# Concise Maths Solutions Circles Chapter-17 ICSE Class 10

## Selina Publishers Concise Maths Solutions Chapter-17 Circles

**Concise Maths Solutions Circles** Chapter-17 ICSE Class 10.** Solutions** of Exercise – 17 (A), Exercise – 17 (B), Exercise – 17 (C)for** Concise **Selina Maths of ICSE Board Class 10th. **Concise Maths Solutions Circles **Chapter-17 for ICSE Maths Class 10 is available here. All **Solutions **of **Concise** Selina **Maths** of **Circles **Chapter-17 has been solved according instruction given by council. This is the **Solutions **of **Circles **Chapter-17 for ICSE Class 10th. ICSE **Maths** text book of **Concise** is In series of famous ICSE writer in maths publications. **Concise** is most famous among students

**Concise Solutions Circles Chapter-17 Selina Maths ICSE Class 10**

The **Solutions** of **Concise** **Mathematics** **Circles **Chapter-17 for ICSE Class 10 have been solved. Experience teachers Solved Chapter-17 **Circles** to help students of class 10th ICSE board. Therefore the ICSE Class 10th **Maths** **Solutions** of **Concise** Selina Publishers helpful on various topics which are prescribed in most ICSE Maths textbook

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**How to Solve Concise Maths Selina Publications Chapter-17 Circles ICSE Maths Class 10**

Note:- Before viewing **Solutions** of Chapter-17 **Circles**** **of **Concise **Selina **Maths** read the Chapter Carefully then solve all example of your text book**. **The Chapter-17 **Circles**** **is main Chapter in ICSE board

**Concise Maths Solutions Circles Selina Maths ICSE Class 10 EXERCISE – 17(A)**

**Question 1.**

**In the given figure, O is the centre of the circle. ∠OAB and ∠OCB are 30° and 40° respectively. Find ∠AOC. Show your steps of working.**

**Answer 1 **

In circle with centre O, ∠BAO = 30°, ∠BCO = 40°.

Join BO.

OA = OB = OC (Radii of the circle)

∠OBA = ∠OAB = 30° and ∠OBC = ∠OCB = 40°

∠ABC = 30° + 40° = 70°

Now, AOC is at the centre and ∠ABC is on the remaining part of the circle.

∠AOC = 2 ∠ABC = 2 x 70° = 140°.

**Question 2.**

**In the given figure, ∠BAD = 65°, ∠ABD = 70°, ∠BDC = 45°**

**(i) Prove that AC is a diameter of the circle.**

**(ii) Find ∠ACB.**

**Answer 2**

(i) In ΔABD

65° + 70° + ∠ADB = 180°

∠ADB = 180° – 65° – 70° = 45°

∠ADC = 45° + 45° = 90°

AC is diameter [Angle in semi circle is 90°]
(ii) ∠ACB = ∠ADB = 45° [angle in same segment]

**Question 3.**

**Given O is the centre of the circle and ∠AOB = 70°. Calculate the value of:**

**(i) ∠OCA,**

**(ii) ∠OAC.**

**Answer 3**

O is the centre of the circle,

∠AOB = 70°

arc AB subtends ∠AOB at the centre

and ∠ OCA is at the remaining part of circle

∠AOB = 2 ∠OCA

or ∠OCA = ∠AOB = x 70° = 35°

In ΔOAC,

OC = OA (Radii of the same circle)

∠OAC = ∠OCA = 35°

**Question 4.**

**In each of the following figures, O is the centre of the circle. Find the values of a, b, and c.**

**Answer 4**

(i) Arc AB subtends ∠AOB at the centre and ∠ACB at the remaining part

∠AOB = 2 ∠ACB

or ∠ACB = ∠AOB = x 130° = 65°

or b = 65°

But a + b = 180° (Opposite angles of a cyclic quad.)

a = 180° – b = 180° – 65° = 115°

a = 115°, b = 65°

(ii) Arc AB subtends ∠ AOB at the centre and ∠ACB at the remaining part.

Reflex ∠AOB = 2 ∠ACB

or ∠ACB = (reflex ∠AOB) = [360°- 112°]
= x 248° = 124°

Hence, c = 124°.

**Question 5. Concise Maths Solutions Circles**

**In each of the following figures, O is the centre of the circle. Find the value of a, b, c and d.**

**Answer 5**

(i) BOD is a diameter

∠BAD = 90° (Angle in a semi-circle)

∠ADB = 180° – (90° + 35°) = 180° – 125° = 55°

But ∠ACB = ∠ADB = 55° (Angles in the same segment)

a = 55°.

(ii) In ΔEBC.

Ext. 120° = 25° + ∠BCE

∠BCE = 120° – 25° = 95°

But ∠ADB = ∠ACB = 95° (Angles in the same segment)

b = 95°.

(iii) Arc AB subtends ∠AOB at the centre and ∠ACB at the remaining part,

∠ AOB = 2 ∠ ACB = 2 x 50° = 100°

In ΔAOB,

OA = OB (Radii of the same circle)

∠OAB = ∠OBA

But ∠OAB + ∠OBA = 180° – 100° = 80°

c = ∠OAB = ~ x 80° = 40°.

(iv) In the given figure, O is the centre of the circle.

AOB is its diameter and ∠ABP = 45°

Q is any point and BQ, PQ are joined

In ΔABP,

∠APB = 90° (Angle in a semicircle)

∠PAB + ∠PBA = 90°

⇒ ∠PAB + 45° = 90°

⇒ ∠PAB = 90° – 45°

⇒ ∠PAB = 45°

Now ∠PAB = ∠PQB (Angle in the same segment)

BPQB = 45°

⇒ d = 45°

**Question 6. Concise Maths Solutions Circles**

**In the figure, AB is common chord of the two circles. If AC and AD are diameters; prove that D, B and C are in a straight line. O1 and 02 are the centres of two circles.**

**Answer 6**

Given- Two circles with centre O1 and O2 intersect each other at A and B.

AC and AD are the diameters of the circles.

To Prove- D, B, C are in the same straight line.

Construction- Join AB.

Proof- AO1C is diameter.

∠ABC = 90°. (Angle in a semi-circle)

Similarly ∠ABD = 90°,

Adding, we get:

∠ABC + ∠ABD = 90° + 90° = 180°

DBC is a straight line.

or D, B, C are in the same line.

**Question 7.**

**In the figure given beow, find :**

**(i) ∠BCD,**

**(ii) ∠ADC,**

**(iii) ∠ABC.**

Show steps of your workng.

**Answer 7**

ABCD is a cyclic quadrilateral

∠A + ∠C = 180°.

∠C = 180° – ∠A = 180° – 105° = 75°

or ∠BCD = 75°

DC || AB

∠ADC + ∠DAB = 180° (Angles on the same side of the transversal of || lines)

∠ADC = 180° – ∠DAB = 180° – 105° = 75°

But ∠ADC + ∠ABC = 180° (opposite angles of a cyclic quad.)

∠ABC = 180° – ∠ADC = 180° – 75° = 105°

**Question 8.**

**In the given figure, O is centre of the circle. If ∠ AOB = 140° and ∠ OAC = 50°; find :**

**(i) ∠ ACB,**

**(ii) ∠OBC,**

**(iii) ∠OAB,**

**(iv) ∠CBA**

**Answer 8**

O is the centre of circle ∠AOB = 140° and ∠OAC = 50°.

AB is joined

Reflex ∠AOB = 360° – 140° = 220°

But ∠ACB = Reflex ∠AOB = x 220° = 110°

In quad. OACB,

∠AOB + ∠OAC + ∠ACB + ∠OBC = 360°

⇒ 140° + 50° + 110° + ∠OBC = 360°

⇒ 300° + ∠OBC = 360°

⇒ ∠OBC = 360° – 300° = 60°

In ∆OAB,

∠AOB + ∠OAB + ∠OBA = 180°

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

140° + ∠OAB + ∠OAB = 180°

2 ∠OAB = 180° – 140° = 40°

∠OAB = = 20°

∠OAB = ∠OBA = 20°.

⇒ ∠OBC = ∠CBA + ∠ABO

and ⇒ 60° = ∠CBA + 20°

hence ⇒ ∠CBA = 40°

**Question 9. Concise Maths Solutions Circles**

**Calculate :**

**(i) ∠ CDB,**

**(ii) ∠ ABC,**

**(iii) ∠ ACB.**

**Answer 9**

∠CDB = ∠BAC (Angles is the same segment) = 49°

∠ABC = ∠ADC (Angles in the same segment) = 43°

∠ADB = ∠ADC + ∠BDC = 43° + 49° = 92°

and ∠ADB + ∠ACB = 180° (opposite angles of a cyclic quad.)

∠ACB = 180° – ∠ADB = 180° – 92° = 88°.

**Question 10.**

**In the figure given below, ABCD is a eyclic quadrilateral in which ∠BAD = 75°; ∠ABD = 58° and ∠ADC = 77°. Find:**

**(i) ∠BDC,**

**(ii) ∠BCD,**

**(iii) ∠BCA.**

**Answer 10**

In cyclic quad. ABCD,

∠BAD = 75°, ∠ABD = 58° and ∠ADC = 77°.

∠A + ∠C = 180° (opposite angles of a cyclic quad.)

∠C = 180° – ∠A = 180° – 75° = 105° or ∠BCD = 105°

In ΔABD,

∠BAD + ∠ABD + ∠ADB = 180°

⇒ 75° + 58° + ∠ ADB = 180°

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