## RS Aggarwal Class 9 Solutions Chapter 11 Circle Ex 11B

These Solutions are part of RS Aggarwal Solutions Class 9. Here we have given RS Aggarwal Solutions Class 9 Chapter 11 Circle Ex 11B.

**Other Exercises**

- RS Aggarwal Solutions Class 9 Chapter 11 Circle Ex 11A
- RS Aggarwal Solutions Class 9 Chapter 11 Circle Ex 11B
- RS Aggarwal Solutions Class 9 Chapter 11 Circle Ex 11C

**Question 1.**

**Solution:**

(i) O is the centre of the circle

∠OAB = 40°, ∠OCB = 30°

Join OB.

**Question 2.**

**Solution:**

O is the centre of the cirlce and ∠AOB = 70°

∵ Arc AB subtends ∠AOB at the centre and ∠ACB at the remaining part of the circle.

∵ ∠ACB = \(\frac { 1 }{ 2 } \) ∠AOB = \(\frac { 1 }{ 2 } \) x 70°

=> ∠ACB = 35°

or ∠OCA = 35°

In ∆OAC,

OA = OC (radii of the same circle)

∴ ∠OAC = ∠OCA = 35° Ans.

**Question 3.**

**Solution:**

In the figure, O is the centre of the circle. ∠PBC = 25°, ∠APB =110°

∠ APB + ∠ BPC = 180° (Linear pair)

=> 110° + ∠ BPC = 180°

**Question 4.**

**Solution:**

O is the centre of the circle

∠ABD = 35° and ∠B AC = 70°

BOD is the diameter of the circle

∠BAD = 90° (Angle in a semi circle)

But ∠ADB + ∠ABD + ∠BAD = 180° (Angles of a triangle)

=> ∠ADB + 35° + 90° = 180°

=> ∠ADB + 125° = 180°

=> ∠ADB = 180° – 125° = 55°

But ∠ACB = ∠ADB (Angles in the same segment of the circle)

∠ACB = 55° Ans.

**Question 5.**

**Solution:**

O is the centre of a circle and ∠ACB = 50°

∴ arc AB subtends ∠ AOB at the centre and ∠ ACB at the remaining part of the circle.

∴ ∠ AOB = 2 ∠ ACB

= 2 x 50° = 100

∴ OA = OB (radii of the same circle)

∴ ∠ OAB = ∠ OBA (Angles opposite to equal sides)

Now in ∆ OAB,

∠ OAB + ∠ OBA + ∠ AOB = 180°

=> ∠ OAB + ∠ OAB + ∠ AOB = 180° (∠OAB = ∠OBA)

=> 2 ∠ OAB + 100°= 180°

=> 2 ∠ OAB = 180° – 100° = 80°

=> ∠OAB = \(\frac { { 80 }^{ o } }{ 2 } \) = 40°

Hence, OAB = 40° Ans.

**Question 6.**

**Solution:**

(i) In the figure,

∠ABD = 54° and ∠BCD = 43°

∠BAD = ∠BCD (Angles in the same segment of a circle)

∠BAD = 43°

**Question 7.**

**Solution:**

Chord DE || diameter AC of the circle with centre O.

∠CBD = 60°

∠CBD = ∠ CAD

(Angles in the same segment of a circle)

∠CAD = 60°

Now in ∆ ADC,

**Question 8.**

**Solution:**

In the figure,

chord CD || diameter AB of the circle with centre O.

∠ ABC = 25°

Join CD and DO.

AB || CD

∠ ABC = ∠ BCD (alternate angles)

**Question 9.**

**Solution:**

AB and CD are two straight lines passing through O, the centre of the circle and ∠AOC = 80°, ∠CDE = 40°

∠ CED = 90° (Angle in a semi circle)

**Question 10.**

**Solution:**

O is the centre of the circle and ∠AOB = 40°, ∠BDC = 100°

Arc AB subtends ∠AOB at the centre and ∠ ACB at the remaining part of the circle

∠ AOB = 2 ∠ ACB

**Question 11.**

**Solution:**

Chords AC and BD of a circle with centre O, intersect each other at E at right angles.

∠ OAB = 25°. Join OB.

In ∆ OAB,

OA = OB (radii of the same circle)

**Question 12.**

**Solution:**

In the figure, O is the centre of a circle ∠ OAB = 20° and ∠ OCB = 55° .

In ∆ OAB,

**Question 13.**

**Solution:**

Given : A ∆ ABC is inscribed in a circle with centre O and ∠ BAC = 30°

To Prove : BC = radius of the circle

Const. Join OB and OC

**Question 14.**

**Solution:**

In a circle with centre O and PQ is its diameter. ∠PQR = 65°, ∠SPR = 40° and ∠PQM = 50°

(i) ∠PRQ = 90° (Angle in a semicircle) and ∠PQR + ∠RPQ + ∠PQR = 180° (Angles of a triangle)

Hope given RS Aggarwal Solutions Class 9 Chapter 11 Circle Ex 11B are helpful to complete your math homework.

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