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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Exercise – 17 (A),

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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 bookThe 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.

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.

(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.

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 = $\frac { 1 }{ 2 }$ ∠AOB = $\frac { 1 }{ 2 }$ 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.

(i) Arc AB subtends ∠AOB at the centre and ∠ACB at the remaining part
∠AOB = 2 ∠ACB
or ∠ACB = $\frac { 1 }{ 2 }$ ∠AOB = $\frac { 1 }{ 2 }$ 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 = $\frac { 1 }{ 2 }$ (reflex ∠AOB) = $\frac { 1 }{ 2 }$ [360°- 112°] = $\frac { 1 }{ 2 }$ 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.

(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.

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°,
∠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,
(iii) ∠ABC.

∠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

O is the centre of circle ∠AOB = 140° and ∠OAC = 50°.
AB is joined
Reflex ∠AOB = 360° – 140° = 220°
But ∠ACB = $\frac { 1 }{ 2 }$ Reflex ∠AOB = $\frac { 1 }{ 2 }$ x 220° = 110°
∠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 = $\frac { 40 }{ 2 }$ = 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.

∠CDB = ∠BAC (Angles is the same segment) = 49°
∠ABC = ∠ADC (Angles in the same segment) = 43°
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.