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The radius of the circle is 15 cm and the length of one of the chords is 18cm. Find the distance of the chord from the centre.

Answer Verified Verified
Hint: Draw a perpendicular from the center to the chord of the circle and use the property of the circle that says a perpendicular drawn from the centre of the circle to the chord bisects the chord.

Complete step-by-step answer:

Let AB=18cm is the chord of the circle as given in the question
And OB=15cm is the radius of the circle.
Let OM be a perpendicular drawn from the centre of the circle to the chord.
Let OM=x cm
Since the perpendicular drawn on a chord from the centre of the circle bisects the chord (property of a circle)
$ \Rightarrow $ MB=9cm
Thus in the right angled triangle OMB,
Using Pythagoras theorem we get,
$O{B^2} = O{M^2} + M{B^2}$
$ \Rightarrow {(15)^2} = {x^2} + {(9)^2}$
$ \Rightarrow 225 = {x^2} + 81$
$ \Rightarrow {x^2} = 144$
$ \Rightarrow x = 12cm$
Thus the distance of the chord from the center of the circle is 12cm.

Note: It is very important in such questions to realize that we have to draw a perpendicular from the centre to the chord or else the property will not be applicable.


No right angled triangle is formed in such a case and hence is difficult to solve the question.
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