## RS Aggarwal Solutions for Class 9 Maths Exercise 12B PDF

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

## Access RS Aggarwal Solutions for Class 9 Chapter 12: Circles Exercise 12B

### 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).

### Access other exercise solutions of Class 9 Maths Chapter 12: Circles

Exercise 12A Solutions 23 Questions

Exercise 12C Solutions 26 Questions

### RS Aggarwal Solutions Class 9 Maths Chapter 12 – Circles Exercise 12B

RS Aggarwal Solutions Class 9 Maths Chapter 12 Circles Exercise 12B contains the theorems which are based on the results of angles which are subtended by arcs and problems regarding them.

### Key features of RS Aggarwal Solutions for Class 9 Maths Chapter 12: Circles Exercise 12B

- Studying concepts in Mathematics is made simple with the help of RS Aggarwal Solutions in PDF format.
- The solutions are in a stepwise manner which improves problem solving abilities among students.
- Several examples before each exercise mainly help students gain conceptual knowledge before solving the problems.
- Theorems are explained in brief as each of them carry more marks in the board exam according to the CBSE syllabus.