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# 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) 360o

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

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

Hence, xo + yo + zo + po = 360o

### Solution 1(c)

Correct option: (iii) xo + 2yo = 360o

### Solution 1(d)

Correct option: (ii) 88o

Arc AB : Arc BC = 11 : 4

AOB : BOC = 11 : 4

*AOC = AOB + BOC = 88o + 32o = 120o, 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) 66o

AB is the side of a regular pentagon.

BC is the side of a regular hexagon.

AOC = AOB + BOC = 72o + 60o = 132o

AOC = 2APB

132o = 2APB

APB = 66o

### Solution 1(f)

Correct option: (ii) 80o

BOD = 2BAD = 2(80o) = 160o

Chord BC = Chord CD

BOC = COD

Now, BOC + COD = BOD

2BOC = BOD = 160o

BOC = 80o

### 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.

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]

∴ 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°

⇒ OCE + 130° = 180°

⇒ OCE = 180° - 130°

⇒ OCE + 5

⇒ OCE = = 25°

∴ AC || ED  [given]

∴ DEC = OCE  [Alternate angles]

⇒ DEC = 25°

### 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

∴ DP is the bisector of ADB

### 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 = 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

2OBD = 40°

OBD = 20°

## Circles Exercise Ex. 17(A)

### Solution 1(a)

Correct option: (ii) 35o

AB is the diameter of a circle with centre O.

Then, BCA = 90o    (angle in a semi-circle)

In ∆ABC,

A + B + C = 180o

A + 55o + 90o = 180o

A = 35o

### Solution 1(b)

Correct option: (iii) 40o

OA = OB     (radii of circle)

OAB = OBA = 50o

AOB = 180oOAB OBA = 180o – 50o – 50o = 80o

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

APB = 80o/2 = 40o

### Solution 1(c)

Correct option: (ii) 50o

APB = ACB = 50o      (angles in the same segment)

Chord AB = Chord PB

PAB = APB = 50o

### Solution 1(d)

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

ABD = 90o     (angle in a semi-circle)

ABC = 90o     (angle in a semi-circle)

Now, ABD + ABC = 180o

Therefore, C, B and D are collinear.

### Solution 1(e)

Correct option: (i) 122o

AB is parallel to DC and AC is the transversal.

Hence, BAC = ACD = 32o     (alternate angles)

DAC = 90o        (angle in a semi-circle)

Therefore, DAB = DAC + BAC = 90o + 32o = 122o

## Circles Exercise Ex. 17(B)

### 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 =  - =

### Solution 1(a)

Correct option: (i) 85o

AD is parallel to BC and AB is the transversal.

ABC= 180o – 95o = 85o

### Solution 1(b)

Correct option: (iv) 120o

ABC is an equilateral triangle.

ABC = 60o

ADC = 180o – 60o = 120o

### Solution 1(c)

Correct option: (iv) 140oJoin OB.

OA = OB      (radii of the circle)

OBA = OAB = 30o

OC = OB      (radii of the circle)

OBC = OCB = 40o

ABC = OBA + OBC = 30o + 40o = 70o

Therefore, AOC = 2 × ABC = 2 × 70o = 140o

### Solution 1(d)

Correct option: (iv) AC//BD

ACQ + APQ = 180o       …..(i)

BDQ + BPQ = 180o      ….(ii)

ACQ + APQ + BDQ + BPQ = 180o + 180o

ACQ + BDQ + (APQ + BPQ) = 360o

ACQ + BDQ + 180o = 360o    (APB is a straight line)

ACQ + BDQ = 180o

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) 105o

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

BCD = 180o – 1050 = 75o

AB is parallel to DC and BC is the transversal.

ABC + BCD = 180o

ABC = 180o – 75o = 105o

## Circles Exercise Ex. 17(C)

### Solution 1(a)

Correct option: (iv) 100o

Chord AB : Chord CD = 3 : 5

AOB : COD = 3 : 5

### Solution 1(b)

Correct option: (ii) 35o

OA = OB     (radii of the circle)

OAB = OBA = 55o

In triangle AOB,

AOB = 180oOAB OBA = 180o – 55o – 55o = 70o

Also, AOB = 2ACB

70o = 2ACB

ACB = 35o

### Solution 1(c)

Correct option: (ii) 105o

AB is the side of a square.

BC is the side of a regular hexagon.

AOD = DOC = x

Then, AOB + BOC + AOD + DOC = 360o

90o + 60o + x + x = 360o

2x = 210

x = 105o

### Solution 1(d)

Correct option: (i) 36o

AB is the side of a regular pentagon.

AOB = 2ACB

72o = 2ACB

ACB = 36o

### Solution 1(e)

Correct option: (iv) 120o

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 = 360o

x + x + x + y + y + y = 360o

3x + 3y = 360o

x + y = 120o

AOC = AOB + BOC = x + y = 120o