# RS Aggarwal Class 9 Solutions Chapter 11 - Circle Ex 11A (11.1)

## RS Aggarwal Class 9 Chapter 11 – Circle Ex 11A (11.1) Solutions Free PDF

The RS Aggarwal Class 9 Solutions help students to score good marks in the exam. It is considered as one of the best study resources to prepare for Class 9 CBSE exam. Practicing these solutions on a daily basis will help student to analyze their performance and speed up the problem-solving skills. Every solution is solved accurately so that students can refer to these solutions whenever they have any doubt.

These solutions are prepared keeping in mind the understanding level of the students. The RS Aggarwal Class 9 Solutions Chapter 11 – Circle Ex 11A (11.1) are prepared by subject experts as per the CBSE syllabus. These solutions assist students to solve difficult questions with ease.

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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 Solutions Chapter 11 – Circle Ex 11A (11.1)

• The solutions are explained in a proper step by step manner.
• It teaches different methods to solve difficult and tricky questions.
• The RS Aggarwal Maths solution help students to score better marks in the exam.
• It boosts up the studentâ€™s confidence level.