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# Class 9 NCERT Solutions- Chapter 10 Circles – Exercise 10.4

• Last Updated : 03 Jan, 2021

### Question 1. Two circles of radii 5cm and 3cm intersect at two points and the distance between their centers is 4 cm. Find the length of the common chord.

Solution:

Given: OP=4cm, AP=3cm, QR=5cm

To find: In âˆ†APO:

AOÂ²=5Â²=25

OPÂ²=4Â²=16

APÂ²=3Â²=9

OPÂ²+APÂ²=AOÂ²

BY converse of Pythagoras theorem

Î”APO: is a right âˆ D=P

Now, in the bigger circle OP is perpendicular AB

AP=Â½AB          —————-(perpendicular from the center of circle  to a chord bisect the chord )

3=Â½AB

6=AB

âˆ´Therefore the length of common chord is 6cm.

### Question 2. If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the center makes equal angles with the chords.

Solution:

Given: Equal chord AB & CD intersect at P.

To find: AP=PD and PB=PC

Construction: Draw OM perpendicular AB ,ON perpendicular CD and join OP.

Because perpendicular from center bisect the chord

âˆ´AM=MB=Â½AB     also   CN=ND=Â½CD

AM=MB=CN=ND        ——————1

Now, In âˆ†OMP  and  âˆ†ONP

ANGLE M=ANGLE N              [90Â° both]

OP=OP               [COMMON]

ON=OM              [equal chords are equilateral from center]

âˆ´âˆ†OMPâ‰…âˆ†ONP

Therefore MP=PN           (C.P.C.T.)      ——————2

i)from   1 and 2

AM+MP=ND+AN

AP=PD

ii)MB-MP=CN=PN

PB=PC

### 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: Equal chords AB and CD intersect at P.

To prove: angle1=angle=2

Construction: Draw OM perpendicular AB & ON perpendicular CD.

Solution: In âˆ†OMP & âˆ†ONP

Angle M= Angel N      [90 Â°  each]

OP=OP           [common]

OM=ON      —————[ Equal chords are equal distant from center]

âˆ´âˆ†OMPâ‰…âˆ†ONP  ———-[R.H.S]

âˆ´âˆ 1=âˆ 2     ———–[C.P.C.T]

### 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 concentric circle with  O. A line intersect them at A, B, C , and  D

To prove: AB=CD

âˆ´AM=MD              —————-[âŠ¥ from center of circle of a circle bisects  the chord]         __________    1

The smaller circle :

BC is chord  OM âŠ¥ BC

BM=MC         ——————-[âŠ¥ from center of circle of a circle bisects  the chord]             __________    2

subtracting  1-2

AM-BM=MD-MC

AB=CD

### Question 5. Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5m 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 6m each, what is the distance between Reshma and Mandip?

Solution:

To find RM=?

Let Reshma, Salma and Mandip be R,S,M

Construction: Draw OP âŠ¥ RS join OR and OS.

RP=Â½RS   ___________[âŠ¥ from center bisects the chord]

RP=Â½*6=3m

In right Î”ORP

OPÂ²=ORÂ²- PRÂ²

OP= âˆš 5Â² -3Â²

=âˆš259    =âˆš16      =4

Area of Î”ORS=Â½*RS*OP

=Â½*6*4=12mÂ²     —————–1

Now, âˆ N=90Â°

Area of Î”ORS=Â½*SO*RN

=Â½*SO*RN     ——————-2

Above ,1=2

12=Â½*5*RN

12/5*2=RN

RN=4.8

RM=2*RN _________________[âŠ¥ from center bisects the chord]

=2*4.8

9.6m

### Question 6. A circular park of radius 20m is situated in a colony. Three boys Ankur, Syed and David are sitting at equal distance on its boundary. Each boy has toy telephone in his hand to talk with each other. Find the length of the string of each phone.

Solution:

Draw AMâŠ¥SD

âˆ´ASD is the equilateral  Î”

Let each side of Î”-2xm

SM=2x/2=x

Now in Î” DMS, by the Pythagoras theorem

AMÂ²+SMÂ²=ASÂ²

AMÂ²= ASÂ²- SMÂ²

AM=âˆš(2xÂ²+xÂ² )

==âˆš(3xÂ² )

AM =âˆš3x

OM=AM-AO

OM=âˆš3x-20

Now in right Î”OMS

OMÂ²+SMÂ²=SOÂ²

(âˆš3x-20)Â²+2xÂ²+xÂ²=20Â²

20Â²+400-40âˆš3x+x^2=400

4xÂ²=40âˆš3x

4xx=40âˆš3x

X=(40âˆš3)/4

X=10âˆš3x

Length of each string =2x

=2*10âˆš3xm

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