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

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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 = 900, OA = 4 crn and OP = 5 cm.
Applying Pythagoras theorem, we have


PA = \sqrt {{{(5)}^2} - {{(4)}^2}} cm

 = \sqrt {25 - 16} cm

 = \sqrt 9 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 =    = \sqrt {25 - {x^2}}   …….(ii)

In right-angled triangle MTS,

MT =  = \sqrt {36 - {{(5 - x)}^2}}                   …….(iii)

From (ii) and (iii), we get

 = \sqrt {25 - {x^2}} = \sqrt {36 - 25 - {x^2} + 10x}

⇒ 25 – x2 = 11 – x2 + 10x

⇒ 10x  = 14                              ⇒ x = 1.4

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

MT =  = \sqrt {25 - {{(1.4)}^2}} = \sqrt {25 - 1.96} = \sqrt {23.04} = 4.8m.

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,           \angle NDO = {1 \over 2}\angle ADS = 30^\circ


{{DN} \over {OD}} = \cos 30^\circ

⇒ {{DN} \over {20}} = {{\sqrt 3 } \over 2} \Rightarrow DN = 10 \times 1.73 = 17.3m

∴ DS = 2DN = 2 × 17.3 = 34.6 m.


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