RS Aggarwal Solutions Class 9 Ex 11C

RS Aggarwal Class 9 Ex 11C Chapter 11

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

Sol.

Consider the triangle Δ PRQ.

PQ is the diameter.

The angle in a semicircle is a right angle.

⇒ ∠ PRQ = 90°

By the angle sum property in Δ PRQ, we have

∠QPR + ∠PRQ +∠PQR = 180°

⇒ ∠QPR + 90° + 65° = 180°

⇒ ∠QPR = 180° – 155° = 25° …….(1)

https://lh3.googleusercontent.com/YxqEvr9IZiSKtBFPFfpIjc-WPBZRb5rkFRCH_IpFY0Sh-GQPpY5afW2n961x6lxR2r2j3GEzZas3Sw8-Ph89qHRF8YSBWf1gMDymIO-YGNiq93vf1IPVsnOA8nRlX8xryi5-PhIIqZjNo4nt4Q

Now Consider the triangle Δ PQM.

Since PQ is the diameter, ∠ PMQ = 90°’

Again by the angle sum property in Δ PRQ, we have

∠QPM+ ∠PMQ +∠PQM = 180°

⇒ ∠QPM + 90° + 50° = 180°

⇒ ∠QPR = 180° – 140° = 40°

Now in quadrilateral PQRS

∠QPS + ∠ SRQ = 180°

⇒ ∠QPR+ ∠RPS +∠PRQ + ∠ PRS = 180° [from (1)]

⇒ 25° + 40° + 90° + ∠ PRS = 180°

⇒       ∠ PRS = 180° – 155° = 25°

∴      ∠ PRS = 25°

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

  1. ∠ BCD
  2. ∠ CAD

Sol.

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

In Δ BCD, we have

∠ BCD +∠ BDC + ∠ DBC =180°

∴ ∠ BCD +40° + 60° =180°

⇒ ∠ BCD = 180° -100° = 80°

∴ ∠ BCD = 80°

https://lh5.googleusercontent.com/-8ZnlJT2pXTAk7mTcxLB5BnHcjZslQQHb4xJeunJUcqPEOBXzkNL-TY7KVbbeDTV52ZQ3dmFET8zj7SoSsxJm1HsUVhjd4fXlhmcjGyhWVDfGg-YH8c-ZuCpLU42XqFrFxM_CCGFkAYQ0O80zw

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

∠ CAD = 60° [∵ ∠ CBD = 60°]

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

Sol.

In cyclic quadrilateral PQRS

∠ PSR + ∠ PQR = 180°

⇒ 150° + ∠ PQR = 180°

⇒ ∠ PQR = 180° – 150° = 30° …….(1)

Also, ∠ PQR = 90° …….(2) [angles in a semicircle]

https://lh5.googleusercontent.com/0mPiPG8RUpFuA6mqXCjYtXBvIPL4NI0RacbylE_KMCNgJUd4tVJ29OQ7x5cGIUNb9QcSYOMrzd-VSjUNF5pRhRwtu3ydbjTO--kjwtGGnFJkLVAR_9uBgRBCm9VOdZ8qdMGTMwKnJ6AcpoRDsA

Now in Δ PRQ, we have

∠ PQR + ∠ PRQ + ∠ RPQ = 180°

⇒ 30° + 90°+ ∠ RPQ = 180° [from (i) and (ii)]

⇒ ∠ RPQ = 180° – 120° = 60°

∴ ∠ RPQ = 60°

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

  1. ∠ BCD
  2. ∠ ADC
  3. ∠ ABC

Sol.

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

https://lh5.googleusercontent.com/gzU-bSslBsfrw6XhUFFhwMsSuKlN-aByZMzh5qcj1sa3Rw7T45rs11hlM2oeZE0YaFK5RjM_ZnnPSBdh2TENwcbCJgs-39-garrbHLPPFSnVun1Y0LB2qzsxnBWbKHM6HTMgGjXaHutLyNNAVA

(i) ∠ BCD + ∠ BAD = 180°

⇒ ∠ BCD + 100° = 180°

⇒ ∠ BCD = 180° – 100° = 80°

(ii) Also, ∠ ADC = ∠ BCD = 80°

∴ ∠ ACD = 80°

(iii) ∠ ABC = ∠ BAD = 100°

∠ ABC = 100°

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  AB is extended to P , find ∠ PBC

Sol.

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

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

https://lh6.googleusercontent.com/kcweYLjCjgZOmpdigjpAeuFr-vf0tUFsHXQrQyeS19KpTvMOCgWpEyAvVGfEz7YA98mnRkVPwXy2nJjvAeWFjyaXYCMMN-EhpTGfBsTQ0Dd-iLP4EPcvEUKZMdRA3vqEd8XleikNrh17hXkRiA

∴ 130° = 2∠ 1

⇒ ∠ 1 = 65° …….(1)

∠ PBC = ∠1 [∵ Exterior angle of a cyclic quadrilateral interior opposite angle]

∴ ∠ PBC = 65°

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

Sol.

ABCD is acyclic quadrilateral

∴ ∠ABC + ∠ ADC = 180°

⇒ 92° + ∠ ADC = 180°

⇒ ∠ ADC = 180° -92° = 88°

https://lh4.googleusercontent.com/fNpL3OlCLwAO1vnpsoS_jqg8sQuYv6P0PdG9VaB7V98n84LgDRs7dtr6iEWjQ8aSVqQbp0YhRxr3iFoHWUrGP9Ij4qJ4gSKGvWRw-eCiXhCdsv6yVlcKbxZ781GPqUUqcg-Q65s79YEvApsknA

Also, AE || CD

∴ ∠EAD = ∠ ADC = 88°

∴ ∠ BCD = ∠ DAF [∵ Exterior angle of a cyclic quadrilateral = interior opposite angle]

∴ ∠ BCD = ∠ EAD + ∠ EAF

= 88° + 20° [∵∠ FAE = 20° (given)]

= 108°

∴ ∠ BCD = 108°

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

Sol.

BD = DC

∴ ∠ BCD = ∠ CBD = 30°

https://lh3.googleusercontent.com/rb6aF5UcefvOoQwCSz_g8PP5scYJpuNztBRXIcyJ5J4oojTKmQGx7kAgyOzuZQipokrlJA7BMYjUHASHXZSMsUfK_mfi-LyoK_EBrQvpF1DhSxdieWj3bypieOx15wgB8Vvlpk4cZa4a330gxg

In Δ BCD, we have

∠ BCD + ∠ CBD + ∠ CDB = 180°

⇒ 30° + 30° + ∠ CDB = 180°

⇒ ∠ CDB = 180° – 60°

= 120°

The opposite angles of a cyclic quadrilateral are supplementary.

ABCD is a cyclic quadrilateral and thus,

∠ CDB + ∠ BAC = 180°

= 180° -120° [∵∠ CDB = 120°]

= 60°

∴ ∠ BAC = 60°

Question 8: In the given figure , O is the centre of the given circle and measure of arc ABC is 1000. Determine ∠ ADC  and ∠ 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 ∠ AOC = 100° at the center of the circle and ∠ ADC on the circumference of the circle.

∴ ∠ AOC = 2∠ ACD

⇒ ∠ ACD = ½ (∠AOC)

⇒ = ½ × 100° [∠ AOC = 100°]

⇒ ∠ ACD = 50°

https://lh3.googleusercontent.com/7SJKMAbdHzqeo5aHOaXUWs6tRFLODYnWJ00XTVuLKwsYRlRnSvD_kmkV54S6w6cIzZyt1pmHUnFlsNd_5wvIwA-_89yVxPY97L2gov5NEjpSEB8VeCTZ_bj61MaMpT8Uq6Pt3YCGLxIVO4yNDA

The opposite angles of a cyclic quadrilateral are supplementary.

ABCD is a cyclic quadrilateral and thus,

∠ ACD + ∠ ABC = 180°

= 180° – 50° [∵ ∠ ADC = 50°]

= 130°

∴ ∠ ABC = 130°

∴ ∠ ACD = 50° and ∠ ABC = 130°

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

  1. ∠ BDC
  2. ∠ BEC

Sol.

Δ ABC is an equilateral triangle.

∴   Each of its angle is equal to 60°

⇒ ∠ BAC = ∠ ABC = ∠  ∠ ACB = 60°

https://lh3.googleusercontent.com/391o0K55XumyChENN4n2vqhhaYKOIKYw66TxBW4uWccGy5YzJSOOTLBV4ELaXKNWClsmyOG3A4QZluaeElJLWigA1E_adNHriBaC2CIazUaPf03-enqfOlAPz2ju4uObRJ25ezjYy4tGEOZOgg

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

∴ ∠ BDC = ∠ BAC

= 60° [∵ ∠ BAC = 60°]

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

ABCE is a cyclic quadrilateral and thus,

∠ BAC + ∠ BEC = 180°

∠ BEC = 180° – 60° [∵ ∠ BAC = 60°]

= 120°

⇒ ∠ BEC = 120°

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

Sol.

ABCD is a cyclic quadrilateral,

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

⇒ ∠ A + 100°= 180°

⇒ ∠ A = 180° – 100° = 80°

https://lh3.googleusercontent.com/Zgs4UhdlVC0J9j4zxCGCgZ_8yt3VsggRwxAcwrUdmLkjscBndMROANh-3kKx0E8bhvG_LokkGNUBMku1TCljvKWItlWdFQgIOlN4g3Z3WUmAehEEh0hLiCZEHXK06_ubCDlTzlWau8qskaZ5HQ

Now in Δ ABD, we have

∠ A + ∠ ABD + ∠ ADB = 180°

⇒ 80° + 50° + ∠ ADB = 180°

⇒ ∠ ADB = 180° – 130° = 50°

∴ ∠ ADB = 50°

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

Sol.

O is the center of the circle and ∠ BOD = 150°

∴ Reflex   ∠ BOD = (360° – ∠ BOD)

https://lh5.googleusercontent.com/d9tRKP2JZGsed7AOXyatw1jfOOT9fmP11qbgn1fJq9b7fJcKcWmoo6Zho8Z-g8cIWGHLWgejiU8xnyrKqSjjCCq13VxAS5JwoTqsJYY36FHt8d3Vn_UlwT0ob2YfZqcmttSM9J2Pe0zd-7Gg3w = (360° – 150°) = 210°

Now, X = ½(reflex ∠ BOD)

= ½ × 210° = 105°

X = 105°

Again, X + Y = 180°

⇒ 105° + Y = 180°

⇒           Y = 180° – 105° = 75°

∴            Y = 75°

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

Sol.

O is the center of the circle and ∠ DAB= 50°

OA = OB [Radii]

⇒ ∠ OBA = ∠ OAB = 50°

https://lh6.googleusercontent.com/maEjVCylMQ5QgZWfAH8z7hSN3P8IV_dHUYt6U-7EIKFR1bpu_9Of8_SFbOCc-6KpjPvJqR62aAxQQMGR9qis6KqD_iaxDBUlnupGEqYwk1UGW1hkpDGM3B8VCIoII72IF1tVRRqZmwc6gY86Vw

In Δ OAB, we have

∠ OAB + ∠ OBA + ∠ AOB =180°

⇒   50° + 50° + ∠ AOB =180°

⇒    ∠ AOB =180° – 100° = 80°

Since, AOD is a straight line,

X = 180° – ∠ AOB

= 180° – 80° = 100°

∴ X = 100°

The opposite angles of a cyclic quadrilateral are supplementary.

ABCD is a cyclic quadrilateral and thus,

∠ DAB + ∠ BCD = 180°

∠ BCD = 180° – 50° [∵ ∠ DAB = 50°, given]

⇒ Y = 130°

Thus, X = 100° and Y = 130°.

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

Sol.

ABCD is a cyclic quadrilateral

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

∴ ∠ CBF = ∠ CDA = (180° – X)

⇒ 130° = 180° – X

https://lh4.googleusercontent.com/9hQ_i88opZCjzAyOmDaou1QgCJVDGdSBNsvVfAcCOLsgeLIrT_tbuOkO1cPjDGPn4wv0kzbTbSU3OButWjGn2tJd5Zqwrn666H_ubF7fS36n27Dmuwx_et9GBAHshNvFp106DkqYXnoeguIVrg

⇒ X = 180° – 130° = 50°

X = 50°


Practise This Question

Rangappa has a piece of farmland next to the river Krishna, whereas his brother Sangappa has farmland away from the flow of the river. What method of irrigation should Rangappa adopt (A) and what method of irrigation should Sangappa adopt (B)?