RRB Junior Engineer Stage 1 Model Papers 7

Question : 8 of 100

Marks: +1, -0

In the following figure, if OA = 10 and AC = 16, then OB must be

b^2/{ac}

Solution:

In OAB,

OA^2=OB^2+AB^2

[ ∴ AB =

1/2

AC, because line drawn from centre to achord bisect & perpendicular to it]

(10)^2=(OB)^2+(8)^2

100-64=OB^2

OB^2=36

OB=6

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