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A line segment whose end points lie on a circle is called a chord.

In the adjoining figure, each of the line segments PQ, RS, AB and CD is a chord of the circle with centre O. Clearly, an infinite number of chords may be drawn in a circle.

In the adjoining figure, each of the line segments PQ, RS, AB and CD is a chord of the circle with centre O. Clearly, an infinite number of chords may be drawn in a circle.

Proof:

If ‘ r ’ is the radius of the circle and ‘ $\theta$ ‘ is the angle made by the chord at the centre of the circle and length of the arc ‘ a ‘

Arc AB subtends an angle ‘ $\theta$ ’ at the centre. Where ‘ $\theta$ ‘ is the measure of angle in radians.

The rule that involves length of the arc ‘ a ‘ and the central angle ‘ $\theta$ ’ and radius ‘ r ‘ is

a = r$\theta$ - - - - - - - - - - - - - - - - - - - (i)

AB is the chord of the circle, ‘ O ‘ is the centre of the circle, draw OM perpendicular to AB.

M is the midpoint of AB and OM bisects $\angle$ AOB.

Therefore, $\angle$ AOM = $\angle$ BOM = $\frac{\theta}{2}$

In triangle AMO, $\angle$ AMO = 90^{0}

Sin($\frac{\theta}{2}$) = $\frac{AM}{OA}$

Sin($\frac{\theta}{2}$) = $\frac{AM}{r}$

r * Sin($\frac{\theta}{2}$) = AM

Since the length of the chord AB = 2 * AM = 2r * Sin($\frac{\theta}{2}$)

Since the length of the chord AB = 2 * AM = 2r * Sin($\frac{\theta}{2}$)

Therefore, formula to find the length of the chord = 2r * Sin( $\frac{\theta}{2}$ )

Find the length of the chord of a circle if radius of the circle and central angle made by the chord are given.

Proof:

If ‘ r ’ is the radius of the circle and ‘ $\theta$ ‘ is the angle made by the chord at the centre of the circle.

AB is the chord of the circle, ‘ O ‘ is the centre of the circle, draw OM perpendicular to AB.

M is the midpoint of AB and OM bisects $\angle$ AOB.

Therefore, $\angle$ AOM = $\angle$ BOM = $\frac{\theta}{2}$

In triangle AMO, $\angle$ AMO = 90^{0}

Sin($\frac{\theta}{2}$) = $\frac{AM}{OA}$

Sin($\frac{\theta}{2}$) = $\frac{AM}{r}$

r * Sin($\frac{\theta}{2}$) = AM

Since the length of the chord AB = 2 * AM = 2r * Sin($\frac{\theta}{2}$)

Therefore, formula to find the length of the chord = 2r * Sin($\frac{\theta}{2}$)

If ‘ r ‘ is the radius of the circle and ‘ p ‘ is the length of the perpendicular drawn to the chord from the centre of the circle then the length of the chord is given by 2

Proof:

Given, a circle of radius ‘ r ‘ and ‘ p ‘ be the length of the perpendicular drawn from centre ‘O’ on the chord.

AB is the chord of the circle, ‘ O ‘ is the centre of the circle, draw OM perpendicular to AB.

r^{2} - p^{2} = AM^{2}

In triangle AMO, $\angle$ AMO = 90^{0}

Using Pythagoras theorem, we get

OA^{2} = AM^{2} + OM^{2}

r^{2} = AM^{2} + p^{2}

r

AM = 2

Since the length of the chord AB = 2 * AM = 2