Exercise 9.2

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 = cm

OA = cm

OO’ = 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 insideoutsideon the bigger circle.

The perpendicular bisector of AB is OO’.

OA = OB = cm

As O is the midpoint of AB, AB = 3 cm + 3 cm = cm

The length of the common chord is 6 cm.

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

2. 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

Solution:

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, considering triangles ΔOME and ΔONE:

∠OME =≠ ∠ONE (they are perpendicularschords)

OE =≠ OE (it is the side)

OM =≠ ON (AB and CD are and so they are equidistantperpendicular from the centre)

∴ ΔOME ≅ ΔONE by congruency rule.

ME =≠ EN (by CPCT) — (iii)

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

AM + ME = ND +

So, AE =≠ ED

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

MB - = CN -

Thus, EB = (Hence, proved)

3. 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.

Solution:

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 equidistantperpendicular from the centre)

(ii) OE =≠ OE (it is side)

(iii) ∠OME =≠ ∠ONE (they are perpendicularschords)

So, by congruency criterion, ΔOEM ≅ Δ.

Hence, by the CPCT rule, ∠MEO = ∠

∴ ∠BEQ = ∠ (Hence, proved)

4. 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 = ODOC and OB = OCOD as they are radius of respective circles.

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

∴ ∠ OCD = ∠ OBAOBC

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

Since, ∠OCD = ∠OBA and ∠OAD = ∠ODA , we get ∠AOBCOB in ΔOAB is equal to ∠COD in ΔOCD.

∴ From congruency, we can say that ΔOAB ≅ Δ

So, AB = (by CPCT).

5. 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 =62 = m

OR = OS = OM = m (radii of same circle)

In △OAR,

OA2+AR2=OR2

OA2+32=52

OA = 25−9

OA = m

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

So in △RSM, ∠RCS = °

RC =

Area of △ORS = 12× OA × RS = 12× RC × OS = 12 × ×

RC × =

RC = m

RM = RC = × = m

So, distance between Reshma and Mandip is 9.6m

6. 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 triangle.

Let the radius of the circular park be r meters.

∴ OS = r = m.

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

Draw AB ⊥ SD

∴ SB = BD = 122SD = m

In △ABS, ∠B = °

By Pythagoras theorem,

AS2=AB2+BS2

∴ AB2=AS2−BS2=x2−x22= 3x24

∴ AB = 3x22x3 m

Now, AB = AO +

OB = AB − AO

OB = 3x2−203x2+20 m

In △OBS, OS2=OB2+SB2

202=3x2−202+x22

400=3x24+400−2203x2+x24

∴ x = 203203 m

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