# RS Aggarwal Class 9 Solutions Chapter 11 - Circle Ex 11C (11.3)

## RS Aggarwal Class 9 Chapter 11 – Circle Ex 11C (11.3) Solutions Free PDF

Students of Class 9 can clear their concepts by referring to the RS Aggarwal Class 9 solutions and practicing it on a regular basis will help them to prepare effectively for the exams. Scoring good marks in the exam is easy for the students with the help of reference RS Aggarwal Class solutions of Class 9 as it covers all the topics of the subject. Prepared by subject experts in accordance to the CBSE syllabus it also helps in improving the understanding of difficult concepts.

These solutions will make your foundation strong to tackle different variety of questions to practice in the most efficient manner. The RS Aggarwal Class 9 Solutions Chapter 11 – Circle Ex 11C (11.3) are easy to understand and crafted in such a way that all the steps of the solutions are described to match your understanding. Students are advised to solve RS Aggarwal maths solution which are highly useful for students.

## Download PDF ofÂ RS Aggarwal Class 9 Solutions Chapter 11 – Circle Ex 11C (11.3)

Question 1:In the given figure ,PQ is a diameter of a circle with centre O . If âˆ  PRQ = 65Â°,âˆ  SPR = 40Â° and âˆ  PQM = 50Â° Â Â Â , Â Â Â Â find âˆ QPR ,âˆ  QPM and âˆ  PRS.

Sol.

Consider the triangle Î” PRQ.

PQ is the diameter.

The angle in a semicircle is a right angle.

â‡’ âˆ  PRQ = 90Â°

By the angle sum property in Î” PRQ, we have

âˆ QPR + âˆ PRQ +âˆ PQR = 180Â°

â‡’ âˆ QPR + 90Â° + 65Â° = 180Â°

â‡’ âˆ QPR = 180Â° – 155Â° = 25Â° â€¦â€¦.(1)

Now Consider the triangle Î” PQM.

Since PQ is the diameter, âˆ  PMQ = 90Â°â€™

Again by the angle sum property in Î” PRQ, we have

âˆ QPM+ âˆ PMQ +âˆ PQM = 180Â°

â‡’ âˆ QPM + 90Â° + 50Â° = 180Â°

â‡’ âˆ QPR = 180Â° – 140Â° = 40Â°

âˆ QPS + âˆ  SRQ = 180Â°

â‡’ âˆ QPR+ âˆ RPS +âˆ PRQ + âˆ  PRS = 180Â° [from (1)]

â‡’ 25Â° + 40Â° + 90Â° + âˆ  PRS = 180Â°

â‡’ Â Â Â Â Â Â âˆ  PRS = 180Â° – 155Â° = 25Â°

âˆ´ Â Â Â Â Â âˆ  PRS = 25Â°

Question 2: In the given figure ABCD is a cyclic quadrilateral whose diagonals intersect at P such that âˆ  DBC = 60Â° and âˆ  BAC = 40Â° , find

1. âˆ  BCD

Sol.

(i) âˆ  BDC = âˆ  BAC = 40Â° [angles in the same segment]

In Î” BCD, we have

âˆ  BCD +âˆ  BDC + âˆ  DBC =180Â°

âˆ´ âˆ  BCD +40Â° + 60Â° =180Â°

â‡’ âˆ  BCD = 180Â° -100Â° = 80Â°

âˆ´ âˆ  BCD = 80Â°

(ii) Also âˆ  CAD = âˆ  CBD [angles in the same segment]

âˆ  CAD = 60Â° [âˆµ âˆ  CBD = 60Â°]

Question 3: In the given figure, PQR is a diameter and PQRS is a cyclic quadrilateral. If âˆ  PSR = 150Â°, findâˆ  RPQ

Sol.

âˆ  PSR + âˆ  PQR = 180Â°

â‡’ 150Â° + âˆ  PQR = 180Â°

â‡’ âˆ  PQR = 180Â° – 150Â° = 30Â° â€¦â€¦.(1)

Also, âˆ  PQR = 90Â° â€¦â€¦.(2) [angles in a semicircle]

Now in Î” PRQ, we have

âˆ  PQR + âˆ  PRQ + âˆ  RPQ = 180Â°

â‡’ 30Â° + 90Â°+ âˆ  RPQ = 180Â° [from (i) and (ii)]

â‡’ âˆ  RPQ = 180Â° – 120Â° = 60Â°

âˆ´ âˆ  RPQ = 60Â°

Question 4: In the given figure ABCD is a cyclic quadrilateral in which AB || DC . If Â Â Â Â Â Â âˆ  BAD = 100Â° Â , find

1. âˆ  BCD
3. âˆ  ABC

Sol.

In cyclic quadrilateral ABCD, AB || DC and âˆ  BAD = 100Â°

(i) âˆ  BCD + âˆ  BAD = 180Â°

â‡’ âˆ  BCD + 100Â° = 180Â°

â‡’ âˆ  BCD = 180Â° – 100Â° = 80Â°

(ii) Also, âˆ  ADC = âˆ  BCD = 80Â°

âˆ´ âˆ  ACD = 80Â°

(iii) âˆ  ABC = âˆ  BAD = 100Â°

âˆ  ABC = 100Â°

Question 5: In the given figure O is the centre of the circle and arc ABC subtends an angle of 1300 at the centre . If Â AB is extended to P , find âˆ  PBC

Sol.

Take a point D on the major arc CA and join AD and DC

âˆ´ âˆ  2 = 2âˆ  1 [Angle subtended by an arc is twice the angle subtended by it on the circumference in the alternate segment.]

âˆ´ 130Â° = 2âˆ  1

â‡’ âˆ  1 = 65Â° â€¦â€¦.(1)

âˆ  PBC = âˆ 1 [âˆµ Exterior angle of a cyclic quadrilateral interior opposite angle]

âˆ´ âˆ  PBC = 65Â°

Question 6: In the given figure ABCD is a cyclic quadrilateral in which AE is drawn parallel to CD and BA is produced . Ifâˆ ABC=92Â° Â and âˆ  EAF = 20Â°, findâˆ  BCD.

Sol.

âˆ´ âˆ ABC + âˆ  ADC = 180Â°

â‡’ 92Â° + âˆ  ADC = 180Â°

â‡’ âˆ  ADC = 180Â° -92Â° = 88Â°

Also, AE || CD

âˆ´ âˆ  BCD = âˆ  DAF [âˆµ Exterior angle of a cyclic quadrilateral = interior opposite angle]

âˆ´ âˆ  BCD = âˆ  EAD + âˆ  EAF

= 88Â° + 20Â° [âˆµâˆ  FAE = 20Â° (given)]

= 108Â°

âˆ´ âˆ  BCD = 108Â°

Question 7: In the given figure , BD=DC and âˆ  CBD = 30Â°, find m(âˆ  BAC)

Sol.

BD = DC

âˆ´ âˆ  BCD = âˆ  CBD = 30Â°

In Î” BCD, we have

âˆ  BCD + âˆ  CBD + âˆ  CDB = 180Â°

â‡’ 30Â° + 30Â° + âˆ  CDB = 180Â°

â‡’ âˆ  CDB = 180Â° – 60Â°

= 120Â°

The opposite angles of a cyclic quadrilateral are supplementary.

ABCD is a cyclic quadrilateral and thus,

âˆ  CDB + âˆ  BAC = 180Â°

= 180Â° -120Â° [âˆµâˆ  CDB = 120Â°]

= 60Â°

âˆ´ âˆ  BAC = 60Â°

Question 8: In the given figure , O is the centre of the given circle and measure of arc ABC is 1000. Determine âˆ  ADC Â and âˆ  ABC .

Sol.

Angle subtended by an arc is twice the angle subtended by it on the circumference in the alternate segment.

Here arc ABC makes âˆ  AOC = 100Â° at the center of the circle and âˆ  ADC on the circumference of the circle.

âˆ´ âˆ  AOC = 2âˆ  ACD

â‡’ âˆ  ACD = Â½ (âˆ AOC)

â‡’ = Â½ Ã— 100Â° [âˆ  AOC = 100Â°]

â‡’ âˆ  ACD = 50Â°

The opposite angles of a cyclic quadrilateral are supplementary.

ABCD is a cyclic quadrilateral and thus,

âˆ  ACD + âˆ  ABC = 180Â°

= 180Â° – 50Â° [âˆµ âˆ  ADC = 50Â°]

= 130Â°

âˆ´ âˆ  ABC = 130Â°

âˆ´ âˆ  ACD = 50Â° and âˆ  ABC = 130Â°

Question 9: In the given figure , Î” ABC is equilateral . Find

1. âˆ  BDC
2. âˆ  BEC

Sol.

Î” ABC is an equilateral triangle.

âˆ´ Â Â Each of its angle is equal to 60Â°

â‡’ âˆ  BAC = âˆ  ABC = âˆ  Â âˆ  ACB = 60Â°

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

âˆ´ âˆ  BDC = âˆ  BAC

= 60Â° [âˆµ âˆ  BAC = 60Â°]

(ii) The opposite angles of a cyclic quadrilateral are supplementary.

ABCE is a cyclic quadrilateral and thus,

âˆ  BAC + âˆ  BEC = 180Â°

âˆ  BEC = 180Â° – 60Â° [âˆµ âˆ  BAC = 60Â°]

= 120Â°

â‡’ âˆ  BEC = 120Â°

Question 10: In the adjoining figure, ABCD is a cyclic quadrilateral in which âˆ  BCD = 100Â° and âˆ  ABD = 50Â° , find âˆ  ADB

Sol.

âˆ  A + âˆ  C = 180Â° [The opposite angles of a cyclic quadrilateral are supplementary]

â‡’ âˆ  A + 100Â°= 180Â°

â‡’ âˆ  A = 180Â° – 100Â° = 80Â°

Now in Î” ABD, we have

âˆ  A + âˆ  ABD + âˆ  ADB = 180Â°

â‡’ 80Â° + 50Â° + âˆ  ADB = 180Â°

â‡’ âˆ  ADB = 180Â° – 130Â° = 50Â°

Question 11: In the given figure , O is the centre of the circle and âˆ  DAB = 150Â° find value of x and y

Sol.

O is the center of the circle and âˆ  BOD = 150Â°

âˆ´ Reflex Â Â âˆ  BOD = (360Â° – âˆ  BOD)

= (360Â° – 150Â°) = 210Â°

Now, X = Â½(reflex âˆ  BOD)

= Â½ Ã— 210Â° = 105Â°

X = 105Â°

Again, X + Y = 180Â°

â‡’ 105Â° + Y = 180Â°

â‡’ Â Â Â Â Â Â Â Â Â Â Y = 180Â° â€“ 105Â° = 75Â°

âˆ´ Â Â Â Â Â Â Â Â Â Â Â Y = 75Â°

Question 12: In the given figure , O is the centre of the circle and âˆ  DAB = 50Â° . Find value of x and y.

Sol.

O is the center of the circle and âˆ  DAB= 50Â°

â‡’ âˆ  OBA = âˆ  OAB = 50Â°

In Î” OAB, we have

âˆ  OAB + âˆ  OBA + âˆ  AOB =180Â°

â‡’ Â Â 50Â° + 50Â° + âˆ  AOB =180Â°

â‡’ Â Â  âˆ  AOB =180Â° – 100Â° = 80Â°

Since, AOD is a straight line,

X = 180Â° – âˆ  AOB

= 180Â° – 80Â° = 100Â°

âˆ´ X = 100Â°

The opposite angles of a cyclic quadrilateral are supplementary.

ABCD is a cyclic quadrilateral and thus,

âˆ  DAB + âˆ  BCD = 180Â°

âˆ  BCD = 180Â° – 50Â° [âˆµ âˆ  DAB = 50Â°, given]

â‡’ Y = 130Â°

Thus, X = 100Â° and Y = 130Â°.

Question 13: In the given figure sides AD and AB of cyclic quadrilateral ABCD are produced to E and F respectively. If âˆ  CBF=130Â° Â and âˆ  CDE= X0 , find the value of x

Sol.

We know that in a cyclic quadrilateral exterior angle = interior angle.

âˆ´ âˆ  CBF = âˆ  CDA = (180Â° – X)

â‡’ 130Â° = 180Â° – X

â‡’ X = 180Â° – 130Â° = 50Â°

X = 50Â°

### Key Features ofÂ RS Aggarwal Class 9 Solutions Chapter 11 – Circle Ex 11C (11.3)

• It will improve your mathematical problem-solving skills.
• It gives ample techniques to solve difficult and tough questions.
• It is the best reference study material.
• It helps in boosting up your confidence.