# Equal Chords and their Distances from the Centre

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Here, we have a line AB and P is a point positioned above it. We know that there are

Now, let's take some random points lying on the line AB as shown below. If we join these points to P, we will get infinitely many such line segments. (PL1, PL2, PM, PL3, PL4 etc).

Which of these is the distance of AB from P?

Out of these line segments, the perpendicular from P to AB, namely PM in the below figure, 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.

**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.9.9 (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. 9.9(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:

**Theorem 9.5**

**Equal Chords are Equidistant Theorem:** **Equal chords of a circle (or of congruent circles) are equidistant from the centre (or centres)**.

**Given:** AB and CD are equal chords of a circle.

**Construction:** Construct the Perpendiculars: Draw the perpendicular from O to AB and let it intersect AB at point M. Draw the perpendicular from O to CD and let it intersect CD at point N. Also join OA,OB, OC and OD.

**To Prove:** OM = ON

**Proof:** In order to prove- OM = ON: We have to prove that the four triangles: OMA, OMB , ONC and OND are congruent. In order to do this: We will prove △OMA ≅ △OMB , △ONC ≅ △OND , △OMA ≅ △ONC.

In the triangles △OMA and △OMB:

(i) ∠OMA = ∠OMB = 90∘ (as OM is

(ii) OA = OB (they are

(iii) OM = OM (

By the

Similarly, in the triangles △ONC and △OND:

(i) ∠ONC = ∠OND = 90∘ (as ON is

(ii) OC = OD (they are

(iii) ON = ON (

By the

Now, in the triangles △OMA and △ONC:

(i) ∠OMA = ∠ONC = 90∘ (as OM and ON are

(ii) OA = OC (they are

(iii) AM = CN (as

By the

**Result:** Thus, we can conclude : OM = ON since that △OMA ≅ △OMB ≅ △ONC ≅ △OND.

Next, let's check whether the converse of this theorem is true or not. The converse of the Theorem 9.5 which is stated as follows:

**Theorem 9.6**

**Equidistant Chords Equal Theorem:** **Chords equidistant from the centre of a circle are equal in length.**

**Given:** Perpendiculars OM and ON are equal.

**Construction:** Join OA,OB, OC and OD.

**To Prove:** AB CD

**Proof:** In the triangles △OMA and △ONC:

(i) ∠OMA = ∠ONC = 90∘ (as OM and ON are

(ii) OA = OC (they are

(iii) OM = ON (Given)

By the

Thus, by CPCT: AM = CN

Since, perpendiculars drawn from the centre

Therefore, AM = BM = CN = DN. But AM + BM = AB and CN + DN = CD. So, AB = CD.

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

## Example 1 :

**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.**

**Solution**

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.9.11). 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

∠ LOE = ∠ MOE (Proved above)

EO = EO (Common)

Therefore, ∆ OLE ≅ ∆ OME

This gives OL = OM (CPCT)

So, AB = CD

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## Exercise 9.2

## Problem 1:

**If a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D, prove that AB = CD.**

We know that, OA =

In ΔOBC, we know that OB = OC ,so ∠OBC = ∠

∴ ∠ OCD = ∠

In ΔOAD, we know that OA = OD ,so ∠OAD = ∠

Since, ∠OCD = ∠OBA and ∠OAD = ∠ODA , we get ∠

∴ From

So, AB = CD.

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## Problem 2:

**Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip, Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6m each, what is the distance between Reshma and Mandip?**

Draw perpendicular OA and OB on RS and SM respectively.

Let R,S and M be the position of Reshma, Salma and Mandip respectively.

AR = AS =

OR = OS = OM =

In △OAR,

OA =

OA =

We know that in an isosceles triangle altitude divides the base,

So in △RSM, ∠RCS =

RC = CM

Area of △ORS =

RC × 5 = 24

RC =

RM = 2RC = 2 × 4.8 =

So, distance between Reshma and Mandip is 9.6m

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## Problem 3:

**A circular park of radius 20m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary each having a toy telephone in his hands to talk each other. Find the length of the string of each phone.**

Let Ankur be represented as A, Syed as S and David as D.

The boys are sitting at an equal distance.

Hence, △ASD is an

Let the radius of the circular park be *r* meters.

∴ OS = r = 20m.

Let the length of each side of △ASD be *x* meters.

Draw AB ⊥ SD

∴ SB = BD =

In △ABS, ∠B =

By Pythagoras theorem,

∴

∴ AB =

Now, AB = AO + OB

OB = AB − AO

OB= ((√3x)/2 − 20) m

In △OBS,

∴ x =

The length of the string of each phone is 20√3m.

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## Problem 4:

**Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centres is 4 cm. Find the length of the common chord.**

The perpendicular bisector of the common chord passes through the centres of both circles.

As the circles intersect at two points, we can construct the above figure.

Consider AB as the common chord and O and O’ as the centres of the circles.

O’A = 5 cm

OA = 3 cm

OO’ = 4 cm [Distance between centres is 4 cm.]

As the radius of the bigger circle is more than the distance between the two centres, we know that the centre of the smaller circle lies inside the bigger circle.

The perpendicular bisector of AB is OO’.

OA = OB = 3 cm

As O is the midpoint of AB

AB = 3 cm + 3 cm = 6 cm

The length of the common chord is 6 cm.

It is clear that the common chord is the diameter of the smaller circle.

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## Problem 5:

**If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to corresponding segments of the other chord.**

Let AB and CD be two equal cords (i.e., AB = CD). In the above question, it is given that AB and CD intersect at a point, say, E.

It is now to be proven that the line segments AE = DE and CE = BE

**Construction:**

Step 1: From the centre of the circle, draw a perpendicular to AB, i.e., OM ⊥ AB.

Step 2: Similarly, draw ON ⊥ CD.

Step 3: Join OE.

Now, the diagram is as follows:

**Proof:**

From the diagram, it is seen that OM bisects AB, and so OM ⊥ AB

Similarly, ON bisects CD, and so ON ⊥ CD.

It is known that AB = CD. So,

AM = ND — (i)

and MB = CN — (ii)

Now, triangles ΔOME and ΔONE are similar by RHS congruency, since

∠OME = ∠ONE (They are perpendiculars.)

OE = OE (It is the common side.)

OM = ON (AB and CD are equal, and so they are equidistant from the centre.)

∴ ΔOME ≅ ΔONE

ME = EN (by CPCT) — (iii)

Now, from equations (i) and (ii), we get

AM+ME = ND+EN

So, AE = ED

Now from equations (ii) and (iii), we get

MB-ME = CN-EN

So, EB = CE (Hence, proved)

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## Problem 6:

**If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the centre makes equal angles with the chords.**

From the question, we know the following:

(i) AB and CD are 2 chords which are intersecting at point E.

(ii) PQ is the diameter of the circle.

(iii) AB = CD.

Now, we will have to prove that ∠BEQ = ∠CEQ

For this, the following construction has to be done.

**Construction:**

Draw two perpendiculars are drawn as OM ⊥ AB and ON ⊥ D. Now, join OE. The constructed diagram will look as follows:

Now, consider the triangles ΔOEM and ΔOEN.

Here,

(i) OM = ON [The equal chords are always equidistant from the centre.]

(ii) OE = OE [It is the common side.]

(iii) ∠OME = ∠ONE [These are the perpendiculars.]

So, by RHS congruency criterion, ΔOEM ≅ ΔOEN.

Hence, by the CPCT rule, ∠MEO = ∠NEO

∴ ∠BEQ = ∠CEQ (Hence, proved)