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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.

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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.

âˆ´ âˆ BAD = 90Â°

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

âˆ BAD + âˆ ADB + âˆ DBA = 180Â°

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]

âˆ´ Â Â âˆ ADC = 180Â° – (âˆ ADC + âˆ CAD)

= 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´ OA = OB [Radius]

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

- âˆ BOC
- âˆ AOC

Sol.

OB = OC [Radius]

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.

âˆ´ âˆ BOC = 2âˆ BAC

= 2 Ã— 30Â° [âˆ´âˆ BAC = 30Â°]

= 60Â° â€¦â€¦.(1)

Now Consider the triangle Î” BOC.

OB = OC [Radii]

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.