Chapter 17 : Circles - Selina Solutions for Class 10 Maths ICSE

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Chapter 17 - Circles Excercise Ex. 17(A)

Question 1

In the given figure, O is the centre of the circle. respectively. Find angle AOC Show your steps of working.

Solution 1

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.

 

  

Solution 2

Question 3

Given O is the centre of the circle and AOB = 70o.Calculate the value of:

(i) OCA ; (ii) OAC.

Solution 3

Question 4

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

Solution 4

Question 5

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

Solution 5

Question 6

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 O2 are the centres of two circles.

Solution 6

Question 7

In the figure, given below, find:

Show steps of your working.

Solution 7

Question 8

In the figure, given below, O is the centre of the circle. If

Solution 8

Question 9

Solution 9

Question 10

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

(i) BDC, (ii) BCD, (iii) BCA.

Solution 10

Question 11

In the figure given below, O is the centre of the circle and triangle ABC is equilateral. Find:

(i) ADB, (ii) AEB.

Solution 11

Question 12

Given CAB = 75o and CBA = 50o. Find the value of DAB + ABD.

Solution 12

Question 13

ABCD is a cyclic quadrilateral in a circle with centre O. IfADC = 130o, find BAC.

Solution 13

Question 14

In the figure given alongside, AOB is a diameter of the circle and AOC = 110o, find BDC.

Solution 14

Question 15

In the following figure, O is the centre of the circle;AOB = 60o and BDC = 100o, find OBC.

Solution 15

Question 16

In ABCD is a cyclic quadrilateral in whichDAC = 27o,DBA = 50o and ADB = 33o. Calculate (i) DBC, (ii) DCB, (iii)CAB.

Solution 16

Question 17

In the figure given alongside, AB and CD are straight lines through the centre O of a circle. If AOC = 80o and CDE = 40o . Find the number of degrees in: (i) DCE; (ii) ABC.

Solution 17

Question 18

In the figure given below, AC is a diameter of a circle, whose centre is O. A circle is described on AO as diameter. AE, a chord of the larger circle, intersects the smaller circle at B. Prove that AB = BE.

Solution 18

Question 19

In the following figure,

(i) if BAD = 96o, find BCD and BFE.

(ii) Prove that AD is parallel to FE.

Solution 19

Question 20

Prove that:

(i) the parallelogram, inscribed in a circle, is a rectangle.

(ii) the rhombus, inscribed in a circle, is a square.

Solution 20

Question 21

In the following figure, AB = AC. Prove that DECB is an isosceles trapezium.

Solution 21

Question 22

Two circles intersect at P and Q. Through P diameters PA and PB of the two circles are drawn. Show that the points A, Q and B are collinear.

Solution 22

Question 23

The figure given below, shows a circle with centre O. Given: AOC = a and ABC = b.

(i) Find the relationship between a and b

(ii) Find the measure of angle OAB, if OABC is a parallelogram.

Solution 23

Question 24

Two chords AB and CD intersect at P inside the circle. Prove that the sum of the angles subtended by the arcs AC and BD as the centre O is equal to twice the angle APC

Solution 24

 

Given : space Two space chords space AB space and space CD space intersect space each space other space at space straight P space inside space the space circle. space OA comma
OB comma space OC space and space OD space are space joined.
To space prove : space angle AOC plus angle BOD space equals 2 angle APC
Construction : space Join space AD.
Proof : space Arc space AC space subtends space angle AOC space at space the space centre space and space angle ADC space at space the space remaining space
part space of space the space circle.
angle AOC space equals space 2 angle ADC space...... left parenthesis 1 right parenthesis
Similarly comma
angle BOD space equals space 2 angle BAD....... left parenthesis 2 right parenthesis
Adding space left parenthesis 1 right parenthesis space and space left parenthesis 2 right parenthesis comma
angle AOC space plus angle BOD space equals space 2 angle ADC space plus 2 angle BAD
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 2 left parenthesis angle ADC plus angle BAD right parenthesis....... left parenthesis 3 right parenthesis
But space in space increment PAD comma
Ext. space angle APC equals angle PAD plus angle ADC
space space space space space space space space space space space space space space space space space space space space space equals angle BAD space plus angle ADC space space space space space space space space space space space space space space......... left parenthesis 4 right parenthesis
From space left parenthesis 3 right parenthesis space and space left parenthesis 4 right parenthesis comma
angle AOC space plus angle BOD space equals 2 angle APC

Question 25

In the figure given RS is a diameter of the circle. NM is parallel to RS and MRS = 29o

Calculate : (i) RNM ; (ii) NRM.

Solution 25

Question 26

In the figure given alongside, AB || CD and O is the centre of the circle. If ADC = 25o; find the angle AEB. Give reasons in support of your answer.

Solution 26

Question 27

Two circles intersect at P and Q. Through P, a straight line APB is drawn to meet the circles in A and B. Through Q, a straight line is drawn to meet the circles at C and D. Prove that AC is parallel to BD.

Solution 27

Question 28

ABCD is a cyclic quadrilateral in which AB and DC on being produced, meet at P such that PA = PD. Prove that AD is parallel to BC.

Solution 28

Question 29

AB is a diameter of the circle APBR as shown in the figure. APQ and RBQ are straight lines. Find:

Solution 29

Question 30

In the given figure, SP is the bisector of angle RPT and PQRS is a cyclic quadrilateral. Prove that: SQ = SR.

Solution 30

Question 31

In the figure, O is the centre of the circle, AOE = 150o, DAO = 51o. Calculate the sizes of the angles CEB and OCE.

Solution 31

Question 32

In the figure, P and Q are the centres of two circles intersecting at B and C. ACD is a straight line. Calculate the numerical value of x.

Solution 32

Question 33

The figure shows two circles which intersect at A and B. The centre of the smaller circle is O and lies on the circumference of the larger circle. Given that APB = ao. Calculate, in terms of ao, the value of:

Give reasons for your answers clearly.

Solution 33

Question 34

In the given figure, O is the centre of the circle and ABC = 55o. Calculate the values of x and y.

Solution 34

Question 35

In the given figure, A is the centre of the circle, ABCD is a parallelogram and CDE is a straight line. Prove that BCD = 2ABE.

Solution 35

Question 36

ABCD is a cyclic quadrilateral in which AB is parallel to DC and AB is a diameter of the circle. Given BED = 65o; calculate: (i) DAB , (ii) BDC.

Solution 36

Question 37

In the given figure, AB is a diameter of the circle. Chord ED is parallel to AB and EAB = 63o; calculate: (i) EBA , (ii)BCD.

Solution 37



(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


Question 38

In the given figure, AB is a diameter of the circle with centre O. DO is parallel to CB and DCB = 120o; calculate:

(i) DAB, (ii) DBA, (iii) DBC, (iv) ADC.

Also, show that the AOD is an equilateral triangle.

Solution 38

Question 39

In the given figure, I is the incentre of the ABC. BI when produced meets the circumcirle of ABC at D. Given BAC = 55° and ACB = 65o ; calculate: (i) DCA, (ii) DAC, (iii) DCI, (iv) AIC.

Solution 39

Question 40

A triangle ABC is inscribed in a circle. The bisectors of angles BAC, ABC and ACB meet the circumcircle of the triangle at points P, Q and R respectively. Prove that:

Solution 40

Question 41

Calculate the angles x, y and z if:

Solution 41

Question 42

In the given figure, AB = AC = CD and ADC = 38o. Calculate:

(i) Angle ABC

(ii) Angle BEC.

Solution 42

Question 43

In the given figure, AC is the diameter of circle, centre O. Chord BD is perpendicular to AC. Write down the angles p, q and r in terms of x.

Solution 43

Question 44

In the given figure, AC is the diameter of circle, centre O. CD and BE are parallel. Angle AOB = 80o and angle ACE = 10o. Calculate:

(i) Angle BEC ; (ii) Angle BCD ; (iii) Angle CED.

Solution 44

Question 45

In the given figure, AE is the diameter of circle. Write down the numerical value of . Give reasons for your answer.

Solution 45

Question 46

In the given figure, AOC is a diameter and AC is parallel to ED. If , calculate .

Solution 46

Question 47

Use the given figure to find

.

Solution 47

Question 48

In the given figure, AOB is a diameter and DC is parallel to AB. If CAB = xo ; find (in terms of x) the values of:

.

Solution 48

Question 49

In the given figure, AB is the diameter of a circle with centre O. BCD = 130°. Find:

(i) DAB  (ii) DBA

Solution 49
  1. ABCD is a cyclic quadrilateral

mDAB = 180° - DCB

= 180° - 130°

= 50°

  1. In ∆ADB,

mDAB + mADB + mDBA = 180°

50° + 90° + mDBA = 180°

mDBA = 40°

Question 50

In the given figure, PQ is the diameter of the circle whose centre is O. GivenROS = ; calculate RTS.

Solution 50

Question 51

In the given figure, PQ is a diameter. Chord SR is parallel to PQ. Given that PQR = ; calculate

Solution 51

Question 52

AB is the diameter of the circle with centre O. OD is parallel to BC and AOD = ; calculate the numerical values of:

Solution 52

Question 53

In the given figure, the centre O of the small circle lies on the circumference of the bigger circle. If APB = and BCD = ; find:

Solution 53

Question 54

In the given figure, BAD = , ABD = and BDC = ; find:

Hence, show that AC is a diameter.

Solution 54

Question 55

In a cyclic quadrilateral ABCD, A :C = 3 : 1 and B : D = 1 : 5; find each angle of the quadrilateral.

Solution 55

Question 56

The given figure shows a circle with centre O andABP =. Calculate the measure of

Solution 56

Question 57

In the given figure, M is the centre of the circle. Chords AB and CD are perpendicular to each other. If MAD =x and BAC = y,

Solution 57

Chapter 17 - Circles Excercise Ex. 17(B)

Question 1

In a cyclic-trapezium, the non-parallel sides are equal and the diagonals are also equal.

Prove it.

Solution 1

 

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Question 2

In the following figure, AD is the diameter of the circle with centre O. chords AB, BC and CD are equal. If  DEF = 110 , calculate :

(i)  AFE,(ii)  FAB.

Solution 2

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Question 3

If two sides of a cycli-quadrilateral are parallel; prove thet:

(i) its other two side are equal.

(ii) its diagonals are equal.

Solution 3

 

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Question 4

The given figure show a circle with centre O. also, PQ = QR = RS and  PTS = 75°. Calculate:

(i)  POS,

(ii)  QOR,

(iii)  PQR.

Solution 4

 

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Question 5

In the given figure, AB is a side of a regular six-sided polygon and AC is a side of a regular eight-sided polygon inscribed in the circle with centre O. calculate the sizes of :

(i)  AOB,

(ii)  ACB,

(iii)  ABC.

Solution 5

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Question 6

In a regular pentagon ABCDE, inscribed in a circle; find ratio between angle EDA and angel ADC.

Solution 6

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Question 7

In the given figure. AB = BC = CD and  ABC = 132° calculate:

(i)  AEB,

(ii)  AED,

(iii)  COD.

Solution 7

 

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Question 8

In the figure, O is the centre of the circle and the length of arc AB is twice the length of arc BC. If angle AOB = 108° find :

(i)  CAB,

(ii)  ADB.

Solution 8

 

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Question 9

The figure shows a circle with centre O. AB is the side of regular pentagon and AC is the side of regular hexagon. Find the angles of triangle ABC.

Solution 9

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Question 10

In the given figire, BD is a side of a regularhexagon, DC is a side of a regular pentagon and AD is adiameter. Calculate:

(i)  ADC,

(ii)  BAD,

(iii)  ABC,

(iv)  AEC.

Solution 10

 

 

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Chapter 17 - Circles Excercise Ex. 17(C)

Question 1

In the given circle with diametre AB, find the valuv of x.

Solution 1

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Question 2

In the given figure, ABC is a triangle in which BAC = 30. Show that BC is equal to the radius of the circum-circle of the triangle ABC, whose centre is O.

Solution 2

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.

Question 3

Prove that the circle drawn on any one a the equalside of an isoscele triangle as diameter bisects the base.

Solution 3




G i v e n minus I n increment A B C comma A B equals A C space a n d space a space c i r c l e space w i t h space A B space a s space d i a m e t e r space i s space d r a w n
w h i c h space i n t e r s e c t s space t h e space s i d e space B C space a n d space D.
T o space p r o v e minus D space i s space t h e space m i d space p o i n t space o f space B C.
C o n s t r u c t i o n minus J o i n space A D.
P r o o f minus angle 1 equals 90 degree space space space left square bracket A n g l e space i n space a space s e m i space c i r c l e right square bracket
B u t space angle 1 plus angle 2 equals 180 degree space space left square bracket L i n e a r space p a i r right square bracket
therefore angle 2 equals 90 degree
N o w space i n space r i g h t space increment A B D space a n d space increment A C D comma
H y p. space A B equals H y p. A C space space left square bracket G i v e n right curly bracket
S i d e space A D equals A D 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 t h e 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 space w e space h a v e
increment A B D approximately equal to increment A C D space space space space space space space space space left square bracket R H S space c r i t e r i o n space o f space c o n g r u e n c e right square bracket
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 D equals D C space space space space space space space space space space space space space left square bracket c. p. c. t right square bracket
H e n c e space D space i s space t h e space m i d space p o i n t space o f space B C.


Question 4

In the given figure, chord ED is parallel to diameter AC of the circle. Given CBE =, calculateDEC.

Solution 4


 

J o i n space O E.
A r c E C space s u b t e n d s angle E O C space a t space t h e space c e n t r e space a n d angle E B C space a t space t h e space r e m a i n i n g
p a r t space o f space t h e space c i r c l e.
angle E O C equals 2 angle E B C equals 2 cross times 65 degree equals 130 degree.
N o w space i n space increment O E C comma O E equals O C 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
therefore angle O E C equals angle O C E
B u t comma space i n space increment E O C comma
angle O E C plus angle O C E plus angle E 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 C E plus angle O C E plus angle E O C equals 180 degree
rightwards double arrow 2 angle O C E plus 130 degree equals 180 degree
rightwards double arrow 2 angle O C E equals 180 degree minus 130 degree
rightwards double arrow 2 angle O C E equals 50 degree
rightwards double arrow angle O C E equals fraction numerator 50 degree over denominator 2 end fraction equals 25 degree
therefore A C parallel to E D space space space space space space space space space space space left square bracket g i v e n right square bracket
therefore angle D E C equals angle O C E space space space space left square bracket A l t e r n a t e space a n g l e s right square bracket
rightwards double arrow angle D E C equals 25 degree


Question 5

The quadrilateral formed by angle bisectors of a cyclic quadrilateral is also cyclic. Prove it.

Solution 5



G i v e n minus 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 P Q R S space i s space a
q u a d r i l a t e r a l space f o r m e d space b y space t h e space a n g l e
b i s e c t o r s space o f space a n g l e space angle A comma angle B comma angle C space a n d space angle D.
T o space p r o v e minus space P Q R S 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.
P r o o f minus space I n space increment A P D comma
angle P A D plus angle A D P plus angle A P D equals 180 degree space space space space space space... left parenthesis 1 right parenthesis
S i m i l a r l y comma space I N space increment B Q C comma
angle Q B C plus angle B C Q plus angle B Q C equals 180 degree space space space space space space space... left parenthesis 2 right parenthesis
A d d i n g space left parenthesis 1 right parenthesis space a n d space left parenthesis 2 right parenthesis comma space w e space g e t
angle P A D plus angle A D P plus angle A P D plus angle Q B C plus angle B C Q plus angle B Q C equals 180 degree plus 180 degree
rightwards double arrow angle P A D plus angle A D P plus angle Q B C plus angle B C Q plus angle A P D plus angle B Q C equals 360 degree space space space space.... left parenthesis 3 right parenthesis
B u t space angle P A D plus angle A D P plus angle Q B C plus angle B C Q equals 1 half left square bracket angle A plus angle B plus angle C plus angle D right square bracket
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 1 half cross times 360 degree equals 180 degree
therefore angle A P D plus angle B Q C equals 360 degree minus 180 degree equals 180 degree space space space space space space left square bracket f r o m space left parenthesis 3 right parenthesis right square bracket
B u t space t h e s e space a r 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 q u a d r i l a t e r a l space P R Q S.
therefore Q u a d. space P R Q S 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 space space space space









Question 6

In the figure, DBC = 58°. BD is a diameter of the circle. Calculate :

(i) BDC

(ii) BEC

(iii) BAC

  

Solution 6

  

Question 7

D and E are points on equal sides AB and AC of an isosceles triangle ABC such that AD = AE. Provet that the points B, C, E and D are concyclic.

Solution 7

G i v e n minus I n space increment A B C comma space A B equals A C space a n d space D space a n d space E space a r e space p o i n t s space o n space A B space a n d space A C
s u c h space t h a t space A D equals A E. space D E space i s space j o i n e d.
T o space p r o v e space B comma C comma E comma D space a r e space c o n c y c l i c.
P r o o f minus I n increment A B C comma A B equals A C
therefore angle B equals angle C space left square bracket A n g l e s space o p p o s i t e space t o space e q u e a l space s i d e s right square bracket
S i m i l i a r l y comma space I n space increment A D E comma space A D equals A E space space space left square bracket g i v e n right square bracket
therefore angle A D E equals angle A E D space left square bracket A n g l e s space o p p o s i t e space t o space e q u a l space s i d e s right square bracket
I n space increment A B C comma
because fraction numerator A P over denominator A B end fraction equals fraction numerator A E over denominator A C end fraction
therefore D E parallel to B C
therefore angle A D E equals angle B space space space space space left square bracket c o r r e s p o n o d i n g space a n g l e s right square bracket
B u t space angle B equals angle C space space space space space left square bracket p r o v e d right square bracket
therefore E x t. angle A D E equals i t s space i n t e r i o r space o p p o s i t e space angle C
therefore B C E 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.
H e n c e space B comma C comma E space a n d space D space a r e space c o n c y c l i c.





Question 8

In the given rigure, ABCD is a cyclic eqadrilateral. AF is drawn parallel to CB and DA is produced to point E. If ADC =,FAE =; determineBCD. Given reason in support of your answer.

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



Question 9

If I is the incentre of triangle ABC and AI when produced meets the cicrumcircle of triangle ABC in points D. ifBAC =and = .calculate:

(i)DBC (ii)IBC (iii)BIC.

Solution 9



J o i n space D B space a n d space D C comma space I B space a n d space I C comma
angle B A C equals 66 degree comma angle A B C equals 80 degree comma space I space i s space t h e space i n c e n t r e space o f space t h e space increment A B C comma
left parenthesis i right parenthesis space S i n c e space angle D B C space a n d space angle D A C space a r e space i n space t h e space s a m e space s e g m e n t comma
angle D B C equals angle D A C
B u t comma angle D A C equals 1 half angle B A C equals 1 half cross times 66 degree equals 33 degree
therefore angle D B C equals 33 degree
left parenthesis i i right parenthesis space S i n c e space I space i s space t h e space i n c e n t r e space o f space increment A B C comma space I B space b i s e c t s space angle A B C
therefore angle I B C equals 1 half angle A B C equals 1 half cross times 80 degree equals 40 degree
left parenthesis i i i right parenthesis space therefore angle B A C equals 66 degree space a n d space angle A B C equals 80 degree
I n space increment A B C comma angle A C B equals 180 degree minus left parenthesis angle A B C plus angle B A C right parenthesis
rightwards double arrow angle A C B equals 180 degree minus left parenthesis 80 degree plus 66 degree right parenthesis
rightwards double arrow angle A C B equals 180 degree minus left parenthesis 156 degree right parenthesis
rightwards double arrow angle A C B equals 34 degree
S i n c e space I C space b i s e c t s space t h e space angle C comma
therefore angle I C B equals 1 half angle C equals 1 half cross times 34 degree equals 17 degree
N o w space i n space increment I B C comma
angle I B C plus angle I C B plus angle B I C equals 180 degree
rightwards double arrow 40 degree plus 17 degree plus angle B I C equals 180 degree
rightwards double arrow 57 degree plus angle B I C equals 180 degree
rightwards double arrow angle B I C equals 180 degree minus 57 degree
rightwards double arrow angle B I C equals 123 degree



Question 10

In the given figure, AB = AD = DC = PB and DBC = xo. Determine, in terms of x :

(i)ABD, (ii)APB.

Hence or otherwise, prove thet AP is parallel to DB.

Solution 10

G i v e n minus I n space t h e space f i g u r e comma space A B equals A D equals D C equals P B space a n d space D B C equals X degree
J o i n space A C space a n d space B D.
T o space f i n d space : space t h e space m e a s u r e space o f space angle A B D space a n d space angle A P B.
P r o o f space : space angle D A C equals angle D B C equals X space space space 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
B u t space angle D C A equals angle D A C equals X space space space space space space space space space space space space space space space space space space space space space space space space left square bracket because A D equals D C right square bracket
A l s o comma space w e space h a v e comma space angle A B D equals angle D A C space 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
I n space increment A B P comma space e x t. angle A B C equals angle B A P plus angle A P B
B u t comma angle B A P equals angle A P B space space space space space space space space space space space space space space space space space space space space space space space space space space space space left square bracket because A B equals B P right square bracket
2 cross times X equals angle A P B plus angle A P B equals 2 angle A P B
therefore 2 angle A P B equals 2 X
rightwards double arrow angle A P B equals X
therefore angle A P B equals angle D B C equals X comma
B u t space t h e s e space a r e space c o r r e s p o n d i n g space a n g l e s
therefore A P parallel to D B




Question 11

In the given figure; ABC, AEQ and CEP are straight lines. Show thatAPE andCQE are supplementary.

Solution 11




G i v e n minus I n space t h e space f i g u r e comma A B C comma A E Q space a n d space C E P space a r e space s t r a i g h t space l i n e
T o space p r o v e minus angle A P E plus angle C Q E equals 180 degree
C o n s t r u c t i o n minus J o i n space E B
P r o o f minus I n space c y c l i c space q u a d. A B E P comma
angle A P E plus angle A B E equals 180 degree space space....... left parenthesis 1 right parenthesis
S i m i l a r l y comma space i n space c y c l i c space q u a d. B C Q E comma
angle C Q E plus angle C B E equals 180 degree space...... left parenthesis 2 right parenthesis
A d d i n g space left parenthesis 1 right parenthesis space a n d space left parenthesis 2 right parenthesis comma
angle A P E plus angle A B E plus angle C Q E plus angle C B E equals 180 degree plus 180 degree equals 360 degree
rightwards double arrow angle A P E plus angle A B E plus angle C B E equals 360 degree
B u t comma space angle A B E plus angle C B E equals 180 degree space space space space space space space space space space space space space space space space space left square bracket L i n e a r space p a i r right square bracket
therefore angle A P E plus angle C Q E plus 180 degree equals 360 degree
rightwards double arrow angle A P E plus angle C Q E equals 360 degree minus 180 degree equals 180 degree
H e n c e space angle A P E space A N D space angle C Q E space a r e space s u p p l e m e n t a r y.


Question 12

In the given, AB is the diameter of the circle with centre O.

IfADC = , find angle BOC.

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




Question 13

In a cyclic-quadrilateral PQRS, angle PQR =. Sides SP and RQ prouduced meet at point A: whereas sides PQ and SR produced meet at point B.

IfA : B =2 : 1 ; find angles A and B.

Solution 13



P Q R S 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 i n space w h i c h space angle P Q R equals 135 degree
S i d e s space S P space a n d space R Q 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 A space a n d
S i d e s space P Q space a n d space S R 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 B.
angle A equals angle B equals 2 : 1
L e t space angle A equals 2 x comma t h e n angle B minus x
N o w comma space i n space c y c l i c space q u a d. P Q R S comma
S i n c e comma angle P Q R equals 135 degree comma angle S equals 180 degree minus 135 degree equals 45 degree
open square brackets S i n c 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 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 close square brackets
S i n c e comma space angle P Q R space a n d space angle P Q A space a r e space l i n e a r space p a i r comma
angle P Q R plus angle P Q A equals 180 degree
rightwards double arrow 135 degree plus angle P Q A equals 180 degree
rightwards double arrow angle P Q A equals 180 degree minus 135 degree equals 45 degree
N o w comma space i n space increment P B S comma
angle P equals 180 degree minus left parenthesis 45 degree plus x right parenthesis equals 180 degree minus 45 degree minus x equals 135 degree minus x space space space.... left parenthesis 1 right parenthesis
A g a i n comma space i n space increment P Q A comma
E x t. angle P equals angle P Q A plus angle A equals 45 degree plus 2 x space space space space space space space space space space space space space space space space space space space space space space space space.... left parenthesis 2 right parenthesis
F r o m space left parenthesis 1 right parenthesis space a n d space left parenthesis 2 right parenthesis comma
45 degree plus 2 x equals 135 degree minus x
rightwards double arrow 2 x plus x equals 135 degree minus 45 degree
rightwards double arrow 3 x equals 90 degree
rightwards double arrow x equals 30 degree
H e n c e comma angle A equals 2 x equals 2 cross times 30 degree equals 60 degree
a n d space angle B equals x equals 30 degree



Question 14

In the following figure, ABCD is a cyclic quadrilateral in which AD is parallel to BC.

If the bisector of angle A meet BC at point E and the given circle at point F, prove that:

(i) EF = FC (ii) BF = DF

Solution 14


 

G i v e n minus 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 i n space w h i c h space A D parallel to B C
B i s e c t o r space o f space angle A space m e e t s space B C space a t space E space a n d space t h e space g i v e n space c i r c l e space a t space F.
D F space a n d space B F space a r e space j o i n e d.
T o space p r o v e minus
left parenthesis i right parenthesis space E F equals F C
left parenthesis i i right parenthesis space B F equals D F
P r o o f minus 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 parallel to B C
because A F space i s space t h e space b i s e c t o r space o f space angle A comma angle B A F equals angle D A F
A l s o comma space angle D A E equals angle B A E
angle D A E equals angle A E B space left square bracket A l t e r n a t e space a n g l e s right square bracket
left parenthesis i right parenthesis I n space increment space A B E comma space angle A B E equals 180 degree minus 2 angle A E B
angle C E F equals angle A E B space left square bracket 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
angle A D C equals 180 degree minus angle A B C equals 180 degree minus left parenthesis 180 degree minus 2 angle A E B right parenthesis
angle A D C equals 2 angle A E B
angle A F C equals 180 degree minus angle A D C
space space space space space space space space space space space space space equals 180 degree minus 2 angle A E B space left square bracket S i n c e space A D C F 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 right square bracket
angle E C F equals 180 degree minus left parenthesis angle A F C plus angle C E F right parenthesis
space space space space space space space space space space space space equals 180 degree minus left parenthesis 180 degree minus 2 angle A E B plus angle A E B right parenthesis
space space space space space space space space space space space space equals angle A E B
therefore E C space equals space E F
left parenthesis i i right parenthesis therefore A r c space B F equals A r c space D F space space space space space space space space space space space space space space space space space space left square bracket E q u a l space a r c 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
rightwards double arrow B F equals D F space space space space space space space space space space space space space space space space space space space space space space space space space space left square bracket E q u a l space a r c s space h a v e space e q u a l space c h o r d s right square bracket

Question 15

ABCD is a cyclic quadrilateral. Sides AB and DC produced meet at point e; whereas sides BC and AD produced meet at point F.

IfDCF : F : E = 3 : 5 : 4, find the angles of the cyclic quadrilateral ABCD.

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


Question 16

The following figure shows a cicrcle with PR as its diameter. If PQ = 7 cm and QR = 3RS = 6 cm, Find the perimetre of the cyclic quadrilateral PQRS.

Solution 16

I n space t h e space f i g u r e comma space P Q R S 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 i n space w h i c h space P R space i s
a space d i a m e t e r
P Q equals 7 c m
Q R equals 3 R S equals 6 c m
3 R S equals 6 space c m rightwards double arrow R S equals 2 space c m
N o w space i n space increment P Q R comma
angle Q equals 90 degree space space space space space space space space space space left square bracket A n g l e s space i n space a space s e m i space c i r c l e right square bracket
therefore P R squared equals P Q squared plus Q R squared space space space space space left square bracket P y t h a g o r a s space T h e o r e m right square bracket
space space space space space space space space space equals 7 squared plus 6 squared
space space space space space space space space space equals 49 plus 36
space space space space space space space space space equals 85
A g a i n space i n space r i g h t space increment P S Q comma P R squared equals P S squared plus R S squared
rightwards double arrow 85 equals P S squared plus 2 squared
rightwards double arrow P S squared equals 85 minus 4 equals 81 equals left parenthesis 9 right parenthesis squared
therefore P S equals 9 c m
N o w comma p e r i m e t e r space o f space q u a d. P Q R S equals P Q plus Q R plus R S plus S P
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals left parenthesis 7 plus 9 plus 2 plus 6 right parenthesis c m
space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 24



Question 17

In the following figure, AB is the diameter of a circle with centre O. If chord AC = chord AD ,prove that:

(i) arc BC = arc DB

(ii) AB is bisector ofCAD.

Further if the lenghof arc AC is twice the lengthof arc BC find : (a)BAC (b) ABC

Solution 17



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

Question 18

In cyclic quadrilateral ABCD; AD = BC,BAC= andCBD= ; find ;

(i)BCD (ii)BCA

(iii)ABC (iv)ADC

Solution 18



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


Question 19

In the given figure, ACE = and CAF=; find the values of a, b and c.

Solution 19

 

Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '.

Question 20

In the given figure, AB is parallel to DC ,BCE = and BAC =

Find

(i)CAD (ii)CBD (iii)ADC

Solution 20

 

Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '.

Question 21

ABCD is a cyclic quadrilalteral of a circle with centre O such that AB is a diameter of this circle and the length of the chord CD is equal to the radius of the circle,.if AD and BC produced meet at P, show that APB =Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '..

Solution 21


Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '.

Question 22

In the figure, given alongside, CP bisects angle ACB.

Show that DP bisects angle ADB.

Solution 22

Syntax error from line 1 column 49 to line 1 column 73. Unexpected '<mstyle '.

Question 23

In the figure, given below , AD = BC, BAC = and CBD = find:

(i)BCD

(ii)BCA

(iii)ABC

(iv)ADB

Solution 23