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**Q.1Â Â Â Â 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.**

**Sol. **

Let O and O' be the centres of the circles of radii 5 cm and 3 cm respectively and let PQ be their common chord.

We have OP = 5 cm, O'P = 3 cm and OO' = 4 cm

Since

OO'P is a right right angled as O'.

Area of

Also ,

Area of

... (2)

From (1) and (2), we have

2 Ã— PL = 6 PL = 3

We know that when two circles intersect at two points, then their centre lie on the perpendicular bisector of the common chord i.e., OO' is the perpendicular bisector of PQ.

Therefore PQ = 2 Ã— PL = (2 Ã— 3) cm = 6 cm

**Q.2Â Â Â Â Â 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.**

**Sol.**

**Given : **AB and CD are chords of a circle with centre O. AB and CD intersect at P and AB = CD.

** To prove :** (i) AP = PD (ii) PB = CP.

** Construction** : Draw OM AB , ON CD.

Join OP,

AM = MB = AB [Perpendicular from centre bisects the chord]

CN = ND = CD [Perpendicular from centre bisects the chord]

AM = ND and MB = CN [Since AB = CD (given)]

In OMP and ONP , we have

OM = ON Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â [Equal chords of a circle are equidistant from the centre]

[Since Each = 90Âº]

OP = OP [Common]

By RHS' criterion of congruence,

MP = PN ... (2) [C.P.C.T.]

Adding (1) and (2) , we have

AM + MP = ND + PN AP = PD

Subtracting (2) from (1), we have

MB â€“ MP = CN â€“ PN PB = CP

Hence (i) AP = PD and (ii) PB = CP

**Q.3Â Â Â Â 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 th chlords.**

**Sol.**

**Given : **AB and CD are chords of a circle with centre O, AB and CD intersect at P and AB = CD.

** To prove :**

** Construction :** Draw OE AB and OF CD. Join OP.

In OEP and OFP , we have

[Since Each = 90Âº]

OP = OP [Common]

OE = OF [ Equal chords of a circle are equidistant from the centre]

Therefore by RHS criterion of congruence

[C.P.C.T.]

**Q.4Â Â Â Â Â 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 (see figure)**

Let OM be perpendicular from O on line l. We know that the perpendicular from the centre of a circle to a chord, bisects the chord.

Since BC is a chord of the smaller circle and OM BC

Therefore BM = CM ... (1)

Again AD is a chord of the larger circle and OM AD.

Therefore AM = DM ... (2)

Subtracting (1) from (2) we get

AM â€“ BM = DM â€“ CM AB = CD.

**Q.5Â Â Â Â Three girls Reshma , Salma and Mandip are playing a game by standing on a circle of radius 5 m 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 6 m each, what is the distance between Reshma and Mandip?**

**Sol. **

Let the three girls Reshma, Salma and Mandip are standing on the circle of radius 5 cm at points B, A and C respectively.

We know that if AB and AC are two equal chords of a circle, then the centre of the circles lies on the bisector of .

Here AB = AC = 6 cm. So the bisector of passes through the centre O i.e. OA is the bisector of .

Since the internal bisector of an angle divides the opposite sides in the ratio of the sides containing the angle. Therefore, M divides BC in the ratio 6 : 6 = 1 : 1 i.e., M is the middle point of BC.

Now, M is the mid- point of BC OM BC.

In right , we have

... (1)

In the right OBM, we have

... (2)

From (1) and (2) , we get

Putting AM = 3.6 in (1) we get

Hence, the distance between Reshma and Mandip = 9.6 cm

**Q.6Â Â Â Â Â A circular park of radius 20 m 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 th e length of the string of each phone.**

**Sol. **

Let ABC is an equilateral triangle of side 2x metres.

Clearly,

In right

Now , OM = AM â€“ OA

In right , we have

Since Therefore

Now,

Hence, the length of each string

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