# Class 10 SELINA Solutions Maths Chapter 17 - Circles

## Circles Exercise TEST YOURSELF

### Solution 1(a)

Correct option: (ii) A is true, R is true

Angle in a semi-circle is a right angle.

Hence, the assertion (A) is true.

The measure of ∠ADC is calculated correctly in reason statement.

Hence, the reason statement is true.

### Solution 1(b)

Correct option: (iii) 360^{o}

The sum of the measures of angles of any quadrilateral is 360^{o}.

For a cyclic quadrilateral, each exterior angle equals its opposite interior angles.

Hence, x^{o} + y^{o} + z^{o} + p^{o} = 360^{o}

### Solution 1(c)

Correct option: (iii) x^{o} + 2y^{o} = 360^{o}

### Solution 1(d)

Correct option: (ii) 88^{o}

Arc AB : Arc BC = 11 : 4

⇒ ∠AOB : ∠BOC = 11 : 4

*∠AOC = ∠AOB + ∠BOC = 88^{o} + 32^{o} = 120^{o}, which is not given in options. Hence, according to back answers, if option (ii) is correct, then ∠AOB is to be found.

### Solution 1(e)

Correct option: (ii) 66^{o}

AB is the side of a regular pentagon.

BC is the side of a regular hexagon.

∠AOC = ∠AOB + ∠BOC = 72^{o} + 60^{o} = 132^{o}

∠AOC = 2∠APB

132^{o} = 2∠APB

∠APB = 66^{o}

### Solution 1(f)

Correct option: (ii) 80^{o}

∠BOD = 2∠BAD = 2(80^{o}) = 160^{o}

Chord BC = Chord CD

⇒ ∠BOC = ∠COD

Now, ∠BOC + ∠COD = ∠BOD

⇒ 2∠BOC = ∠BOD = 160^{o}

⇒ ∠BOC = 80^{o}

### Solution 2

### Solution 3

### Solution 4

Given - In Δ ABC, AB = AC and a circle with AB as diameter is drawn which intersects the side BC at D.

To prove - D is the midpoint of BC.

Construction - Join AD.

Proof:

∠1 = 90° [Angle in a semi circle]

But ∠1 + ∠2 = 180° [Linear pair]

∴ ∠2 = 90°

Now in right ΔABD and Δ ACD,

Hyp. AB = Hyp. AC [Given]

Side AD = Ad [Common]

∴ By the Right angle - Hypotenuse - Side criterion of congruence, we have

Δ ABD ≅ ΔACD [RHS criterion of congruence]

The corresponding parts of the congruent triangles are congruent.

∴ BD = DC [c.p.c.t]

Hence D is the mid point of BC.

### Solution 5

Join OE.

Arc EC subtends ∠EOC at the centre and ∠EBC at the remaining part of the circle.

∠EOC = 2 ∠EBC = 2 × 65° = 130°.

Now in Δ OEC, OE = OC [Radii of the same circle]

∴ ∠OEC = ∠OCE

But, in Δ EOC,

∠OEC + ∠OCE + ∠EOC = 180° [Angles of a triangle]

⇒ ∠OCE + ∠OCE + ∠EOC = 180°

⇒ 2 ∠OCE + 130° = 180°

⇒ 2 ∠OCE = 180° - 130°

⇒ 2 ∠OCE + 50°

⇒ ∠OCE = = 25°

∴ AC || ED [given]

∴ ∠DEC = ∠OCE [Alternate angles]

⇒ ∠DEC = 25°

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### Solution 20

Given - In the figure, CP is the bisector of ∠ABC

To prove - DP is the bisector of ∠ADB

Proof - Since CP is the bisector of ∠ACB

∴ ∠ACP = ∠BCP

But ∠ACP = ∠ADP [Angles in the same segment of the circle]

and ∠BCP = ∠BDP

But ∠ACP = ∠BCP

∴ ∠ADP = ∠BDP

∴ DP is the bisector of ∠ADB

### Solution 21

### Solution 22

i. AD is parallel to BC, i.e., OD is parallel to BC and BD is transversal.

### Solution 23

∠DAE and ∠DAB are linear pair

So,

∠DAE + ∠DAB = 180°

∴∠DAB = 110°

Also,

∠BCD + ∠DAB = 180°……Opp. Angles of cyclic quadrilateral BADC

∴∠BCD = 70°

∠BCD = ∠BOD…angles subtended by an arc on the center and on the circle

∴∠BOD = 140°

In ΔBOD,

OB = OD……radii of same circle

So,

∠OBD =∠ODB……isosceles triangle theorem

∠OBD + ∠ODB + ∠BOD = 180°……sum of angles of triangle

2∠OBD = 40°

∠OBD = 20°

## Circles Exercise Ex. 17(A)

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### Solution 1(a)

Correct option: (ii) 35^{o}

AB is the diameter of a circle with centre O.

Then, ∠BCA = 90^{o} (angle in a semi-circle)

In ∆ABC,

∠A + ∠B + ∠C = 180^{o}

∠A + 55^{o} + 90^{o} = 180^{o}

∠A = 35^{o}

### Solution 1(b)

Correct option: (iii) 40^{o}

OA = OB (radii of circle)

⇒ ∠OAB = ∠OBA = 50^{o}

⇒ ∠AOB = 180^{o} – ∠OAB – ∠OBA = 180^{o} – 50^{o} – 50^{o}_{ = }80^{o}

The angle at the centre is twice the angle at remaining circumference.

∴ ∠APB = 80^{o}/2 = 40^{o}

### Solution 1(c)

Correct option: (ii) 50^{o}

∠APB = ∠ACB = 50^{o} (angles in the same segment)

Chord AB = Chord PB

⇒ ∠PAB = ∠APB = 50^{o}

Note: Back answer is incorrect

### Solution 1(d)

Correct option: (iv) C, B and D are collinear

∠ABD = 90^{o} (angle in a semi-circle)

∠ABC = 90^{o} (angle in a semi-circle)

Now, ∠ABD + ∠ABC = 180^{o}

Therefore, C, B and D are collinear.

### Solution 1(e)

Correct option: (i) 122^{o}

AB is parallel to DC and AC is the transversal.

Hence, ∠BAC = ∠ACD = 32^{o} (alternate angles)

∠DAC = 90^{o} (angle in a semi-circle)

Therefore, ∠DAB = ∠DAC + ∠BAC = 90^{o} + 32^{o} = 122^{o}

## Circles Exercise Ex. 17(B)

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### Solution 17

(i) AEB =

(Angle in a semicircle is a right angle)

Therefore EBA = - EAB = - =

(ii) AB ED

Therefore DEB = EBA = (Alternate angles)

Therefore BCDE is a cyclic quadrilateral

Therefore DEB + BCD =

[Pair of opposite angles in a cyclic quadrilateral are supplementary]

Therefore BCD = - =

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### Solution 1(a)

Correct option: (i) 85^{o}

∠DCE = ∠BAD (Exterior angle property of a cyclic quadrilateral)

⇒ ∠BAD = 95^{o}

AD is parallel to BC and AB is the transversal.

⇒ ∠BAD + ∠ABC = 180^{o}

⇒ ∠ABC= 180^{o} – 95^{o} = 85^{o}

### Solution 1(b)

Correct option: (iv) 120^{o}

ABC is an equilateral triangle.

⇒ ∠ABC = 60^{o}

ABCD is a cyclic quadrilateral.

⇒ ∠ABC + ∠ADC = 180^{o}

⇒ ∠ADC = 180^{o} – 60^{o} = 120^{o}

### Solution 1(c)

Correct option: (iv) 140^{o}Join OB.

OA = OB (radii of the circle)

⇒ ∠OBA = ∠OAB = 30^{o}

OC = OB (radii of the circle)

⇒ ∠OBC = ∠OCB = 40^{o}

∠ABC = ∠OBA + ∠OBC = 30^{o} + 40^{o} = 70^{o}

Therefore, ∠AOC = 2 × ∠ABC = 2 × 70^{o} = 140^{o}

### Solution 1(d)

Correct option: (iv) AC//BD

ACQP is a cyclic quadrilateral.

∠ACQ + ∠APQ = 180^{o} …..(i)

BDQP is a cyclic quadrilateral.

∠BDQ + ∠BPQ = 180^{o} ….(ii)

Adding (i) and (ii),

∠ACQ + ∠APQ + ∠BDQ + ∠BPQ = 180^{o} + 180^{o}

∠ACQ + ∠BDQ + (∠APQ + ∠BPQ) = 360^{o}

∠ACQ + ∠BDQ + 180^{o} = 360^{o} (APB is a straight line)

∠ACQ + ∠BDQ = 180^{o}

Since, the interior angles ACQ and BDQ on the same side of the transversal CD are supplementary, AC is parallel to BD.

### Solution 1(e)

Correct option: (i) 105^{o}

ABCD is a cyclic quadrilateral.

∠DAC + ∠BCD = 180^{o } (Opposite angles of a cyclic quadrilateral are supplementary)

∠BCD = 180^{o} – 105^{0} = 75^{o}

AB is parallel to DC and BC is the transversal.

⇒ ∠ABC + ∠BCD = 180^{o}

⇒ ∠ABC = 180^{o} – 75^{o} = 105^{o}

## Circles Exercise Ex. 17(C)

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### Solution 1(a)

Correct option: (iv) 100^{o}

Chord AB : Chord CD = 3 : 5

⇒ ∠AOB : ∠COD = 3 : 5

### Solution 1(b)

Correct option: (ii) 35^{o}

OA = OB (radii of the circle)

⇒ ∠OAB = ∠OBA = 55^{o}

In triangle AOB,

∠AOB = 180^{o} – ∠OAB – ∠OBA = 180^{o} – 55^{o} – 55^{o} = 70^{o}

Also, ∠AOB = 2∠ACB

70^{o} = 2∠ACB

∠ACB = 35^{o}

### Solution 1(c)

Correct option: (ii) 105^{o}

AB is the side of a square.

BC is the side of a regular hexagon.

Arc AD = Arc CD

⇒ ∠AOD = ∠DOC = x

Then, ∠AOB + ∠BOC + ∠AOD + ∠DOC = 360^{o}

90^{o} + 60^{o} + x + x = 360^{o}

2x = 210

x = 105^{o}

### Solution 1(d)

Correct option: (i) 36^{o}

AB is the side of a regular pentagon.

∠AOB = 2∠ACB

72^{o} = 2∠ACB

∠ACB = 36^{o}

### Solution 1(e)

Correct option: (iv) 120^{o}

Chord AB = Chord CD = Chord EF

⇒ ∠AOB = ∠COD = ∠EOF = x

Chord BC = Chord DE = Chord FA

⇒ ∠BOC = ∠DOE = ∠AOF = y

Now,

∠AOB + ∠COD + ∠EOF + ∠BOC + ∠DOE + ∠AOF = 360^{o}

x + x + x + y + y + y = 360^{o}

3x + 3y = 360^{o}

x + y = 120^{o}

∠AOC = ∠AOB + ∠BOC = x + y = 120^{o}