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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 2

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Solution 3

G i v e n space minus space I n space t h e space f i g u r e space A B C space i s space a space t r i a n g l e space i n space w h i c h space angle A equals 30 degree.
T o space p r o v e minus B C space i s space t h e space r a d i u s space o f space c i r c u m c i r c l e space o f space increment A B C space w h o s e space c e n t r e
i s space O.
C o n s t r u c t i o n minus J o i n space O B space a n d space O C.
P r o o f :
angle B O C equals 2 angle B A C equals 2 cross times 30 degree equals 60 degree
N o w space i n space increment O B C comma
O B equals O C space space space space space space space space space space space space space left square bracket R a d i i space o f space t h e space s a m e space c i r c l e right square bracket
angle O B C equals angle O C B
B u t comma space i n space increment B O C comma
angle O B C plus angle O C B plus angle B O C equals 180 degree space space left square bracket A n g l e s space o f space a space t r i a n g l e right square bracket
rightwards double arrow angle O B C plus angle O B C plus 60 degree equals 180 degree
rightwards double arrow 2 angle O B C plus 60 degree equals 180 degree
rightwards double arrow 2 angle O B C equals 180 degree minus 60 degree
rightwards double arrow 2 angle O B C equals 120 degree
rightwards double arrow angle O B C equals fraction numerator 120 degree over denominator 2 end fraction equals 60 degree
rightwards double arrow angle O B C equals angle O C B equals angle B O C equals 60 degree
rightwards double arrow increment B O C space i s space a n space e q u i l a t e r a l space t r i a n g l e.
rightwards double arrow B C equals O B equals O C
B u t comma space O B space a n d space O C space a r e space t h e space r a d i i space o f space t h e space c i r c u m minus c i r c l e.
therefore B C space i s space a l s o space t h e space r a d i u s space o f space t h e space c i r c u m minus c i r c l e.

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° 

⇒ OCE + 130° = 180° 

⇒ OCE = 180° - 130° 

⇒ OCE + 5 

⇒ OCE = begin mathsize 12px style fraction numerator 50 degree over denominator 2 end fraction end style= 25° 

∴ AC || ED  [given]

∴ DEC = OCE  [Alternate angles]

⇒ DEC = 25°

Solution 6

  

Solution 7

Solution 8

I n space c y c l i c space q u a d. space A B C D comma
A F parallel to C B space a n d space D A space i s space p r o d u c e d space t o space E space s u c h space t h a t space angle A D C equals 92 degree space a n d space angle F A E equals 20 degree
N o w space w e space n e e d space t o space f i n d space t h e space m e a s u r e space o f space angle B C D
I n space c y c l i c space q u a d. space A B C D comma
angle B plus angle D equals 180 degree
rightwards double arrow angle B plus 92 degree equals 180 degree
rightwards double arrow angle B equals 180 degree minus 92 degree
rightwards double arrow angle B equals 88 degree
S i n c e space A F parallel to C B comma space angle F A B equals angle B equals 88 degree
B u t comma space angle F A E equals 20 degree space space space space space left square bracket g i v e n right square bracket
E x t. angle B A E equals angle B A F plus angle F A E
equals 88 degree plus 22 degree equals 108 degree
B u t comma space E x t. angle B A E equals angle B C D
therefore angle B C D equals 108 degree

Solution 9

Solution 10

Solution 11

Solution 12

A r c space A C space s u b t e n d s space angle A O C space a t space t h e space c e n t r e space a n d space angle A D C space a t space t h e space r e m a i n i n g space p a r t
o f space t h e space c i r c l e
therefore angle A O C equals 2 angle A D C
rightwards double arrow angle A O C equals 2 cross times 32 degree equals 64 degree
S i n c e space angle A O C space a n d space angle B O C space a r e space l i n e a r space p a i r comma space w e space h a v e
angle A O C plus angle B O C equals 180 degree
rightwards double arrow 64 degree plus angle B O C equals 180 degree
rightwards double arrow angle B O C equals 180 degree
rightwards double arrow angle B O C equals 180 degree minus 64 degree
rightwards double arrow angle B O C equals 116 degree

Solution 13

Solution 14

 

Solution 15

G i v e n minus I n space a space c i r c l e comma space A B C D space i s space a space c y c l i c space q u a d r i l a t e r a l space A B space a n d space D C
a r e space p r o d u c e d space t o space m e e t space a t space E space a n d space B C space a n d space A D space a r e space p r o d u c e d space t o space m e e t space a t space F.
angle D C F : angle F : angle E equals 3 : 5 : 4
L e t space angle D C F equals 3 X comma angle F equals 5 x comma angle E equals 4 x
N o w comma space w e space h a v e space t o space f i n d comma angle A comma angle B comma angle C space A N D space angle D
I n space c y c l i c space q u a d. space A B C D comma space B C space i s space p r o d u c e d.
therefore angle A equals angle D C F equals 3 x
I n space increment C D F comma
E x t. angle C D A equals angle D C F plus angle F equals 3 x plus 5 x equals 8 x
I n space increment B C E comma
E x t. angle A B C equals angle B C E plus angle E space space space space space space space space left square bracket angle B C E equals angle D C F comma v e r t i c a l l y space o p p o s i t e space a n g l e s right square bracket
equals angle D C F plus angle E
equals 3 x plus 4 x equals 7 x
N o w comma space i n space c y c l i c space q u a d. A B C D comma
sin c e comma angle B plus angle D equals 180 degree
space space space space space space space space space space space space space space space space space space space space space left square bracket S i n c e space s u m space o f space o p p o s i t e space o f space a space c y c l i c space q u a d r i l a t e r a l space a r e space s u p p l e m e n t a r y right square bracket
rightwards double arrow 7 x plus 8 x equals 180 degree
rightwards double arrow 15 x equals 180 degree
rightwards double arrow x equals fraction numerator 180 degree over denominator 15 end fraction equals 12 degree
therefore angle A equals 3 x equals 3 cross times 12 degree equals 36 degree
angle B equals 7 x equals 7 cross times 12 degree equals 84 degree
angle C equals 180 degree minus angle A equals 180 degree minus 36 degree equals 144 degree
angle D equals 8 x equals 8 cross times 12 degree equals 96 degree

Solution 16


G i v e n minus I n space a space c i r c l e space w i t h space c e n t r e space O comma space A B space i s space t h e space d i a m e t e r space a n d space A C space a n d space
A D space a r e space t w o space c h o r d s space s u c h space t h a t space A C equals A D.
T o space p r o v e : left parenthesis i right parenthesis space a r c space B C equals a r c space D B
left parenthesis i i right parenthesis space A B space i s space t h e space b i s e c t o r space o f space angle C A D
left parenthesis i i i right parenthesis space I f space a r c space A C equals 2 a r c space B C comma space t h e n space f i n d
space space space space space space space space space space space left parenthesis a right parenthesis angle B A C space space left parenthesis b right parenthesis angle A B C
C o n s t r u c t i o n space : space J o i n space B C space a n d space B D
P r o o f : space I n space r i g h t space a n g l e d space increment A B C space a n d space increment A B D
S i d e space A C equals A D space space space space space space space space space space space space space space left square bracket g i v e n right square bracket
H y p. space A B equals A B space space space space space space space space space space space space space space space left square bracket c o m m o n right square bracket
therefore B y space R i g h t space A n g l e minus H y p o t e n u s e minus S i d e space c r i t e r i o n space o f space c o n g r u e n c e comma
increment A B C approximately equal to increment A B D
left parenthesis i right parenthesis space T h e space c o r r e s p o n d i n g space p a r t s space o f space t h e space c o n g r u e n t space t r i a n g l e s space a r e space c o n g r u e n t.
therefore B C equals B D space space space space space space space space space space space space space space space space left square bracket c. p. c. t right square bracket
therefore A r c space B C equals A r c B D space space space space space space left square bracket e q u a l space c h o r d s space h a v e space e q u a l space a r c s right square bracket
left parenthesis i i right parenthesis space angle B A C equals angle B A D
therefore A B space i s space t h e space b i s e c t o r space o f space angle C A D
left parenthesis i i i right parenthesis space I f space A r c space A C equals 2 space a r c space B C comma
t h e n space angle A B C equals 2 angle B A C
B u t space angle A B C plus angle B A C equals 90 degree
rightwards double arrow 2 angle B A C plus angle B A C equals 90 degree
rightwards double arrow 3 angle B A C equals 90 degree
rightwards double arrow angle B A C equals fraction numerator 90 degree over denominator 3 end fraction equals 30 degree
angle A B C equals 2 angle B A C rightwards double arrow angle A B C equals 2 cross times 30 degree equals 60 degree

Solution 17

A B C D space i s space a space c y c l i c space q u a d r i l a t e r a l space a n d space A D equals B C
angle B A C equals 30 degree comma angle C B D equals 70 degree
W e space h a v e
angle D A C equals angle C B D space space space space space space space space space space space space space left square bracket a n g l e s space i n space t h e space s a m e space s e g m e n t right square bracket
rightwards double arrow angle D A C equals 70 degree space space space space space space space space space space space space space space left square bracket because angle C B D equals 70 degree right square bracket
rightwards double arrow angle B A D equals angle B A C plus angle D A C equals 30 degree plus 70 degree equals 100 degree space space..... left parenthesis 1 right parenthesis
S i n c e space t h e space s u m space o f space o p p o s i t e space a n g l e s space o f space c y c l i c space q u a d r i l a t e r a l space i s space s u p p l e m e n t a r y
angle B A D plus angle B C D equals 180 degree
rightwards double arrow 100 degree plus angle B C D equals 180 degree space space space space space space space space left square bracket f r o m space left parenthesis 1 right parenthesis right square bracket
rightwards double arrow angle B C D equals 180 degree minus 100 degree equals 80 degree
S i n c e space A D equals B C comma angle A C D equals angle B D C space space space left square bracket E q u a l space c h o r d s space s u b t e n d s space e q u a l space a n g l e s right square bracket
B u t space angle A C B equals angle A D B space space space space left square bracket a n g l e s space i n space t h e space s a m e space s e g m e n t right square bracket
therefore angle A C D plus angle A C B equals angle B D C plus angle A D B
rightwards double arrow angle B C D equals angle A D C equals 80 degree
B u t space i n space increment B C D comma
angle C B D plus angle B C D plus angle B D C equals 180 degree space space space space space space space space left square bracket a n g l e s space o a f space a space t r i a n g l e right square bracket
rightwards double arrow 70 degree plus 80 degree plus angle B D C equals 180 degree
rightwards double arrow 150 degree plus angle B D C equals 180 degree
therefore angle B D C equals 180 degree minus 150 degree equals 30 degree
rightwards double arrow angle A C D equals 30 degree space space space space space space space space space space space space space space space space space space left square bracket because angle A C D equals angle B D C right square bracket
therefore angle B C A equals angle B C D minus angle A C D equals 80 degree minus 30 degree equals 50 degree
S i n c e space t h e space s u m space o f space o p p o s i t e space a n g l e s space o f space c y c l i c space q u a d r i l a t e r a l space i s space s u p p l e m e n t a r y comma
angle A D C plus angle A B C equals 180 degree
rightwards double arrow 80 degree plus angle A B C equals 180 degree
rightwards double arrow angle A B C equals 180 degree minus 80 degree equals 100 degree

Solution 18

 

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Solution 19

 

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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

2OBD = 40° 

OBD = 20° 

Circles Exercise Ex. 17(A)

Solution 2

Solution 3

 

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

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

Note: Back answer is incorrect

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 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17



(i) angleAEB = 90 to the power of 0

(Angle in a semicircle is a right angle)

Therefore angle EBA = 90 to the power of 0 - angleEAB = 90 to the power of 0 - 63 to the power of 0= 27 to the power of 0


(ii) AB parallel to ED

Therefore angle DEB = EBA = 27 to the power of 0                        (Alternate angles)

Therefore BCDE is a cyclic quadrilateral

Therefore angle DEB  angleBCD = 180 to the power of 0

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

Therefore angle BCD = 180 to the power of 0 - 27 to the power of 0 = 153 to the power of 0


Solution 18

Solution 19

Solution 20

Solution 21

Solution 22

Solution 23

Solution 24

Solution 25

Solution 26

Solution 27

Solution 28

Solution 1(a)

Correct option: (i) 85o

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

BAD = 95o

AD is parallel to BC and AB is the transversal.

BAD + ABC = 180o

ABC= 180o – 95o = 85o

Solution 1(b)

Correct option: (iv) 120o

ABC is an equilateral triangle.

ABC = 60o

ABCD is a cyclic quadrilateral.

ABC + ADC = 180o

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

ACQP is a cyclic quadrilateral.

ACQ + APQ = 180o       …..(i)

BDQP is a cyclic quadrilateral.

BDQ + BPQ = 180o      ….(ii)

Adding (i) and (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

ABCD is a cyclic quadrilateral.

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 2

 

Solution 3

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Solution 4

 

Solution 5

 

Solution 6

 

Solution 7

Solution 8

 

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Solution 9

 

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Solution 10

Solution 11

 

 

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.

Arc AD = Arc CD

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

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