RS Aggarwal Solutions for Class 9 Chapter 12: Circles

The students can boost their exam preparation by downloading the RS Aggarwal Solutions which are available in PDF format. These solutions are prepared by experts at BYJU’S with the objective of helping students score well in the exams. The answers mainly contain step by step explanations in understandable language to help students master the subject. By solving the exercise wise problems on a regular basis, the students can recognise the areas which require improvement.

The experts provide solutions which are comfortable for the students to solve and clear the exams with a good score. All the solutions contain various methods which can be used for solving them based on the CBSE syllabus. RS Aggarwal Solutions for Class 9 Chapter 12 Circles are provided here.

RS Aggarwal Solutions for Class 9 Chapter 12: Circles Download PDF

 

RS Aggarwal Solutions Class 9 Maths Chapter 12
RS Aggarwal Solutions Class 9 Maths Chapter 12
RS Aggarwal Solutions Class 9 Maths Chapter 12
RS Aggarwal Solutions Class 9 Maths Chapter 12
RS Aggarwal Solutions Class 9 Maths Chapter 12
RS Aggarwal Solutions Class 9 Maths Chapter 12
RS Aggarwal Solutions Class 9 Maths Chapter 12
RS Aggarwal Solutions Class 9 Maths Chapter 12
RS Aggarwal Solutions Class 9 Maths Chapter 12
RS Aggarwal Solutions Class 9 Maths Chapter 12
RS Aggarwal Solutions Class 9 Maths Chapter 12
RS Aggarwal Solutions Class 9 Maths Chapter 12
RS Aggarwal Solutions Class 9 Maths Chapter 12
RS Aggarwal Solutions Class 9 Maths Chapter 12
RS Aggarwal Solutions Class 9 Maths Chapter 12
RS Aggarwal Solutions Class 9 Maths Chapter 12
RS Aggarwal Solutions Class 9 Maths Chapter 12
RS Aggarwal Solutions Class 9 Maths Chapter 12
RS Aggarwal Solutions Class 9 Maths Chapter 12
RS Aggarwal Solutions Class 9 Maths Chapter 12
RS Aggarwal Solutions Class 9 Maths Chapter 12

RS Aggarwal Solutions Class 9 Maths Chapter 12
RS Aggarwal Solutions Class 9 Maths Chapter 12
RS Aggarwal Solutions Class 9 Maths Chapter 12
RS Aggarwal Solutions Class 9 Maths Chapter 12
RS Aggarwal Solutions Class 9 Maths Chapter 12
RS Aggarwal Solutions Class 9 Maths Chapter 12
RS Aggarwal Solutions Class 9 Maths Chapter 12
RS Aggarwal Solutions Class 9 Maths Chapter 12
RS Aggarwal Solutions Class 9 Maths Chapter 12
RS Aggarwal Solutions Class 9 Maths Chapter 12
RS Aggarwal Solutions Class 9 Maths Chapter 12
RS Aggarwal Solutions Class 9 Maths Chapter 12
RS Aggarwal Solutions Class 9 Maths Chapter 12
RS Aggarwal Solutions Class 9 Maths Chapter 12
RS Aggarwal Solutions Class 9 Maths Chapter 12
RS Aggarwal Solutions Class 9 Maths Chapter 12
RS Aggarwal Solutions Class 9 Maths Chapter 12
RS Aggarwal Solutions Class 9 Maths Chapter 12
RS Aggarwal Solutions Class 9 Maths Chapter 12

RS Aggarwal Solutions Class 9 Maths Chapter 12
RS Aggarwal Solutions Class 9 Maths Chapter 12
RS Aggarwal Solutions Class 9 Maths Chapter 12
RS Aggarwal Solutions Class 9 Maths Chapter 12
RS Aggarwal Solutions Class 9 Maths Chapter 12
RS Aggarwal Solutions Class 9 Maths Chapter 12
RS Aggarwal Solutions Class 9 Maths Chapter 12
RS Aggarwal Solutions Class 9 Maths Chapter 12
RS Aggarwal Solutions Class 9 Maths Chapter 12
RS Aggarwal Solutions Class 9 Maths Chapter 12
RS Aggarwal Solutions Class 9 Maths Chapter 12
RS Aggarwal Solutions Class 9 Maths Chapter 12
RS Aggarwal Solutions Class 9 Maths Chapter 12
RS Aggarwal Solutions Class 9 Maths Chapter 12
RS Aggarwal Solutions Class 9 Maths Chapter 12
RS Aggarwal Solutions Class 9 Maths Chapter 12
RS Aggarwal Solutions Class 9 Maths Chapter 12
RS Aggarwal Solutions Class 9 Maths Chapter 12
RS Aggarwal Solutions Class 9 Maths Chapter 12
RS Aggarwal Solutions Class 9 Maths Chapter 12

 

Access RS Aggarwal Solutions for Class 9 Chapter 12: Circles

Exercise 12(A) page: 436

1. A chord of length 16cm is drawn in a circle of radius 10cm. Find the distance of the chord from the centre of the circle.

Solution:

Consider AB as the chord with O as the centre and radius 10cm

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12A 1

So we get

OA = 10 cm and AB = 16cm

Construct OL ⊥ AB

Perpendicular from the centre of a circle to a chord bisects the chord

So we get

AL = ½ × AB

By substituting the values

AL = ½ × 16

So we get

AL = 8 cm

Consider △ OLA

Using the Pythagoras theorem it can be written as

OA2 = OL2 + AL2

By substituting the values we get

102 = OL2 + 82

On further calculation

OL2 = 102 – 82

So we get

OL2 = 100 – 64

By subtraction

OL2 = 36

By taking the square root

OL = √36

So we get

OL = 6cm

Therefore, the distance of the chord from the centre of the circle is 6cm.

2. Find the length of a chord which is at a distance of 3cm from the centre of a circle of radius 5cm.

Solution:

Consider AB as the chord of the circle with O as the centre and radius 5cm.

Construct OL ⊥ AB

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12A 2

It is given that OA = 5cm and OL = 3cm

Perpendicular from the centre of a circle to a chord bisects the chord

Consider △ OLA

Using the Pythagoras theorem it can be written as

OA2 = OL2 + AL2

By substituting the values we get

52 = OL2 + 32

On further calculation

OL2 = 52 – 32

So we get

OL2 = 25 – 9

By subtraction

OL2 = 16

By taking the square root

OL = √16

So we get

OL = 4cm

We know that

AB = 2AL

By substituting the values

AB = 2 × 4

So we get

AB = 8cm

Therefore, length of the chord is 8cm.

3. A chord of length 30cm is drawn at a distance of 8cm from the centre of a circle. Find out the radius of the circle.

Solution:

Consider AB as as the chord of the circle with O as the centre

Construct OL ⊥ AB

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12A 3

From the figure we know that OL is the distance from the centre of chord

It is given that AB = 30cm and OL = 8cm

Perpendicular from the centre of a circle to a chord bisects the chord

So we get

AL = ½ × AB

By substituting the values

AL = ½ × 30

By division

AL = 15cm

Consider △ OLA

Using the Pythagoras theorem it can be written as

OA2 = OL2 + AL2

By substituting the values we get

OA2 = 82 + 152

On further calculation

OA2 = 64 + 225

By addition

OA2 = 289

By taking the square root

OA = √289

So we get

OA = 17cm

Therefore, the radius of the circle is 17cm.

4. In a circle of radius 5cm, AB and CD are two parallel chords of lengths 8cm and 6cm respectively. Calculate the distance between the chords if they are

(i) on the same side of the centre,

(ii) on the opposite sides of the centre.

Solution:

(i) Consider AB and CD as the two chords of the circle where AB || CD lying on the same side of the circle

It is given that AB = 8cm and OB = OD = 5cm

Draw a line to join the points OL and LM

Perpendicular from the centre of a circle to a chord bisects the chord

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12A 4

We know that

LB = ½ × AB

By substituting the values we get

LB = ½ × 8

So we get

LB = 4cm

We know that

MD = ½ × CD

By substituting the values we get

MD = ½ × 6

So we get

MD = 3cm

Consider △ BLO

Using the Pythagoras theorem it can be written as

OB2 = LB2 + LO2

By substituting the values we get

52 = 42 + LO2

On further calculation

LO2 = 52 – 42

So we get

LO2 = 25 – 16

By subtraction

LO2 = 9

By taking the square root

LO = √9

LO = 3cm

Consider △ DMO

Using the Pythagoras theorem it can be written as

OD2 = MD2 + MO2

By substituting the values we get

52 = 32 + MO2

On further calculation

MO2 = 52 – 32

So we get

MO2 = 25 – 9

By subtraction

MO2 = 16

By taking the square root

MO = √16

MO = 4cm

So the distance between the chords = MO – LO

By substituting the values

Distance between the chords = 4 – 3 = 1cm

Therefore, the distance between the chords on the same side of the centre is 1cm.

(ii) Consider AB and CD as the chords of the circle and AB || CD on the opposite sides of the centre

It is given that AB = 8cm and CD = 6cm.

Construct OL ⊥ AB and OM ⊥ CD

Join the diagonals OA and OC

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12A 5

We know that OA = OC = 5cm

Perpendicular from the centre of a circle to a chord bisects the chord

We know that AL = ½ × AB

By substituting the values we get

AL = ½ × 8

So we get

AL = 4cm

We know that CM = ½ × CD

By substituting the values we get

CM = ½ × 6

So we get

CM = 3cm

Consider △ OLA

Using the Pythagoras theorem it can be written as

OA2 = AL2 + OL2

By substituting the values

52 = 42 + OL2

On further calculation

OL2 = 52 – 42

So we get

OL2 = 25 – 16

By subtraction

OL2 = 9

By taking the square root

OL = √9

OL = 3cm

Consider △ OMC

Using the Pythagoras theorem it can be written as

OC2 = OM2 + CM2

By substituting the values

52 = OM2 +32

On further calculation

OM2 = 52 – 32

OM2 = 25 – 9

By subtraction

OM2 = 16

By taking the square root

OM = √16

OM = 4cm

Distance between the chords = OM + OL

So we get

Distance between the chords = 4 + 3 = 7cm

Therefore, the distance between the chords on the opposite sides of the centre is 7 cm.

5. Two parallel chords of lengths 30cm and 16cm are drawn on the opposite sides of the centre of a circle of radius 17cm. Find the distance between the chords.

Solution:

Consider AB and CD as the chords of circle with centre O

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12A 6

It is given that AB = 30cm and CD = 16cm

Join the lines OA and OC

We know that AO = 17cm and CO = 17cm

Construct OM ⊥ CD and OL ⊥ AB

Perpendicular from the centre of a circle to a chord bisects the chord

We know that

AL = ½ × AB

By substituting the values

AL = ½ × 30

So we get

AL = 15cm

We know that

CM = ½ × CD

By substituting the values

CM = ½ × 16

So we get

CM = 8cm

Consider △ ALO

Using the Pythagoras theorem it can be written as

AO2 = OL2 + AL2

By substituting the values

172 = OL2 + 152

So we get

OL2 = 172 – 152

On further calculation

OL2 = 289 – 225

By subtraction

OL2 = 64

By taking the square root

OL = √64

OL = 8cm

Consider △ CMO

Using the Pythagoras theorem it can be written as

CO2 = CM2 + OM2

By substituting the values

172 = 82 + OM2

So we get

OM2 = 172 – 82

On further calculation

OM2 = 289 – 64

By subtraction

OM2 = 225

By taking the square root

OM = √225

OM = 15cm

So the distance between the chords = OM + OL

By substituting the values

Distance between the chords = 8 + 15 = 23cm

Therefore, the distance between the chords is 23 cm.

6. In the given figure, the diameter CD of a circle with centre O is perpendicular to chord AB. If AB = 12 cm and CE = 3 cm, calculate the radius of the circle.

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12A 7

Solution:

From the figure we know that CD is the diameter of the circle with centre O which is perpendicular to chord AB.

Draw the line OA.

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12A 8

It is given that AB = 12cm and CE = 3cm

Consider OA = OC = r cm

It can be written as

OE = (r – 3) cm

Perpendicular from the centre of a circle to a chord bisects the chord

We know that

AE = ½ × AB

By substituting the values

AE = ½ × 12

So we get

AE = 6 cm

Consider △ OEA

By using the Pythagoras theorem

OA2 = OE2 + AE2

By substituting the values

r2 = (r – 3)2 + 62

So we get

r2 = r2 – 6r + 9 + 36

On further calculation

r2 – r2 + 6r = 45

So we get

6r = 45

By division

r = 45/6

r = 7.5 cm

Therefore, the radius of the circle is 7.5 cm.

7. In the given figure, a circle with centre O is given in which a diameter AB bisects the chord CD at a point E such that CE = ED = 8cm and EB = 4cm. Find the radius of the circle.

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12A 9

Solution:

Consider AB as the diameter with centre O which bisects the chord CD at E

It is given that

CE = ED = 8cm and EB = 4cm

Join the diagonal OC

Consider OC = OB = r cm

It can be written as

OE = (r – 4) cm

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12A 10

Consider △ OEC

By using the Pythagoras theorem

OC2 = OE2 + EC2

By substituting the values we get

r2 = (r – 4)2 + 82

On further calculation

r2 = r2 – 8r + 16 + 64

So we get

r2 = r2 – 8r + 80

It can be written as

r2 – r2 + 8r = 80

We get

8r = 80

By division

r = 10 cm

Therefore, the radius of the circle is 10 cm.

8. In the adjoining figure, OD is perpendicular to the chord AB of a circle with centre O. If BC is a diameter, show that AC || DO and AC = 2 × OD.

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12A 11

Solution:

Perpendicular from the centre of a circle to a chord bisects the chord

We know that OD ⊥ AB

From the figure we know that D is the midpoint of AB

We get

AD = BD

We also know that O is the midpoint of BC

We get

OC = OB

Consider △ ABC

Using the midpoint theorem

We get OD || AC and

OD = ½ × AC

By cross multiplication

AC = 2 × OD

Therefore, it is proved that AC || DO and AC = 2 × OD.

9. In the given figure, O is the centre of a circle in which chords AB and CD intersect at P such that PO bisects ∠BPD. Prove that AB = CD.

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12A 12

Solution:

Consider △ OEP and △ OFP

We know that

∠OEP = ∠OFP = 90o

OP is common i.e. OP = OP

From the figure we know that OP bisects ∠BPD

It can be written as

∠OPE = ∠OPF

By ASA congruence criterion

△ OEP ≅ △ OFP

OE = OF (c. p. c. t)

We know that AB and CD are equidistant from the centre

So we get

AB = CD

Therefore, it is proved that AB = CD.

10. Prove that the diameter of a circle perpendicular to one of the two parallel chords of a circle is perpendicular to the other and bisects it.

Solution:

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12A 13

Consider AB || CD and POQ as the diameter

It is given that ∠PEB = 90o

From the figure we know that AB || CD and ∠PFD and ∠PEB are corresponding angles

So we get

∠PFD = ∠PEB

It can be written as

PF ⊥ CD

In the same way

OF ⊥ CD

Perpendicular from the centre of a circle to a chord bisects the chord

So we get

CF = FD

Therefore, it is proved that the diameter of a circle perpendicular to one of the two parallel chords of a circle is perpendicular to the other and bisects it.

11. Prove that two different circles cannot intersect each other at more than two points.

Solution:

Consider two different circles intersecting at three point A, B and C

We know that these points are non collinear and a unique circle can be drawn using these points

This shows that our assumption is wrong

Therefore, it is proved that two different circles cannot intersect each other at more than two points.

12. Two circles of radii 10 cm and 8 cm intersect each other, and the length of the common chord is 12 cm. Find the distance between their centres.

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12A 14

Solution:

It is given that

OA = 10cm and AB = 12 cm

So we get

AD = ½ × AB

By substituting the values

AD = ½ × 12

By division

AD = 6cm

Consider △ ADO

Using the Pythagoras theorem

OA2 = AD2 + OD2

By substituting the values we get

102 = 62 + OD2

On further calculation

OD2 = 102 – 62

So we get

OD2 = 100 – 36

By subtraction

OD2 = 64

By taking the square root

OD = √64

So we get

OD = 8cm

We know that O’A = 8cm

Consider △ ADO’

Using the Pythagoras theorem

O’A2 = AD2 + O’D2

By substituting the values we get

82 = 62 + O’D2

On further calculation

O’D2 = 82 – 62

So we get

O’D2 = 64 – 36

By subtraction

O’D2 = 28

By taking the square root

O’D = √28

We get

O’D = 2 √7

We know that

OO’ = OD + O’D

By substituting the values

OO’ = (8 + 2 √7) cm

Therefore, the distance between their centres is (8 + 2 √7) cm.

13. Two equal circles intersect in P and Q. A straight line through P meets the circles in A and B. Prove that QA = QB.

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12A 15

Solution:

We know that two circles will be congruent if they have equal radii

From the figure we know that if the two chords are equal then the corresponding arcs are congruent

We know that PQ is the common chord in both the circles

So their corresponding arcs are equal

It can be written as

Arc PCQ = arc PDQ

We know that the congruent arcs have the same degree

So we get

∠QAP = ∠QBP

We know that the base angles of isosceles triangle are equal

So we get

QA = QB

Therefore, it is proved that QA = QB.

14. If a diameter of a circle bisects each of the two chords of a circle then prove that the chords are parallel.

Solution:

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12A 16

It is given that AB and CD are the two chords of the circle having O as the centre

We know that POQ bisects them at the points L and M

So we get

OL ⊥ AB and OM ⊥ CD

We know that the alternate angles are equal

∠ALM = ∠LMD

We get

AB || CD

Therefore, it is proved that the chords are parallel.

15. In the adjoining figure, two circles with centres at A and B, and of radii 5cm and 3cm touch each other internally. If the perpendicular bisector of AB meets the bigger circle in P and Q, find the length of PQ.

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12A 17

Solution:

It is given that two circles with centres at A and B, and of radii 5cm and 3cm touch each other internally.

The perpendicular bisector of AB meets the bigger circle in P

Join the line AP

From the figure we know that PQ intersects the line AB at L

So we get

AB = (5 – 3)

AB = 2cm

We know that that PQ is the perpendicular bisector of AB

So we get

AL = ½ × AB

Substituting the values

AL = ½ × 2

So we get

AL = 1 cm

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12A 18

Consider △ PLA

Using the Pythagoras theorem

AP2 = AL2 + PL2

By substituting the values we get

52 = 12 + PL2

So we get

PL2 = 52 – 12

On further calculation

PL2 = 25 – 1

PL2 = 24

By taking the square root

PL = √24

So we get

PL = 2 √6 cm

We know that

PQ = 2 × PL

By substituting the values

PQ = 2 × 2 √6

PQ = 4 √6 cm

Therefore, the length of PQ = 4 √6 cm.

16. In the given figure, AB is a chord of a circle with centre O and AB is produced to C such that BC = OB. Also, CO is joined and produced to meet the circle in D. If ∠ACD = yo and ∠AOD = xo, prove that x = 3y.

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12A 19

Solution:

It is given that AB is a chord of a circle with centre O and AB is produced to C such that BC = OB

We know that ∠BOC and ∠BCO form isosceles triangle

∠BOC = ∠BCO = yo

We know that OA and OB are the radii of same circle

OA = OB

We know that ∠OAB and ∠OBA form isosceles triangle

∠OAB = ∠OBA = 2yo

From the figure we know that

Exterior ∠AOD = ∠OAC + ∠ACO

It can be written as

Exterior ∠AOD = ∠OAB + ∠BCO

So we get

Exterior ∠AOD = 3yo

It is given that

∠AOD = xo

So we get

xo = 3yo

Therefore, it is proved that x = 3y.

17. AB and AC are two chords of a circle of radius r such that AB = 2AC. If p and q are the distances of AB and AC from the centre then prove that 4q2 = p2 + 3r2.

Solution:

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12A 20

Consider O as the centre of the circle with radius r

So we get

OB = OC = r

Consider AC = x and AB = 2x

We know that OM ⊥ AB

So we get

OM = p

We know that ON ⊥ AC

So we get

ON = q

Consider △ OMB

Using the Pythagoras theorem

OB2 = OM2 + BM2

We know that the perpendicular from the centre of the circle bisects the chord

So we get

r2 = p2 + ((1/2) AB) 2

It can be written as

r2 = p2 + ¼ × 4x2

So we get

r2 = p2 + x2 ……… (1)

Consider △ ONC

Using the Pythagoras theorem

OC2 = ON2 + CN2

We know that the perpendicular from the centre of the circle bisects the chord

So we get

r2 = q2 + ((1/2) AC) 2

It can be written as

r2 = q2 + x2/4

We get

q2 = r2 – x2/4

Multiplying the equation by 4

4q2 = 4r2 – x2

Substituting equation (1)

4q2 = 4r2 – (r2 – p2)

So we get

4q2 = 3r2 + p2

Therefore, it is proved that 4q2 = p2 + 3r2.

18. In the adjoining figure, O is the centre of a circle. If AB and AC are chords of the circle such that AB = AC, OP ⊥ AB and OQ ⊥ AC, prove that PB = QC.

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12A 21

Solution:

It is given that AB = AC

Dividing the equation by 2

We get

½ AB = ½ AC

Perpendicular from the centre of a circle to a chord bisects the chord

MB = NC ……. (1)

We know that the equal chords are equidistant from the centre of the circle

OM = ON and OP = OQ

Subtracting both the equations

OP – OM = OQ – ON

So we get

PM = QN ……… (2)

Consider △ MPB and △ NQC

We know that

∠PMB = ∠QNC = 90o

By SAS congruence criterion

△ MPB ≅ △ NQC

PB = QC (c. p. c. t)

Therefore, it is proved that PB = QC.

19. In the adjoining figure, BC is a diameter of a circle with centre O. If AB and CD are two chords such that AB || CD, prove that AB = CD.

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12A 22

Solution:

Construct OL ⊥ AB and OM ⊥ CD

Consider △ OLB and △ OMC

We know that ∠OLB and ∠OMC are perpendicular bisector

∠OLB = ∠OMC = 90o

We know that AB || CD and BC is a transversal

From the figure we know that ∠OBL and ∠OCD are alternate interior angles

∠OBL = ∠OCD

So we get OB = OC which is the radii

By AAS congruence criterion

△ OLB ≅ △ OMC

OL = CM (c. p. c. t)

We know that the chords equidistant from the centre are equal

So we get

AB = CD

Therefore, it is proved that AB = CD.

20. An equilateral triangle of side 9cm is inscribed in a circle. Find the radius of the circle.

Solution:

Consider △ ABC as an equilateral triangle with side 9 cm

Take AD as one of its median

We know that

AD ⊥ BC

It can be written as

BD = ½ × BC

By substituting the values

BD = ½ × 9

So we get

BD = 4.5 cm

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12A 23

Consider △ ADB

Using the Pythagoras theorem

AB2 = AD2 + BD2

Substituting the values

92 = AD2 + (9/2)2

On further calculation

AD2 = 92 – (9/2)2

So we get

AD2 = 81 – 81/4

By taking out the square root

AD = 9√3/ 2 cm

We know that the centroid and circumcenter coincide in a equilateral triangle

AG: GD = 2: 1

The radius can be written as

AG = 2/3 AD

By substituting the values

AG = (2/3) × (9√3/ 2)

So we get

AG = 3√3 cm

Therefore, the radius of the circle is 3√3 cm.

21. In the adjoining figure, AB and AC are two equal chords of a circle with centre O. Show that O lies on the bisector of ∠BAC.

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12A 24

Solution:

Consider △ OAB and △ OAC

It is given that AB = AC

OA is common i.e. OA = OA

From the figure we know that OB = OC which is the radii

By SSS congruence criterion

△ OAB ≅ △ OAC

∠OAB = ∠OAC (c. p. c. t)

Therefore, it is proved that O lies on the bisector of ∠BAC.

22. In the adjoining figure, OPQR is a square. A circle drawn with centre O cuts the square in X and Y. Prove that QX = QY.

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12A 25

Solution:

Consider △ OXP and △ OYR

We know that ∠OPX and ∠ORY are right angles

So we get

∠OPX = ∠ORY = 90o

We know that OX and OY are the radii

OX = OY

From the figure we know that the sides of a square are equal

OP = OR

By RHS congruence criterion

△ OXP ≅ △ OYR

PX = RY (c. p. c. t)

We know that PQ = QR

So we get

PQ – PX = QR – RY

QX = QY

Therefore, it is proved that QX = QY.

23. Two circles with centres O and O’ intersect at two points A and B. A line PQ is drawn parallel to OO’ through A or B, intersecting the circles at P and Q. Prove that PQ = 2OO’.

Solution:

Construct OM ⊥ PQ and O’N ⊥ PQ

So we get

OM ⊥ AP

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12A 26

We know that the perpendicular from the centre of a circle bisects the chord

AM = PM

It can be written as

AP = 2AM ………. (1)

We know that

O’N ⊥ AQ

We know that the perpendicular from the centre of a circle bisects the chord

AN = QN

It can be written as

AQ = 2AN ……. (2)

So we get

PQ = AP + PQ

By substituting the values

PQ = 2 (AM + AN)

We get

PQ = 2MN

From the figure we know that MNO’O is a rectangle

PQ = 2OO’

Therefore, it is proved that PQ = 2OO’.


Exercise 12(B) page: 456

1. (i) In Figure (1), O is the centre of the circle. If ∠OAB = 40o and ∠OCB = 30o, find ∠AOC.

(ii) In Figure (2), A, B and C are three points on the circle with centre O such that ∠AOB = 90o and

∠AOC = 110o. Find ∠BAC.

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12B 1

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12B 2

Solution:

(i) Join the line BO

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12B 3

Consider △ BOC

We know that the sides are equal to the radius

So we get

OC = OB

From the figure we know that the base angles of an isosceles triangle are equal

∠OBC = ∠OCB

It is given that

∠OCB = 30o

So we get

∠OBC = ∠OCB = 30o

So we get ∠OBC = 30o ……. (1)

Consider △ BOA

We know that the sides are equal to the radius

So we get

OB = OC

From the figure we know that the base angles of an isosceles triangle are equal

∠OAB = ∠OBA

It is given that

∠OAB = 40o

So we get

∠OAB = ∠OBA = 40o

So we get ∠OBA = 40o ……. (2)

We know that

∠ABC = ∠OBC + ∠OBA

By substituting the values

∠ABC = 30o + 40o

So we get

∠ABC = 70o

We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at any point on the circumference.

It can be written as

∠AOC = 2 × ∠ABC

So we get

∠AOC = 2 × 70o

By multiplication

∠AOC = 140o

Therefore, ∠AOC = 140o

(ii) From the figure we know that

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12B 4

∠AOB + ∠AOC + ∠BOC = 360o

By substituting the values

90o + 110o + ∠BOC = 360o

On further calculation

∠BOC = 360o – 90o – 110o

By subtraction

∠BOC = 360o – 200o

So we get

∠BOC = 160o

We know that

∠BOC = 2 × ∠BAC

It is given that ∠BOC = 160o

∠BAC = 160o/2

By division

∠BAC = 80o

Therefore, ∠BAC = 80o.

2. In the given figure, O is the centre of the circle and ∠AOB = 70o. Calculate the values of

(i) ∠OCA,

(ii) ∠OAC.

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12B 5

Solution:

(i) We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at any point on the circumference.

So we get

∠AOB = 2 ∠OCA

It is given that ∠AOB = 70o

We can write it as

∠OCA = 70o/2

By division

∠OCA = 35o

Therefore, ∠OCA = 35o.

(ii) From the figure we know that the radius is

OA = OC

We know that the base angles of an isosceles triangle are equal

∠OAC = ∠OCA

It is given that ∠OCA = 35o

So we get

∠OAC = 35o

Therefore, ∠OAC = 35o.

3. In the given figure, O is the centre of the circle. If ∠PBC = 25o and ∠APB = 110o, find the value of ∠ADB.

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12B 6

Solution:

We know that ∠ACB = ∠PCB

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12B 7

In △ PCB

Using the angle sum property

∠PCB + ∠BPC + ∠PBC = 180o

We know that ∠APB and ∠BPC are linear pair

By substituting the values

∠PCB + (180o – 110o) + 25o = 180o

On further calculation

∠PCB + 70o + 25o = 180o

∠PCB + 95o = 180o

By subtraction

∠PCB = 180o – 95o

So we get

∠PCB = 85o

We know that the angles in the same segment of a circle are equal

∠ADB = ∠ACB = 85o

Therefore, the value of ∠ADB is 85o.

4. In the given figure, O is the centre of the circle. If ∠ABD = 35o and ∠BAC = 70o, find ∠ACB.

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12B 8

Solution:

We know that BD is the diameter of the circle

Angle in a semicircle is a right angle

∠BAD = 90o

Consider △ BAD

Using the angle sum property

∠ADB + ∠BAD + ∠ABD = 180o

By substituting the values

∠ADB + 90o + 35o = 180o

On further calculation

∠ADB = 180o – 90o – 35o

By subtraction

∠ADB = 180o – 125o

So we get

∠ADB = 55o

We know that the angles in the same segment of a circle are equal

∠ACB = ∠ADB = 55o

So we get

∠ACB = 55o

Therefore, ∠ACB = 55o.

5. In the given figure, O is the centre of the circle. If ∠ACB = 50o, find ∠OAB.

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12B 9

Solution:

We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at any point on the circumference.

So we get

∠AOB = 2 ∠ACB

It is given that ∠ACB = 50o

By substituting

∠AOB = 2 × 50o

By multiplication

∠AOB = 100o …… (1)

Consider △ OAB

We know that the radius of the circle are equal

OA = OB

Base angles of an isosceles triangle are equal

So we get

∠OAB = ∠OBA ……. (2)

Using the angle sum property

∠AOB + ∠OAB + ∠OBA = 180o

Using equations (1) and (2) we get

100o + 2 ∠OAB = 180o

By subtraction

2 ∠OAB = 180o – 100o

So we get

2 ∠OAB = 80o

By division

∠OAB = 40o

Therefore, ∠OAB = 40o.

6. In the given figure, ∠ABD = 54o and ∠BCD = 43o, calculate

(i) ∠ACD,

(ii) ∠BAD,

(iii) ∠BDA.

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12B 10

Solution:

(i) We know that the angles in the same segment of a circle are equal

From the figure we know that ∠ABD and ∠ACD are in the segment AD

∠ABD = ∠ACD = 54o

(ii) We know that the angles in the same segment of a circle are equal

From the figure we know that ∠BAD and ∠BCD are in the segment BD

∠BAD = ∠BCD = 43o

(iii) In △ ABD

Using the angle sum property

∠BAD + ∠ADB + ∠DBA = 180o

By substituting the values

43o + ∠ADB + 54o = 180o

On further calculation

∠ADB = 180o – 43o – 54o

By subtraction

∠ADB = 180o – 97o

So we get

∠ADB = 83o

It can be written as

∠BDA = 83o

7. In the adjoining figure, DE is a chord parallel to diameter AC of the circle with centre O. If ∠CBD = 60o, calculate ∠CDE.

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12B 11

Solution:

We know that the angles in the same segment of a circle are equal

From the figure we know that ∠CAD and ∠CBD are in the segment CD

∠CAD = ∠CBD = 60o

An angle in a semi-circle is a right angle

So we get

∠ADC = 90o

Using the angle sum property

∠ACD + ∠ADC + ∠CAD = 180o

By substituting the values

∠ACD + 90o + 60o = 180o

On further calculation

∠ACD = 180o – 90o – 60o

By subtraction

∠ACD = 180o – 150o

So we get

∠ACD = 30o

We know that AC || DE and CD is a transversal

From the figure we know that ∠CDE and ∠ACD are alternate angles

So we get

∠CDE = ∠ACD = 30o

Therefore, ∠CDE = 30o.

8. In the adjoining figure, O is the centre of a circle. Chord CD is parallel to diameter AB. If ∠ABC = 25o, calculate ∠CED.

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12B 12

Solution:

We know that ∠BCD and ∠ABC are alternate interior angles

∠BCD = ∠ABC = 25o

We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at any point on the circumference.

∠BOD = 2 ∠BCD

It is given that ∠BCD = 25o

So we get

∠BOD = 2 (25o)

By multiplication

∠BOD = 50o

In the same way

∠AOC = 2 ∠ABC

So we get

∠AOC = 50o

From the figure we know that AB is a straight line passing through the centre

Using the angle sum property

∠AOC + ∠COD + ∠BOD = 180o

By substituting the values

50o + ∠COD + 50o = 180o

On further calculation

∠COD + 100o = 180o

By subtraction

∠COD = 180o – 100o

So we get

∠COD = 80o

We know that

∠CED = ½ ∠COD

So we get

∠CED = 80o/2

By division

∠CED = 40o

Therefore, ∠CED = 40o.

9. In the given figure, AB and CD are straight lines through the centre O of a circle. If ∠AOC = 80o and ∠CDE = 40o, find

(i) ∠DCE,

(ii) ∠ABC.

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12B 13

Solution:

(i) From the figure we know that ∠CED = 90o

Consider △ CED

Using the angle sum property

∠CED + ∠EDC + ∠DCE = 180o

By substituting the values

90o + 40o + ∠DCE = 180o

On further calculation

∠DCE = 180o – 90o – 40o

By subtraction

∠DCE = 180o – 130o

So we get

∠DCE = 50o

(ii) We know that ∠AOC and ∠BOC form a linear pair

It can be written as

∠BOC = 180o – 80o

By subtraction

∠BOC = 100o

Using the angle sum property

∠ABC + ∠BOC + ∠DCE = 180o

By substituting the values

∠ABC + 100o + 50o = 180o

On further calculation

∠ABC = 180o – 100o – 50o

By subtraction

∠ABC = 180o – 150o

So we get

∠ABC = 30o

10. In the given figure, O is the centre of a circle, ∠AOB = 40o and ∠BDC = 100o, find ∠OBC.

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12B 14

Solution:

We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at any point on the circumference.

So we get

∠AOB = 2 ∠ACB

From the figure we know that

∠ACB = ∠DCB

It can be written as

∠AOB = 2 ∠DCB

We also know that

∠DCB = ½ ∠AOB

By substituting the values

∠DCB = 40o/2

By division

∠DCB = 20o

In △ DBC

Using the angle sum property

∠BDC + ∠DCB + ∠DBC = 180o

By substituting the values we get

100o + 20o + ∠DBC = 180o

On further calculation

∠DBC = 180o – 100o – 20o

By subtraction

∠DBC = 180o – 120o

So we get

∠DBC = 60o

From the figure we know that

∠OBC = ∠DBC = 60o

So we get

∠OBC = 60o

Therefore, ∠OBC = 60o.

11. In the adjoining figure, chords AC and BD of a circle with centre O, intersect at right angles at E. If ∠OAB = 25o, calculate ∠EBC.

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12B 15

Solution:

We know that OA and OB are the radius

Base angles of an isosceles triangle are equal

So we get

∠OBA = ∠OAB = 25o

Consider △ OAB

Using the angle sum property

∠OAB + ∠OBA + ∠AOB = 180o

By substituting the values

25o + 25o + ∠AOB = 180o

On further calculation

∠AOB = 180o – 25o – 25o

By subtraction

∠AOB = 180o – 50o

So we get

∠AOB = 130o

We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at any point on the circumference

So we get

∠AOB = 2 ∠ACB

It can be written as

∠ACB = ½ ∠AOB

By substituting the values

∠ACB = 130/2

By division

∠ACB = 65o

So we get

∠ECB = 65o

In △ BEC

Using the angle sum property

∠EBC + ∠BEC + ∠ECB = 180o

By substituting the values

∠EBC + 90o + 65o = 180o

On further calculation

∠EBC = 180o – 90o – 65o

By subtraction

∠EBC = 180o – 155o

So we get

∠EBC = 25o

Therefore, ∠EBC = 25o.

12. In the given figure, O is the centre of a circle in which ∠OAB = 20o and ∠OCB = 55o. Find

(i) ∠BOC,

(ii) ∠AOC.

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12B 16

Solution:

(i) We know that OB = OC which is the radius

The base angles of an isosceles triangle are equal

So we get

∠OBC = ∠OCB = 55o

In △ BOC

Using the angle sum property

∠BOC + ∠OCB + ∠OBC = 180o

By substituting the values

∠BOC + 55o + 55o = 180o

On further calculation

∠BOC = 180o – 55o – 55o

By subtraction

∠BOC = 180o – 110o

So we get

∠BOC = 70o

(ii) We know that OA = OB which is the radius

The base angles of an isosceles triangle are equal

So we get

∠OBA= ∠OAB = 20o

In △ AOB

Using the angle sum property

∠AOB + ∠OAB + ∠OBA = 180o

By substituting the values

∠AOB + 20o + 20o = 180o

On further calculation

∠AOB = 180o – 20o – 20o

By subtraction

∠AOB = 180o – 40o

So we get

∠AOB = 140o

We know that

∠AOC = ∠AOB – ∠BOC

By substituting the values

∠AOC = 140o – 70o

So we get

∠AOC = 70o

13. In the given figure, O is the centre of the circle and ∠BCO = 30o. Find x and y.

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12B 17

Solution:

From the figure we know that ∠AOD and ∠OEC form right angles

So we get

∠AOD = ∠OEC = 90o

We know that OD || BC and OC is a transversal

From the figure we know that ∠AOD and ∠OEC are corresponding angles

∠AOD = ∠OEC

We know that ∠DOC and ∠OCE are alternate angles

∠DOC = ∠OCE = 30o

Angle subtended by an arc of a circle at its centre is twice the angle subtended by the same arc at a point on the circumference.

So we get

∠DOC = 2 ∠DBC

It can be written as

∠DBC = ½ ∠DOC

By substituting the values

∠DBC = 30/2

By division

y = ∠DBC = 15o

In the same way

∠ABD = ½ ∠AOD

By substituting the values

∠ABD = 90/2

By division

∠ABD = 45o

We know that

∠ABE = ∠ABC = ∠ABD + ∠DBC

So we get

∠ABE = ∠ABC = 45o + 15o

By addition

∠ABE = ∠ABC = 60o

Consider △ ABE

Using the angle sum property

∠BAE + ∠AEB + ∠ABE = 180o

By substituting the values

x + 90o + 60o = 180o

On further calculation

x = 180o – 90o – 60o

By subtraction

x = 180o – 150o

So we get

x = 30o

Therefore, the value of x is 30o and y is 15o.

14. In the given figure, O is the centre of the circle, BD = OD and CD ⊥ AB. Find ∠CAB.

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12B 18

Solution:

Join the points AC

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12B 19

It is given that BD = OD

We know that the radii of same circle are equal

OD = OB

It can be written as

BD = OD = OB

Consider △ ODB as equilateral triangle

We know that

∠ODB = 60o

The altitude of an equilateral triangle bisects the vertical angle

So we get

∠BDE = ∠ODE = ½ ∠ODB

By substituting the values

∠BDE = ∠ODE = 60/2

By division

∠BDE = ∠ODE = 30o

We know that the angles in the same segment are equal

∠CAB = ∠BDE = 30o

Therefore, ∠CAB = 30o.

15. In the given figure, PQ is a diameter of a circle with centre O. If ∠PQR = 65o, ∠SPR = 40o and ∠PQM = 50o, find ∠QPR, ∠QPM and ∠PRS.

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12B 20

Solution:

In △ PQR

We know that PQ is the diameter

So we get

∠PRQ = 90o as the angle in a semicircle is a right angle

Using the angle sum property

∠QPR + ∠PRQ + ∠PQR = 180o

By substituting the values

∠QPR + 90o + 65o = 180o

On further calculation

∠QPR = 180o – 90o – 65o

By subtraction

∠QPR = 180o – 155o

So we get

∠QPR = 25o

In △ PQM

We know that PQ is the diameter

So we get

∠PMQ = 90o as the angle in a semicircle is a right angle

Using the angle sum property

∠QPM + ∠PMQ + ∠PQM = 180o

By substituting the values

∠QPM + 90o + 50o = 180o

On further calculation

∠QPM = 180o – 90o – 50o

By subtraction

∠QPM = 180o – 140o

So we get

∠QPM = 40o

Consider the quadrilateral PQRS

We know that

∠QPS + ∠SRQ = 180o

It can be written as

∠QPR + ∠RPS + ∠PRQ + ∠PRS = 180o

By substituting the values

25o + 40o + 90o + ∠PRS = 180o

On further calculation

∠PRS = 180o – 25o – 40o – 90o

By subtraction

∠PRS = 180o – 155o

So we get

∠PRS = 25o

16. In the figure given below, P and Q are centres of two circles, intersecting at B and C, and ACD is a straight line. If ∠APB = 150o and ∠BQD = xo, find the value of x.

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12B 21

Solution:

The angle subtended by an arc of a circle at its centre is twice the angle subtended by the same arc at a point on the circumference

So we get

∠APB = 2 ∠ACB

It can be written as

∠ACB = ½ ∠APB

By substituting the values

∠ACB = 150/2

So we get

∠ACB = 75o

We know that ACD is a straight line

It can be written as

∠ACB + ∠DCB = 180o

By substituting the values

75o + ∠DCB = 180o

On further calculation

∠DCB = 180o – 75o

By subtraction

∠DCB = 105o

We know that

∠DCB = ½ × reflex ∠BQD

By substituting the values

105o = ½ × (360o – x)

On further calculation

210o = 36o – x

By subtraction

x = 150o

Therefore, the value of x is 150o.

17. In the given figure, ∠BAC = 30o. Show that BC is equal to the radius of the circumcircle of △ ABC whose centre is O.

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12B 22

Solution:

Join the lines OB and OC

We know that the angle subtended by an arc of a circle at the centre is double the angle subtended by the arc at any point on the circumference

So we get

∠BOC = 2 ∠BAC

It is given that ∠BAC = 30o

∠BOC = 2 × 30o

By multiplication

∠BOC = 60o

In △ BOC

We know that the radii are equal

OB = OC

The base angles of an isosceles triangle are equal

∠OBC = ∠OCB

Consider △ BOC

Using the angle sum property

∠BOC + ∠OBC + ∠OCB = 180o

By substituting the values we get

60o + ∠OCB + ∠OCB = 180o

So we get

2 ∠OCB = 180o – 60o

By subtraction

2 ∠OCB = 120o

By division

∠OBC = 60o

We get ∠OBC = ∠OCB = ∠BOC = 60o

We know that △ BOC is an equilateral triangle

So we get OB = OC = BC

Therefore, it is proved that BC is equal to the radius of the circumcircle of △ ABC whose centre is O.

18. In the given figure, AB and CD are two chords of a circle, intersecting each other at a point E. Prove that ∠AEC = ½ (angle subtended by arc CXA. At the centre + angle subtended by arc DYB at the centre).

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12B 23

Solution:

Construct the lines AC and BC

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12B 24

The angle subtended by an arc of a circle at its centre is twice the angle subtended by the same arc at a point on the circumference.

So we get

∠AOC = 2 ∠ABC …… (1)

In the same way

∠BOD = 2 ∠BCD …… (2)

By adding both the equations

∠AOC + ∠BOD = 2 ∠ABC + 2 ∠BCD

By taking 2 as common

∠AOC + ∠BOD = 2 (∠ABC + ∠BCD)

It can be written as

∠AOC + ∠BOD = 2 (∠EBC + ∠BCE)

So we get

∠AOC + ∠BOD = 2 (180o – ∠CEB)

We can write it as

∠AOC + ∠BOD = 2 (180o – (180o – ∠AEC))

We get

∠AOC + ∠BOD = 2 ∠AEC

Dividing the equation by 2

∠AEC = ½ (∠AOC + ∠BOD)

∠AEC = ½ (angle subtended by arc CXA at the centre + angle subtended by arc DYB at the centre)

Therefore, it is proved that ∠AEC = ½ (angle subtended by arc CXA. At the centre + angle subtended by arc DYB at the centre).


Exercise 12(C) PAGE: 482

1. In the given figure, ABCD is a cyclic quadrilateral whose diagonals intersect at P such that ∠DBC = 60o and ∠BAC = 40o. Find

(i) ∠BCD,

(ii) ∠CAD.

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12C 1

Solution:

(i) We know that the angles in the same segment are equal

So we get

∠BDC = ∠BAC = 40o

Consider △ BCD

Using the angle sum property

∠BCD + ∠BDC + ∠DBC = 180o

By substituting the values

∠BCD + 40o + 60o = 180o

On further calculation

∠BCD = 180o – 40o – 60o

By subtraction

∠BCD = 180o – 100o

So we get

∠BCD = 80o

(ii) We know that the angles in the same segment are equal

So we get

∠CAD = ∠CBD

So ∠CAD = 60o

2. In the given figure, POQ is a diameter and PQRS is a cyclic quadrilateral. If ∠PSR = 150o, find ∠RPQ.

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12C 2

Solution:

We know that PQRS is a cyclic quadrilateral

It can be written as

∠PSR+ ∠PQR = 180o

By substituting the values

150o + ∠PQR = 180o

On further calculation

∠PQR = 180o – 150o

By subtraction

∠PQR = 30o

We know that the angle in semi-circle is a right angle

∠PRQ = 90o

Consider △ PRQ

Using the angle sum property

∠PQR + ∠PRQ + ∠RPQ = 180o

By substituting the values

30o + 90o + ∠RPQ = 180o

On further calculation

∠RPQ = 180o – 30o – 90o

By subtraction

∠RPQ = 180o – 120o

So we get

∠RPQ = 60o

Therefore, ∠RPQ = 60o.

3. In the given figure, O is the centre of the circle and arc ABC subtends an angle of 130o at the centre. If AB is extended to P, find ∠PBC.

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12C 3

Solution:

Consider a point D on the arc CA and join DC and AD

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12C 4

We know that the angle subtended by an arc is twice the angle subtended by it on the circumference in the alternate segment

So we get

∠2 = 2 ∠1

By substituting the values

130o = 2 ∠1

So we get

∠1 = 65o

From the figure we know that the exterior angle of a cyclic quadrilateral = interior opposite angle

∠PBC = ∠1

So we get ∠PBC = 65o

Therefore, ∠PBC = 65o.

4. In the given figure, ABCD is a cyclic quadrilateral in which AE is drawn parallel to CD, and BA is produced to F. If ∠ABC = 92o and ∠FAE = 20o, find ∠BCD.

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12C 5

Solution:

We know that ABCD is a cyclic quadrilateral

So we get

∠ABC + ∠ADC = 180o

By substituting the values

92o + ∠ADC = 180o

On further calculation

∠ADC = 180o – 92o

By subtraction

∠ADC = 88o

We know that AE || CD

From the figure we know that

∠EAD = ∠ADC = 88o

We know that the exterior angle of a cyclic quadrilateral = interior opposite angle

So we get

∠BCD = ∠DAF

We know that

∠BCD = ∠EAD + ∠EAF

It is given that ∠FAE = 20o

By substituting the values

∠BCD = 88o + 20o

By addition

∠BCD = 108o

Therefore, ∠BCD = 108o.

5. In the given figure, BD = DC and ∠CBD = 30o, find ∠BAC.

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12C 6

Solution:

It is given that BD = DC

From the figure we know that

∠BCD = ∠CBD = 30o

Consider △ BCD

Using the angle sum property

∠BCD + ∠CBD + ∠CDB = 180o

By substituting the values

30o + 30o + ∠CDB = 180o

On further calculation

∠CDB = 180o – 30o – 30o

By subtraction

∠CDB = 180o – 60o

So we get

∠CDB = 120o

We know that the opposite angles of a cyclic quadrilateral are supplementary

It can be written as

∠CDB + ∠BAC = 180o

By substituting the values

120o + ∠BAC = 180o

On further calculation

∠BAC = 180o – 120o

By subtraction

∠BAC = 60o

Therefore, ∠BAC = 60o.

6. In the given figure, O is the centre of the given circle and measure of arc ABC is 100o. Determine ∠ADC and ∠ABC.

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12C 7

Solution:

We know that the angle subtended by an arc is twice the angle subtended by it on the circumference

From the figure we know that

∠AOC = 100o

So we get

∠AOC = 2 ∠ADC

It can be written as

∠ADC = ½ ∠AOC

By substituting the values

∠ADC = ½ (100o)

So we get

∠ADC = 50o

We know that the opposite angles of a cyclic quadrilateral are supplementary

It can be written as

∠ADC + ∠ABC = 180o

By substituting the values

50o + ∠ABC = 180o

On further calculation

∠ABC = 180o – 50o

By subtraction

∠ABC = 130o

Therefore, ∠ADC = 50o and ∠ABC = 130o.

7. In the given figure, △ ABC is equilateral. Find

(i) ∠BDC,

(ii) ∠BEC.

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12C 8

Solution:

It is given that △ ABC is equilateral

We know that

∠BAC = ∠ABC = ∠ACB = 60o

(i) We know that the angles in the same segment of a circle are equal

∠BDC = ∠BAC = 60o

So we get

∠BDC = 60o

(ii) We know that the opposite angles of a cyclic quadrilateral are supplementary

So we get

∠BAC + ∠BEC = 180o

By substituting the values

60o + ∠BEC = 180o

On further calculation

∠BEC = 180o – 60o

By subtraction

∠BEC = 120o

8. In the adjoining figure, ABCD is a cyclic quadrilateral in which ∠BCD = 100o and ∠ABD = 50o. Find ∠ADB.

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12C 9

Solution:

It is given that ABCD is a cyclic quadrilateral

We know that the opposite angles of a cyclic quadrilateral are supplementary

So we get

∠A + ∠C = 180o

By substituting the values

∠A + 100o = 180o

On further calculation

∠A = 180o – 100o

By subtraction

∠A = 80o

Consider △ ABD

Using the angle sum property

∠A + ∠ABD + ∠ADB = 180o

By substituting the values

80o + 50o + ∠ADB = 180o

On further calculation

∠ADB = 180o – 80o – 50o

By subtraction

∠ADB = 180o – 130o

So we get

∠ADB = 50o

Therefore, ∠ADB = 50o.

9. In the given figure, O is the centre of a circle and ∠BOD =150o. Find the values of x and y.

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12C 10

Solution:

It is given that O is the centre of a circle and ∠BOD =150o

We know that

Reflex ∠BOD = (360o – ∠BOD)

By substituting the values

Reflex ∠BOD = (360o – 150o)

By subtraction

Reflex ∠BOD = 210o

Consider x = ½ (reflex ∠BOD)

By substituting the value

x = 210/2

So we get

x = 105o

We know that

x + y = 180o

By substituting the values

105o + y = 180o

On further calculation

y = 180o – 105o

By subtraction

y = 75o

Therefore, the value of x is 105o and y is 75o.

10. In the given figure, O is the centre of the circle and ∠DAB = 50o. Calculate the values of x and y.

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12C 11

Solution:

It is given that O is the centre of the circle and ∠DAB = 50o

We know that the radii of the circle are equal

OA = OB

From the figure we know that

∠OBA = ∠OAB = 50o

Consider △ OAB

Using the angle sum property

∠OAB + ∠OBA + ∠AOB = 180o

By substituting the values

50o + 50o + ∠AOB = 180o

On further calculation

∠AOB = 180o – 50o – 50o

By subtraction

∠AOB = 180o – 100o

So we get

∠AOB = 80o

From the figure we know that AOD is a straight line

It can be written as

x = 180o – ∠AOB

By substituting the values

x = 180o – 80o

By subtraction

x = 100o

We know that the opposite angles of a cyclic quadrilateral are supplementary

So we get

∠DAB + ∠BCD = 180o

By substituting the values

50o + ∠BCD = 180o

On further calculation

∠BCD = 180o – 50o

By subtraction

y = ∠BCD = 130o

Therefore, the value of x is 100o and y is 130o.

11. In the given figure, sides AD and AB of cyclic quadrilateral ABCD are produced to E and F respectively. If ∠CBF = 130o and ∠CDE = xo, find the value of x.

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12C 12

Solution:

In a cyclic quadrilateral we know that the exterior angle is equal to the interior opposite angle

So we get

∠CBF = ∠CDA

It can be written as

130o = 180o – x

On further calculation

x = 180o – 130o

By subtraction

x = 50o

12. In the given figure, AB is a diameter of a circle with centre O and DO || CB. If ∠BCD = 120o, calculate

(i) ∠BAD,

(ii) ∠ABD,

(iii) ∠CBD,

(iv) ∠ADC.

Also, show that △ AOD is an equilateral triangle.

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12C 13

Solution:

It is given that AB is a diameter of a circle with centre O and DO || CB

(i) We know that ABCD is a cyclic quadrilateral

It can be written as

∠BCD + ∠BAD = 180o

By substituting the values

120o + ∠BAD = 180o

On further calculation

∠BAD = 180o – 120o

By subtraction

∠BAD = 60o

(ii) We know that the angle in a semi-circle is right angle

∠BDA = 90o

Consider △ ABD

By using the angle sum property

∠BDA + ∠BAD + ∠ABD = 180o

By substituting the values

90o + 60o + ∠ABD = 180o

On further calculation

∠ABD = 180o – 90o – 60o

By subtraction

∠ABD = 180o – 150o

So we get

∠ABD = 30o

(iii) We know that OD = OA

So we get ∠ODA = ∠OAD = ∠BAD = 60o

From the figure we know that

∠ODB + ∠ODA = 90o

By substituting the values

∠ODB + 60o = 90o

On further calculation

∠ODB = 90o – 60o

By subtraction

∠ODB = 30o

It is given that DO || CB

We know that the alternate angles are equal

∠CDB = ∠ODB = 30o

(iv) From the figure we know that

∠ADC = ∠ADB + ∠CDB

By substituting the values

∠ADC = 90o + 30o

By addition

∠ADC = 120o

Consider △ AOD

By using the angle sum property

∠ODA + ∠OAD + ∠AOD = 180o

By substituting the values

60o + 60o + ∠AOD = 180o

On further calculation

∠AOD = 180o – 60o – 60o

By subtraction

∠AOD = 180o – 120o

So we get

∠AOD = 60o

We know that all the angles of the △ AOD is 60o

Therefore, it is proved that △ AOD is an equilateral triangle.

13. Two chords AB and CD of a circle intersect each other at P outside the circle. If AB = 6cm, BP = 2cm and PD = 2.5cm, find CD.

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12C 14

Solution:

It is given that AB and CD of a circle intersect each other at P outside the circle. If AB = 6cm, BP = 2cm and PD = 2.5cm

So we get

AP × BP = CP × DP

From the figure we know that CP = CD + DP

By substituting the values

8 × 2 = (CD + 2.5) × 2.5 cm

Consider x = CD

So we get

8 × 2 = (x + 2.5) × 2.5

On further calculation

16 = 2.5x + 6.25

It can be written as

2.5x = 16 – 6.25

By subtraction

2.5x = 9.75

By division

x = 9.75/2.5

So we get

x = 3.9cm

Therefore, CD = 3.9cm.

14. In the given figure, O is the centre of a circle. If ∠AOD = 140o and ∠CAB = 50o, calculate

(i) ∠EDB,

(ii) ∠EBD.

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12C 15

Solution:

(i) We know that

∠BOD + ∠AOD = 180o

By substituting the values

∠BOD + 140o = 180o

On further calculation

∠BOD = 180o – 140o

By subtraction

∠BOD = 40o

We know that OB = OD

So we get ∠OBD = ∠ODB

Consider △ OBD

By using the angle sum property

∠BOD + ∠OBD + ∠ODB = 180o

We know that ∠OBD = ∠ODB

So we get

40o + 2 ∠OBD = 180o

On further calculation

2 ∠OBD = 180o – 40o

By subtraction

2 ∠OBD = 140o

By division

∠OBD = 70o

We know that ABCD is a cyclic quadrilateral

∠CAB + ∠BDC = 180o

∠CAB + ∠ODB + ∠ODC = 180o

By substituting the values

50o + 70o + ∠ODC = 180o

On further calculation

∠ODC = 180o – 50o – 70o

By subtraction

∠ODC = 180o – 120o

So we get

∠ODC = 60o

Using the angle sum property

∠EDB + ∠ODC + ∠ODB = 180o

By substituting the values

∠EDB + 60o + 70o = 180o

On further calculation

∠EDB = 180o – 60o – 70o

By subtraction

∠EDB = 180o – 130o

So we get

∠EDB = 50o

(ii) We know that

∠EDB + ∠OBD = 180o

By substituting the values

∠EDB + 70o = 180o

On further calculation

∠EDB = 180o – 70o

By subtraction

∠EDB = 110o

15. In the given figure, △ ABC is an isosceles triangle in which AB = AC and a circle passing through B and C intersects AB and AC at D and E respectively. Prove that DE || BC.

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12C 16

Solution:

We know that △ ABC is an isosceles triangle in which AB = AC and a circle passing through B and C intersects AB and AC at D and E

So AB = AC

We get

∠ACB = ∠ABC

It can be written as

∠ADE = ∠ACB = ∠ABC

So we get

∠ADE = ∠ABC

DE || BC

Therefore, it is proved that DE || BC.

16. In the given figure, AB and CD are two parallel chords of a circle. If BDE and ACE are straight lines, intersecting at E, prove that △ AEB is isosceles.

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12C 17

Solution:

It is given that AB and CD are two parallel chords of a circle. If BDE and ACE are straight lines, intersecting at E

We know that in a cyclic quadrilateral exterior angle is equal to the interior opposite angle.

It can be written as

Exterior ∠EDC = ∠A

Exterior ∠DCE = ∠B

We know that AB || CD

So we get

∠EDC = ∠B and ∠DCE = ∠A

We get

∠A = ∠B

Therefore, it is proved that △ AEB is isosceles.

17. In the given figure, ∠BAD = 75o, ∠DCF = xo and ∠DEF = yo. Find the values of x and y.

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12C 18

Solution:

In a cyclic quadrilateral we know that the exterior angle is equal to the interior opposite angle

So we get

∠BAD = ∠DCF = 75o

It can be written as

∠DCF = x = 75o

We get x = 75o

We know that the opposite angles of a cyclic quadrilateral is 180o

So we get

∠DCF + ∠DEF = 180o

By substituting the values

75o + ∠DEF = 180o

On further calculation

∠DEF = 180o – 75o

By subtraction

∠DEF = y = 105o

Therefore, the values of x and y is 75o and 105o.

18. In the given figure, ABCD is a quadrilateral in which AD = BC and ∠ADC = ∠BCD. Show that the points A, B, C, D lie on a circle.

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12C 19

Solution:

It is given that ABCD is a quadrilateral in which AD = BC and ∠ADC = ∠BCD

Construct DE ⊥ AB and CF ⊥ AB

Consider △ ADE and △ BCF

We know that

∠AED + ∠BFC = 90o

From the figure it can be written as

∠ADE = ∠ADC – 90o = ∠BCD – 90o = ∠BCF

It is given that

AD = BC

By AAS congruence criterion

△ ADE ≅ △ BCF

∠A = ∠B (c. p. c. t)

We know that the sum of all the angles of a quadrilateral is 360o

∠A + ∠B + ∠C + ∠D = 360o

By substituting the values

2 ∠B + 2 ∠D = 360o

By taking 2 as common

2 (∠B + ∠D) = 360o

By division

∠B + ∠D = 360/2

So we get

∠B + ∠D = 180o

So, ABCD is a cyclic quadrilateral.

Therefore, it is proved that the points A, B, C and D lie on a circle.

19. Prove that the perpendicular bisectors of the sides of a cyclic quadrilateral are concurrent.

Solution:

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12C 20

Consider ABCD as a cyclic quadrilateral with centre O passing through the points A, B, C, D

We know that AB, BC, CD and DA are the chords of the circle and its right bisector passing through the centre O

So we know that the right bisectors of AB, BC, CD and DA pass through the centre O and are concurrent.

Therefore, it is proved that the perpendicular bisectors of the sides of a cyclic quadrilateral are concurrent.

20. Prove that the circles described with the four sides of a rhombus as diameters pass through the point of intersection of its diagonals.

Solution:

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12C 21

Consider ABCD as a rhombus

We know that the diagonals AC and BD intersect at the point O

From the figure we know that the diagonals of the rhombus bisect at right angles

It can be written as

∠BOC = 90o

Thus, ∠BOC lies in the circle

We know that the circle can be drawn with BC as the diameter having the centre O

In the same way, all the circles with AB, AD and CD as diameters will pass through the centre O.

Therefore, it is proved that the circles described with the four sides of a rhombus as diameters pass through the point of intersection of its diagonals.

21. ABCD is a rectangle. Prove that the centre of the circle through A, B, C, D is the point of intersection of its diagonals.

Solution:

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12C 22

Consider ABCD as a rectangle

We know that O is the point of intersection of the diagonals AC and BD

The diagonals of a rectangle are equal and bisect each other

So we get

OA = OB = OC = OD

We get O as the centre of the circle through A, B, C and D

Therefore, it is proved that the centre of the circle through A, B, C, D is the point of intersection of its diagonals.

22. Give a geometrical construction for finding the fourth point lying on a circle passing through three given points, without finding the centre of the circle. Justify the construction.

Solution:

Consider A, B, C as the points

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12C 23

Taking B as the centre and radius equal to AC construct an arc

Taking C as the centre and radius equal to AB construct another arc which cuts the arc at D

So we get D as the required point BD and CD

Consider △ ABC and △ DCB

We know that

AB = DC and AC = DB

BC and CB are common i.e. BC = CB

By SSS congruence criterion

△ ABC ≅ △ DCB

∠BAC = ∠CDB (c. p. c. t)

We know that BC subtends equal angles ∠BAC and ∠CDB on the same side

So we get A, B, C, D are cyclic.

Therefore, the points A, B, C, D are cyclic.

23. In a cyclic quadrilateral ABCD, if (∠B – ∠D) = 60o, show that the smaller of the two is 60o.

Solution:

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12C 24

It is given that ABCD is a cyclic quadrilateral

(∠B – ∠D) = 60o …….. (1)

We know that

(∠B + ∠D) = 180o ……. (2)

By adding both the equations

∠B – ∠D + ∠B + ∠D = 60o + 180o

So we get

2 ∠B = 240o

By division

∠B = 120o

By substituting equation it in equation (1)

(∠B – ∠D) = 60o

120o – ∠D = 60o

On further calculation

∠D = 120o – 60o

By subtraction

∠D = 60o

Therefore, the smaller of the two angles ∠D = 60o.

24. The diagonals of a cyclic quadrilateral are at right angles. Prove that the perpendicular from the point of their intersection on any side when produced backwards, bisects the opposite side.

Solution:

Consider ABCD as a cyclic quadrilateral with diagonals AC and BD intersecting at right angles

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12C 25

We know that OL ⊥ AB so that ∠O meets the line CD at the point M

From the figure we know that the angles in the same segment are equal

∠1 = ∠2

We know that ∠OLB = 90o so we get

∠2 + ∠3 = 90o …… (1)

We know that OLM is a straight line and ∠BOC = 90o

So we get

∠3 + ∠4 = 90o ……… (2)

It can be written as

∠2 + ∠3 = ∠3 + ∠4

On further calculation

∠2 = ∠4

So we get

∠1 = ∠2 and ∠2 = ∠4

It can be written as

∠1 = ∠4

We get

OM = CM

In the same way

OM = MD

So CM = MD

Therefore, it is proved that the perpendicular from the point of their intersection on any side when produced backwards, bisects the opposite side.

25. On a common hypotenuse AB, two right triangles ACB and ADB are situated on opposite sides. Prove that ∠BAC = ∠BDC.

Solution:

We know that AB is the common hypotenuse of △ ACB and △ ADB

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12C 26

So we get

∠ACB = 90o

∠BDC = 90o

It can be written as

∠ACB + ∠BDC = 180o

We know that the opposite angles of a quadrilateral ABCD are supplementary

So we get ABCD as a cyclic quadrilateral which means that a circle passes through the points A, C, B and D

From the figure we know that the angles in the same segment are equal

∠BAC = ∠BDC

Therefore, it is proved that ∠BAC = ∠BDC.

26. ABCD is a quadrilateral such that A is the centre of the circle passing through B, C and D. Prove that ∠CBD + ∠CDB = ½ ∠BAD.

Solution:

Consider a point E on the circle and join BE, DE and BD

RS Aggarwal Solutions for Class 9 Maths Chapter 12 Ex 12C 27

The angle subtended by an arc of a circle at its centre is twice the angle subtended by the same arc at a point on the circumference

So we get

∠BAD = 2 ∠BED

It can be written as

∠BED = ½ ∠BAD …… (1)

Consider EBCD as a cyclic quadrilateral

So we get

∠BED + ∠BCD = 180o

We can write it as

∠BCD = 180o – ∠BED

Substituting equation (1)

∠BCD = 180o – ½ ∠BAD …… (2)

Consider △ BCD

Using the angle sum property

∠CBD + ∠CDB + ∠BCD = 180o

By using the equation (2)

∠CBD + ∠CDB + 180o – ½ ∠BAD = 180o

So we get

∠CDB + ∠CDB – ½ ∠BAD = 180o – 180o

On further calculation

∠CDB + ∠CDB = ½ ∠BAD

Therefore, it is proved that ∠CDB + ∠CDB = ½ ∠BAD.


RS Aggarwal Solutions for Class 9 Maths Chapter 12: Circles

Chapter 12, Circles, has 3 exercises containing answers for all the problems in PDF format which can be easily accessed by the students. The major concepts which are explained based on the RS Aggarwal Solutions Chapter 12 are as follows:

  • Terms related to a circle
  • Position of a point with respect to a circle
  • Degree measure of an arc
  • Minor and major arcs of a circle
  • Congruence of circles
  • Chord properties of circles
  • Some results on congruent circles
  • Results on angles subtended by arcs
  • Results on cyclic quadrilaterals

RS Aggarwal Solutions Class 9 Maths Chapter 12 – Exercise list

Exercise 12A Solutions 23 Questions

Exercise 12B Solutions 18 Questions

Exercise 12C Solutions 26 Questions

RS Aggarwal Solutions Class 9 Maths Chapter 12 – Circles

The solutions prepared by experienced faculties majorly help students revise the concepts from exam point of view. The problems are solved based on the CBSE guidelines and evaluation process which help students to score better marks. This chapter contains problems which are based on the circle and theorems.

Circles are mainly used in construction of high rise buildings and in creating maps. The examples of circles in real life are pizzas, camera lenses, steering wheels, buttons and satellite’s orbit etc. PDF of RS Aggarwal Solutions contain exercise wise answers of the entire chapter which are solved based on the understanding capacity of the students.