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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

### 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

So we get

OA = 10 cm and AB = 16cm

Construct OL âŠ¥ AB

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

So we get

AL = Â½ Ã— AB

By substituting the values

AL = Â½ Ã— 16

So we get

AL = 8 cm

Consider â–³ OLA

Using the Pythagoras theorem it can be written as

OA^{2} = OL^{2} + AL^{2}

By substituting the values we get

10^{2} = OL^{2} + 8^{2}

On further calculation

OL^{2} = 10^{2} – 8^{2}

So we get

OL^{2} = 100 â€“ 64

By subtraction

OL^{2} = 36

By taking the square root

OL = âˆš36

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 âŠ¥ AB

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

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

Consider â–³ OLA

Using the Pythagoras theorem it can be written as

OA^{2} = OL^{2} + AL^{2}

By substituting the values we get

5^{2} = OL^{2} + 3^{2}

On further calculation

OL^{2} = 5^{2} – 3^{2}

So we get

OL^{2} = 25 â€“ 9

By subtraction

OL^{2} = 16

By taking the square root

OL = âˆš16

So we get

OL = 4cm

We know that

AB = 2AL

By substituting the values

AB = 2 Ã— 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 âŠ¥ AB

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 = Â½ Ã— AB

By substituting the values

AL = Â½ Ã— 30

By division

AL = 15cm

Consider â–³ OLA

Using the Pythagoras theorem it can be written as

OA^{2} = OL^{2} + AL^{2}

By substituting the values we get

OA^{2} = 8^{2} + 15^{2}

On further calculation

OA^{2} = 64 + 225

By addition

OA^{2} = 289

By taking the square root

OA = âˆš289

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

We know that

LB = Â½ Ã— AB

By substituting the values we get

LB = Â½ Ã— 8

So we get

LB = 4cm

We know that

MD = Â½ Ã— CD

By substituting the values we get

MD = Â½ Ã— 6

So we get

MD = 3cm

Consider â–³ BLO

Using the Pythagoras theorem it can be written as

OB^{2} = LB^{2} + LO^{2}

By substituting the values we get

5^{2} = 4^{2} + LO^{2}

On further calculation

LO^{2} = 5^{2} – 4^{2}

So we get

LO^{2} = 25 â€“ 16

By subtraction

LO^{2} = 9

By taking the square root

LO = âˆš9

LO = 3cm

Consider â–³ DMO

Using the Pythagoras theorem it can be written as

OD^{2} = MD^{2} + MO^{2}

By substituting the values we get

5^{2} = 3^{2} + MO^{2}

On further calculation

MO^{2} = 5^{2} – 3^{2}

So we get

MO^{2} = 25 â€“ 9

By subtraction

MO^{2} = 16

By taking the square root

MO = âˆš16

MO = 4cm

So the distance between the chords = MO â€“ LO

By substituting the values

Distance between the chords = 4 â€“ 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 âŠ¥ AB and OM âŠ¥ CD

Join the diagonals OA and OC

We know that OA = OC = 5cm

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

We know that AL = Â½ Ã— AB

By substituting the values we get

AL = Â½ Ã— 8

So we get

AL = 4cm

We know that CM = Â½ Ã— CD

By substituting the values we get

CM = Â½ Ã— 6

So we get

CM = 3cm

Consider â–³ OLA

Using the Pythagoras theorem it can be written as

OA^{2} = AL^{2} + OL^{2}

By substituting the values

5^{2} = 4^{2} + OL^{2}

On further calculation

OL^{2} = 5^{2} – 4^{2}

So we get

OL^{2} = 25 â€“ 16

By subtraction

OL^{2} = 9

By taking the square root

OL = âˆš9

OL = 3cm

Consider â–³ OMC

Using the Pythagoras theorem it can be written as

OC^{2} = OM^{2} + CM^{2}

By substituting the values

5^{2} = OM^{2} +3^{2}

On further calculation

OM^{2} = 5^{2} – 3^{2}

OM^{2} = 25 â€“ 9

By subtraction

OM^{2} = 16

By taking the square root

OM = âˆš16

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

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 âŠ¥ CD and OL âŠ¥ AB

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

We know that

AL = Â½ Ã— AB

By substituting the values

AL = Â½ Ã— 30

So we get

AL = 15cm

We know that

CM = Â½ Ã— CD

By substituting the values

CM = Â½ Ã— 16

So we get

CM = 8cm

Consider â–³ ALO

Using the Pythagoras theorem it can be written as

AO^{2} = OL^{2} + AL^{2}

By substituting the values

17^{2} = OL^{2} + 15^{2}

So we get

OL^{2} = 17^{2} – 15^{2}

On further calculation

OL^{2} = 289 â€“ 225

By subtraction

OL^{2} = 64

By taking the square root

OL = âˆš64

OL = 8cm

Consider â–³ CMO

Using the Pythagoras theorem it can be written as

CO^{2} = CM^{2} + OM^{2}

By substituting the values

17^{2} = 8^{2} + OM^{2}

So we get

OM^{2} = 17^{2} – 8^{2}

On further calculation

OM^{2} = 289 â€“ 64

By subtraction

OM^{2} = 225

By taking the square root

OM = âˆš225

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.**

**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.

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

Consider OA = OC = r cm

It can be written as

OE = (r â€“ 3) cm

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

We know that

AE = Â½ Ã— AB

By substituting the values

AE = Â½ Ã— 12

So we get

AE = 6 cm

Consider â–³ OEA

By using the Pythagoras theorem

OA^{2} = OE^{2} + AE^{2}

By substituting the values

r^{2} = (r â€“ 3)^{2} + 6^{2}

So we get

r^{2} = r^{2} â€“ 6r + 9 + 36

On further calculation

r^{2} – r^{2} + 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.**

**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 â€“ 4) cm

Consider â–³ OEC

By using the Pythagoras theorem

OC^{2} = OE^{2} + EC^{2}

By substituting the values we get

r^{2} = (r â€“ 4)^{2} + 8^{2}

On further calculation

r^{2} = r^{2} â€“ 8r + 16 + 64

So we get

r^{2} = r^{2} â€“ 8r + 80

It can be written as

r^{2} – r^{2} + 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 Ã— OD.**

**Solution:**

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

We know that OD âŠ¥ 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 â–³ ABC

Using the midpoint theorem

We get OD || AC and

OD = Â½ Ã— AC

By cross multiplication

AC = 2 Ã— OD

Therefore, it is proved that AC || DO and AC = 2 Ã— 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 âˆ BPD. Prove that AB = CD.**

**Solution:**

Consider â–³ OEP and â–³ OFP

We know that

âˆ OEP = âˆ OFP = 90^{o}

OP is common i.e. OP = OP

From the figure we know that OP bisects âˆ BPD

It can be written as

âˆ OPE = âˆ OPF

By ASA congruence criterion

â–³ OEP â‰… â–³ 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:**

Consider AB || CD and POQ as the diameter

It is given that âˆ PEB = 90^{o}

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

So we get

âˆ PFD = âˆ PEB

It can be written as

PF âŠ¥ CD

In the same way

OF âŠ¥ 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.**

**Solution:**

It is given that

OA = 10cm and AB = 12 cm

So we get

AD = Â½ Ã— AB

By substituting the values

AD = Â½ Ã— 12

By division

AD = 6cm

Consider â–³ ADO

Using the Pythagoras theorem

OA^{2} = AD^{2} + OD^{2}

By substituting the values we get

10^{2} = 6^{2} + OD^{2}

On further calculation

OD^{2} = 10^{2} – 6^{2}

So we get

OD^{2} = 100 â€“ 36

By subtraction

OD^{2} = 64

By taking the square root

OD = âˆš64

So we get

OD = 8cm

We know that Oâ€™A = 8cm

Consider â–³ ADOâ€™

Using the Pythagoras theorem

Oâ€™A^{2} = AD^{2} + Oâ€™D^{2}

By substituting the values we get

8^{2} = 6^{2} + Oâ€™D^{2}

On further calculation

Oâ€™D^{2} = 8^{2} – 6^{2}

So we get

Oâ€™D^{2} = 64 â€“ 36

By subtraction

Oâ€™D^{2} = 28

By taking the square root

Oâ€™D = âˆš28

We get

Oâ€™D = 2 âˆš7

We know that

OOâ€™ = OD + Oâ€™D

By substituting the values

OOâ€™ = (8 + 2 âˆš7) cm

Therefore, the distance between their centres is (8 + 2 âˆš7) 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.**

**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

âˆ QAP = âˆ 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:**

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 âŠ¥ AB and OM âŠ¥ CD

We know that the alternate angles are equal

âˆ ALM = âˆ 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.**

**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 â€“ 3)

AB = 2cm

We know that that PQ is the perpendicular bisector of AB

So we get

AL = Â½ Ã— AB

Substituting the values

AL = Â½ Ã— 2

So we get

AL = 1 cm

Consider â–³ PLA

Using the Pythagoras theorem

AP^{2} = AL^{2} + PL^{2}

By substituting the values we get

5^{2} = 1^{2} + PL^{2}

So we get

PL^{2} = 5^{2} – 1^{2}

On further calculation

PL^{2} = 25 â€“ 1

PL^{2} = 24

By taking the square root

PL = âˆš24

So we get

PL = 2 âˆš6 cm

We know that

PQ = 2 Ã— PL

By substituting the values

PQ = 2 Ã— 2 âˆš6

PQ = 4 âˆš6 cm

Therefore, the length of PQ = 4 âˆš6 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 âˆ ACD = y ^{o} and âˆ AOD = x^{o}, prove that x = 3y.**

**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 âˆ BOC and âˆ BCO form isosceles triangle

âˆ BOC = âˆ BCO = y^{o}

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

OA = OB

We know that âˆ OAB and âˆ OBA form isosceles triangle

âˆ OAB = âˆ OBA = 2y^{o}

From the figure we know that

Exterior âˆ AOD = âˆ OAC + âˆ ACO

It can be written as

Exterior âˆ AOD = âˆ OAB + âˆ BCO

So we get

Exterior âˆ AOD = 3y^{o}

It is given that

âˆ AOD = x^{o}

So we get

x^{o} = 3y^{o}

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 4q ^{2} = p^{2 }+ 3r^{2}.**

**Solution:**

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 âŠ¥ AB

So we get

OM = p

We know that ON âŠ¥ AC

So we get

ON = q

Consider â–³ OMB

Using the Pythagoras theorem

OB^{2} = OM^{2} + BM^{2}

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

So we get

r^{2} = p^{2} + ((1/2) AB) ^{2}

It can be written as

r^{2} = p^{2} + Â¼ Ã— 4x^{2}

So we get

r^{2} = p^{2} + x^{2} â€¦â€¦â€¦ (1)

Consider â–³ ONC

Using the Pythagoras theorem

OC^{2} = ON^{2} + CN^{2}

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

So we get

r^{2} = q^{2} + ((1/2) AC) ^{2}

It can be written as

r^{2} = q^{2} + x^{2}/4

We get

q^{2} = r^{2} – x^{2}/4

Multiplying the equation by 4

4q^{2} = 4r^{2} – x^{2}

Substituting equation (1)

4q^{2} = 4r^{2} â€“ (r^{2} – p^{2})

So we get

4q^{2} = 3r^{2} + p^{2}

Therefore, it is proved that 4q^{2} = p^{2 }+ 3r^{2}.

**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 âŠ¥ AB and OQ âŠ¥ AC, prove that PB = QC.**

**Solution:**

It is given that AB = AC

Dividing the equation by 2

We get

Â½ AB = Â½ AC

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

MB = NC â€¦â€¦. (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 â€“ OM = OQ – ON

So we get

PM = QN â€¦â€¦â€¦ (2)

Consider â–³ MPB and â–³ NQC

We know that

âˆ PMB = âˆ QNC = 90^{o}

By SAS congruence criterion

â–³ MPB â‰… â–³ 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.**

**Solution:**

Construct OL âŠ¥ AB and OM âŠ¥ CD

Consider â–³ OLB and â–³ OMC

We know that âˆ OLB and âˆ OMC are perpendicular bisector

âˆ OLB = âˆ OMC = 90^{o}

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

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

âˆ OBL = âˆ OCD

So we get OB = OC which is the radii

By AAS congruence criterion

â–³ OLB â‰… â–³ 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 â–³ ABC as an equilateral triangle with side 9 cm

Take AD as one of its median

We know that

AD âŠ¥ BC

It can be written as

BD = Â½ Ã— BC

By substituting the values

BD = Â½ Ã— 9

So we get

BD = 4.5 cm

Consider â–³ ADB

Using the Pythagoras theorem

AB^{2} = AD^{2} + BD^{2}

Substituting the values

9^{2} = AD^{2} + (9/2)^{2}

On further calculation

AD^{2} = 9^{2} â€“ (9/2)^{2}

So we get

AD^{2} = 81 â€“ 81/4

By taking out the square root

AD = 9âˆš3/ 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) Ã— (9âˆš3/ 2)

So we get

AG = 3âˆš3 cm

Therefore, the radius of the circle is 3âˆš3 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 âˆ BAC.**

**Solution:**

Consider â–³ OAB and â–³ 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

â–³ OAB â‰… â–³ OAC

âˆ OAB = âˆ OAC (c. p. c. t)

Therefore, it is proved that O lies on the bisector of âˆ 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.**

**Solution:**

Consider â–³ OXP and â–³ OYR

We know that âˆ OPX and âˆ ORY are right angles

So we get

âˆ OPX = âˆ ORY = 90^{o}

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

â–³ OXP â‰… â–³ OYR

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

We know that PQ = QR

So we get

PQ â€“ PX = QR â€“ RY

QX = QY

Therefore, it is proved that QX = QY.

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

**Solution:**

Construct OM âŠ¥ PQ and Oâ€™N âŠ¥ PQ

So we get

OM âŠ¥ AP

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

AM = PM

It can be written as

AP = 2AM â€¦â€¦â€¦. (1)

We know that

Oâ€™N âŠ¥ AQ

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

AN = QN

It can be written as

AQ = 2AN â€¦â€¦. (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 MNOâ€™O is a rectangle

PQ = 2OOâ€™

Therefore, it is proved that PQ = 2OOâ€™.

## Exercise 12(B) page: 456

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

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

** âˆ AOC = 110 ^{o}. Find âˆ BAC.**

**Solution:**

(i) Join the line BO

Consider â–³ 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

âˆ OBC = âˆ OCB

It is given that

âˆ OCB = 30^{o}

So we get

âˆ OBC = âˆ OCB = 30^{o}

So we get âˆ OBC = 30^{o} â€¦â€¦. (1)

Consider â–³ 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

âˆ OAB = âˆ OBA

It is given that

âˆ OAB = 40^{o}

So we get

âˆ OAB = âˆ OBA = 40^{o}

So we get âˆ OBA = 40^{o} â€¦â€¦. (2)

We know that

âˆ ABC = âˆ OBC + âˆ OBA

By substituting the values

âˆ ABC = 30^{o} + 40^{o}

So we get

âˆ ABC = 70^{o}

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

âˆ AOC = 2 Ã— âˆ ABC

So we get

âˆ AOC = 2 Ã— 70^{o}

By multiplication

âˆ AOC = 140^{o}

Therefore, âˆ AOC = 140^{o}

(ii) From the figure we know that

âˆ AOB + âˆ AOC + âˆ BOC = 360^{o}

By substituting the values

90^{o} + 110^{o} + âˆ BOC = 360^{o}

On further calculation

âˆ BOC = 360^{o} – 90^{o} – 110^{o}

By subtraction

âˆ BOC = 360^{o} – 200^{o}

So we get

âˆ BOC = 160^{o}

We know that

âˆ BOC = 2 Ã— âˆ BAC

It is given that âˆ BOC = 160^{o}

âˆ BAC = 160^{o}/2

By division

âˆ BAC = 80^{o}

Therefore, âˆ BAC = 80^{o}.

**2. In the given figure, O is the centre of the circle and âˆ AOB = 70 ^{o}. Calculate the values of **

**(i) âˆ OCA, **

**(ii) âˆ OAC.**

**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

âˆ AOB = 2 âˆ OCA

It is given that âˆ AOB = 70^{o}

We can write it as

âˆ OCA = 70^{o}/2

By division

âˆ OCA = 35^{o}

Therefore, âˆ OCA = 35^{o}.

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

OA = OC

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

âˆ OAC = âˆ OCA

It is given that âˆ OCA = 35^{o}

So we get

âˆ OAC = 35^{o}

Therefore, âˆ OAC = 35^{o}.

**3. In the given figure, O is the centre of the circle. If âˆ PBC = 25 ^{o} and âˆ APB = 110^{o}, find the value of âˆ ADB.**

**Solution:**

We know that âˆ ACB = âˆ PCB

In â–³ PCB

Using the angle sum property

âˆ PCB + âˆ BPC + âˆ PBC = 180^{o}

We know that âˆ APB and âˆ BPC are linear pair

By substituting the values

âˆ PCB + (180^{o} â€“ 110^{o}) + 25^{o }= 180^{o}

On further calculation

âˆ PCB + 70^{o} + 25^{o }= 180^{o}

âˆ PCB + 95^{o} = 180^{o}

By subtraction

âˆ PCB = 180^{o} – 95^{o}

So we get

âˆ PCB = 85^{o}

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

âˆ ADB = âˆ ACB = 85^{o}

Therefore, the value of âˆ ADB is 85^{o}.

**4. In the given figure, O is the centre of the circle. If âˆ ABD = 35 ^{o} and âˆ BAC = 70^{o}, find âˆ ACB.**

**Solution:**

We know that BD is the diameter of the circle

Angle in a semicircle is a right angle

âˆ BAD = 90^{o}

Consider â–³ BAD

Using the angle sum property

âˆ ADB + âˆ BAD + âˆ ABD = 180^{o}

By substituting the values

âˆ ADB + 90^{o} + 35^{o} = 180^{o}

On further calculation

âˆ ADB = 180^{o} – 90^{o} – 35^{o}

By subtraction

âˆ ADB = 180^{o} – 125^{o}

So we get

âˆ ADB = 55^{o}

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

âˆ ACB = âˆ ADB = 55^{o}

So we get

âˆ ACB = 55^{o}

Therefore, âˆ ACB = 55^{o}.

**5. In the given figure, O is the centre of the circle. If âˆ ACB = 50 ^{o}, find âˆ OAB.**

**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

âˆ AOB = 2 âˆ ACB

It is given that âˆ ACB = 50^{o}

By substituting

âˆ AOB = 2 Ã— 50^{o}

By multiplication

âˆ AOB = 100^{o} â€¦â€¦ (1)

Consider â–³ OAB

We know that the radius of the circle are equal

OA = OB

Base angles of an isosceles triangle are equal

So we get

âˆ OAB = âˆ OBA â€¦â€¦. (2)

Using the angle sum property

âˆ AOB + âˆ OAB + âˆ OBA = 180^{o}

Using equations (1) and (2) we get

100^{o} + 2 âˆ OAB = 180^{o}

By subtraction

2 âˆ OAB = 180^{o} – 100^{o}

So we get

2 âˆ OAB = 80^{o}

By division

âˆ OAB = 40^{o}

Therefore, âˆ OAB = 40^{o}.

**6. In the given figure, âˆ ABD = 54 ^{o} and âˆ BCD = 43^{o}, calculate**

**(i) âˆ ACD,**

**(ii) âˆ BAD,**

**(iii) âˆ BDA.**

**Solution:**

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

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

âˆ ABD = âˆ ACD = 54^{o}

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

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

âˆ BAD = âˆ BCD = 43^{o}

(iii) In â–³ ABD

Using the angle sum property

âˆ BAD + âˆ ADB + âˆ DBA = 180^{o}

By substituting the values

43^{o} + âˆ ADB + 54^{o} = 180^{o}

On further calculation

âˆ ADB = 180^{o} – 43^{o} – 54^{o}

By subtraction

âˆ ADB = 180^{o} – 97^{o}

So we get

âˆ ADB = 83^{o}

It can be written as

âˆ BDA = 83^{o}

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

**Solution:**

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

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

âˆ CAD = âˆ CBD = 60^{o}

An angle in a semi-circle is a right angle

So we get

âˆ ADC = 90^{o}

Using the angle sum property

âˆ ACD + âˆ ADC + âˆ CAD = 180^{o}

By substituting the values

âˆ ACD + 90^{o} + 60^{o} = 180^{o}

On further calculation

âˆ ACD = 180^{o} – 90^{o} – 60^{o}

By subtraction

âˆ ACD = 180^{o} – 150^{o}

So we get

âˆ ACD = 30^{o}

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

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

So we get

âˆ CDE = âˆ ACD = 30^{o}

Therefore, âˆ CDE = 30^{o}.

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

**Solution:**

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

âˆ BCD = âˆ ABC = 25^{o}

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.

âˆ BOD = 2 âˆ BCD

It is given that âˆ BCD = 25^{o}

So we get

âˆ BOD = 2 (25^{o})

By multiplication

âˆ BOD = 50^{o}

In the same way

âˆ AOC = 2 âˆ ABC

So we get

âˆ AOC = 50^{o}

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

Using the angle sum property

âˆ AOC + âˆ COD + âˆ BOD = 180^{o}

By substituting the values

50^{o} + âˆ COD + 50^{o} = 180^{o}

On further calculation

âˆ COD + 100^{o} = 180^{o}

By subtraction

âˆ COD = 180^{o} – 100^{o}

So we get

âˆ COD = 80^{o}

We know that

âˆ CED = Â½ âˆ COD

So we get

âˆ CED = 80^{o}/2

By division

âˆ CED = 40^{o}

Therefore, âˆ CED = 40^{o}.

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

**(i) âˆ DCE,**

**(ii) âˆ ABC.**

**Solution:**

(i) From the figure we know that âˆ CED = 90^{o}

Consider â–³ CED

Using the angle sum property

âˆ CED + âˆ EDC + âˆ DCE = 180^{o}

By substituting the values

90^{o} + 40^{o} + âˆ DCE = 180^{o}

On further calculation

âˆ DCE = 180^{o} – 90^{o} – 40^{o}

By subtraction

âˆ DCE = 180^{o} – 130^{o}

So we get

âˆ DCE = 50^{o}

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

It can be written as

âˆ BOC = 180^{o} – 80^{o}

By subtraction

âˆ BOC = 100^{o}

Using the angle sum property

âˆ ABC + âˆ BOC + âˆ DCE = 180^{o}

By substituting the values

âˆ ABC + 100^{o} + 50^{o} = 180^{o}

On further calculation

âˆ ABC = 180^{o} – 100^{o} – 50^{o}

By subtraction

âˆ ABC = 180^{o} – 150^{o}

So we get

âˆ ABC = 30^{o}

**10. In the given figure, O is the centre of a circle, âˆ AOB = 40 ^{o} and âˆ BDC = 100^{o}, find âˆ OBC.**

**Solution:**

So we get

âˆ AOB = 2 âˆ ACB

From the figure we know that

âˆ ACB = âˆ DCB

It can be written as

âˆ AOB = 2 âˆ DCB

We also know that

âˆ DCB = Â½ âˆ AOB

By substituting the values

âˆ DCB = 40^{o}/2

By division

âˆ DCB = 20^{o}

In â–³ DBC

Using the angle sum property

âˆ BDC + âˆ DCB + âˆ DBC = 180^{o}

By substituting the values we get

100^{o} + 20^{o} + âˆ DBC = 180^{o}

On further calculation

âˆ DBC = 180^{o} – 100^{o} – 20^{o}

By subtraction

âˆ DBC = 180^{o} – 120^{o}

So we get

âˆ DBC = 60^{o}

From the figure we know that

âˆ OBC = âˆ DBC = 60^{o}

So we get

âˆ OBC = 60^{o}

Therefore, âˆ OBC = 60^{o}.

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

**Solution:**

We know that OA and OB are the radius

Base angles of an isosceles triangle are equal

So we get

âˆ OBA = âˆ OAB = 25^{o}

Consider â–³ OAB

Using the angle sum property

âˆ OAB + âˆ OBA + âˆ AOB = 180^{o}

By substituting the values

25^{o} + 25^{o} + âˆ AOB = 180^{o}

On further calculation

âˆ AOB = 180^{o} – 25^{o} – 25^{o}

By subtraction

âˆ AOB = 180^{o} – 50^{o}

So we get

âˆ AOB = 130^{o}

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

âˆ AOB = 2 âˆ ACB

It can be written as

âˆ ACB = Â½ âˆ AOB

By substituting the values

âˆ ACB = 130/2

By division

âˆ ACB = 65^{o}

So we get

âˆ ECB = 65^{o}

In â–³ BEC

Using the angle sum property

âˆ EBC + âˆ BEC + âˆ ECB = 180^{o}

By substituting the values

âˆ EBC + 90^{o} + 65^{o} = 180^{o}

On further calculation

âˆ EBC = 180^{o} – 90^{o} – 65^{o}

By subtraction

âˆ EBC = 180^{o} – 155^{o}

So we get

âˆ EBC = 25^{o}

Therefore, âˆ EBC = 25^{o}.

**12. In the given figure, O is the centre of a circle in which âˆ OAB = 20 ^{o} and âˆ OCB = 55^{o}. Find **

**(i) âˆ BOC,**

**(ii) âˆ AOC.**

**Solution:**

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

The base angles of an isosceles triangle are equal

So we get

âˆ OBC = âˆ OCB = 55^{o}

In â–³ BOC

Using the angle sum property

âˆ BOC + âˆ OCB + âˆ OBC = 180^{o}

By substituting the values

âˆ BOC + 55^{o} + 55^{o} = 180^{o}

On further calculation

âˆ BOC = 180^{o} – 55^{o} – 55^{o}

By subtraction

âˆ BOC = 180^{o} – 110^{o}

So we get

âˆ BOC = 70^{o}

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

The base angles of an isosceles triangle are equal

So we get

âˆ OBA= âˆ OAB = 20^{o}

In â–³ AOB

Using the angle sum property

âˆ AOB + âˆ OAB + âˆ OBA = 180^{o}

By substituting the values

âˆ AOB + 20^{o} + 20^{o} = 180^{o}

On further calculation

âˆ AOB = 180^{o} – 20^{o} – 20^{o}

By subtraction

âˆ AOB = 180^{o} – 40^{o}

So we get

âˆ AOB = 140^{o}

We know that

âˆ AOC = âˆ AOB – âˆ BOC

By substituting the values

âˆ AOC = 140^{o} – 70^{o}

So we get

âˆ AOC = 70^{o}

**13. In the given figure, O is the centre of the circle and âˆ BCO = 30 ^{o}. Find x and y.**

**Solution:**

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

So we get

âˆ AOD = âˆ OEC = 90^{o}

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

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

âˆ AOD = âˆ OEC

We know that âˆ DOC and âˆ OCE are alternate angles

âˆ DOC = âˆ OCE = 30^{o}

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

âˆ DOC = 2 âˆ DBC

It can be written as

âˆ DBC = Â½ âˆ DOC

By substituting the values

âˆ DBC = 30/2

By division

y = âˆ DBC = 15^{o}

In the same way

âˆ ABD = Â½ âˆ AOD

By substituting the values

âˆ ABD = 90/2

By division

âˆ ABD = 45^{o}

We know that

âˆ ABE = âˆ ABC = âˆ ABD + âˆ DBC

So we get

âˆ ABE = âˆ ABC = 45^{o} + 15^{o}

By addition

âˆ ABE = âˆ ABC = 60^{o}

Consider â–³ ABE

Using the angle sum property

âˆ BAE + âˆ AEB + âˆ ABE = 180^{o}

By substituting the values

x + 90^{o} + 60^{o} = 180^{o}

On further calculation

x = 180^{o} â€“ 90^{o} â€“ 60^{o}

By subtraction

x = 180^{o} â€“ 150^{o}

So we get

x = 30^{o}

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

**14. In the given figure, O is the centre of the circle, BD = OD and CD âŠ¥ AB. Find âˆ CAB.**

**Solution:**

Join the points AC

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 â–³ ODB as equilateral triangle

We know that

âˆ ODB = 60^{o}

The altitude of an equilateral triangle bisects the vertical angle

So we get

âˆ BDE = âˆ ODE = Â½ âˆ ODB

By substituting the values

âˆ BDE = âˆ ODE = 60/2

By division

âˆ BDE = âˆ ODE = 30^{o}

We know that the angles in the same segment are equal

âˆ CAB = âˆ BDE = 30^{o}

Therefore, âˆ CAB = 30^{o}.

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

**Solution:**

In â–³ PQR

We know that PQ is the diameter

So we get

âˆ PRQ = 90^{o} as the angle in a semicircle is a right angle

Using the angle sum property

âˆ QPR + âˆ PRQ + âˆ PQR = 180^{o}

By substituting the values

âˆ QPR + 90^{o} + 65^{o} = 180^{o}

On further calculation

âˆ QPR = 180^{o} – 90^{o} – 65^{o}

By subtraction

âˆ QPR = 180^{o} – 155^{o}

So we get

âˆ QPR = 25^{o}

In â–³ PQM

We know that PQ is the diameter

So we get

âˆ PMQ = 90^{o} as the angle in a semicircle is a right angle

Using the angle sum property

âˆ QPM + âˆ PMQ + âˆ PQM = 180^{o}

By substituting the values

âˆ QPM + 90^{o} + 50^{o} = 180^{o}

On further calculation

âˆ QPM = 180^{o} – 90^{o} – 50^{o}

By subtraction

âˆ QPM = 180^{o} – 140^{o}

So we get

âˆ QPM = 40^{o}

Consider the quadrilateral PQRS

We know that

âˆ QPS + âˆ SRQ = 180^{o}

It can be written as

âˆ QPR + âˆ RPS + âˆ PRQ + âˆ PRS = 180^{o}

By substituting the values

25^{o} + 40^{o} + 90^{o} + âˆ PRS = 180^{o}

On further calculation

âˆ PRS = 180^{o} – 25^{o} – 40^{o} – 90^{o}

By subtraction

âˆ PRS = 180^{o} – 155^{o}

So we get

âˆ PRS = 25^{o}

**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 âˆ APB = 150 ^{o} and âˆ BQD = x^{o}, find the value of x.**

**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

âˆ APB = 2 âˆ ACB

It can be written as

âˆ ACB = Â½ âˆ APB

By substituting the values

âˆ ACB = 150/2

So we get

âˆ ACB = 75^{o}

We know that ACD is a straight line

It can be written as

âˆ ACB + âˆ DCB = 180^{o}

By substituting the values

75^{o} + âˆ DCB = 180^{o}

On further calculation

âˆ DCB = 180^{o} – 75^{o}

By subtraction

âˆ DCB = 105^{o}

We know that

âˆ DCB = Â½ Ã— reflex âˆ BQD

By substituting the values

105^{o} = Â½ Ã— (360^{o} â€“ x)

On further calculation

210^{o} = 36^{o} – x

By subtraction

x = 150^{o}

Therefore, the value of x is 150^{o}.

**17. In the given figure, âˆ BAC = 30 ^{o}. Show that BC is equal to the radius of the circumcircle of â–³ ABC whose centre is O.**

**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

âˆ BOC = 2 âˆ BAC

It is given that âˆ BAC = 30^{o}

âˆ BOC = 2 Ã— 30^{o}

By multiplication

âˆ BOC = 60^{o}

In â–³ BOC

We know that the radii are equal

OB = OC

The base angles of an isosceles triangle are equal

âˆ OBC = âˆ OCB

Consider â–³ BOC

Using the angle sum property

âˆ BOC + âˆ OBC + âˆ OCB = 180^{o}

By substituting the values we get

60^{o} + âˆ OCB + âˆ OCB = 180^{o}

So we get

2 âˆ OCB = 180^{o} – 60^{o}

By subtraction

2 âˆ OCB = 120^{o}

By division

âˆ OBC = 60^{o}

We get âˆ OBC = âˆ OCB = âˆ BOC = 60^{o}

We know that â–³ 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 â–³ 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 âˆ AEC = Â½ (angle subtended by arc CXA. At the centre + angle subtended by arc DYB at the centre).**

**Solution:**

Construct the lines AC and BC

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

âˆ AOC = 2 âˆ ABC â€¦â€¦ (1)

In the same way

âˆ BOD = 2 âˆ BCD â€¦â€¦ (2)

By adding both the equations

âˆ AOC + âˆ BOD = 2 âˆ ABC + 2 âˆ BCD

By taking 2 as common

âˆ AOC + âˆ BOD = 2 (âˆ ABC + âˆ BCD)

It can be written as

âˆ AOC + âˆ BOD = 2 (âˆ EBC + âˆ BCE)

So we get

âˆ AOC + âˆ BOD = 2 (180^{o} – âˆ CEB)

We can write it as

âˆ AOC + âˆ BOD = 2 (180^{o} â€“ (180^{o} – âˆ AEC))

We get

âˆ AOC + âˆ BOD = 2 âˆ AEC

Dividing the equation by 2

âˆ AEC = Â½ (âˆ AOC + âˆ BOD)

âˆ AEC = Â½ (angle subtended by arc CXA at the centre + angle subtended by arc DYB at the centre)

Therefore, it is proved that âˆ AEC = Â½ (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 âˆ DBC = 60 ^{o} and âˆ BAC = 40^{o}. Find **

**(i) âˆ BCD,**

**(ii) âˆ CAD.**

**Solution:**

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

So we get

âˆ BDC = âˆ BAC = 40^{o}

Consider â–³ BCD

Using the angle sum property

âˆ BCD + âˆ BDC + âˆ DBC = 180^{o}

By substituting the values

âˆ BCD + 40^{o} + 60^{o} = 180^{o}

On further calculation

âˆ BCD = 180^{o} – 40^{o} – 60^{o}

By subtraction

âˆ BCD = 180^{o} – 100^{o}

So we get

âˆ BCD = 80^{o}

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

So we get

âˆ CAD = âˆ CBD

So âˆ CAD = 60^{o}

**2. In the given figure, POQ is a diameter and PQRS is a cyclic quadrilateral. If âˆ PSR = 150 ^{o}, find âˆ RPQ.**

**Solution:**

We know that PQRS is a cyclic quadrilateral

It can be written as

âˆ PSR+ âˆ PQR = 180^{o}

By substituting the values

150^{o} + âˆ PQR = 180^{o}

On further calculation

âˆ PQR = 180^{o} – 150^{o}

By subtraction

âˆ PQR = 30^{o}

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

âˆ PRQ = 90^{o}

Consider â–³ PRQ

Using the angle sum property

âˆ PQR + âˆ PRQ + âˆ RPQ = 180^{o}

By substituting the values

30^{o} + 90^{o} + âˆ RPQ = 180^{o}

On further calculation

âˆ RPQ = 180^{o} – 30^{o} – 90^{o}

By subtraction

âˆ RPQ = 180^{o} – 120^{o}

So we get

âˆ RPQ = 60^{o}

Therefore, âˆ RPQ = 60^{o}.

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

**Solution:**

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

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

âˆ 2 = 2 âˆ 1

By substituting the values

130^{o} = 2 âˆ 1

So we get

âˆ 1 = 65^{o}

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

âˆ PBC = âˆ 1

So we get âˆ PBC = 65^{o}

Therefore, âˆ PBC = 65^{o}.

**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 âˆ ABC = 92 ^{o} and âˆ FAE = 20^{o}, find âˆ BCD.**

**Solution:**

We know that ABCD is a cyclic quadrilateral

So we get

âˆ ABC + âˆ ADC = 180^{o}

By substituting the values

92^{o} + âˆ ADC = 180^{o}

On further calculation

âˆ ADC = 180^{o} – 92^{o}

By subtraction

âˆ ADC = 88^{o}

We know that AE || CD

From the figure we know that

âˆ EAD = âˆ ADC = 88^{o}

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

So we get

âˆ BCD = âˆ DAF

We know that

âˆ BCD = âˆ EAD + âˆ EAF

It is given that âˆ FAE = 20^{o}

By substituting the values

âˆ BCD = 88^{o} + 20^{o}

By addition

âˆ BCD = 108^{o}

Therefore, âˆ BCD = 108^{o}.

**5. In the given figure, BD = DC and âˆ CBD = 30 ^{o}, find âˆ BAC.**

**Solution:**

It is given that BD = DC

From the figure we know that

âˆ BCD = âˆ CBD = 30^{o}

Consider â–³ BCD

Using the angle sum property

âˆ BCD + âˆ CBD + âˆ CDB = 180^{o}

By substituting the values

30^{o} + 30^{o} + âˆ CDB = 180^{o}

On further calculation

âˆ CDB = 180^{o} – 30^{o} – 30^{o}

By subtraction

âˆ CDB = 180^{o} – 60^{o}

So we get

âˆ CDB = 120^{o}

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

It can be written as

âˆ CDB + âˆ BAC = 180^{o}

By substituting the values

120^{o} + âˆ BAC = 180^{o}

On further calculation

âˆ BAC = 180^{o} – 120^{o}

By subtraction

âˆ BAC = 60^{o}

Therefore, âˆ BAC = 60^{o}.

**6. In the given figure, O is the centre of the given circle and measure of arc ABC is 100 ^{o}. Determine âˆ ADC and âˆ ABC.**

**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

âˆ AOC = 100^{o}

So we get

âˆ AOC = 2 âˆ ADC

It can be written as

âˆ ADC = Â½ âˆ AOC

By substituting the values

âˆ ADC = Â½ (100^{o})

So we get

âˆ ADC = 50^{o}

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

It can be written as

âˆ ADC + âˆ ABC = 180^{o}

By substituting the values

50^{o} + âˆ ABC = 180^{o}

On further calculation

âˆ ABC = 180^{o} – 50^{o}

By subtraction

âˆ ABC = 130^{o}

Therefore, âˆ ADC = 50^{o} and âˆ ABC = 130^{o}.

**7. In the given figure, â–³ ABC is equilateral. Find **

**(i) âˆ BDC,**

**(ii) âˆ BEC.**

**Solution:**

It is given that â–³ ABC is equilateral

We know that

âˆ BAC = âˆ ABC = âˆ ACB = 60^{o}

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

âˆ BDC = âˆ BAC = 60^{o}

So we get

âˆ BDC = 60^{o}

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

So we get

âˆ BAC + âˆ BEC = 180^{o}

By substituting the values

60^{o} + âˆ BEC = 180^{o}

On further calculation

âˆ BEC = 180^{o} – 60^{o}

By subtraction

âˆ BEC = 120^{o}

**8. In the adjoining figure, ABCD is a cyclic quadrilateral in which âˆ BCD = 100 ^{o} and âˆ ABD = 50^{o}. Find âˆ ADB.**

**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

âˆ A + âˆ C = 180^{o}

By substituting the values

âˆ A + 100^{o} = 180^{o}

On further calculation

âˆ A = 180^{o} – 100^{o}

By subtraction

âˆ A = 80^{o}

Consider â–³ ABD

Using the angle sum property

âˆ A + âˆ ABD + âˆ ADB = 180^{o}

By substituting the values

80^{o} + 50^{o} + âˆ ADB = 180^{o}

On further calculation

âˆ ADB = 180^{o} – 80^{o} – 50^{o}

By subtraction

âˆ ADB = 180^{o} – 130^{o}

So we get

âˆ ADB = 50^{o}

Therefore, âˆ ADB = 50^{o}.

**9. In the given figure, O is the centre of a circle and âˆ BOD =150 ^{o}. Find the values of x and y.**

**Solution:**

It is given that O is the centre of a circle and âˆ BOD =150^{o}

We know that

Reflex âˆ BOD = (360^{o} – âˆ BOD)

By substituting the values

Reflex âˆ BOD = (360^{o} â€“ 150^{o})

By subtraction

Reflex âˆ BOD = 210^{o}

Consider x = Â½ (reflex âˆ BOD)

By substituting the value

x = 210/2

So we get

x = 105^{o}

We know that

x + y = 180^{o}

By substituting the values

105^{o} + y = 180^{o}

On further calculation

y = 180^{o} – 105^{o}

By subtraction

y = 75^{o}

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

**10. In the given figure, O is the centre of the circle and âˆ DAB = 50 ^{o}. Calculate the values of x and y.**

**Solution:**

It is given that O is the centre of the circle and âˆ DAB = 50^{o}

We know that the radii of the circle are equal

OA = OB

From the figure we know that

âˆ OBA = âˆ OAB = 50^{o}

Consider â–³ OAB

Using the angle sum property

âˆ OAB + âˆ OBA + âˆ AOB = 180^{o}

By substituting the values

50^{o} + 50^{o} + âˆ AOB = 180^{o}

On further calculation

âˆ AOB = 180^{o} – 50^{o} – 50^{o}

By subtraction

âˆ AOB = 180^{o} – 100^{o}

So we get

âˆ AOB = 80^{o}

From the figure we know that AOD is a straight line

It can be written as

x = 180^{o} – âˆ AOB

By substituting the values

x = 180^{o} – 80^{o}

By subtraction

x = 100^{o}

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

So we get

âˆ DAB + âˆ BCD = 180^{o}

By substituting the values

50^{o} + âˆ BCD = 180^{o}

On further calculation

âˆ BCD = 180^{o} – 50^{o}

By subtraction

y = âˆ BCD = 130^{o}

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

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

**Solution:**

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

So we get

âˆ CBF = âˆ CDA

It can be written as

130^{o} = 180^{o} – x

On further calculation

x = 180^{o} – 130^{o}

By subtraction

x = 50^{o}

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

**(i) âˆ BAD,**

**(ii) âˆ ABD,**

**(iii) âˆ CBD,**

**(iv) âˆ ADC.**

**Also, show that â–³ AOD is an equilateral triangle.**

**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

âˆ BCD + âˆ BAD = 180^{o}

By substituting the values

120^{o} + âˆ BAD = 180^{o}

On further calculation

âˆ BAD = 180^{o} – 120^{o}

By subtraction

âˆ BAD = 60^{o}

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

âˆ BDA = 90^{o}

Consider â–³ ABD

By using the angle sum property

âˆ BDA + âˆ BAD + âˆ ABD = 180^{o}

By substituting the values

90^{o} + 60^{o} + âˆ ABD = 180^{o}

On further calculation

âˆ ABD = 180^{o} – 90^{o} – 60^{o}

By subtraction

âˆ ABD = 180^{o} – 150^{o}

So we get

âˆ ABD = 30^{o}

(iii) We know that OD = OA

So we get âˆ ODA = âˆ OAD = âˆ BAD = 60^{o}

From the figure we know that

âˆ ODB + âˆ ODA = 90^{o}

By substituting the values

âˆ ODB + 60^{o} = 90^{o}

On further calculation

âˆ ODB = 90^{o} – 60^{o}

By subtraction

âˆ ODB = 30^{o}

It is given that DO || CB

We know that the alternate angles are equal

âˆ CDB = âˆ ODB = 30^{o}

(iv) From the figure we know that

âˆ ADC = âˆ ADB + âˆ CDB

By substituting the values

âˆ ADC = 90^{o} + 30^{o}

By addition

âˆ ADC = 120^{o}

Consider â–³ AOD

By using the angle sum property

âˆ ODA + âˆ OAD + âˆ AOD = 180^{o}

By substituting the values

60^{o} + 60^{o} + âˆ AOD = 180^{o}

On further calculation

âˆ AOD = 180^{o} – 60^{o} – 60^{o}

By subtraction

âˆ AOD = 180^{o} – 120^{o}

So we get

âˆ AOD = 60^{o}

We know that all the angles of the â–³ AOD is 60^{o}

Therefore, it is proved that â–³ 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.**

**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 Ã— BP = CP Ã— DP

From the figure we know that CP = CD + DP

By substituting the values

8 Ã— 2 = (CD + 2.5) Ã— 2.5 cm

Consider x = CD

So we get

8 Ã— 2 = (x + 2.5) Ã— 2.5

On further calculation

16 = 2.5x + 6.25

It can be written as

2.5x = 16 â€“ 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 âˆ AOD = 140 ^{o} and âˆ CAB = 50^{o}, calculate**

**(i) âˆ EDB,**

**(ii) âˆ EBD.**

**Solution:**

(i) We know that

âˆ BOD + âˆ AOD = 180^{o}

By substituting the values

âˆ BOD + 140^{o} = 180^{o}

On further calculation

âˆ BOD = 180^{o} – 140^{o}

By subtraction

âˆ BOD = 40^{o}

We know that OB = OD

So we get âˆ OBD = âˆ ODB

Consider â–³ OBD

By using the angle sum property

âˆ BOD + âˆ OBD + âˆ ODB = 180^{o}

We know that âˆ OBD = âˆ ODB

So we get

40^{o} + 2 âˆ OBD = 180^{o}

On further calculation

2 âˆ OBD = 180^{o} – 40^{o}

By subtraction

2 âˆ OBD = 140^{o}

By division

âˆ OBD = 70^{o}

We know that ABCD is a cyclic quadrilateral

âˆ CAB + âˆ BDC = 180^{o}

âˆ CAB + âˆ ODB + âˆ ODC = 180^{o}

By substituting the values

50^{o} + 70^{o} + âˆ ODC = 180^{o}

On further calculation

âˆ ODC = 180^{o} – 50^{o} – 70^{o}

By subtraction

âˆ ODC = 180^{o} – 120^{o}

So we get

âˆ ODC = 60^{o}

Using the angle sum property

âˆ EDB + âˆ ODC + âˆ ODB = 180^{o}

By substituting the values

âˆ EDB + 60^{o} + 70^{o} = 180^{o}

On further calculation

âˆ EDB = 180^{o} – 60^{o} – 70^{o}

By subtraction

âˆ EDB = 180^{o} – 130^{o}

So we get

âˆ EDB = 50^{o}

(ii) We know that

âˆ EDB + âˆ OBD = 180^{o}

By substituting the values

âˆ EDB + 70^{o} = 180^{o}

On further calculation

âˆ EDB = 180^{o} – 70^{o}

By subtraction

âˆ EDB = 110^{o}

**15. In the given figure, â–³ 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.**

**Solution:**

We know that â–³ 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

âˆ ACB = âˆ ABC

It can be written as

âˆ ADE = âˆ ACB = âˆ ABC

So we get

âˆ ADE = âˆ 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 â–³ AEB is isosceles.**

**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 âˆ EDC = âˆ A

Exterior âˆ DCE = âˆ B

We know that AB || CD

So we get

âˆ EDC = âˆ B and âˆ DCE = âˆ A

We get

âˆ A = âˆ B

Therefore, it is proved that â–³ AEB is isosceles.

**17. In the given figure, âˆ BAD = 75 ^{o}, âˆ DCF = x^{o} and âˆ DEF = y^{o}. Find the values of x and y.**

**Solution:**

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

So we get

âˆ BAD = âˆ DCF = 75^{o}

It can be written as

âˆ DCF = x = 75^{o}

We get x = 75^{o}

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

So we get

âˆ DCF + âˆ DEF = 180^{o}

By substituting the values

75^{o} + âˆ DEF = 180^{o}

On further calculation

âˆ DEF = 180^{o} – 75^{o}

By subtraction

âˆ DEF = y = 105^{o}

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

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

**Solution:**

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

Construct DE âŠ¥Â AB and CF âŠ¥Â AB

Consider â–³ ADE and â–³ BCF

We know that

âˆ AED + âˆ BFC = 90^{o}

From the figure it can be written as

âˆ ADE = âˆ ADC – 90^{o} = âˆ BCD – 90^{o} = âˆ BCF

It is given that

AD = BC

By AAS congruence criterion

â–³ ADE â‰… â–³ BCF

âˆ A = âˆ B (c. p. c. t)

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

âˆ A + âˆ B + âˆ C + âˆ D = 360^{o}

By substituting the values

2 âˆ B + 2 âˆ D = 360^{o}

By taking 2 as common

2 (âˆ B + âˆ D) = 360^{o}

By division

âˆ B + âˆ D = 360/2

So we get

âˆ B + âˆ D = 180^{o}

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:**

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:**

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

âˆ BOC = 90^{o}

Thus, âˆ 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:**

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

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 â–³ ABC and â–³ DCB

We know that

AB = DC and AC = DB

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

By SSS congruence criterion

â–³ ABC â‰… â–³ DCB

âˆ BAC = âˆ CDB (c. p. c. t)

We know that BC subtends equal angles âˆ BAC and âˆ 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 (âˆ B – âˆ D) = 60 ^{o}, show that the smaller of the two is 60^{o}.**

**Solution:**

It is given that ABCD is a cyclic quadrilateral

(âˆ B – âˆ D) = 60^{o} â€¦â€¦.. (1)

We know that

(âˆ B + âˆ D) = 180^{o} â€¦â€¦. (2)

By adding both the equations

âˆ B – âˆ D + âˆ B + âˆ D = 60^{o} + 180^{o}

So we get

2 âˆ B = 240^{o}

By division

âˆ B = 120^{o}

By substituting equation it in equation (1)

(âˆ B – âˆ D) = 60^{o}

120^{o} – âˆ D = 60^{o}

On further calculation

âˆ D = 120^{o} – 60^{o}

By subtraction

âˆ D = 60^{o}

Therefore, the smaller of the two angles âˆ D = 60^{o}.

**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

We know that OL âŠ¥ AB so that âˆ O meets the line CD at the point M

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

âˆ 1 = âˆ 2

We know that âˆ OLB = 90^{o} so we get

âˆ 2 + âˆ 3 = 90^{o} â€¦â€¦ (1)

We know that OLM is a straight line and âˆ BOC = 90^{o}

So we get

âˆ 3 + âˆ 4 = 90^{o} â€¦â€¦â€¦ (2)

It can be written as

âˆ 2 + âˆ 3 = âˆ 3 + âˆ 4

On further calculation

âˆ 2 = âˆ 4

So we get

âˆ 1 = âˆ 2 and âˆ 2 = âˆ 4

It can be written as

âˆ 1 = âˆ 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 âˆ BAC = âˆ BDC.**

**Solution:**

We know that AB is the common hypotenuse of â–³ ACB and â–³ ADB

So we get

âˆ ACB = 90^{o}

âˆ BDC = 90^{o}

It can be written as

âˆ ACB + âˆ BDC = 180^{o}

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

âˆ BAC = âˆ BDC

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

**26. ABCD is a quadrilateral such that A is the centre of the circle passing through B, C and D. Prove that âˆ CBD + âˆ CDB = Â½ âˆ BAD.**

**Solution:**

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

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

âˆ BAD = 2 âˆ BED

It can be written as

âˆ BED = Â½ âˆ BAD â€¦â€¦ (1)

Consider EBCD as a cyclic quadrilateral

So we get

âˆ BED + âˆ BCD = 180^{o}

We can write it as

âˆ BCD = 180^{o} – âˆ BED

Substituting equation (1)

âˆ BCD = 180^{o} – Â½ âˆ BAD â€¦â€¦ (2)

Consider â–³ BCD

Using the angle sum property

âˆ CBD + âˆ CDB + âˆ BCD = 180^{o}

By using the equation (2)

âˆ CBD + âˆ CDB + 180^{o} – Â½ âˆ BAD = 180^{o}

So we get

âˆ CDB + âˆ CDB – Â½ âˆ BAD = 180^{o} – 180^{o}

On further calculation

âˆ CDB + âˆ CDB = Â½ âˆ BAD

Therefore, it is proved that âˆ CDB + âˆ CDB = Â½ âˆ 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 satelliteâ€™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.