SSC CPO SI and ASI 2015 Paper

Question : 139 of 199

Marks: +1, -0

In the given circle, the chord PQ is of length 18 cm. OB is the perpendicular bisector of PQ at M. If MB = 3 cm, then the length of AB is

Solution:

(2)
In ΔOPM,

\ {OP}^{2}=9^{2}+(\ {OP}-3)^{2}

⇒ 6 {OP}=90 \⇒{OP}=15 \ {cm}

Now,

\ {AB}

is the diameter, therefore

\ {AB}=2 \×{OP}=30 \ {cm}

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