SSC CGL 12 Jun 2019 Shift 1 Paper

Question : 60 of 100

Marks: +1, -0

Chords AB and CD of a circle intersect at a point P inside the circle. If AB = 10 cm, AP = 4 cm and PC = 5 cm, then CD is equal to:

Solution:

\AP \×\ {\BP}=\ {\CP} \×\ {\DP}

4 \×6=5 \×\ {\DP}

\ {\DP}=\ {24}/{5}=4.8

\ {\CD}=\ {\CP}+\ {\PD}

=5+4.8=9.8 \ {\cm}

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