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NCERT Solutions for Class 9 Maths Chapter 10 Circles

NCERT Solutions for Class 9 Maths Chapter 10 Circles

Edited By Ramraj Saini | Updated on May 11, 2023 03:25 PM IST

NCERT Solutions for Class 9 Maths Chapter 10 Circles

Circles Class 9 Questions And Answers are provided here. These NCERT solutions are prepared by experts at Careers360 considering latest CBSE syllabus 2023. All NCERT problems are discussed in simple, easy to understand and comprehensive way. Thus, these will help students to score well in the exams. Many objects that we come across in our daily life is circular in shape, such as ring, bangle, wheels of the vehicle, clock, etc. NCERT solutions for class 9 maths chapter 10 Circles will help in solving all problems given in the book related to circular shapes.

This Story also Contains
  1. NCERT Solutions for Class 9 Maths Chapter 10 Circles
  2. Circles Class 9 Questions And Answers PDF Free Download
  3. Circles Class 9 Solutions - Important Formulae And Points
  4. Circles Class 9 NCERT Solutions (Intext Questions and Exercise)
  5. Summary Of NCERT Maths Circles Class 9
  6. NCERT Solutions For Class 9 Maths Chapter Wise
  7. NCERT Solutions For Class 9 Subject Wise
  8. NCERT Books and NCERT Syllabus
NCERT Solutions for Class 9 Maths Chapter 10 Circles
NCERT Solutions for Class 9 Maths Chapter 10 Circles

The circle is an integral part of unit geometry. A circle divides the plane in which the circle lies into three parts as shown in figure 1, which are the interior of the circle, the exterior of the circle and the circle. In this particular chapter, you will learn the proof of theorems and problems based on those theorems. NCERT class 9 maths chapter 10 question answer are designed in such a way that a student can fetch 100% marks in a particular question. Here you will get NCERT solutions for class 9 Maths also.

Also Read :

Circles Class 9 Questions And Answers PDF Free Download

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Circles Class 9 Solutions - Important Formulae And Points

Concentric Circles: Concentric circles are circles that share the same centre but have different radii.

Arc: An arc of a circle is a continuous portion of the circle.

Chord of a Circle: The chord of a circle is a line segment that connects any two points on the circle.

>> Some Important Properties Of Circle Chords

  • The diameter of a circle is a chord that passes through its centre.

  • A circle's diameter divides it into two equal arcs, forming a semicircle.

  • Congruent arcs have the same degree measure.

  • Equal arcs have associated chords of the same length.

  • A perpendicular drawn from the centre to a chord bisects the chord, and vice versa.

  • Three non-collinear points define one and only one circle.

  • Chords equidistant from the centre are equal in length.

  • The line connecting the centres of two intersecting circles and their common chord are perpendicular.

  • The central angle of an arc is twice the angle it subtends on the circumference.

  • Any two angles in the same circle segment are equal.

  • Equal chords of a circle create equal central angles at the centre.

  • The larger chord of a circle is closer to the centre than the smaller chord.

  • A semicircle contains a right angle.

  • Equal chords in a circle subtend equal angles at the centre.

Cyclic Quadrilateral:

  • A quadrilateral is termed cyclic if all of its vertices lie on the circumference of a circle.

  • The sum of opposite angles in a cyclic quadrilateral is 180°, and vice versa.

  • An exterior angle of a cyclic quadrilateral is equal to its opposite inner angle.

Tangent and Radius:

  • The tangent and radius of a circle intersect at a right angle.

Free download NCERT Solutions for Class 9 Maths Chapter 10 Circles for CBSE Exam.

Circles Class 9 NCERT Solutions (Intext Questions and Exercise)

Class 9 maths chapter 10 NCERT solutions - Exercise: 10.1

Fill in the blanks:

Q1 (i) The centre of a circle lies in _____________ of the circle. (exterior/ interior)

Answer:

The centre of a circle lies in the interior of the circle.

Fill in the blanks:

Q1 (ii) A point, whose distance from the centre of a circle is greater than its radius lies in_____________ of the circle. (exterior/ interior)

Answer:

A point, whose distance from the centre of a circle is greater than its radius lies in exterior of the circle.

Fill in the blanks:

Q1 (iii) The longest chord of a circle is a ___________ of the circle.

Answer:

The longest chord of a circle is a diameter of the circle.

Fill in the blanks:

Q1 (iv) An arc is a ________________ when its ends are the ends of a diameter.

Answer:

An arc is a semi- circle when its ends are the ends of a diameter.

Fill in the blanks:

Q1 (v) Segment of a circle is the region between an arc and ________________ of the circle.

Answer:

Segment of a circle is the region between an arc and chord of the circle.

Fill in the blanks:

Q1 (vi) A circle divides the plane, on which it lies, in _______________ parts.

Answer:

A circle divides the plane, on which it lies, in two parts.

Write True or False: Give reasons for your answers.

Q2 (i) Line segment joining the centre to any point on the circle is a radius of the circle.

Answer:

True. As line segment joining the centre to any point on the circle is a radius of the circle.

Write True or False: Give reasons for your answers.

Q2 (ii) A circle has only finite number of equal chords.

Answer:

False . As a circle has infinite number of equal chords.

Write True or False: Give reasons for your answers.

Q2 (iii) If a circle is divided into three equal arcs, each is a major arc.

Answer:

False. If a circle is divided into three equal arcs, each arc makes angle of 120 degrees whereas major arc makes angle greater than 180 degree at centre.

Write True or False: Give reasons for your answers.

Q2 (iv) A chord of a circle, which is twice as long as its radius, is a diameter of the circle.

Answer:

True.A chord of a circle, which is twice as long as its radius, is a diameter of the circle.

Write True or False: Give reasons for your answers.

Q2 (v) Sector is the region between the chord and its corresponding arc.

Answer:

False. As the sector is the region between the radii and arc.

Write True or False: Give reasons for your answers.

Q2 (vi) A circle is a plane figure.

Answer:

True. A circle is a plane figure.

Circles class 9 NCERT solutions - Exercise: 10.2

Q1 Recall that two circles are congruent if they have the same radii. Prove that equal chords of congruent circles subtend equal angles at their centres.

Answer:

Given: The two circles are congruent if they have the same radii.

To prove: The equal chords of congruent circles subtend equal angles at their centres i.e. BAC= QPR

Proof :

1640237118228

In ABC and PQR,

BC = QR (Given)

AB = PQ (Radii of congruent circle)

AC = PR (Radii of congruent circle)

Thus, ABC PQR (By SSS rule)

BAC= QPR (CPCT)

Q2 Prove that if chords of congruent circles subtend equal angles at their centres, then the chords are equal.

Answer:

Given : chords of congruent circles subtend equal angles at their centres,

To prove : BC = QR

Proof : 1651649779761

In ABC and PQR,

BAC= QPR (Given)

AB = PQ (Radii of congruent circle)

AC = PR (Radii of congruent circle)

Thus, ABC PQR (By SAS rule)

BC = QR (CPCT)

Class 9 circles ncert solutions - Exercise: 10.3

Q1 Draw different pairs of circles. How many points does each pair have in common? What ii the maximum number of common points?

Answer:

1640237144301

In (i) we do not have any common point.

In (ii) we have 1 common point.

In (iii) we have 1 common point.

In (iv) we have 2 common points.

The maximum number of common points is 2.

Q2 Suppose you are given a circle. Give a construction to find its centre.
Answer:

1640237176161

Given : Points P,Q,R lies on circle.

Construction :

1. Join PR and QR

2. Draw perpendicular bisector of PR and QR which intersects at point O.

3. Taking O as centre and OP as radius draw a circle.

4. The circle obtained is required.

Q3 If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord.

Answer:

Given: Two circles intersect at two points.

To prove: their centres lie on the perpendicular bisector of the common chord.

1640237204682

Construction: Joinpoint P and Q to midpoint M of chord AB.

Proof: AB is a chord of circle C(Q,r) and QM is the bisector of chord AB.

PMAB

Hence, PMA=90

Similarly, AB is a chord of circle(Q,r' ) and QM is the bisector of chord AB.

QMAB

Hence, QMA=90

Now, QMA+PMA=90+90=180

PMA and QMA are forming linear pairs so PMQ is a straight line.

Hence, P and Q lie on the perpendicular bisector of common chord AB.

NCERT Solutions for Class 9 Maths Chapter 10 Circles - Exercise: 10.4

Q1 Two circles of radii 5cm and 3cm intersect at two points and the distance between their centres is 4cm . Find the length of the common chord.
Answer:

Given: Two circles of radii 5cm and 3cm intersect at two points and the distance between their centres is 4cm .

To find the length of the common chord.

Construction: Join OP and draw OMABandONCD.

1640237228523

Proof: AB is a chord of circle C(P,3) and PM is the bisector of chord AB.

PMAB

PMA=90

Let, PM = x , so QM=4-x

In APM, using Pythagoras theorem

AM2=AP2PM2 ...........................1

Also,

In AQM, using Pythagoras theorem

AM2=AQ2MQ2 ...........................2

From 1 and 2, we get

AP2PM2=AQ2MQ2

32x2=52(4x)2

9x2=2516x2+8x

9=9+8x

8x=0

x=0

Put,x=0 in equation 1

AM2=3202=9

AM=3

AB=2AM=6

Q2 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.
Answer:

Given: two equal chords of a circle intersect within the circle

To prove: Segments of one chord are equal to corresponding segments of the other chord i.e. AP = CP and BP=DP.

Construction : Join OP and draw OMABandONCD.

Proof :

1640237267424

In OMP and ONP,

AP = AP (Common)

OM = ON (Equal chords of a circle are equidistant from the centre)

OMP = ONP (Both are right angled)

Thus, OMP ONP (By SAS rule)

PM = PN..........................1 (CPCT)

AB = CD ............................2(Given )

12AB=12CD

AM=CN ......................3

Adding 1 and 3, we have

AM + PM = CN + PN

AP=CP

Subtract 4 from 2, we get

AB-AP = CD - CP

PB=PD

Q3 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.
Answer:

Given: two equal chords of a circle intersect within the circle.

To prove: the line joining the point of intersection to the centre makes equal angles with the chords.
i.e. OPM= OPN

Proof :

Construction: Join OP and draw OMABandONCD.

In OMP and ONP,

AP = AP (Common)

OM = ON (Equal chords of a circle are equidistant from the centre)

OMP = ONP (Both are right-angled)

Thus, OMP ONP (By RHS rule)

OPM= OPN (CPCT)

Q4 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 Fig. 10.25 ).

Answer:

Given: a line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D.

To prove : AB = CD

Construction: Draw OMAD

Proof :

1640237314606

BC is a chord of the inner circle and OMBC

So, BM = CM .................1

(Perpendicular OM bisect BC)

Similarly,

AD is a chord of the outer circle and OMAD

So, AM = DM .................2

(Perpendicular OM bisect AD )

Subtracting 1 from 2, we get

AM-BM = DM - CM

AB=CD


Q5 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?

Answer:

Given: From the figure, R, S, M are the position of Reshma, Salma, Mandip respectively.

So, RS = SM = 6 cm

Construction : Join OR,OS,RS,RM and OM.Draw OLRS .

Proof:

1640237339300 In ORS,

OS = OR and OLRS (by construction )

So, RL = LS = 3cm (RS = 6 cm )

In OLS, by pytagoras theorem,

OL2=OS2SL2

OL2=5232=259=16

OL=4

In ORK and OMK,

OR = OM (Radii)

ROK = MOK (Equal chords subtend equal angle at centre)

OK = OK (Common)

ORK OMK (By SAS)

RK = MK (CPCT)

Thus, OKRM

area of ORS = 12×RS×OL ...............................1

area of ORS = 12×OS×KR .............................2

From 1 and 2, we get

12×RS×OL =12×OS×KR

RS×OL=OS×KR

6×4=5×KR

KR=4.8cm

Thus, RM=2KR=2×4.8cm=9.6cm

Q6 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 having a toy telephone in his hands to talk each other. Find the length of the string of each phone.

Answer:

Given: In the figure, A, S, D are positioned Ankur, Syed and David respectively.

So, AS = SD = AD

Radius of circular park = 20 m

so, AO=SO=DO=20 m

Construction: AP SD

Proof :

1640237370579

Let AS = SD = AD = 2x cm

In ASD,

AS = AD and AP SD

So, SP = PD = x cm

In OPD, by Pythagoras,

OP2=OD2PD2

OP2=202x2=400x2

OP=400x2

In APD, by Pythagoras,

AP2=AD2PD2

(AO+OP)2+x2=(2x)2

(20+400x2)2+x2=4x2

400+400x2+40400x2+x2=4x2

800+40400x2=4x2

200+10400x2=x2

10400x2=x2200

Squaring both sides,

100(400x2)=(x2200)2

40000100x2=x440000400x2

x4300x2=0

x2(x2300)=0

x2=300

x=103

Hence, length of string of each phone =2x=203 m

NCERT Solutions for Class 9 Maths Chapter 10 Circles - Exercise: 10.5

Q1 In Fig. 10.36 , A,B and C are three points on a circle with centre O such that BOC=30 and AOB=60 . If D is a point on the circle other than the arc ABC, find ADC .

1640237406418

Answer:

AOC = AOB + BOC= 60+30=90

AOC = 2 ADC (angle subtended by an arc at the centre is double the angle subtended by it at any)

ADC=12AOC

ADC=1290=45

Q2 A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.

Answer:

Given: A chord of a circle is equal to the radius of the circle i.e. OA=OB.

To find: ADB and ACB.

Solution :

1640237429779 In OAB,

OA = AB (Given )

OA = OB (Radii of circle)

So, OA=OB=AB

ABC is a equilateral triangle.

So, AOB = 60

AOB = 2 ADB

ADB=12AOB

ADB=1260=30

ACBD is a cyclic quadrilateral .

So, ACB+ ADB = 180

ACB+30=180

ACB=18030=150

Q3 In Fig. 10.37 , PQR=100 , where P, Q and R are points on a circle with centre O. Find OPR .

1640237467378

Answer:

Construction: Join PS and RS.

PQRS is a cyclic quadrilateral.

So, PSR + PQR = 180

PSR+100=180

PSR=180100=80

Here, POR = 2 PSR

POR=2×80=160

In OPR ,

OP=OR (Radii )

ORP = OPR (the angles opposite to equal sides)

In OPR ,

OPR+ ORP+ POR= 180

2OPR+160=180

2OPR=180160

2OPR=20

OPR=10


Q4 In Fig. 10.38 , ABC=69,ACB=31, find BDC


1640237504699

Answer:

In ABC,

A+ ABC+ ACB= 180

A+69+31=180

A+100=180

A=180100

A=80

A = BDC = 80 (Angles in same segment)

Q5 In Fig. 10.39 , A, B, C and D are four points on a circle. AC and BD intersect at a point E such that BEC=130 and ECD=20 . Find BAC

1640237534860 .

Answer:

DEC+ BEC = 180 (linear pairs)

DEC+ 130 = 180 ( BEC = 130 )

DEC = 180 - 130

DEC = 50

In DEC,

D+ DEC+ DCE = 180

D+50+20=180

D+70=180

D=18070=110

D = BAC (angles in same segment are equal )

BAC = 110


Q6 ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If DBC=70 , BAC is 30 , find BCD . Further, if AB=BC , find ECD .

Answer:

1640237557595

BDC=BAC (angles in the same segment are equal )

BDC=30

In BDC,

BCD+BDC+DBC=180

BCD+30+70=180

BCD+100=180

BCD=180100=80

If AB = BC ,then

BCA=BAC

BCA=30

Here, ECD+BCE=BCD

ECD+30=80

ECD=8030=50


Q7 If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.

Answer:

1640237583763

AC is the diameter of the circle.

Thus, ADC=90 and ABC=90 ............................1(Angle in a semi-circle is a right angle)

Similarly, BD is the diameter of the circle.

Thus, BAD=90 and BCD=90 ............................2(Angle in a semi-circle is a right angle)

From 1 and 2, we get

BCD=ADC=ABC=BAD=90

Hence, ABCD is a rectangle.

Q8 If the non-parallel sides of a trapezium are equal, prove that it is cyclic.

Answer:

1640237665409

Given: ABCD is a trapezium.

Construction: Draw AD || BE.

Proof: In quadrilateral ABED,

AB || DE (Given )

AD || BE ( By construction )

Thus, ABED is a parallelogram.

AD = BE (Opposite sides of parallelogram )

AD = BC (Given )

so, BE = BC

In EBC,

BE = BC (Proved above )

Thus, C=2 ...........1(angles opposite to equal sides )

A=1 ...............2(Opposite angles of the parallelogram )

From 1 and 2, we get

1+2=180 (linear pair)

A+C=180

Thus, ABED is a cyclic quadrilateral.


Q9 Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see Fig. 10.40 ). Prove that ACP=QCD .

1640237697041

Answer:

1640237709420

ABP=QBD ................1(vertically opposite angles)

ACP=ABP ..................2(Angles in the same segment are equal)

QBD=QCD .................3(angles in the same segment are equal)

From 1,2,3 ,we get

ACP=QCD

Q10 If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.

Answer:

Given: circles are drawn taking two sides of a triangle as diameters.

Construction: Join AD.

1640237735195

Proof: AB is the diameter of the circle and ADB is formed in a semi-circle.

ADB = 90 ........................1(angle in a semi-circle)

Similarly,

AC is the diameter of the circle and ADC is formed in a semi-circle.

ADC = 90 ........................2(angle in a semi-circle)

From 1 and 2, we have

ADB+ ADC= 90 + 90 = 180

ADB and ADC are forming a linear pair. So, BDC is a straight line.

Hence, point D lies on this side.


Q11 ABC and ADC are two right triangles with common hypotenuse AC. Prove that
CAD=CBD .

Answer:

Given: ABC and ADC are two right triangles with common hypotenuse AC.

To prove : CAD=CBD

Proof :

1640237758812

Triangle ABC and ADC are on common base BC and BAC = BDC.

Thus, point A,B,C,D lie in the same circle.

(If a line segment joining two points subtends equal angles at two other points lying on the same side of line containing line segment, four points lie on the circle.)

CAD = CBD (Angles in same segment are equal)


Q12 Prove that a cyclic parallelogram is a rectangle.
Answer:

Given: ABCD is a cyclic quadrilateral.

To prove: ABCD is a rectangle.

Proof :

1640237806403

In cyclic quadrilateral ABCD.

A+C=180 .......................1(sum of either pair of opposite angles of a cyclic quadrilateral)

A=C ........................................2(opposite angles of a parallelogram are equal )

From 1 and 2,

A+A=180

2A=180

A=90

We know that a parallelogram with one angle right angle is a rectangle.

Hence, ABCD is a rectangle.

NCERT maths chapter 10 class 9 - Exercise: 10.6

Q1 Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.

Answer:

Given: Circle C(P,r) and circle C(Q,r') intersect each other at A and B.

To prove : PAQ = PBQ

Proof : In APQ and BPQ,

PA = PB (radii of same circle)

PQ = PQ (Common)

QA = QB (radii of same circle)

So, APQ BPQ (By SSS)

PAQ = PBQ (CPCT)


Q2 Two chords AB and CD of lengths 5cm and 11cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is 6cm , find the radius of the circle.

Answer:

Given : AB = 5 cm, CD = 11 cm and AB || CD.

To find Radius (OA).

Construction: Draw OMCDandONAB

Proof :

1640237834315

Proof: CD is a chord of circle and OMCD

Thus, CM = MD = 5.5 cm (perpendicular from centre bisects chord)

and AN = NB = 2.5 cm

Let OM be x.

So, ON = 6 - x (MN = 6 cm )

In OCM , using Pythagoras,

OC2=CM2+OM2 .............................1

and

In OAN , using Pythagoras,

OA2=AN2+ON2 .............................2

From 1 and 2,

CM2+OM2=AN2+ON2 (OC=OA =radii)

5.52+x2=2.52+(6x)2

30.25+x2=6.25+36+x212x

30.2542.25=12x

12=12x

x=1

From 2, we get

OC2=5.52+12=30.25+1=31.25

OC=525cm

OA = OC

Thus, the radius of the circle is 525cm

Q3 The lengths of two parallel chords of a circle are 6cm and 8cm . If the smaller chord is at distance 4cm from the centre, what is the distance of the other chord from the centre?

Answer:

Given : AB = 8 cm, CD = 6 cm , OM = 4 cm and AB || CD.

To find: Length of ON

Construction: Draw OMCDandONAB

Proof :

1640237861219

Proof: CD is a chord of circle and OMCD

Thus, CM = MD = 3 cm (perpendicular from centre bisects chord)

and AN = NB = 4 cm

Let MN be x.

So, ON = 4 - x (MN = 4 cm )

In OCM , using Pythagoras,

OC2=CM2+OM2 .............................1

and

In OAN , using Pythagoras,

OA2=AN2+ON2 .............................2

From 1 and 2,

CM2+OM2=AN2+ON2 (OC=OA =radii)

32+42=42+(4x)2

9+16=16+16+x28x

9=16+x28x

x28x+7=0

x27xx+7=0

x(x7)1(x7)=0

(x1)(x7)=0

x=1,7

So, x=1 (since x7>OM )

ON =4-x =4-1=3 cm

Hence, second chord is 3 cm away from centre.

Q4 Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.

Answer:

Given : AD = CE

To prove : ABC=12(AOCDOE)

Construction: Join AC and DE.

Proof :

1640237883312

Let ADC = x , DOE = y and AOD = z

So, EOC = z (each chord subtends equal angle at centre)

AOC + DOE + AOD + EOC = 360

x+y+z+z=360

x+y+2z=360 .........................................1

In OAD ,

OA = OD (Radii of the circle)

OAD = ODA (angles opposite to equal sides )

OAD + ODA + AOD = 180

2OAD+z=180

2OAD=180z

OAD=180z2

OAD=90z2 .............................................................2

Similarly,

OCE=90x2 .............................................................3

OED=90y2 ..............................................................4

ODB is exterior of triangle OAD . So,

ODB = OAD + ODA

ODB=90z2+z (from 2)

ODB=90+z2 .................................................................5

similarly,

OBE is exterior of triangle OCE . So,

OBE = OCE + OEC

OEB=90z2+z (from 3)

OEB=90+z2 .................................................................6

From 4,5,6 ;we get

BDE = BED = OEB - OED

BDE=BED=90+z2(90y2)=y+z2

BDE+BED=y+z ..................................................7

In BDE ,

DBE + BDE + BED = 180

DBE+y+z=180

DBE=180(y+z)

ABC=180(y+z) ...................................................8

Here, from equation 1,

xy2=360y2xy2

xy2=3602y2x2

xy2=180yx ...................................9

From 8 and 9,we have

ABC=xy2=12(AOCDOE)

Q5 Prove that the circle drawn with any side of a rhombus as diameter, passes through the point of intersection of its diagonals.

Answer:

Given : ABCD is a rhombus.

To prove: the circle drawn with AB as diameter passes through the point O.

Proof :

1640237931704

ABCD is rhombus.

Thus, AOC=90 (diagonals of a rhombus bisect each other at 90 )

So, a circle drawn AB as diameter will pass through point O.

Thus, the circle is drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.

Q6 ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that AE=AD .

Answer:

Given: ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E.

To prove : AE = AD

Proof :

1640237957576

ADC = 3 , ABC = 4, ADE = 1 and AED = 2

3+1=180 .................1(linear pair)

2+4=180 ....................2(sum of opposite angles of cyclic quadrilateral)

3 = 4 (oppsoite angles of parallelogram )

From 1 and 2,

3+ 1 = 2 + 4

From 3, 1 = 2

From 4, AQB, 1 = 2

Therefore, AE = AD (In an isosceles triangle ,angles oppsoite to equal sides are equal)

Q7 (i) AC and BD are chords of a circle which bisect each other. Prove that AC and BD are diameters

Answer:

Given: AC and BD are chords of a circle which bisect each other.

To prove: AC and BD are diameters.

Construction : Join AB,BC,CD,DA.

Proof :

1640237982446

In ABD and CDO,

AO = OC (Given )

AOB = COD (Vertically opposite angles )

BO = DO (Given )

So, ABD CDO (By SAS)

BAO = DCO (CPCT)

BAO and DCO are alternate angle and are equal .

So, AB || DC ..............1

Also AD || BC ...............2

From 1 and 2,

A+C=180 ......................3(sum of opposite angles)

A = C ................................4(Opposite angles of the parallelogram )

From 3 and 4,

A+A=180

2A=180

A=90

BD is a diameter of the circle.

Similarly, AC is a diameter.

Q7 (ii) AC and BD are chords of a circle which bisect each other. Prove that ABCD is a rectangle.

Answer:

Given: AC and BD are chords of a circle which bisect each other.

To prove: ABCD is a rectangle.

Construction : Join AB,BC,CD,DA.

Proof :

1651650039652

ABCD is a parallelogram. (proved in (i))

A=90 (proved in (i))

A parallelogram with one angle 90 , is a rectangle )

Thus, ABCD is rectangle.

Q8 Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively. Prove that the angles of the triangle DEF are 9012C , 9012B and 9012A

Answer:

Given : Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively.

To prove : the angles of the triangle DEF are 9012C , 9012B and 9012A

Proof :

1640238007328

1 and 3 are angles in same segment.therefore,

1 = 3 ................1(angles in same segment are equal )

and 2 = 4 ..................2

Adding 1 and 2,we have

1+ 2= 3+ 4

D=12B+12C ,

D=12(B+C)

D=12(180+C)

and D=12(180A)

D=9012A

Similarly, E=9012B and F=9012C

Q9 Two congruent circles intersect each other at points A and B. Through A any line segment PAQ is drawn so that P, Q lie on the two circles. Prove that BP=BQ .

Answer:

Given: Two congruent circles intersect each other at points A and B. Through A any line segment PAQ is drawn so that P, Q lie on the two circles.

To prove : BP = BQ

Proof :

1640238028954

AB is a common chord in both congruent circles.

APB=AQB

In BPQ,

APB=AQB

BQ=BP (Sides opposite to equal of the triangle are equal )

Q10 In any triangle ABC, if the angle bisector of A and perpendicular bisector of BC intersect, prove that they intersect on the circumcircle of the triangle ABC.

Answer:

Given :In any triangle ABC, if the angle bisector of A and perpendicular bisector of BC intersect.

To prove : D lies on perpendicular bisector BC.

Construction: Join BD and DC.

Proof :

1640238078439

Let ABD = 1 , ADC = 2 , DCB = 3 , CBD = 4

1 and 3 lies in same segment.So,

1 = 3 ..........................1(angles in same segment)

similarly, 2 = 4 ......................2

also, 1= 2 ..............3(given)

From 1,2,3 , we get

3 = 4

Hence, BD = DC (angles opposite to equal sides are equal )

All points lying on perpendicular bisector BC will be equidistant from B and C.

Thus, point D also lies on perpendicular bisector BC.

Summary Of NCERT Maths Circles Class 9

  • A circle is a closed figure in which all points on the boundary are equidistant from a fixed point called the center of the circle.
  • The distance between the center and any point on the circle is called the radius of the circle.
  • The diameter of a circle is twice the radius.
  • The circumference of a circle is the distance around its boundary and is given by the formula C = 2πr, where r is the radius of the circle and π is the mathematical constant pi (approximately equal to 3.14).
  • The area of a circle is given by the formula A = πr², where r is the radius of the circle.
  • The chord of a circle is a line segment joining any two points on the circle.
  • The diameter is the longest chord of a circle and passes through the center.
  • The tangent to a circle is a line that touches the circle at only one point.
  • A line drawn perpendicular to a tangent at the point of contact is called the normal to the circle at that point.
  • The angle between a tangent and a chord drawn from the point of contact is equal to the angle in the alternate segment.
  • If a line intersects two chords of a circle, then the product of the segments of one chord is equal to the product of the segments of the other chord.
  • The angle subtended by an arc at the center of a circle is double the angle subtended by it at any point on the remaining part of the circle.
  • The angle in a semicircle is a right angle.

Interested students can practice these class 9 maths ch 10 question answer using the exercise solutions provided below.

NCERT Solutions For Class 9 Maths Chapter Wise

Chapter No. Chapter Name
Chapter 1 Number Systems
Chapter 2 Polynomials
Chapter 3 Coordinate Geometry
Chapter 4 Linear Equations In Two Variables
Chapter 5 Introduction to Euclid's Geometry
Chapter 6 Lines And Angles
Chapter 7 Triangles
Chapter 8 Quadrilaterals
Chapter 9 Areas of Parallelograms and Triangles
Chapter 10 Circles
Chapter 11 Constructions
Chapter 12 Heron’s Formula
Chapter 13 Surface Area and Volumes
Chapter 14 Statistics
Chapter 15 Probability

NCERT Solutions For Class 9 Subject Wise

How To Use NCERT Solutions For Class 9 Maths Chapter 10 Circles

  • Learn and memorize some theorems related to circles
  • Learn the application of those theorems in the problems
  • Start applying the theorems and concepts on the practice exercises.
  • During the practice exercises, you can take the help of NCERT solutions for class 9 maths chapter 10 Circles.
  • After doing all the above-written things, you can practice more using the previous year questions papers.

NCERT Books and NCERT Syllabus

Keep working hard & happy learning!

Frequently Asked Questions (FAQs)

1. What are the important topics in chapter ch 10 maths class 9 Circles ?

Circles and the related terms, angle subtended by a chord at a point, perpendicular from the centre to a chord, circle through three points, equal chords and their distances from the centre, angle subtended by an arc of a circle are the important topics of this chapter.

2. What benefits do Class 9 students derive from studying NCERT Solutions for Maths Chapter 10?

Students in Class 9 can benefit from class 9 chapter 10 maths by gaining a thorough understanding of all the concepts within the subject, which can serve as the basis for their future academic pursuits. The solutions provided by Careers360 experts are designed in a clear and comprehensive manner, making it easier for students to solve complex problems with greater efficiency. By mastering these solutions, CBSE Class 9 students can establish a strong foundation in the fundamentals and achieve excellent scores in their final exams.

3. What is the rationale behind adhering to circles class 9 NCERT solutions?

The use of class 9th circles NCERT solutions  is a reliable and effective approach to help students develop mastery of the subject's concepts. Reviewing these solutions, in conjunction with the textbooks, can aid in solving any problems that may arise on the board exams. These solutions also enhance students' problem-solving skills and logical reasoning abilities. They are among the most widely used study materials for CBSE exams. Consistent practice with these solutions can improve students' performance and help them excel in the subject.

4. Where can I find the complete solutions of NCERT for class 9 ?

Here you will get the detailed NCERT solutions for class 9. you can practice these maths ch 10 class 9 solutions and problems to command the concepts discussed in the chapter. after practicing you will get confidence to solve any kind of problem related to circle given in class 9th.

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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

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2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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