SSC CHSL 15 Nov 2015 Shift 2 Paper

Question : 111 of 200

Marks: +1, -0

In a circle with center at O (0,0) and radius 5cm, AB is a chord of length 8 cm. If OM is perpendicular to AB, then the length of OM is:

Solution:

Given : AB = 8 cm and OB = 5 cm
To find : OM = ?
Solution : The line from the centre of the circle to the chord bisects it at right angle.
⇒

AM = BM = 1/2AB

⇒

BM=8/2 = 4cm

In ∆OBM,
⇒

(OM)^ 2 = (OB) ^2 − (BM) ^2

⇒

(OM)^2 = (5)^ 2 − (4)^ 2

⇒

(OM)^2 = 25 − 16 = 9

⇒

OM = √9 = 3

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