6 basic properties of circles (i) non-foundation · basic properties of circles (i) non-foundation...

6

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1

NON-FOUNDATIONBasic Properties of Circles (I)

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6.2 Chords of a Circle

Key Concepts and Formulae

1. If ON ⊥ AB, then AN = BN.

[Abbreviation: line from centre ⊥ chord bisects chord]O

A BN

2. If AN = BN, then ON ⊥ AB.

[Abbreviation: line joining centre to mid-pt. of chord⊥ chord]

O

A BN

O

C DN

A

BM

O

C DN

A

BM

3. If OM ⊥ AB, ON ⊥ CD and AB = CD, thenOM = ON.

[Abbreviation: equal chords, equidistant from centre]

4. If OM ⊥ AB, ON ⊥ CD and OM = ON, thenAB = CD.

[Abbreviation: chords equidistant from centre areequal]

2

Measures, Shape and SpaceMeasures, Shape and Space

In this exercise, unless otherwise specified, O is the centre of a circle.

1. In the figure, N is the mid-point of the chord AB and ∠ = °NOB 55 .Find ∠ OAN.

Solution

ON ⊥ ( ) (line joining centre to mid-pt. of chord ⊥ chord)

OA = ( ) (radius)

∠ OAN = ∠ ( ) (base ∠ s, isos. ¡ )

But ∠ OBN = 180° − ( ) − ( ) (∠ sum of ¡ )

= ( )

∴ ∠ OAN = ( )

2. In the figure, OC = 5 cm, ON = 3 cm and ON ⊥ CD. Find CD.

Solution

CN = ND (line from centre ⊥ chord bisects chord)

By Pythagoras’ theorem,

CN

ND

CD

= −

=

=

=

5 3

4

4

8

2 2 cm

cm

cm

cm

∴

∴

O

A

B

N

55o

5 cm

3 cm

O

NC D

AB

OB

OBN

90° 55°

35°

35°

3

6 Basic Properties of Circles (I)

3. In the figure, AB = 8 cm, OM = ON = 2 cm. Find

(a) the chord CD,

(b) OD.

(Give your answers in surd form if necessary.)

Solution

(a) CD = AB (chords equidistant from centre are equal)

∴ CD = 8 cm

(b) DN = NC (line from centre ⊥ chord bisects chord)

∴

DN =

=

8

2

4

cm

cm


OD ON ND= +

= +

=

=

2 2

2 22 4

20

2 5

cm

cm

cm

4. In the figure, OM = ON, OM ⊥ AB, ON ⊥ CD, ∠ MBO = 30°and OB = 5 cm. Find

(a) OM,

(b) CD.

Solution

(a)

OM = °

=

5 30

2 5

sin

.

cm

cm

O

A

B

N

C

D

M2 cm

8 cm

2 cm

5 cm

30o

O

A

B

N

C

DM

4


O

A BN

5 cm

5. In the figure, AB = 24 cm, ON = 5 cm and ON ⊥ AB. Find theradius of the circle.

Solution

AN NB

AB

=

=

=2

12 cm

(line from centre ⊥ chord bisects chord)


OB ON NB= +

= +

=

2 2

2 25 12

13

cm

cm

∴ The radius of the circle is 13 cm.

(b)

MB = °

=

5 30

5 3

2

cos cm

cm

AM MB= (line from centre ⊥ chord bisects chord)

∴

AB = ×

=

2

5 3

5 3

2 cm

cm

CD AB= (chords equidistant from centre are equal)

∴ CD = 5 3 cm

5


2 cm 4 cm

O

A

BM

6. In the figure, OB = 4 cm, OM = 2 cm and OM ⊥ AB. Find thelength of AB.

Solution

7. In the figure, OM = ON, AM = MC, BN = NC and ∠ ACB = 60°.Find ∠ MON.

Solution

OM ⊥ AC (line joining centre to mid-pt. of chord ⊥ chord)

∴ ∠ OMC = 90°

ON ⊥ BC (line joining centre to mid-pt. of chord ⊥ chord)

∴ ∠ ONC = 90°

∠ MON = 360° − 90° − 90° − 60°

= 120°

60oO

A

C

N

B

M


MB OB MO= −

= −

=

2 2

2 24 2

2 3

cm

cm

AM = MB (line from centre ⊥ chord bisects chord)

∴

AB = ×

=

(2 2 3

4 3

) cm

cm

6


8. In the figure, BN = NC and ∠ BAN = 30°. Find ∠ OCN.

Solution

AN ⊥ BC (line joining mid-pt. of chord ⊥ chord)

∴ ∠ ANB = 90°

∠ ABN = 360° − 30° − 90° (∠ sum of ¡ )

= 60°

OA = OB (radius)

∠ OBA = 30° (base ∠ s, isos. ¡ )

OB = OC (radius)

∠ OBN = ∠ OCN (base ∠ s, isos. ¡ )

Q ∠ OBN = ∠ ABN − ∠ OBA

= 60° − 30°

= 30°

∠ OCN = 30°

A B

C

D

M

9. In the figure, CD is the common chord of two identicalcircles of centres A and B. M is the mid-point of CD. IfAM = (x + 1)2 cm and MB = (4x + 4) cm, find thevalue of x.

Solution

AM ⊥ CD (line joining centre to mid-pt. of chord ⊥ chord)

BM ⊥ CD (line joining centre to mid-pt. of chord ⊥ chord)

AM = MB (equal chord, equidistant from centre)

(x + 1)2 = 4x + 4

x 2 + 2x + 1 = 4x + 4

x 2 − 2x − 3 = 0

(x − 3)(x + 1) = 0

x = 3 or x = −1 (rejected)

30o

O

A

B N C

7


10. In the figure, AB = 48 cm, BC = 56 cm and OC = 82 cm.Find

(a) OM,

(b) the area of ¡ OAC.

Solution

(a) AM = MB (line from centre ⊥ chord bisects chord)

∴

MB =

=

48

2

24

cm

cm


OM OC MC= −

= − +

=

=

2 2

224 56

18

82 cm

324 cm

cm

2 ( )

(b) The area of ¡ OAC

=

=

1

2

9

( )( )

(

AC OM

= 48 + 56)(18) cm

36 cm

1

22

2

48 cm 56 cm

82 cm

O

A BM

C

8


11. In the figure, the radius of the circle is 13 cm. AB and CD are perpendicularchords intersecting at P. If AB = CD = 24 cm, find the length of BC.

Solution

Let M and N be the mid-points of AB and CD respectively.

OM ⊥ AB (line joining centre to mid-pt. of chord ⊥ chord)

∴

MB = 24

2

1

cm

= 2 cm


OM OB MB= −

= −

=

2 2

23 121 cm

5 cm

2

ON = OM (equal chords, equidistant from centre)

∴ ON = 5 cm

∴ PB = PC

= (12 + 5) cm

= 17 cm

BC PB PC= +

=

2 2

17 2 cm

O

A

N

M B

C

D

P

9


12. In the figure, ABCD is a rectangle with vertices lying on a circle ofradius 15 cm.

(a) Prove that AOC is a straight line.

(b) If AB = 24 cm, find the area of ABCD.

Solution

(a) Join QN and PM as shown in the figure.

DA = CB property of rectangle

OQ = ON equal chords, equidistant from centre

DC = AB property of rectangle

OP = OM equal chords, equidistant from centre

Let OQ = ON = a and OP = OM = b.

OC = a b2 2+ and OA = a b2 2+

Q ON = PC and

PC = PD (line from centre ⊥ chord bisects chord)

∴ DC = 2a

Similarly, AD = 2b


AC b a

b

AO OC

= +

= +

= +

( ) ( )2 2

2

2 2

2a cm2

∴ AOC is a straight line.

(b) Consider ¡ AMO

AMAB=2

1= 2 cm


OM OA AM= −

= −

=

2 2

2 213 12

5

cm

cm

∴ CB = 10 cm

∴ The area of ABCD = (10 × 24) cm2 = 240 cm2

O

A B

CD

M

P

NQ

6

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Basic Properties of Circles (I)

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6.3 Angles in a Circle


1. ∠ = ∠AOB APB2

[Abbreviation: ∠ at centre twice ∠ at ¡ ce]

2. ∠ = °APB 90

[Abbreviation: ∠ in semi-circle]

3. x y=

[Abbreviation: ∠ s in the same segment]

O

AB

Q

P

OA B

P

x

y

A B

P

Q

11


In this exercise, unless otherwise specified,O is the centre of a circle. Find the unknownsin the following figures. (1 – 12)

1.

x

70o

O

A

B

C

x

100o

O

A

B

C

Solution

∠ = ∠

=

ABC

x

1

2( )

( )

2.

Solution

reflex ∠ AOC = 2∠ ABC(∠ at centre twice ∠ at ¡ ce)

x = 2 × 100°

x = 200°

x

120oO

A

B

C3.

Solution

reflex ∠ AOC = 360° − 120° (∠ s at a pt.)

= 240°

reflex ∠ AOC = 2∠ ABC(∠ at centre twice ∠ at ¡ ce)

= 240°

240° = 2x

x = 120°

4.

x 30o

OA

B

C

Solution

∠ ABC = 90° (∠ in the semi-circle)

x + 90° + 30° = 180° (∠ sum of ¡ )

x = 60°

AOC

35°

12


5.x60o

A

B

C

D

Solution

x = 60° (∠ s in the same segment)

x

70o

A

B

C

D6.

Solution

∠ ABC = 70° (base ∠ s, isos. ¡ )

∠ ACB = 180° − 70° − 70° (∠ sum of ¡ )

= 40°

x = 40° (∠ s in the same segment )

x

OA B

C7.

Solution

∠ CAB = x (base ∠ s, isos. ¡ )

∠ ACB = 90° (∠ in the semi-circle)

x + 90° + x = 180° (∠ sum of ¡ )

x = 45°

8.

x

30o

A B

E

C

D

Solution

∠ EBA = x (∠ s in the same segment)

∠ BEC + ∠ ECB = ∠ EBA (ext. ∠ of ¡ )

∴ 30° + 25° = x

∴ x = 55°

13


9.

x

135o

O

A

B

C

D

Solution

∠ CBD = 180° − 135° (adj. ∠ s on st. line)

= 45°

∠ COD = 2∠ CBD(∠ at centre twice ∠ at ¡ ce)

x = 2 × 45°

= 90°

10.x

50o

110o

A

BE

C

D

Solution

∠ EDC = ∠ BAE(∠ s in the same segment)

= 50°

∠ CDE + ∠ DCE = ∠ CEB (ext. ∠ of ¡ )

50° + x = 110°

x = 60°

x

35o

OA B

C

D

x

60o

O

A

B

C

11.

Solution


∠ CBA = 180° − 90° − 35° (∠ sum of ¡ )

= 55°

Consider ¡ BCD

x + 90° + 55° = 180° (∠ sum of ¡ )

x = 35°

12.

Solution

∠ OAB = x (base ∠ s, isos. ¡ )

∠ AOB = 180° − x − x (∠ sum of ¡ )

= 180° − 2x

∠ AOB = 2∠ ACB

180° − 2x = 2 × 60°

x = 30°

14


13. In the figure, AB is a diameter of the circle. If AB is producedto C such that DA = DC and ∠ ABD = 60°, find ∠ ADC.

Solution

∠ ADB = 90° (∠ in the semi-circle)

∠ DAB = 180° − 60° − 90° (∠ sum of ¡ )

= 30°

∠ DCA = 30° (base ∠ s, isos. ¡ )

∠ BDC = 60° − 30° (ext. ∠ of ¡ )

= 30°

∴ ∠ ADC = ∠ ADB + ∠ BDC

= 90° + 30°

= 120°

60o

AB

C

D

14. In the figure, AB intersects OC at K. If ∠ ACK = 70° and∠ BOC = 110°, find ∠ KBO.

Solution

∠ CAB = 1

2× 110° (∠ at centre twice ∠ at ¡ ce)

= 55°

∠ CKB = 70° + 55° (ext. ∠ of ¡ )

= 125°

∠ KBO + ∠ BOK = ∠ CKB (ext. ∠ of ¡ )

∠ KBO = 125° − 110°

= 15°

70o

110oOA

B

K

C

15


30oA

BC

D15. In the figure, diameter AB is produced to C. If BC = BD and∠ BCD = 30°, find ∠ DAB.

Solution

∠ BDC = 30° (base ∠ s, isos. ¡ )

∠ DBA = 30° + 30° (ext. ∠ of ¡ )

= 60°

∠ ADB = 90° (∠ in the semi-circle)

∠ DAB = 180° − ∠ ADB − ∠ DBA (∠ sum of ¡ )

= 180° − 90° − 60°

= 30°

6

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6.4 Relationships among Arcs, Chords and Angles


1. If ∠ AOB = ∠ COD, then

(a)¡_ ¡_AB = CD

[Abbreviation: equal ∠ s, equal arcs]

(b) AB = CD

[Abbreviation: equal ∠ s, equal chords]

2.¡_ ¡_

If AB = CD, then ∠ AOB = ∠ COD.

[Abbreviation: equal arcs, equal ∠ s]

3. If AB = CD, then ∠ AOB = ∠ COD.

[Abbreviation: equal chords, equal ∠ s]

O

A

B C

D

O

A

B C

D

O

A

B C

D

O

A

B C

D

17



5.¡_ ¡_

If AB = CD, then AB = CD.

[Abbreviation: equal arcs, equal chords]

6.¡_ ¡_AB : CD = x : y

[Abbreviation: arcs prop. to ∠ s at centre]

7.¡_ ¡_AB : CD = m : n

[Abbreviation: arcs prop. to ∠ s at ¡ ce]

A

BC

D

A

BC

D

x

y

O

AB

C

D

m

n

A

B

Q

C

D

P

4.¡_ ¡_

If AB = CD then AB = CD.

[Abbreviation: equal chords, equal arcs]

18


In this exercise, unless otherwise specified, Ois the centre of a circle. Find the unknowns inthe following figures (1 – 4)

1.

Solution

= x

x

30°=

=

( )

( )

(arcs prop. to ∠ s at ¡ ce)

4

630

20

× °

= °

Solution

x

y

x

y

7 5

25

125

125

3

7 5

1 5

50

.

.

.

=

=

=

= °°

(arcs prop. to ∠ s at centre)

(arcs prop. to ∠ s at centre)

x cm

y

25o

125o

3 cm

7.5 cm

OA

B C

DE

6 cm

x30o

4 cm Solution

∠ ACB = 180° − 60° − 70° (∠ sum of ¡ )

= 50°

x

x

12

50

60

10

=

=

°°


4.

x cm

60o 70o

12 cm

AB

C

x

12 cm 6 cm

OA

B

C

3.

Solution


∠ CBA = 180° − 90° − x (∠ sum of ¡ )

= 90° − x

12

6

90

30

=

= °

° − x

x

x


2.

4

6

19


5.¡_ ¡_

In the figure, if AB : BC = 3 : 4, ∠ AOB = 45°, OB and AC intersectat K, find

(a) ∠ BOC, (b) ∠ CAB, (c) ∠ OKC.

Solution

(a)¡_ ¡_AB : BC = ∠ AOB : ∠ BOC (arcs prop. to ∠ s at ¡ ce)

3

4=

45°

∠ BOC

∴ ∠ BOC = 60°

(b) ∠ CAB = 1

2∠ BOC (∠ at centre twice ∠ at ¡ ce)

= 1

2× 60°

= 30°

(c) ∠ AOC = ∠ AOB + ∠ BOC

= 45° + 60°

= 105°

OA = OC (radius)

∠ OAC = ∠ OCA (base ∠ s, isos. ¡ )

= 180 105

2

° − °(∠ sum of ¡ )

= 37.5°

∠ OKC = 45° + 37.5° (ext. ∠ of ¡ )

= 82 5. °

45o

O

A B

C

K

20


6. In the figure, ABC and AED are straight lines. If AD = AC,¡_ ¡_

find EB : BC.

Solution

∠ BDC = 110° − 65° (ext. ∠ of ¡ )

= 45°

Q ∠ ADC = 65° (base ∠ s, isos. ¡ )

∴ ∠ EDB = 65° − 45°

= 20°¡_ ¡_

∴ EB : BC = 20 : 45

= 4 : 9

E

65o110o

AB C

D

7.¡_ ¡_

In the figure, AB is a diameter of the circle and BC = CD.¡_ ¡_

If ∠ CAB = 15°, find BD : DA.

Solution

Q¡_BC

¡_= CD

∴ ∠ DAC = 15° (arcs prop. to ∠ s at ¡ ce)

Join OD.

∠ ADO = 30° (base ∠ s, isos. ¡ )

∠ DOB = 30° + 30° (ext. ∠ of ¡ )

= 60°

∠ DOA = 180° − ∠ DOB (adj. ∠ s on st. line)

= 180° − 60°

= 120°

∴¡_ ¡_BD : DA = 60 : 120 (arcs prop. to ∠ s at centre)

= 1: 2

15o OA B

C

D

21


8.¡_ ¡_

In the figure, ∠ PQS = 30°, ∠ QRS = 75° and SR = 2PS. If PR and

QS intersect at K, find ∠ PKS.

Solution¡_

Q SR¡_

= 2PS

∴ ∠ SQR = 2 × 30° (arcs prop. to ∠ s at ¡ ce)

= 60°

∠ SRP = 30° (∠ s in the same segment)

∴ ∠ KRQ = 75° − 30°

= 45°

∠ QKR = 180° − 60° − 45° (∠ sum of ¡ )

= 75°

∴ ∠ PKS = 75° (vert. opp. ∠ s)

30o

75o

K

PS

Q

R

9. In the figure, AB is a diameter of the circle. If BC = CD,prove that OC // AD.

Solution

Let ∠ DAC = a.

∠ CAB = a arcs prop. to ∠ s at ¡ ce

OA = OC radius

∠ ACO = ∠ CAO base ∠ s, isos. ¡

= a

∴ OC // AD alt. ∠ s equal

O

A

B

C

D

22


Solution

(a) AE = AE common side

BE = ED line joining centre ⊥ chord bisects chord

∠ AEB = 180° − 90° adj. ∠ s on st. line

= 90°

∴ ∠ AEB = ∠ AED

∴ ¡ ABE ≅ ¡ ADE S.A.S.

10. In the figure, diameter AC of the circle is perpendicular to chord BD.

(a) Prove that ¡ ABE ≅ ¡ ADE.

(b)¡_ ¡_

If ∠ EAD = 30°, prove that AD = 2CD.

(b) ∠ ADE = 180° − 90° − 30° ∠ sum of ¡

= 60°

∠ ABE = 60° corr. ∠ s, ≅ ¡ s

= = =∠

∠ABE

EAD

60

30

2

1¡_ ¡_

∴ AD = 2CD

O

A

B

C

DE

¡_CD

¡_AD

6

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NON-FOUNDATIONBasic Properties of Circles (I)

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6.5 Cyclic Quadrilaterals


1. ∠ A + ∠ C = 180° and ∠ B + ∠ D = 180°

[Abbreviation: opp. ∠ s, cyclic quad.]

2. ∠ DCE = ∠ DAB

[Abbreviation: ext. ∠ , cyclic quad.]

A

B

C

D

A

B

CE

D

In this exercise, unless otherwise specified, O is the centre of a circle. Find the unknowns in thefollowing figures (1 – 11)

1.

Solution

x + ( ) = 180° (opp. ∠ s, cyclic quad.)

x = ( )

y = ( ) (ext. ∠ , cyclic quad.)

x

y

85o

115o

85°

105°

115°

24


2.

Solution

∠ DAC = 25° (base ∠ , isos. ¡ )

∠ ADC = 180° − 25° − 25° (∠ sum of ¡ )

= 130°

∠ ADC + x = 180° (opp. ∠ s, cyclic quad.)

x = 180° − 130°

= 50°

x

25o

OA B

CD

3.

x

y

55o45o

A

B C

D

Solution

x = 45° (∠ in the same segment)

∠ ADC + ∠ ABC = 180° (opp. ∠ s, cyclic quad.)

x + 55° + y = 180°

45° + 55° + y = 180°

y = 80°

25


4.

x

D

C

B

A

Solution

∠ ADC = 60° (prop. of equil. ¡ )

∠ ADC + ∠ ABC = 180° (opp. ∠ s, cyclic quad.)

∴ ∠ ABC = 120°

∠ BCA = x (base ∠ s, isos. ¡ )

x + x + ∠ ABC = 180° (∠ sum of ¡ )

2x + 120° = 180°

x = 30°

5. In the figure, ABCDE is a pentagon inscribed in the circle. If ∠ EBC = 70°,∠ BAE = 120° and DE = DC, find ∠ BED.

Solution

∠ = ° − °

= °

∠ = ∠

∠ =

= °

∠ = ° − °

= °

∠ = ° − ° − °

= °

° − °

EDC

DEC DCE

DEC

ECB

BEC

180 70

110

35

180 120

60

180 70 60

50

180 110

2

Q ∠ = ∠ + ∠

∴ ∠ = ° + °

= °

BED BEC CED

BED 50 35

85

120o

70o

A

B C

D

E

(opp. ∠ s. cyclic quad.)

(base ∠ s, isos. ¡ )

(∠ sum of ¡ )

(opp. ∠ s, cyclic quad.)

(∠ sum of ¡ )

26


7. In the figure, ∠ ADB = 40°, CB = CD and AD // BC, find ∠ BAD.

6. In the figure, BCDF is a cyclic quadrilateral. ABC, AFD, EFB and

EDC are all straight lines. If ∠ FDE = 96° and ∠ BFD = 120°,find

(a) ∠ FED, (b) ∠ ABF.

Solution

(a)

∠ = ° − °

= °

FED 120 96

24

(ext. ∠ of ¡ )

(b) ∠ FDC = 180° − 96° (adj. ∠ s on st. line)

= 84°

∠ ABF = ∠ FDC (ext. ∠ , cyclic quad.)

= 84°

120o

96o

A

B

F

C DE

40oA

B C

D

(alt. ∠ s, AD // BC)

(base ∠ s, isos. ¡ )

(∠ sum of ¡ )

(opp. ∠ s, cyclic quad.)

Solution

∠ = °

∠ = °

∠ = ° − ∠ − ∠

= ° − ° − °

= °

∠ = ° − ∠

= ° − °

= °

DBC

BDC

BCD DBC BDC

BAD BCD

40

40

180

180 40 40

100

180

180 100

80

27


8.¡_ ¡_

In the figure, ABCD is a cyclic quadrilateral. If BC = 3BA and

∠ BDC = 60°, find

(a) ∠ ADB, (b) ∠ ABD.

Solution

(a)

∠∠

∠°

= =

∠ = °

ADB

BDC

ADB

ADB

60

1

3

20


∴

(b) ∠ BAD = 180° − ∠ BCD (opp. ∠ s, cyclic quad.)

= 180° − 50°

= 130°

∠ ABD = 180° − ∠ ADB − ∠ BAD (∠ sum of ¡ )

= 180° − 20° − 130°

= 30°

60o

50o

A

B

C

D

O

A

B C

D

9.¡_ ¡_ ¡_ ¡_

In the figure, AB : BC : CD : DA = 6 : 5 : 4 : 3.

(a) Find

(i) ∠ BCA, (ii) ∠ ACD.

(b) Hence, determine whether BA ⊥ AD.

Solution

(a) (i)¡_ ¡_ ¡_ ¡_AB + BC + CD + DA = circumference of the circle

∠ BOA + ∠ BOC + ∠ COD + ∠ DOA = 360°

∠ BCA + ∠ BAC + ∠ CAD + ∠ ACD = 180° (∠ at centre twice ∠ at ¡ ce)

But ∠ BCA : ∠ BAC : ∠ CAD : ∠ ACD¡_ ¡_ ¡_ ¡_

= AB : BC : CD : DA (arcs prop. to ∠ s at ¡ ce)

= 6 : 5 : 4 : 3

∴ ∠ BCA

=

= °+ + +

6

6 5 4 3

60

× 180°

28


A

B

C D

E

10. In the figure, AC = AD, prove that ∠ ABC = ∠ AED.

Solution

(ii) ∴ ∠ ACD

=

= °+ + +

3

6 5 4 3

30

× 180°

(b) ∠ BCD = ∠ BCA + ∠ ACD

= 60° + 30°

= 90°

∠ BAD = 180° − ∠ BCD (opp. ∠ s, cyclic quad.)

= 180° − 90°

= 90°

∴ BA ⊥ AD

∠ ACD = ∠ ADC base ∠ s, isos. ¡

∠ ABC = 180° − ∠ ADC opp. ∠ , cyclic quad.

∠ AED = 180° − ∠ ACD opp. ∠ , cyclic quad.

∴ ∠ ABC = ∠ AED

29


O

A BC

E

D

11. In the figure, ABC is a straight line, AO // BD and AO = AB.

(a) Show that ∠ AOD = 2∠ DBC.

(b) If ∠ CBD = 60°, prove that OABD is a parallelogram.

Solution

(a) ∠ DBC = ∠ AED ext. ∠ , cyclic quad.

∠ AOD = 2∠ AED ∠ at centre twice ∠ at ¡ ce

∴ ∠ AOD = 2∠ DBC

(b) ∠ AOD = 2 × 60°

= 120°

∠ OAB = ∠ CBD corr. ∠ s, AO // BD

∠ OAB + ∠ AOD = 60° + 120°

= 180°

∴ OD // AB int. ∠ s, supp.

Since OD // AB and AO // BD, OABD is a parallelogram.

6

30

NON-FOUNDATION

Name :

Date :

Mark :

6E


Multiple Choice Questions

In this exercise, unless otherwise specified,O is the centre of a circle.

1. In the figure, POQ is a straight line,OC = 10 cm, AB = CD = 16 cm.Find PQ.

3. In the figure, BC is a diameter of thecircle and AB = AD = DC. Find∠ ABD.

16 cm

16 cm

O

A B

QC D

P

A. 11 cm B. 12 cm

C. 13 cm D. 14 cm

2. In the figure, AM = MB. Which of thefollowing is / are true?

O

A BM

I. OM ⊥ AB

II. AO ⊥ OB

III. ¡ OAM ≅ ¡ OBM

IV. ¡ OBM ≅ ¡ OBA

A. I

B. I and II

C. I and III

D. I, III and IV

A

B C

D

A. 27°B. 28°C. 29°D. 30°

4. In the figure, AC and BD intersect at K.Find x.

x

42o

45o

58o

A

B

C

D

K

A. 32°B. 35°C. 45°D. 50° B

C

DB

31


5. In the figure, find x. 8. In the figure, AB is a diameter of thecircle and AD // OC. Find x.

x

35o 23o

O

A

B

C

A. 115° B. 116°C. 117° D. 118°

6. In the figure, AB is a diameter of thecircle. Find x.

x

44o

A B

C

D

A. 19° B. 21°C. 23° D. 25°

7.¡_ ¡_

In the figure, AD = DC. Find x.

x

100o

45o

A

B

C

D

A. 65° B. 67°C. 69° D. 80°

x

35o

OA B

C

D

A. 100° B. 105°C. 107.5° D. 110°

9. In the figure, find x + y.

xO

A

B

C D

Ey

A. 215° B. 220°C. 225° D. 230°

10. In the figure, BD is a diameter of thecircle. Find ∠ DBC.

53o

A

B C

D

A. 30° B. 35°C. 36° D. 37°

C

C

A

B

C

D

32


11. In the figure, ABC and AED are straightlines. Find ∠ EDB.

36o48o

A

B

C

D

E

A. 11° B. 12°C. 18° D. 22°

12.¡_ ¡_ ¡_

In the figure, BCD : CDA : DAB =3 : 4 : 6. Find ∠ ADC.

A

B

C

D

A. 105° B. 110°C. 115° D. 121°

13. In the figure, AOB is a diameter of thecircle and ∠ DAO = °50 . If DC CB= ,find the value of x y+ .

xy

50o

OA B

CD

14. In the figure, COD is a diameter of thecircle and AB ⊥ CD. If AB = 8 cm andCM = 2 cm, find the radius of the circle.

O

A B

D

C

M

A. 2.5 cm B. 3 cm

C. 4 cm D. 5 cm

15. In the figure, chord ED and AB areproduced to meet at C. If ED = DB,AB = AE and ∠ DCB = 30°, find ∠ EAB.

30o

O

A

BC

D

E

A. 90° B. 98°C. 105° D. 112°

A. 50° B. 55°C. 58° D. 60°

16. In the figure, ∠ ABC = 105°, ∠ BCE = 70°and AB = BC, find x.

A. 100° B. 102.5°C. 105° D. 107.5°

x

70o

105o

A B

EC

D

B

D

C

B

D

D

33


17. In the figure, ABDF is inscribed in thecircle. AFE, BDE, FDC and ABC are thestraight lines. If ∠ E = 22° and ∠ C = 30°,find ∠ A.

30o

22o

F

A BC

D

E

A. b + c = 180°B. a + d = 180°C. a = x + y

D. d = b + c − 90°

19. In the figure, AB and BC are two chordsof the circle. If reflex ∠ COA = 285°,find x + y.

x y

a

b

c

d

AB

F

E

C

D

A. 30°B. 37.5°C. 52.5°D. 75°

20. In the figure, AB is a diameter of thecircle. Find ∠ DOC.

x

y

285o

O

A

B

C

125o

OA B

CD

K

A. 60°B. 64°C. 68°D. 78°

18. In the figure, which of the followingMUST be true?

A. 70°B. 75°C. 78°

D. 82°

B

A

B

C

6 basic properties of circles (i) non-foundation · basic properties of circles (i) non-foundation...

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