10)=0

since PB cannot be negative, Therefore, PB = 10

\ {cm} .

So, PA

=10-6=4 {cm}

Go to Question:

Prev Question

Next Question