NCERT Solutions for Class 9 Maths Chapter 9 Circles Ex 9.2

# NCERT Solutions for Class 9 Maths Chapter 9 Circles Ex 9.2

## NCERT Solutions for Class 9 Maths Chapter 9 Circles Ex 9.2

NCERT Solutions for Class 9 Maths Chapter 9 Circles Ex 9.2 are the part of NCERT Solutions for Class 9 Maths. In this post, you will find the NCERT Solutions for Chapter 9 Circles Ex 9.2.

Ex 9.2 Class 9 Maths Question 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.

Solution:
We have two intersecting circles with centres at O and O’ respectively. Let PQ be the common chord of both the circles.
Since in two intersecting circles, the line joining their centres is perpendicular bisector of the common chord.

Therefore, OLP = OLQ = 90° and PL = LQ
Now, in right-angled ∆OLP, we have
PL2 + OL2 = OP2
PL2 + (4 – x)2 = 52
PL2 = 52 – (4 – x)2
PL2 = 25 – 16 – x2 + 8x
PL2 = 9 – x2 + 8x                    …..…(i)
Again, in right-angled ∆O’LP, we have
PL2 = PO‘2 – LO‘= 32 – x2 = 9 – x2          ….…(ii)
From (i) and (ii), we have
9 – x2 + 8x = 9 – x2
8x = 0
x = 0
L and O’ coincide.
PQ is a diameter of the smaller circle.
PL = 3 cm
But PL = LQ
LQ = 3 cm
PQ = PL + LQ = 3 cm + 3 cm = 6 cm
Thus, the required length of the common chord = 6 cm.

Ex 9.2 Class 9 Maths Question 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.

Solution:
Given: A circle with centre O and equal chords AB and CD intersect at E.
To prove: AE = DE and CE = BE
Construction: Draw OM
AB and ON  CD. Join OE.

Proof: Since AB = CD         [Given]
OM = ON                         [Equal chords are equidistant from the centre]
Now, in ∆OME and ∆ONE, we have
OME = ONE                  [Each equal to 90°]
OM = ON                            [Proved above]
OE = OE                              [Common hypotenuse]
∆OME  ∆ONE              [By RHS congruence rule]
ME = NE                        [C.P.C.T.]

Adding AM on both sides, we get

AM + ME = AM + NE
AE = DN + NE = DE       [ AB = CD  ½ AB = ½ DC  AM = DN]
AE = DE                          ….…(i)
Now, AB – AE = CD – DE
BE = CE                          …….(ii)
From (i) and (ii), we have
AE = DE and CE = BE

Ex 9.2 Class 9 Maths Question 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.

Solution:
Given: A circle with centre O and equal chords AB and CD are intersecting at E.
To prove:
OEA = OED
Construction: Draw OM
AB and ON  CD. Join OE.
Proof: In ∆OME and ∆ONE, we have

OM = ON                         [Equal chords are equidistant from the centre]
OE = OE                           [Common hypotenuse]
OME = ONE              [Each equal to 90°]
∆OME  ∆ONE          [By RHS congruence rule]
OEM = OEN         [C.P.C.T.]
OEA = OED

Ex 9.2 Class 9 Maths Question 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).

Solution:
Given: Two circles with the common centre O. A line intersects the outer circle at A and D and the inner circle at B and C.
To prove: AB = CD.
Construction: Draw OM

Proof: For the outer circle,
OM
AM = MD                 ….…(i)
[Perpendicular from the centre to the chord bisects the chord]
For the inner circle,

OM  BC                      [By construction]
BM = MC                  ….…(ii)
[Perpendicular from the centre to the chord bisects the chord]
Subtracting eq. (ii) from (i), we have
AM – BM = MD – MC
AB = CD

Ex 9.2 Class 9 Maths Question 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?

Solution:
Let the three girls Reshma, Salma and Mandip be positioned at R, S and M, respectively on the circle with centre O and radius 5 m such that
RS = SM = 6 m          [Given]

Equal chords of a circle subtend equal angles at the centre.
1 = 2
In ∆POR and ∆POM, we have
OP = OP                    [Common]
OR = OM                  [Radii of the same circle]
1 =                    [Proved above]
∆POR  ∆POM    [By SAS congruence rule]
PR = PM and OPR = OPM             [C.P.C.T.]
∵ ∠OPR + OPM = 180°                        [Linear pair]
∴ ∠OPR = OPM = 90°
OP  RM
Now, in ∆RSP and ∆MSP, we have
RS = MS                 [Each 6 cm]
SP = SP                  [Common]
PR = PM                [Proved above]
∆RSP  ∆MSP   [By SSS congruence rule]
RPS = MPS   [C.P.C.T.]
But
RPS + MPS = 180°         [Linear pair]
RPS = MPS = 90°
SP passes through O.
Let OP = x m
SP = (5 – x) m
Now, in right-angled ∆OPR, we have
x2 + RP2 = 52
RP2 = 52 – x2                                            ……..(i)
In right-angled ∆SPR, we have
(5 – x)2 + RP2 = 62
RP2 = 62 – (5 – x)2               ……..(ii)
From eq. (i) and (ii), we get
52 – x2 = 62 – (5 – x)2
25 – x2 = 36 – [25 – 10x + x2]

25 – x2 = 36 – 25 + 10x – x2

25 + 25 – 36 = 10x

10x = 14
x = 14/10 = 1.4
Now, RP2 = 52 – x2
RP2 = 25 – (1.4)2
RP2 = 25 – 1.96 = 23.04
RP = 23.04 = 4.8
RM = 2RP = 2 × 4.8 = 9.6
Thus, distance between Reshma and Mandip is 9.6 m.

Ex 9.2 Class 9 Maths Question 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 the length of the string of each phone.

Solution:
Let Ankur, Syed and David are sitting at A, S and D respectively in the circular park with centre O such that AS = SD = DA
i.e., ∆ASD is an equilateral triangle.
Let the length of each side of the equilateral triangle be 2x.
Draw AM
SD.
Since ∆ASD is an equilateral triangle.
AM passes through O.
SM = ½ SD = ½ (2x)
SM = x

Now, in right-angled ∆ASM, we have
AM2 + SM2 = AS2            [Using Pythagoras theorem]

AM= AS2 – SM= (2x)2 – x2 = 4x2 – x2 = 3x2
AM = 3x m
Now, OM = AM – OA = (3x – 20) m
Again, in right-angled ∆OSM, we have
OS2 = SM2 + OM2            [Using Pythagoras theorem]
202 = x2 + (3x – 20)2
400 = x2 + 3x2 – 403x + 400
4x2 = 403x
x = 10√3 m
Now, SD = 2x = 2 × 10√3 m = 20√3 m
Thus, the length of the string of each phone = 20√3 m