# Circles – Exercise 10.4 – (MATHEMATICS) – 9th Class

**Exercise 10.4**

**1. 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.**

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**Sol.**We know that line joining the centres is perpendicular bisector of the common chord.

The common chord passes through the centre of the smaller circle.

∴ ∠PAO = 90^{0}, OA = 4 crn and OP = 5 cm.

Applying Pythagoras theorem, we have

cm

cm

cm = 3 cm.

Further, PQ = 2PA = 2 × 3 = 6 cm.

**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.**

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**Sol. Construction:**Draw OL and OM perpendiculars to chords AB and CD respectively. Join OP.

**To prove:**AP = DP and PB = CP.

**Proof:**Consider triangles OLP and OMP,

OL = OM [Equal chords AB and CD are equidistant from the centre of the circle]

OP is common.

∠OLP = ∠OMP [90° each]

∴ ΔOLP ≅ ΔOMP [RHS]

∴ LP= PM …(i) [CPCT]

Also, AL= LB …(ii)

[Perpendicular from centre to the chord bisects the chord]

CM = DM …(iii) [Reason same as above]

AL + LP = DM + MP [From (i), (ii), (iii)]

AP= DP …(iv)

Now, AB = CD

⇒ AP + PB = CP + PD

⇒ AP + PB = CP + AP [From (iv)]

⇒ PB = CP

**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.**

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**Sol.**Construction: Draw OL and OM perpendiculars to chords AB and CD respectively. Join OP.

**To prove:** ∠OPL = ∠OPM

**Proof:** Consider triangles OLP and OMP,

OL = OM [Equal chords AB and CD are equidistant from the centre of the circle]

OP is common.

∠OLP = ∠UMP

ΔOLP ≅ ΔOMP [RHS]

⇒ ∠OPL = ∠OPM. [CPCT]

**4. ****If a line intersects two concentric circles (circles with the same centre) with centre 0 at A, B, C and D, prove that AB = CD (see figure).**

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**Sol**.

**Construction:**Draw

perpendicular OL, from centre O, to the line

*l*.

**Proof:** AD is the chord of a bigger circle and OL ⊥ AD.

∴ AL = DL …(i)

[Perpendicular from centre of the circle to the chord bisects the chord]

Also, BC is the chord of a smaller circle and OL ⊥ BC.

BL = CL …(ii) [Reason same as above]

⇒ AL – BL = DL – CL [From (i) and (ii)]

⇒ AB = CD.

**5. 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 Sal ma, 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?**

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**Sol.**∠MOS = ∠ROS

[Angles subtended by equal chords are equal]

OM = OR [Radii]

OT is common.

∴ ΔOMT ≅ ΔORT [SAS]

∴ MT = TR …(i)

∠OTM = ∠OTR = 90°

Let OT = *x*

In right-angled triangle OTM,

MT = …….(ii)

In right-angled triangle MTS,

MT = …….(iii)

From (ii) and (iii), we get

⇒ 25 – x^{2} = 11 – x^{2} + 10x

⇒ 10x = 14 ⇒ x = 1.4

Substituting this value of x in (ii), we get

MT =

From (i), MR = 2MT = 2 × 4.8 m = 9.6 m.

**6. A circular park of radius 20 in 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.**

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**Sol.**Let Ankur, Syed and David are sitting at A, S and D respectively and so, DAS is an equilateral triangle, as if arc are equal then corresponding chords are equal.

∴ ∠ADS = 60°

Also,

⇒

∴ DS = 2DN = 2 × 17.3 = 34.6 m.