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Circle : Exercise 10.4 (Mathematics NCERT Class 9th)


Q.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.

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 O{P^2} = PO{'^2} + O'{O^2} \left[ {\,\,\,{5^2} = {3^2} + {4^2}} \right]
 \Rightarrow OO'P is a right \angle d\,\Delta , right angled as O'.

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Area of \Delta \,OO'P = {1 \over 2} \times O'P \times OO'
 = {1 \over 2} \times 3 \times 4
 = 6\,sq.\,units\,\,\,\,\,\,\,\,\,...\left( 1 \right)
Also ,
Area of \Delta OO'P = {1 \over 2} \times OO' \times PL
 = {1 \over 2} \times 4 \times PL = 2\,PL ... (2)
From (1) and (2), we have
2 × PL = 6  \Rightarrow 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 × PL = (2 × 3) cm = 6 cm


Q.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.

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  \bot AB , ON  \bot CD.
Join OP,

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AM = MB = {1 \over 2} AB [Perpendicular from centre bisects the chord]
CN = ND = {1 \over 2} CD [Perpendicular from centre bisects the chord]
AM = ND and MB = CN [Since AB = CD (given)]
In \Delta s OMP and ONP , we have
OM = ON                                                [Equal chords of a circle are equidistant from the centre]
\angle OMP = \angle ONP [Since Each = 90º]
OP = OP [Common]
By RHS' criterion of congruence,
\Delta \,OMP \cong \Delta ONP
 \Rightarrow MP = PN ... (2) [C.P.C.T.]
Adding (1) and (2) , we have
AM + MP = ND + PN  \Rightarrow AP = PD
Subtracting (2) from (1), we have
MB – MP = CN – PN  \Rightarrow PB = CP
Hence (i) AP = PD and (ii) PB = CP


Q.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 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 : \angle OPE = \angle OPF

3
Construction : Draw OE  \bot AB and OF  \bot CD. Join OP.
In \Delta s OEP and OFP , we have
\angle OEP = \angle OFP [Since Each = 90º]
OP = OP [Common]
OE = OF [ Equal chords of a circle are equidistant from the centre]
Therefore by RHS criterion of congruence
\Delta OEP\, \cong \Delta OFP
 \Rightarrow \angle OPE = \angle OPF [C.P.C.T.]


Q.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 (see figure)

33
Sol.

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  \bot BC

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Therefore BM = CM ... (1)
Again AD is a chord of the larger circle and OM  \bot AD.
Therefore AM = DM ... (2)
Subtracting (1) from (2) we get
AM – BM = DM – CM  \Rightarrow AB = CD.


Q.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 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.

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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 \angle BAC.
Here AB = AC = 6 cm. So the bisector of \angle BAC passes through the centre O i.e. OA is the bisector of \angle BAC.
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  \Rightarrow OM  \bot BC.
In right \angle d\,\Delta ABM , we have
A{B^2} = A{M^2} + B{M^2}
 \Rightarrow 36 = A{M^2} + B{M^2}
 \Rightarrow B{M^2} = 36 - A{M^2} ... (1)
In the right \Delta OBM, we have
O{B^2} = O{M^2} + B{M^2}
 \Rightarrow 25 = {\left( {OA - AM} \right)^2} + B{M^2}
 \Rightarrow B{M^2} = 25 - {\left( {OA - AM} \right)^2}
 \Rightarrow B{M^2} = 25 - {\left( {5 - AM} \right)^2} ... (2)
From (1) and (2) , we get
36 - A{M^2} = 25 - {\left( {5 - AM} \right)^2}
 \Rightarrow 11 - A{M^2} + {\left( {5 - AM} \right)^2} = 0
 \Rightarrow 11 - A{M^2} + 25 - 10AM + A{M^2} = 0
 \Rightarrow 10AM = 36
 \Rightarrow AM = 3.6
Putting AM = 3.6 in (1) we get
B{M^2} = 3.6 - {\left( {3.6} \right)^2} = 36 - 12.96
 \Rightarrow BM = \sqrt {36 - 12.96} = \sqrt {23.04} = 4.8\,cm
 \Rightarrow BC = 2BM = 2 \times 4.8\, = 9.6\,cm
Hence, the distance between Reshma and Mandip = 9.6 cm


Q.6      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, BM = {{BC} \over 2} = {{2x} \over 2} = x\,metres
In right \angle d\,\Delta ABM
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A{M^2} = A{B^2} - B{M^2}
 = {\left( {2x} \right)^2} - {x^2} = 4{x^2} - {x^2} = 3{x^2}
 \Rightarrow AM = \sqrt 3 \,x
Now , OM = AM – OA \left( {\sqrt {3x} - 20} \right)m
In right \angle d\,\Delta OBM, we have
O{B^2} = B{M^2} + O{M^2}
 \Rightarrow {20^2} = {x^2} + {\left( {\sqrt {3x} - 20} \right)^2}
 \Rightarrow 400 = {x^2} + 3{x^2} - 40\sqrt {3x} + 400
 \Rightarrow 4{x^2} - 40\sqrt {3x} = 0
 \Rightarrow 4x\left( {x - 10\sqrt 3 } \right) = 0
Since x \ne 0 Therefore x - 10\sqrt 3 = 0
 \Rightarrow x = 10\sqrt 3

Now, BC = 2BM = 2x = 20\sqrt 3
Hence, the length of each string  = 20\sqrt 3 \,\,m



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