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• Equal Chords and Their Distances from the Centre

Let AB be a line and P be a point. Since there are infinite numbers of points on a line, if you join these points to P, you will get infinitely many line segments `color{blue}(PL_1, PL_2, PM, PL_3, PL_4,)` etc.

Which of these is the distance of AB from P? You may think a while and get the answer. Out of these line segments, the perpendicular from P to AB, namely PM in Fig. 10.21, will be the least.

In Mathematics, we define this least length PM to be the distance of AB from P. So you may say that:

The length of the perpendicular from a point to a line is the distance of the line from the point.

Note that if the point lies on the line, the distance of the line from the point is zero.

A circle can have infinitely many chords. You may observe by drawing chords of a circle that longer chord is nearer to the centre than the smaller chord.

You may observe it by drawing several chords of a circle of different lengths and measuring their distances from the centre. What is the distance of the diameter, which is the longest chord from the centre?

Since the centre lies on it, the distance is zero. Do you think that there is some relationship between the length of chords and their distances from the centre? Let us see if this is so.

`color{orange}("Activity :")` Draw a circle of any radius on a tracing paper. Draw two equal chords AB and CD of it and also the perpendiculars OM and ON on them from the centre O. Fold the figure so that D falls on B and C falls on A [see Fig.10.22 (i)].

You may observe that O lies on the crease and N falls on M. Therefore, OM = ON. Repeat the activity by drawing congruent circles with centres O and O′ and taking equal chords AB and CD one on each.

Draw perpendiculars OM and O′N on them [see Fig. 10.22(ii)]. Cut one circular disc and put it on the other so that AB coincides with CD. Then you will find that O coincides with O′ and M coincides with N. In this way you verified the following:

`color{green}("Theorem 10.6 :")`

`"Equal chords of a circle (or of congruent circles) are equidistant from the centre (or centres)."`

Next, it will be seen whether the converse of this theorem is true or not. For this, draw a circle with centre O. From the centre O, draw two line segments OL and OM of equal length and lying inside the circle [see Fig. 10.23(i)].

Then draw chords PQ and RS of the circle perpendicular to OL and OM respectively [see Fig 10.23(ii)]. Measure the lengths of PQ and RS. Are these different? No, both are equal.

Repeat the activity for more equal line segments and drawing the chords perpendicular to them. This verifies the converse of the Theorem 10.6 which is stated as follows:

`color{red}("Theorem 10.7 :")`

`"Chords equidistant from the centre of a circle are equal in length."`

We now take an example to illustrate the use of the above results:

Which of these is the distance of AB from P? You may think a while and get the answer. Out of these line segments, the perpendicular from P to AB, namely PM in Fig. 10.21, will be the least.

In Mathematics, we define this least length PM to be the distance of AB from P. So you may say that:

The length of the perpendicular from a point to a line is the distance of the line from the point.

Note that if the point lies on the line, the distance of the line from the point is zero.

A circle can have infinitely many chords. You may observe by drawing chords of a circle that longer chord is nearer to the centre than the smaller chord.

You may observe it by drawing several chords of a circle of different lengths and measuring their distances from the centre. What is the distance of the diameter, which is the longest chord from the centre?

Since the centre lies on it, the distance is zero. Do you think that there is some relationship between the length of chords and their distances from the centre? Let us see if this is so.

`color{orange}("Activity :")` Draw a circle of any radius on a tracing paper. Draw two equal chords AB and CD of it and also the perpendiculars OM and ON on them from the centre O. Fold the figure so that D falls on B and C falls on A [see Fig.10.22 (i)].

You may observe that O lies on the crease and N falls on M. Therefore, OM = ON. Repeat the activity by drawing congruent circles with centres O and O′ and taking equal chords AB and CD one on each.

Draw perpendiculars OM and O′N on them [see Fig. 10.22(ii)]. Cut one circular disc and put it on the other so that AB coincides with CD. Then you will find that O coincides with O′ and M coincides with N. In this way you verified the following:

`color{green}("Theorem 10.6 :")`

`"Equal chords of a circle (or of congruent circles) are equidistant from the centre (or centres)."`

Next, it will be seen whether the converse of this theorem is true or not. For this, draw a circle with centre O. From the centre O, draw two line segments OL and OM of equal length and lying inside the circle [see Fig. 10.23(i)].

Then draw chords PQ and RS of the circle perpendicular to OL and OM respectively [see Fig 10.23(ii)]. Measure the lengths of PQ and RS. Are these different? No, both are equal.

Repeat the activity for more equal line segments and drawing the chords perpendicular to them. This verifies the converse of the Theorem 10.6 which is stated as follows:

`color{red}("Theorem 10.7 :")`

`"Chords equidistant from the centre of a circle are equal in length."`

We now take an example to illustrate the use of the above results:

Q 3210478310

If two intersecting chords of a circle make equal angles with the diameter passing through their point of intersection, prove that the chords are equal.

Class 9 Chapter 10 Example 2

Class 9 Chapter 10 Example 2

Given that AB and CD are two chords of a circle, with centre O intersecting at a point E. PQ is a diameter through E, such that ∠ AEQ = ∠ DEQ (see Fig.10.24). You have to prove that AB = CD. Draw perpendiculars OL and OM on chords AB and CD, respectively. Now

`∠ LOE = 180° – 90° – ∠ LEO = 90° – ∠ LEO`

`"(Angle sum property of a triangle)"`

`= 90° – ∠ AEQ = 90° – ∠ DEQ`

`= 90° – ∠ MEO = ∠ MOE`

In triangles OLE and OME,

`∠ LEO = ∠ MEO \ \ \ \ \ \ ("Why ?")`

`∠ LOE = ∠ MOE\ \ \ \ \ \ ("Proved above")`

`EO = EO\ \ \ \ \ \ \ \ \ \ ("Common")`

`"Therefore," \ \ \ \ \ D OLE ≅ D OME\ \ \ \ ("Why" ?)`

`"This gives" \ \ \ \ \ \ OL = OM \ \ \ \ \ \ "(CPCT)"`

`So, \ \ \ \ \ \ \ \ \ \ \ \ AB = CD \ \ \ \ \ \ ("Why ?")`