RRB NTPC Mains or Stage 2 Model Paper 7 with solutions

Question : 16 of 120

Marks: +1, -0

In the figure given above, AD is a straight line, OP perpendicular to AD and O is the centre of both circles. If OA = 20 cm, OB = 15 cm and OP = 12 cm, then what is AB equal to ?

7 cm
8 cm
10 cm
12 cm

Validate

Solution:

(a) Given O(a) = 20 cm O(b) = 15 cm and OP = 12 cm

(a)P

= √{AO^2 -OP^2 }

= √{20^2 -12^2}

√{400 -144}

= √256 = 16 cm

(b)P

= √{15^2 -12^2 }

= √{225 -144}

√81 = 9 cm

∴(a)(b) = (a)P – (b)P = 16 – 9 = 7 cm

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