# RS Aggarwal Class 9 Solutions Chapter 11 - Circle Ex 11A (11.2)

## RS Aggarwal Class 9 Chapter 11 - Circle Ex 11A (11.2) Solutions Free PDF

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These solutions cover the entire syllabus and provide step by step solutions to the questions asked in RS Aggarwal maths textbook in a concise and accurate manner. Scoring good marks is easy with the access of these Class 10 maths solutions of RS Aggarwal. The well-structured solutions ofÂ RS Aggarwal Class 9 Solutions Chapter 11 – Circle Ex 11BÂ (11.2) help students in fetching good marks in the exam.

## Download PDF ofÂ RS Aggarwal Class 9 Solutions Chapter 11 – Circle Ex 11B (11.2)

Question 1: In the given figure, O is the centre of the circle and âˆ  AOB = 70Â° Â calculate the values of i) âˆ  OCA Â ii)âˆ  OAC

Sol.

(i) The angle subtended by an arc of a circle at the center is doubled the angle subtended by the arc at any point on the circumference.

âˆ´ âˆ  AOB = 2 Ã— âˆ  OCA

â‡’ âˆ  OCA = 70Â°/2 = 35Â° [âˆµ âˆ  AOB =70Â°]

(ii) The radius of the circle is

OA = OC

â‡’ âˆ  OAC = âˆ  OCA [âˆµ base angles of an isosceles triangle are equal]

â‡’ âˆ  OAC = 35Â° [as âˆ  OCA = 35Â°]

Question 2: Â In the given figure , O is the centre of the circle . If âˆ  PBC and âˆ  B calculate the values of âˆ  ADB.

Sol.

It is clear that âˆ  ACB = âˆ  PCB

Consider the triangle Î” PCB

Applying the sum property, we have

âˆ  PCB = 180Â° – (âˆ  BPC + âˆ  PCB)

= 180Â° – (180Â° – 110Â° + 25Â°)

âˆ  PCB = 180Â° – 95Â° = 85Â°

Angles in the same segment of a circle are equal.

âˆ´ âˆ  ADB = âˆ  ACB = 85Â°

Question 3: In the given figure, O is the centre of the circle.If âˆ  ABD = 35Â° and âˆ  BAC = 70Â°, find âˆ  ACB.

Sol.

It is clear that, BD is the diameter of the circle.

Also we know that, the angle in a semicircle is a right angle.

Now consider the triangle Î” BAD

â‡’ Â Â âˆ  ADB = 180Â° – (âˆ  BAD + âˆ ABD) [Angle sum property]

â‡’ Â Â Â Â Â = 180Â° – (90Â° + 35Â°) [âˆ  BAD = 90Â° and âˆ ABD= 35Â°]

â‡’ Â Â âˆ  ADB = 55Â°

Angles in the same segment of a circle are equal.

âˆ´ âˆ ACB = âˆ  ADB = 55Â°

âˆ´ âˆ ACB = 55Â°

Question 4: In the given figure, O is the centre of the circle. If âˆ ACB = 50Â°, Find âˆ OAB .

Sol.

The angle subtended by an arc of a circle at the center is doubled the angle subtended by the arc at any point on the circumference.

âˆ´ âˆ  AOB = 2 Ã— âˆ  ACB

= 2 Ã— 50Â° [Given]

â‡’ âˆ AOB = 100Â° â€¦â€¦.(1)

Now, in Î” OAB, we have

OA = OB [radius of the circle]

â‡’ âˆ  OAB = âˆ  OBA [âˆµ base angles of an isosceles triangle are equal]

Thus we have

âˆ  OAB = âˆ  OBA â€¦â€¦.(2)

By angle sum property, we have

Now Â Â âˆ AOB + âˆ  OAB + âˆ  OBA = 180Â°

â‡’ 100Â° + 2âˆ  OAB = 180Â°

â‡’ 2âˆ  OAB = 180Â° – 100Â° [from (1) and (2)]

â‡’ âˆ  OAB = 80Â°/2 = 40Â°

âˆ´ âˆ  OAB = 40Â°

Question 5: In the given figure, âˆ ABD = 54Â° Â and âˆ ABD = 43Â° . Calculate

i) âˆ ACD

ii) âˆ BDA

iii) âˆ BDA

Sol.

(i) Angles in the same segment of a circle are equal.

âˆ  ABD and âˆ  ACD are in the segment AD.

âˆ´ âˆ  ACD = âˆ  ABD

= 54Â° [Given]

(ii) Angles in the same segment of a circle are equal.

âˆ  BAD and âˆ  BCD are in the segment BD.

âˆ´ âˆ  BAD = âˆ  BCD

= 43Â° [Given]

(iii) Consider the Î” ABD.

By angle sum property, we have

â‡’ 43Â° + âˆ  ADB + 54Â° = 180Â°

â‡’ âˆ  ADB = 180Â° -97Â° = 83Â°

â‡’ âˆ  BDA = 83Â°

Question 6: In the adjoining figure ,DE is the chord parallel to diameter AC of the circle with Centre O. If âˆ  CBD = 60Â°, calculate âˆ  CDE.

Sol.

Angles in the same segment of a circle are equal.

âˆ´ Â Â âˆ  CAD =âˆ  CBD = 60Â° [Given]

We know that an angle in a semi-circle is a right angle.

âˆ´ Â Â âˆ  ADC = 90Â° [angle in a semi-circle]

= 180Â° – (90Â° + 60Â°)

= 180Â° -150Â° =30Â°

âˆ´ Â Â âˆ  CDE = âˆ  ADC =30Â°

Question 7: In the adjoining figure, O is the centre of the circle. Chord CD is parallel to diameter AB. Ifâˆ  ABC = 25Â°, calculate âˆ  CED.

Sol.

Join CO and DO, âˆ  BCD = âˆ  ABC = 25Â° [alternate interior angles]

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´ Â Â âˆ BOD = 2âˆ  BCD

= 50Â° [âˆ  BCD =25Â°]

Similarly,

âˆ AOC = 2âˆ  ABC

= 50Â°

AB is a straight line passing through the center.

âˆ´ Â Â Â âˆ AOC + âˆ COD + âˆ BOD = 180Â°

â‡’ 50Â° + âˆ  COD + 50Â° = 180Â°

â‡’ âˆ  COD = 180Â° – 100Â° = 80Â°

âˆ´ âˆ  CED = Â½ âˆ  COD

= 80Â°/2 = 40Â°

âˆ´ âˆ  CED = 40Â°

Question 8: In the given figure AB and CD are straight lines to the centre O of a circle .If âˆ  AOC= 80Â° and âˆ  CDE = 40Â° find

I – âˆ  DCE

II- âˆ  ABC

Sol.

(i) âˆ  CED = 90Â°

In CED, We have

âˆ  CED + âˆ  EDC + âˆ  DCE =180Â°

â‡’ 90Â° + 50Â° + âˆ  DCE = 180Â°

âˆ´ âˆ  DCE = 180Â° -130Â°

âˆ  DCE = 50Â° Â Â â€¦.(1)

(ii) âˆ  AOC and âˆ  BOC are linear Pair.

âˆ´ âˆ  BOC = (180Â° – 80Â°) = 100Â° â€¦.(2)

âˆ´ âˆ ABC = 180Â° – (âˆ  BOC +âˆ  DCE)

= 180Â° – (100Â° + 50Â°) [from (1) and (2)]

= 180Â° – 150Â° = 30Â°

Question 9: In the given figure, O is the centre of the circle , âˆ  AOB = 40Â° and âˆ  BDC = 100Â° , find âˆ  OBC = 60Â°

Sol.

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´ Â Â âˆ AOB = 2âˆ  ACB

â‡’ Â Â Â = 2âˆ  ACB [âˆµ âˆ  ACB = âˆ  DCB]

$\Rightarrow \angle DCB = \frac{1}{2}\angle AOB$

= $\left ( \frac{1}{2}\times 40 \right )$ = 20Â°

Consider the Î” DBC;

By angle sum property, we have

âˆ  BDC + âˆ DCB + âˆ  DBC = 180Â°

â‡’ 100Â° + 20Â° + âˆ DCB = 180Â°

â‡’ âˆ  DCB = 180Â° – 120Â° = 60Â°

â‡’ âˆ  OBC = âˆ  DBC = 60Â°

âˆ´ âˆ  OBC = 60Â°

Question 10: In the adjoining figure ,choprd AC and BD of a circle with centre O , intersect at right angle at E . If Â âˆ  OAB = 25Â°, Calculate âˆ  EBC

Sol.

Join OB,

âˆ´ âˆ  OBA = âˆ  OAB = 25Â° [base angles are equal in isosceles triangle]

Now in Î” OAB, WE have

â‡’Â âˆ OAB + âˆ  OBA + âˆ  AOB = 180Â°

â‡’Â  25 + 25 + âˆ AOB =180Â°

â‡’Â  âˆ AOB = 180Â° â€“ 50Â° = 130Â°

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´ âˆ  AOB = 2âˆ  ACB

â‡’Â  âˆ ACB = Â½ âˆ  AOB = Â½ Ã— 130Â° = 65Â°

â‡’Â  âˆ ECB = 65Â°

In rt Â angled Î” BEC,

We know that sum of three angles in a triangle is 180Â°.

â‡’Â  âˆ EBC + âˆ BEC +âˆ ECB = 180Â°

â‡’ âˆ EBC + 90Â° + 65Â° = 180Â°

â‡’ âˆ EBC = 180Â° – 155Â° =25Â°

âˆ´ âˆ EBC =25Â°

Question 11:In the fgiven figure , O is the centre of the circle in which âˆ  OAB = 20Â° and âˆ  OCB = 55Â° . Find

1. âˆ  BOC
2. âˆ  AOC

Sol.

â‡’ âˆ  OBC = âˆ  OCB =55Â° [base angles are equal in isosceles triangle]

Consider the triangle Î” BOC.

By angle sum property, we have

âˆ  BOC = 180Â° – (âˆ  OCB + âˆ  OBC)

= 180Â° – (55Â° + 55Â°)

= 180Â° – 110Â° = 70Â°

âˆ´ âˆ  BOC = 70Â°

Again, OA = OB

â‡’ âˆ  OBA = âˆ  OAB =20Â° [base angles are equal in isosceles triangle]

Consider the triangle Î” AOB.

By angle sum property, we have

â‡’ âˆ  AOB = 180Â° – (âˆ  OAB + âˆ  OBA)

= 180Â° – (20Â° + 20Â°)

= 180Â° – 40Â° = 140Â°

âˆ´ âˆ  AOC = âˆ  AOB + âˆ  BOC

= 140Â° – 70Â° = 70Â°

âˆ´ âˆ  AOC = 70Â°

Question 12: In the given figure , âˆ  OAB = 25Â°. Show that BC is equal to the radius of the circumcircle of Î” ABC whose centre is O.

Sol.

Join OB and OC.

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´ âˆ  BOC = 2âˆ  BAC

= 2 Ã— 30Â° [âˆ´âˆ  BAC = 30Â°]

= 60Â° â€¦â€¦.(1)

Now Consider the triangle Î” BOC.

â‡’ âˆ OBC = âˆ  OCB â€¦â€¦.(2) Â  [base angles are equal in isosceles triangle]

Now in Î” BOC, we have

â‡’Â  âˆ BOC + âˆ OBC +âˆ OCB = 180Â°

â‡’Â  60Â° + âˆ OBC +âˆ OCB = 180Â° [from (1) and (2)]

â‡’ Â 2âˆ  OBC = 180Â° â€“ 60Â° = 120Â°

â‡’ âˆ  OBC = 120Â°/2 = 60Â°

â‡’ âˆ  OBC = 60Â° [from (2)]

Thus, we have

âˆ BOC = âˆ OBC = âˆ OCB = 60Â°

So, BOC is an equilateral triangle

â‡’ OB = OC = BC

âˆ´ Â Â BC is the radius of the circumference.’

### Key Features ofÂ RS Aggarwal Class 9 Chapter 11 – Circle Ex 11B (11.2)

• It is the best revision study material for Class 9 students.
• It provides different methods to solve tricky and difficult questions.
• It helps you analyze your overall performance, speed and accuracy.
• It boosts your confidence while attempting the final paper.

#### Practise This Question

In the following triangle, if CD = BE and EC = 3cm, then, BD is equal to ______cm.