## 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°

Now in quadrilateral PQRS

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

- âˆ BCD
- âˆ CAD

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.

In cyclic quadrilateral PQRS

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

- âˆ BCD
- âˆ ADC
- 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 130^{0 }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.

ABCD is acyclic quadrilateral

âˆ´ âˆ ABC + âˆ ADC = 180Â°

â‡’ 92Â° + âˆ ADC = 180Â°

â‡’ âˆ ADC = 180Â° -92Â° = 88Â°

Also, AE || CD

âˆ´ âˆ EAD = âˆ ADC = 88Â°

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 100^{0}. 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

- âˆ BDC
- 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.

ABCD is a cyclic quadrilateral,

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°

âˆ´ âˆ ADB = 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°

OA = OB [Radii]

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= X^{0} , find the value of x

Sol.

ABCD is a cyclic quadrilateral

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.