NCERT Solutions for Class 9 Maths Exercise 10.4

NCERT Solutions for Class 9 Maths Exercise 10.4 book solutions are available in PDF format for free download. These ncert book chapter wise questions and answers are very helpful for CBSE board exam. CBSE recommends NCERT books and most of the questions in CBSE exam are asked from NCERT text books. Class 9 Maths chapter wise NCERT solution for Maths Book for all the chapters can be downloaded from our website and myCBSEguide mobile app for free.

NCERT solutions for Class 9 Maths Circles Download as PDF

NCERT Solutions for Class 9 Mathematics Circles

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

Ans. Let two circles with centres O and O’ intersect each other at points A and B. On joining A and B, AB is a common chord.

Radius OA = 5 cm, Radius O’A = 3 cm,

Distance between their centers OO’ = 4 cm

In triangle AOO’,

5² = 4² + 3²

 25 = 16 + 9

 25 = 25

Hence AOO’ is a right triangle, right angled at O’.

Since, perpendicular drawn from the center of the circle bisects the chord.

Hence O’ is the mid-point of the chord AB. Also O’ is the centre of the circle II.

Therefore length of chord AB = Diameter of circle II

 Length of chord AB = 2 x 3 = 6 cm.

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NCERT Solutions for Class 9 Maths Exercise 10.4

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.

Ans. Given: Let AB and CD are two equal chords of a circle of centers O intersecting each other at point E within the circle.

To prove: (a) AE = CE

(b) BE = DE

Construction: Draw OM  AB, ON  CD. Also join OE.

Proof: In right triangles OME and ONE,

OME = ONE =

OM = ON

[Equal chords are equidistance from the centre]

OE = OE [Common]

 OMEONE [RHS rule of congruency]

 ME = NE [By CPCT] ……….(i)

Now, O is the centre of circle and OM  AB

 AM = AB

[Perpendicular from the centre bisects the chord] …..(ii)

Similarly, NC = CD ……….(iii)

But AB = CD [Given]

From eq. (ii) and (iii), AM = NC ……….(iv)

Also MB = DN …….…(v)

Adding (i) and (iv), we get,

AM + ME = NC + NE

 AE = CE [Proved part (a)]

Now AB = CD [Given]

AE = CE [Proved]

 AB – AE = CD – CE

 BE = DE [Proved part (b)]

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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 chord.

Ans. Given: AB and CD be two equal chords of a circle with centre O intersecting each other with in the circle at point E. OE is joined.

To prove: OEM = OEN

Construction: Draw OM  AB and

ON  CD.

Proof: In right angled triangles OME and ONE,

OME = ONE [Each ]

OM = ON [Equal chords are equidistant from the centre]

OE = OE [Common]

 OMEONE [RHS rule of congruency]

 OEM = OEN [By CPCT]

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NCERT Solutions for Class 9 Maths Exercise 10.4

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)

Ans. Given: Line  intersects two concentric circles with centre O at points A, B, C and D.

To prove: AB = CD

Construction: Draw OL

Proof: AD is a chord of outer circle and OL AD.

 AL = LD ………(i) [Perpendicular drawn from the centre bisects the chord]

Now,   BC is a chord of inner circle and

OL BC

 BL = LC …(ii) [Perpendicular drawn from the centre bisects the chord]

Subtracting (ii) from (i), we get,

AL – BL = LD – LC

 AB = CD

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NCERT Solutions for Class 9 Maths Exercise 10.4

5. Three girls Reshma, Salma and Mandip are 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?

Ans. Let Reshma, Salma and Mandip takes the position C, A and B on the circle. Since AB = AC

The centre lies on the bisector of BAC.

Let M be the point of intersection of BC and OA.

Again, since AB = AC and AM bisects

CAB.

 AM CB and M is the mid-point of CB.

Let OM =  then MA =

From right angled triangle OMB,

OB² = OM² + MB²

 5² =  + MB² ……….(i)

Again, in right angled triangle AMB,

AB² = AM² + MB²

 6² =  + MB² ……….(ii)

Equating the value of MB² from eq. (i) and (ii),

 

 

 

 

Hence, from eq. (i),

MB² =  =

=  =

 MB =  = 4.8 cm

 BC = 2MB = 2 x 4.8 = 9.6 cm

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NCERT Solutions for Class 9 Maths Exercise 10.4

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.

Ans. Let position of three boys Ankur, Syed and David are denoted by the points A, B and C respectively.

A = B = C =  [say]

Since equal sides of equilateral triangle are as equal chords and perpendicular distances of equal chords of a circle are equidistant from the centre.

 OD = OE = OF =  cm [say]

Join OA, OB and OC.

 Area of AOB

= Area of BOC = Area of AOC

And Area of ABC

= Area of AOB + Area of BOC + Area of AOC

 And Area of ABC = 3 x Area of BOC

  = 3 ( BC x OE)

  = 3 ( )

 

  ……….(i)

Now, CE  BC

 BE = EC =  BC    [ Perpendicular drawn from the centre bisects the chord]

 BE = EC =

 BE = EC =  [Using eq. (i)]

 BE = EC =

Now in right angled triangle BEO,

OE² + BE² = OB² [Using Pythagoras theorem]

 

 

 

 

  = 10 m

And  =  =  m

Thus distance between any two boys is  m.

NCERT Solutions for Class 9 Maths Exercise 10.4

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