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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.

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.

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.

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).

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?

Sol.      ∠MOS = ∠ROS

[Angles subtended by equal chords are equal]

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.

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.

Also,           $\angle NDO = {1 \over 2}\angle ADS = 30^\circ$
${{DN} \over {OD}} = \cos 30^\circ$
⇒ ${{DN} \over {20}} = {{\sqrt 3 } \over 2} \Rightarrow DN = 10 \times 1.73 = 17.3m$