Question

# 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.

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.

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.