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# 1_7_10_37

### 1_7_10_37

Question Bank In Mathematics Class IX (Term II)

10 CIRCLES

A. SUMMATIVE ASSESSMENT

10.1 CIRCLES AND ITS RELATED through the centre of the circle. In the given
TERMS : A REVIEW figure, AOB is the diameter of the circle. A
diameter is the longest chord of a circle.
1. The collection of all the points in a plane,
which are at a fixed distance from a fixed point Diameter = 2 × radius
in the plane, is called a circle. 6. A piece of a circle between two points is
called an arc. Look at the pieces of the circle
2. The fixed point is called the centre of the between two points P and Q in the given figure.
circle and the fixed distance is called the radius You find that there are two pieces, one longer
of the circle. and the other smaller. The longer one is called
the major arc PQ and the shorter one is called
the minor arc PQ.
GOYAL BROTHERS PRAKASHAN
In the given figure, O is the centre and the 7. The length of the complete circle is called
length OP is the radius of the circle. its circumference. The
region between a chord and
3. A circle divides the plane on which it lies either of its arcs is called a
into three parts. They are : (i) inside the circle, segment of the circular
which is also called the interior of the circle; region or simply a segment
(ii) the circle and (iii) outside the circle, which is of the circle. You will find
also called the exterior of the circle. The circle that there are two types of
and its interior make up the circular region. segments also, which are the major segment and
the minor segment.

4. A chord of a circle 8. The region between an arc and the two
is a line segment joining radii, joining the centre to the end points of the
any two points on the arc is called a sector. Like
circle. In the given figure segments, you find that the
PQ, RS and AOB are the minor arc corresponds to
chords of a circle. the minor sector and the
major arc corresponds to
5. A diameter is a chord of a circle passing the major sector.

1

TEXTBOOK’S EXERCISE 10.1

Q.1. Fill in the blanks : (ii) A circle has only finite number of equal
(i) The centre of a circle lies in chords.
___________ of the circle. (exterior/interior)
(ii) A point, whose distance from the centre (iii) If a circle is divided into three equal
of a circle is greater than its radius lies in arcs, each is a major arc.
__________ of the circle. (exterior/interior)
(iii) The longest chord of a circle is a (iv) A chord of a circle, which is twice as
__________ of the circle. long as its radius, is a diameter of the circle.
(iv) An arc is a __________ when its ends
are the ends of a diameter. (v) Sector is the region between the chord
(v) Segment of a circle is the region between and its corresponding arc.
an arc and __________ of the circle.
(vi)A circle divides the plane, on which it (vi) A circle is a plane figure.
lies, in __________ parts.
Sol. (i) interior (ii) exterior (iii) diameter Sol. (i) True, because all points on the circle
(iv) semicircle (v) the chord (vi) three are equidistant from its centre.

Q.2. Write True or False: Give reasons for (ii) False, because there are infinitely many

(i) Line segment joining the centre to any (iii) False, because for each arc, the
point on the circle is a radius of the circle. remaining arc will have greater length.
GOYAL BROTHERS PRAKASHAN
(iv) True, because of definition of diameter.

(v) False by virtue of its definition.

(vi) True as it is a part of a plane.

10.2 ANGLE SUBTENDED BY A CHORD 2. If the angles subtended by the chords of a
AT A POINT circle at the centre are equal, then the chords are
equal.
1. Equal chords of a circle subtend equal
angles at the centre.

TEXTBOOK’S EXERCISE 10.2

Q.1. Recall that two circles are congruent if Proof : In triangles AOB and COD,

they have the same radii. Prove that equal AB = CD [Given]

chords of congruent circles subtend equal angles AO = CO

at their centres [V. Imp] BO = DO [Radii of congruent circles]

Sol.  AOB  COD [SSS axiom]

 AOB = COD [CPCT] Proved.

Q.2. Prove that if chords of congruent

circles subtend equal angles at their centres,

then the chords are equal. [2010]

Sol.

Given : Two congruent circles with centres
O and O. AB and CD are equal chords of the
circles with centres O and O respectively.

To Prove : AOB = COD

2

Given : Two congruent circles with centres O AO = CO
and O. AB and CD are chords of circles with
centre O and O respectively such that AOB = BO = DO [Radii of congruent circle]
COD
 AOB = COD [Given]
To Prove : AB = CD
Proof : In triangles AOB and COD,  AOB  COD [SAS axiom]

 AB = CD [CPCT] Proved.

10.3 PERPENDICULAR FROM THE 2. The line drawn through the centre of a circle
CENTRE TO A CHORD to bisect a chord is perpendicular to the chord.

1. The perpendicular from the centre of a 3. There is one and only one circle passing
circle to a chord bisects the chord. through three given non-collinear points.
GOYAL BROTHERS PRAKASHAN
TEXTBOOK’S EXERCISE 10.3

Q.1. Draw different pairs of circles. How Sol. Given : AB is the common chord of two
many points does each pair have in common? intersecting circles (O, r) and (O, r).
What is the maximum number of common points?
To Prove : Centres of both circles lie on the
Sol. perpendicular bisector of chord AB, i.e., AB is
bisected at right angle by OO.

Construction : Join AO, BO, AO and BO.

Maximum number of common points = 2 Proof : In AOO and BOO
AO = OB [Radii of the circle (O, r)]
Q.2. Suppose you are given a circle. Give a AO = BO [Radii of the circle (O, r)]
construction to find its centre. OO = OO [Common]

Sol. Steps of Construction :  AOO BOO [SSS congruency]
1. Take arc PQ of the given circle.  AOO= BOO [CPCT]
2. Take a point R on the arc PQ and draw Now in AOC and BOC,
chords PR and RQ.
3. Draw perpendicular AOC = BOC [AOO = BOO]
bisectors of PR and RQ. AO = BO [Radii of the circle (O, r)]
These perpendicular bisectors
intersect at point O. OC = OC [Common]
Hence, point O is the  AOC  BOC [SAS congruency]
centre of the given circle. AC = BC and ACO = BCO ...(i) [CPCT]
ACO + BCO = 180° ..(ii) [Linear pair]
Q.3. If two circles intersect at two points, ACO = BCO = 90° [From (i) and (ii)]
prove that their centres lie on the perpendicular Hence, OO lie on the perpendicular bisector
bisector of the common chord. [2011 (T-II)] of AB. Proved.

3

10.4 EQUAL CHORDS AND THEIR circles) are equidistant from the centre (or centres).
DISTANCES FROM THE CENTRE
2. Chords equidistant from the centre of a
1. Equal chords of a circle (or of congruent circle are equal in length.

TEXTBOOK’S EXERCISE 10.4

Q.1. Two circles of radii 5 cm and 3 cm Now, AB = CD [Given]

intersect at two points and the distance between 11

their centres is 4 cm. Find the length of the  2 AB = 2 CD

common chord. [2011 (T-II)]  AM = CN ...(ii) [Perpendicular from
GOYAL BROTHERS PRAKASHAN
Sol. In AOO, centre bisects the chord]
AO2 = 52 = 25
AO2 = 32 = 9 Adding (i) and (ii), we get ...(iii)
OO2 = 42 = 16 EM + AM = EN + CN ...(iv)
AO2 + OO2 [From (iii)]
 AE = CE Proved.
= 9 + 16 = 25 = AO2 Now, AB = CD
 AB – AE = CD – AE
 AOO= 90° [By converse of  BE = CD – CE

Pythagoras theorem] Q.3. If two equal chords of a circle intersect

Similarly, BOO = 90°. within the circle, prove that the line joining the
 AOB= 90° + 90° = 180°
AOB is a straight line, whose mid-point point of intersection to the centre makes equal
is O.
angles with the chords. [2011 (T-II)]

 AB = (3 + 3) cm = 6 cm Sol. Given : AB and
CD are two equal chords
Q.2. If two equal chords of a circle intersect of a circle which meet at
E within the circle and a
within the circle, prove that the segments of one line PQ joining the point
of intersection to the
chord are equal to corresponding segments of centre.

the other chord. [V. Imp.] To Prove : AEQ = DEQ
Construction : Draw OL  AB and
Sol. Given : AB and CD are two equal OM  CD.
Proof : In OLE and OME, we have
chords of a circle which meet at E.

To prove : AE = CE and BE = DE

Construction : Draw OM  AB and
ON  CD and join OE.

Proof : In OME and OL = OM [Equal chords are equidistant]

ONE, OM = ON OE = OE [Common]

[Equal chords OLE = OME [Each = 90°]

are equidistant]  OLE  OME [RHS congruence]
OE = OE [Common]
OME = ONE  LEO = MEO [CPCT]

Q.4. If a line intersects two concentric

[Each equal to 90°] circles (circles with the same centre) with centre

 OME  ONE [RHS axiom] O at A, B, C and D, prove that AB = CD (See

 EM = EN...(i) [CPCT] fig.) [2011 (T-II)]

4

Sol. Given : A line  KR = 12  2  24 = 4.8 m  RM = 2KR
AD intersects two 5 5
concentric circles at A,
B, C and D, where O is  RM = 2 × 4.8 = 9.6 m
the centre of these
circles. Hence, distance between Reshma and

To prove : AB = CD Mandip is 9.6 m.
Construction : Draw OM  AD.
Proof : AD is the chord of larger circle. Q.6. A circular park of radius 20 m is
AM = DM ..(i) [OM bisects the chord]
situated in a colony. Three boys Ankur, Syed and
BC is the chord of
smaller circle David are sitting at equal distance on its

 BM = CM ...(ii)
[OM bisects the chord]

Subtracting (ii) from
(i), we get

AM – BM = DM – CM  AB = CD Proved.
GOYAL BROTHERS PRAKASHAN boundary each having a toy telephone in his

hands to talk each other. Find the length of the

string of each phone. [HOTS]

Sol. Let Ankur, Syed and David be
represented by A, S and D respectively.

Q.5. Three girls Reshma, Salma and Mandip

are playing a game by standing on a circle of

radius 5 m drawn in a park. Reshma throws a

ball to Salma, Salma to Mandip, Mandip to

Reshma. If the distance between Reshma and Let PD = SP = SQ = QA = AR = RD = x
In OPD, OP2 = 400 – x2
Salma and between Salma and Mandip is 6 m

each, what is the distance between Reshma and  OP = 400  x2

Mandip? [HOTS]

Sol. Let Reshma, Salma and Mandip be  AP = 2 400  x2  400  x2
[ centroid divides the
represented by R, S and M respectively.
median in the ratio 2 : 1]
Draw OL  RS, OL2 = OR2 – RL2

OL2 = 52 – 32 [RL = 3 m, because OL  RS] = 3 400  x2
Now, in APD, PD2 = AD2 – AP2
= 25 – 9 = 16
 2
OL = 16 = 4 m
 x2 = (2x)2 – 3 400  x2
Now, area of

triangle ORS  x2 = 4x2 – 9(400 – x2)

1  x2 = 4x2 – 3600 + 9x2  12x2 = 3600

= 2 × KR × OS  x2 = 3600 = 300  x = 10 3
12
1
Now, SD = 2x = 2 × 10 3 = 20 3
= 2 × KR × 5  ASD is an equilateral triangle.

Also, area of ORS = 1 × RS × OL
2

= 1 × 6 × 4 = 12 m2  SD = AS = AD = 20 3
2 Hence, length of the string of each phone is

1 20 3 m.

 2 × KR × 5 = 12

5

OTHER IMPORTANT QUESTIONS

Q.1. In the given figure, O is the centre of Q.5. There are three non-collinear points.

the circle. If OA = 5 cm and OC = 3 cm, then the The number of circles passing through them

length of AB is : [2011 (T-II)] are : [2010]

(a) 2 (b) 1 (c) 3 (d) 4
GOYAL BROTHERS PRAKASHAN
Sol. (b) There is one and only one circle

passing through three given non-collinear points.

Q.6. In the given

figure, OM  to the

chord AB of the circle

with centre O. If OA =

(a) 4 cm (b) 6 cm (c) 8 cm (d) 15 cm 13 cm and AB = 24 cm,

Sol. (c) AC = AO2 – OC2 = 25 – 9 cm then OM equals : [2010]
= 4 cm
(a) 3 cm (b) 4 cm

AB = 2 × AC = 2 × 4 cm = 8 cm. (c) 5 cm (d) 4.7 cm

Q.2. Three chords AB, CD and EF of a circle Sol. (c) In AMO, M = 90°
 OA2 = AM2 + OM2

are respectively 3 cm, 3.5 cm and 3.8 cm away  OM = OA2  AM2
from the centre. Then which of the following

relations is correct ? [HOTS]  OM = (13)2  (12)2 = 5 cm

(a) AB > CD > EF (b) AB < CD < EF Q.7. In the given figure,

(c) AB = CD = EF (d) none of these in a circle with centre O, a
Sol. (a) We know that longer the chord,
chord AB is drawn and C is
shorter is its distance from the centre.
its mid-point ACO will be :
Q.3. In a circle, chord AB of length 6 cm is
at a distance of 4 cm from the centre O. The [2010]
length of another chord CD which is also 4 cm
away from the centre is : (a) more than 90° (b) less than 90°

(a) 6 cm (b) 4 cm (c) 8 cm (d) 3 cm (c) 90° (d) none of these
Sol. (a) Chords equidistant from the centre
are equal. Sol. (c) The line drawn through the centre of a

circle to bisect a chord is perpendicular to the chord.

Q.8. Two chords AB and CD subtend x° each

Q.4. In the figure, at the centre of the circle. If chord AB = 8 cm,
chord AB is greater than
chord CD. OL and OM then chord CD is : [2011 (T-II)]
are the perpendiculars
from the centre O on these (a) 4 cm (b) 8 cm (c) 16 cm (d) 12 cm
two chords as shown in
the figure. The correct [HOTS]
releation between OL and OM is :

(a) OL = OM (b) OL < OM

(c) OL > OM (d) none of these Sol. (b) Equal chords subtend equal angles at
the centre.
Sol. (b) Longer the chord, shorter is its

distance from the centre.

6

Q.9. In the given (a) 5 (b) 53 cm
figure, a circle with 2
centre O is shown, 2 cm
where ON > OM. Then
which of the following [HOTS] 53
relations is true
between the chord AB (b) AB > CD (c) 5 cm (d) 4 cm
and chord CD ?
Sol. (c) We have, OA = 5 cm = OB
(a) AB = CD [Radii of the circle]

Clearly A = B = 60°
[Opposite angle of equal sides]

 AB = 5 cm
[ AOB is an equilateral triangle]

Q.13. Find the length
of a chord which is at a
distance of 5 cm from the
centre of a circle whose
radius is 13 cm.

Sol. AB = 2 AO2 – OC2

= 2 132 – 52 cm
(c) AB < CD (d) none of these B R O T H -ERS
PRAKSHAN
Sol. (b) Longer the chord, shorter is its

distance from the centre.

Q.10. In the figure,
O is the centre of the
circle. If OA = 5 cm, AB
= 8 cm and OD is
perpendicular to AB,
then CD is equal to :

(a) 2 cm (b) 3 cm (c) 4 cm (d) 5 cm

Sol. (a) OC = AO2 – AC2 = 2 × 12 cm = 24 cm.

= 25 – 16 cm = 3 cm Q.14. Two concentric circles with centre O

have A, B, C and D as

Since, OD = OA = 5 cm points of intersection

 CD = OD – OC = (5 – 3) cm = 2 cm. with a line l as shown

Q.11. In the given figure, in the figure. If AD =

O is the centre of the circle 12 cm and BC = 8 cm,

of radius 5 cm. OP ⊥ AB, find the length of AB

OQ ⊥ CD, AB | | CD, AB = and CD. [2011 (T-II)]
GOYAL
8 cm and CD = 6 cm. The Sol. Since OM  BC

length of PQ is : [2011 (T-II)] 1
 BM = CM = 2 BC = 4 cm
(a) 8 cm (b) 1 cm Similarly, OM  AD

(c) 6 cm (d) none of these

Sol. (b) OQ = OC2 – CQ2 1
 AM = DM = 2 AD = 6 cm
= 25  9 cm = 4 cm Now, AB = AM – BM = (6 – 4) cm = 2 cm
Also, CD = DM – CM = (6 – 4) cm = 2 cm
OP = OA2 – AP2 = 25 – 16 cm = 3 cm Hence, AB = CD = 2 cm
Q.15. Two circles of radii 10 cm and 8 cm
 PQ = OQ – OP = 1 cm. intersect and the length of the common chord is
Q.12. In the given figure, ∠AOB chord AB 12 cm. Find the distance between their centres.
subtends angle equal to 60° at the centre of the
circle. If OA = 5 cm, then length of AB (in cm) [2011 (T-II)]
is : [2010]

7

Sol. Let O and O  AB2 = AC2 + BC2 [Pythagoras theorem]
be the centres of the
circles of radii 10 cm Q.18. Two chords of a circle of lengths
and 8 cm respectively 10 cm and 8 cm are at the distances 8 cm and
and let PQ be their 3.5 cm respectively from the centre. Check
common chord. whether the above statement is true or not.

We have, Sol. False, because larger the chord, shorter
OP = 10 cm, OP = 8 cm and PQ = 12 cm is its distance from the centre.

1 Q.19. If the perpendicular bisector of a
 PL = 2 PQ = 6 cm
In right OLP, we have OP2 = OL2 + LP2 chord AB of a circle PXAQBY intersects the

 OL = OP2  LP2 = 102  62 circle at P and Q, prove that arc PXA  arc

= 64 cm = 8 cm PYB. [HOTS]
In right OLP,
we have OP2 = OL2 + LP2 Sol.

 OL = OP2  LP2  82  62  28 cm In PAO and PBO,
= 5.29 cm
BROTHERS AO = BO [Given]
 OO = OL + LO = (8 + 5.29) cm
= 13.29 cm. POA = POB

Q.16. Two congruent circles with centres O = 90°
and O intersect at two points A and B. Check
whether ∠AOB = ∠AO B or not. [V. Imp.] [Given]

Sol. OA = OB = OA = OB PO = PO

[Common]

 PAO  PBO [SAS]

 PA = PB [CPCT]

 arc PXA = arc PYB

 arc PXA  arc PYB Proved.

Q.20. Show that two circles cannot intersect

at more than two points. [Imp.]

Sol.Let us assume that two circles intersect

PRGAOKYAASLHAN at three points say A, B and C. Then clearly, A,

B and C are not collinear. But, through three

In  AOB and  AOB, we have non-collinear points we can draw one and only

AO = AO one circle. Therefore, we cannot have two circles

OB = OB passing A, B and C. Or two circles cannot

AB = AB [Common] intersect at more than two points.

 AOB  AO B [SSS congruency axiom] Q.21. AB and AC are two chords of a circle

 AOB = AO B [CPCT] of radius r such that AB = 2AC. If p and q are

Q.17. AOB is a diameter of a circle and C is the distances of AB and AC from the centre,

a point on the circle. prove that 4q2 = p2 + 3r2. [2011 (T-II)]

Check whether AC2 + BC2 Sol. Draw OD  AB,

= AB2 is true or OE  AC and join AO.
Let AC = 2x, then
not. [Imp.]

Sol. True. We know AE = CE = x
that ACB = 90° So, AB = 4x and
AD = BD = 2x.
[Angle made in semi-circle]

8

In AOD, OM = ON [From (i)]

AO2 = AD2 + OD2 [By Pythagoras theorem] OME = ONE [each equal to 90°]

 AD2 = AO2 – OD2 OE = OE [common]

 4x2 = r2 – p2 … (i)  OME  ONE [by SAS congruence]

In AOE, we have  ME = NE [CPCT]

AO2 = OE2 + AE2  AE2 = AO2 – OE2 Thus, in quad. OMEN, we have

 x2 = r2 – q2 OM = ON, ME = NE and

 4x2 = 4r2 – 4q2 … (ii) OME = ONE = 90°

From (i) and (ii), we have Hence, OMEN is a square. Proved.

4r2 – 4q2 = r2 – p2  4q2 = p2 + 3r2. Q.23. In the given

Proved. figure, OD is perpendi-

cular to the chord AB of a

Q.22. In the given circle whose centre is O. If

figure, equal chords AB BC is a diameter, show

and CD of a circle cut that CA = 2OD.

at right angles at E. If [2011 (T-II)]

M and N are the mid- Sol. Since OD  AB and the perpendicular

points of AB and CD drawn from the centre to a chord bisects the

respectively. Prove that chord.

OMEN is a square. [2011 (T-II)]  D is the mid-point of AB
GOYAL
Sol. Join OE.PRABKRAOSTHHAENRS Also, O being the centre, is the mid-point of

Since the line joining the centre of a circle to BC.

the mid-point of a chord is perpendicular to the Thus, in ABC, D and O are mid-points of

chord, we have OM  AB and ON  CD AB and BC respectively.

 OMB = 90° and OND = 90° 1

 OME = 90° and ONE = 90°  OD || AC and OD = CA
2
Also, equal chords of a circle are equidistant
[ segment joining the mid-points of two
from the centre.
sides of a triangle is half of the third side.]
 OM = ON ..... (i)
 CA = 2OD. Proved.
Now, in OME and ONE, we have

PRACTICE EXERCISE 10A

1 Mark Questions (a) 10 cm (b) 8 cm

1. In the given figure, O (c) 12 cm (d) 16 cm
is the centre and AB = BC. If
BOC = 80°, then AOB is 3. In the given figure, O is the centre of the
circle, OB = 5 cm and AB = 8 cm. The distance of
(a) 80° (b) 70° AB from the centre is

(c) 85° (d) 90° (a) 4 cm
(b) 3 cm
2. In the given figure, O is
the centre of the circle, OA = (c) 89 cm
10 cm and OC = 6 cm. The (d) 6 cm
length of AB is

9

2 Marks Questions 10. In a circle of radius 5 cm, there are two
parallel chords of length 6 cm and 4 cm. Find the
4. Calculate the length of a chord which is at a distance between them, when they are on (i) oppo-
distance 5 cm from the centre of a circle whose site sides of the centre (ii) same side of the centre.
radius is 13 cm.
11. Two concentric circles are of radii 7 cm,
5. Find the distance from the centre to a chord 4 cm. A line PQRS cuts one circle at P, S and other
70 cm in length in a circle whose diameter is 74 cm. at Q, R. If QR = 6 cm, find the length of PS.

6. Two points P and Q are 9 cm apart. A circle 12. If in the figure, AN
of radius 5.1 cm passes through P and Q. Calculate = NB = l and ND = h,
the distance of its centre from the chord PQ. prove that the diameter of

7. Two circles of radii 26 cm and 25 cm inter- the circle is l2  h2 .
sect at two points which are 48 cm apart. Find the
distance between their centres. h

8. Two parallel chords of a circle whose diam- [HOTS]
eter is 13 cm are respectively 5 cm and 12 cm.
Find the distance between them if they lie on op- 13. AB and CD are two equal chords of a circle.
posite sides of the centre. M and N are their mid-points respectively. Prove
that MN makes equal angles with AB and CD.
3/4 Marks Questions
[HOTS]
9. PQ is a variable chord of a circle of radius
7.5 cm. If PQ = 9 cm, find the radius of the circle 14. A chord AB of a circle (O, r) is produced
which is the locus of the mid-point of PQ.
PRGAOKYAASLHBARNOTHERS to P so that BP = 2 AB. Prove that

OP2 = OA2 + 6AB2. [HOTS]

10.5 ANGLE SUBTENDED BY AN ARC OF subtends equal angles at two other points lying
A CIRCLE AND CYCLIC QUADRILATERAL on the same side of the line containing the line
segment, the four points lie on a circle (i.e., they
1. The angle subtended by an arc at the are concyclic).
centre is double the angle subtended by it at any
point on the remaining part of the circle. 4. The sum of either pair of opposite angles
of a cyclic quadrilateral is 180°.
2. Angles in the same segment of a circle are
equal. 5. If the sum of a pair of opposite angles of
a quadrilateral is 180°, the quadrilateral is cyclic.
3. If a line segment joining two points

TEXTBOOK’S EXERCISE 10.5

Q.1. In the figure, A, B and C are three the centre of a circle is double the angle
points on a circle with centre
O such that  BOC = 30° subtended by the same arc on the remaining part
and  AOB = 60°. If D is a
point on the circle other than of the circle.
the arc ABC, find  ADC.  2ADC = AOC

[2010] 11

Sol. We have, BOC = 30° and AOB = 60°  ADC = 2 AOC = 2 × 90°
AOC = AOB + BOC = 60° + 30° = 90°  ADC = 45°.

We know that angle subtended by an arc at Q.2. A chord of a circle is equal to the

radius of the circle. Find the angle subtended by

the chord at a point on the minor arc and also

at a point on the major arc. [Imp.]

10

Sol. We have, OA = OB = AB  69° + 31° + BAC = 180°
 BAC = 180° – 100° = 80°
Therefore, OAB is a equilateral triangle. Also, BAC = BDC

 AOB = 60° [Angles in the same segment]
 BDC = 80°
We know that angle Q.5. In the figrue, A, B,
C and D are four points on
subtended by an arc at the a circle. AC and BD
intersect at a point E such
centre of a circle is double that BEC = 130° and
ECD = 20°. Find BAC.
the angle subtended by the
[2011 (T-II)]
same arc on the remaining
Sol. BEC + DEC = 180° [Linear pair]
part of the circle.  130° + DEC = 180°
 DEC= 180° – 130° = 50°
    AOB = 2ACB Now, in DEC,
 DEC + DCE + CDE = 180°
    ACB = 1 AOB = 1
2 [Angle sum property of a triangle]
    ACB = 30° 2 × 60°
 50° + 20° + CDE = 180°
1 G  CDE = 180° – 70° = 110°
O Also, CDE = BAC
Also, ADB = 2 reflex AOB YA
L [Angles in same segment]
11 PRBARKOATSHHEARNS
 BAC = 110°
= 2 (360° – 60°) = 2 × 300° = 150°

Hence, angle subtended by the chord at a

point on the minor arc is 150° and at a point on

the major arc is 30°

Q.3. In the figure, Q.6. ABCD is a cyclic
PQR = 100°, where P, Q
and R are points on a circle quadrilateral whose dia-
with centre O. Find OPR.
gonals intersect at a point
[2011 (T-II)]
E. If DBC = 70°,
Sol. Reflex angle
POR = 2PQR BAC = 30°, find BCD.

= 2 × 100° = 200°

Now, angle POR = 360° – 200° = 160° Further, if AB = BC, find

Also, PO = OR [Radii of a circle] ECD. [V. Imp.]

 OPR = ORP [Opposite angles of Sol. CAD = DBC = 70° [Angles in the

isosceles triangle] same segment]

In OPR, POR = 160° Therefore, DAB = CAD + BAC

 OPR = ORP = 10° = 70° + 30° = 100°

[Angle sum property of a triangle] But, DAB + BCD = 180° [Opposite

Q.4. In the figure, ABC = 69°, ACB = 31°, angles of a cyclic quadrilateral]
find  BDC. [2011 (T-II)]
So, BCD = 180° – 100° = 80°
Sol. In ABC, we have
Now, we have AB = BC
ABC + ACB + BAC
= 180° Therefore, BCA = 30° [Opposite angles

of an isosceles triangle]

Again, DAB + BCD = 180° [Opposite

[Angle sum property angles of a cyclic quadrilateral]
of a triangle]
 100° + BCA + ECD = 180°

[ BCD = BCA + ECD]

11

 100° + 30° + ECD = 180° Sol. Given : A trapezium ABCD in which
 130° + ECD = 180° AB || CD and AD = BC.
 ECD = 180° – 130° = 50°
To Prove : ABCD is a cyclic trapezium.

Hence, BCD = 80° and ECD = 50°. Construction : Draw DE  AB and CF  AB.
Proof : In DEA and CFB, we have
Q.7. If diagonals of a cyclic quadrilateral

are diameters of the circle through the vertices AD = BC [Given]
of the quadrilateral, prove that it is a rectangle.
DEA = CFB = 90°
[2010]
[DE  AB and CF  AB]
Sol. Given : ABCD is a
DE = CF [Distance between parallel
diagonals AC and BD are lines remains constant]
GOYAL BROTHERS PRAKASHAN
diameter of the circle passing  DEA  CFB [RHS axiom]
through A, B, C and D.
 A = B ...(i) [CPCT]

and, ADE = BCF ...(ii) [CPCT]

To Prove : ABCD is a Since, ADE = BCF [From (ii)]

rectangle.  ADE + 90° = BCF + 90°

Proof : In AOD and COB  ADE + CDE = BCF + DCF

AO = CO [Radii of a circle]  D = C ...(iii)

OD = OB [Radii of a circle] [ADE + CDE = D,

AOD = COB [Vertically BCF + DCF = C]

opposite angles]  A = B and C = D …(iv)

[From (i) and (iii)]

 AOD  COB [SAS axiom] A + B + C + D = 360°

 OAD = OCB [CPCT] [Sum of the angles
But these are alternate interior angles made
of a quadrilateral is 360°]

by the transversal AC, intersecting AD and BC.  2(B + D) = 360° [Using (iv)]
 AD || BC
Similarly, AB || CD.  B + D = 180°

Hence, quadrilateral ABCD is a parallelogram.  Sum of a pair of opposite angles of
Also, ABC = ADC ..(i)
quadrilateral ABCD is 180°.
[Opposite angles of a ||gm are equal]
And, ABC + ADC = 180° ...(ii)  ABCD is a cyclic trapezium Proved.

[Sum of opposite angles of a Q.9. Two circles intersect at two points B
and C. Through B, two line segments ABD and
cyclic quadrilateral is 180°] PBQ are drawn to intersect the circles at A, D
ABC = ADC = 90° [From (i) and (ii)] and P, Q respectively (see Fig.). [2011 (T-II)]

Prove that ACP = QCD.

 ABCD is a rectangle. [A ||gm one of
whose angles is 90° is a rectangle] Proved.

Q.8. If the non-parallel sides of a trapezium

are equal, prove that it is cyclic.
[2010, 2011 (T-II)]

Sol. Given : Two circles intersect at two
points B and C. Through B, two line segments
ABD and PBQ are drawn to intersect the circles
at A, D and P, Q respectively.

12

To Prove : ACP = QCD. Proof : Let O be the

Proof : ACP = ABP ...(i) [Angles in mid-point of AC.

the same segment] Then OA = OB = OC

QCD = QBD ..(ii) [Angles in = OD

the same segment] Mid point of the hypo-

But, ABP = QBD ..(iii) [Vertically tenuse of a right triangle is

opposite angles] equidistant from its vertices

By (i), (ii) and (iii), we get with O as centre and radius equal to OA, draw

ACP = QCD. Proved. a circle to pass through A, B, C and D.

Q.10. If circles are drawn taking two sides We know that angles in the same segment of
GOYAL BROTHERS PRAKASHAN
of a triangle as diameters, prove that the point a circle are equal.

of intersection of these circles lie on the third Since, CAD and CBD are angles of the

side. [2011 (T-II)] same segment.

Sol. Given : Sides AB Therefore, CAD = CBD. Proved.

and AC of a triangle ABC Q.12. Prove that a cyclic parallelogram is a

are diameters of two circles rectangle. [2010]

which intersect at D. Or

To Prove : D lies on BC. Prove that a parallelogram inscribed in a

Proof : Join AD circle is a rectangle. [2011 (T-II)]

ADB = 90° ...(i) Sol. Given : ABCD is a

[Angle in a semicircle] cyclic parallelogram.

Also, ADC = 90° ...(ii) To prove : ABCD is a

Adding (i) and (ii), we get rectangle.

ADB + ADC = 90° + 90° Proof :

 ADB + ADC = 180° ABC = ADC ...(i)

 BDC is a straight line. [Opposite angles of a ||gm are equal]

 D lies on BC But, ABC + ADC = 180° ... (ii)

Hence, point of intersection of circles lie on [Sum of opposite angles of a cyclic

the third side BC. Proved. quadrilateral is 180°]

Q.11. ABC and ADC are two right triangles  ABC = ADC = 90° [From (i) and (ii)]

with common hypotenuse AC. Prove that  ABCD is a rectangle

CAD = CBD. [2011 (T-II)] [A ||gm one of whose angles is

Sol. Given : ABC and ADC are two right 90° is a rectangle]

triangles with common hypotenuse AC. Hence, a cyclic parallelogram is a rectangle.

To Prove : CAD = CBD Proved.

OTHER IMPORTANT QUESTIONS

Q.1. In the figure, O (c) 70°
is the centre of the circle (d) 80°
with AB as diameter. If Sol. (c) OA = OC  OAC = OCA
AOC = 40°, the value  Now, OAC + OCA + AOC = 180°
of x is equal to : [Imp.]
180  40
(a) 50° 2x + 40° = 180° x = 2 = 70°
(b) 60°

13

Q.2. For what value 1
Sol. (a) x = 2 × 70°
of x in the figure, points
= 35° [Angle at the
A, B, C and D are centre is double the angle
at the circumference]
concyclic ? [2011 (T-II)]
y = x = 35°
(a) 9° [Angles in the same segment are equal]

(b) 10° Q.7. In the given
figure, if POQ is a
(c) 11° diameter of the circle
and PR = QR, then
(d) 12° RPQ is equal to :

Sol. (b) Since, opposite angles of a cyclic [2011 (T-II)]

quadrilateral are supplementary. (a) 30° (b) 60°
(c) 90° (d) 45°
 81° + x + 89° = 180° Sol. (d) PRQ = 90°
[ angle in a semicircle is a right angle]
 x = 180° – 170° = 10°.  PR = PQ
 P = Q = 45°
Q.3. In the given PRAKASHANBROTH- Q.8. ABCE is a cyclic
quadrilateral. 'O' is the
figure, O is the centre centre of the circle and
AOC = 150°, then
of the circle. If CAB CBD is : [2011 (T-II)]
(a) 225° (b) 128°
= 40° and CBA = (c) 150° (d) 75°
Sol.(d) Since the angle subtended by an arc
110°, the value of x is : at the centre of a circle is twice the angle
sustended at a point on the remaining part of the
(a) 50° circumference, we have

(b) 80° 11
AEC = 2  AOC = 2 × 150° = 75°
(c) 55° Now, ABCE is a cyclic quadrilateral whose
side AB is produce to D
(d) 60°  CBD = AEC = 75°

Sol. (d) ACB = 180° – (110° + 40°) = 30° [ ext.  of cyclic quad. = int. opp. ]
Q.9. 'O' is the centre
AOB = 2ACB [Angle at the centre is of the circle QPS = 65°;
PRS = 33°, PSQ is
twice the angle at the circumference] equal to : [2011 (T-II)]

 x = 2 × 30° = 60°. (a) 90° (b) 82°
Q.4. Angle inscribed in a semicircle is :
(c) 102° (d) 42°
[2010]

(a) 60° (b) 75° (c) 90° (d) 120°

Sol.(c) Angle in a semicircle is a right angle.

Q.5. In the given

figure, O is the centre of
L
the circle. If QPR is 50°,

A
then QOR is : [2010]

(a) 130°
Y
(b) 40°S

O
(c) 100°

(d) 50°
G R
Sol. (c) QOR = 2QPR

[The angle subtended by an arc at the
E
centre is double the angle subtended by it at

any point on the remaining part of the circle.]

= 2 × 50° = 100°

Q.6. In the given figure the value of y is :

[2011 (T-II)]

(a) 35° (b) 70° + x

(c) 70° – x (d) 140°

14

Sol. (b) R = Q = 33° [Angles in the (c) 140° (d) 110°
same segment are equal]
Sol. (c) OA = OB
Now, in PQS, P + Q + S = 180°
 PSQ = 82° OBA = OAB = 30°
Q.10. In the given
OC = OB  OBC = OCB = 40°
figure, AB is a diameter of
ABC = OBA + OBC = 30° + 40°
the circle. CD || AB and
BAD = 40°, then ACD = 70°
is : [2011 (T-II)]
Now, AOC = 2ABC = 2 × 70° = 140°
(a) 40° (b) 90°
Q.14. ABCD is a cyclic
(c) 130° (d) 140°
Sol. (c) ADB = 90° [Angle in a semicircle quadrilateral as shown in

is a right angle] the figure. The value of
BAD + ADB + ABD = 180°
 ABD = 180° – (40° + 90°) = 50° (x + y) is : [2011 (T-II)]
ACD = 130° [opposite angles of a cyclic
(a) 200° (b) 100°
E RGOSYAL(c) 180° (d) 160°
Q.11. In the given PRAKASHANBROTH-
Sol. (d) x + 90° = 180°
figure the values of x and y
 x = 90°
is : [2011 (T-II)]
y + 110° = 180°  y = 70°
(a) 20°, 30°
 x + y = 90° + 70° = 160°
(b) 36°, 60°
Q.15. Arc ABC subtends an angle of 130° at
(c) 15°, 30°
the centre O of the circle. AB is extended to P.
(d) 25°, 30°
Sol. (b) 2x + 3x = 180°  x = 36° Then CBP equals : [2010]
y + 2y = 180°  y = 60° [opp. angles of a
(a) 60° (b) 65° (c) 70° (d) 130°
cyclic quad. are supplementary]
Sol. (b) Take a point E on the remaining part
Q.12. In the given
of the circumference. Join EA and EC.
figure, if AOB is the
Since the angle subtended by an arc at the
diameter of the circle and
B = 35°, then x is equal centre of a circle is twice the angle subtended at
to : [2011 (T-II)]
a point on the remaining part of the
(a) 90° (b) 55°
circumference, we have
(c) 75° (d) 45°
Sol. (b) BCA = 90° AEC = 1 AOC = 1 × 130° = 65°
2 2
[Angle in a semicircle is a right angle]
Now, BCA + CBA + CAB = 180° Now, ABCE is a cyclic quadrilateral whose
 x = 180° – (90° + 35°) = 55°
Q.13. In the given side AB is produced to P.

figure, O is the centre of  CBP = AEC = 65°
the circle. If OAB = 30°
and OCB = 40°, then [ ext.  of a cyclic quad. = int. opp. ]
measure of AOC is :
Hence, CBP = 65°.
[2011 (T-II)]
Q.16. In the figure, O is the centre of the
(a) 70° (b) 220°
circle and ∠AOB = 80°. The value of x is :

[Imp.]

15

ADB = ACB [Angles in the

same segment are equal]

 ACB = 70°.

Q.19. In the figure , O is the centre of the

circle. If ∠ABC = 20°, then ∠AOC is equal to :

(a) 30° (b) 40° (c) 60° (d) 160°

Sol. (b) x = 1 AOB [Angle at the centre
GOYAL BROTHERS PRAKASHAN2
is double the angle at the circumference]

1 (a) 20° (b) 40° (c) 60° (d) 10°

 2 × 80° = 40°.

Q.17. In the figure, O is the centre of the Sol. (b) AOC = 2ABC [Angle at the centre
circle. If ∠OAB = 40°, then ∠ACB is equal to :
is twice the angle at the circumference]
[Imp.]
 AOC = 40°.

Q.20. In the given figure, a circle is centred

at O. The value of x is : [2010]

(a) 50° (b) 40° (c) 60° (d) 70°

Sol. (a) OA = OB  OAB = OBA = 40°

 AOB = 180° – (40° + 40°) = 100°
1
ACB = 2 AOB [Angle at the centre is (a) 55° (b) 70° (c) 110° (d) 125°

double the angle at the circumference] Sol. (c) OA = OC  OAC = OCA = 20°

 ACB = 50°. OC = OB  OCB = OBC = 35°

BCA = OCA + OCB = 20° + 35° = 55°

Q.18. In the figure , if ∠DAB = 60°, ∠ABD Now, x° = AOB = 2BCA = 2 × 55° = 110°
= 50°, then ∠ACB is equal to :
 x = 110°

Q.21. In the given figure, O is the centre of

circle, BAO = 68°, AC is diameter of circle,

then measure of BCO is : [2010]

(a) 60° (b) 50° (c) 70° (d) 80°
Sol. (c) ADB = 180° – (60° + 50°) = 70°

16

(a) 22° (b) 33° (c) 44° (d) 68° Q.24. In the figure, O is the centre of the

Sol. (a) We have, ABC = 90° circle with ∠AOB = 85° and ∠AOC = 115°.

( angle in a semicircle is a right angle) Then ∠BAC is : [Imp.]

Now, in ABC,

ABC + BAO + BCO = 180°

 90° + 68° + BCO = 180°

 158° + BCO = 180°  BCO = 22°

Q.22. In the figure, if ∠SPR = 73°, ∠SRP =

42°, then ∠PQR is equal to : [V. Imp.]

GOYAL BROTHERS PRAKASHAN(a) 65° (b) 70° (c) 74° (d) 76° (a) 115° (b) 85° (c) 80° (d) 100°
Sol. (a) PSR = 180° – (73° + 42°) = 65° Sol. (c) BOC = 360° – (85° + 115°) = 160°
PSR = PQR
1
[Angles in the same segment are equal]
 PQR = 65°. BAC = 2 BOC [Angle at the centre is
Q.23. In the figure, O is the cnetre of the
circle. If ∠OPQ = 25° and ∠ORQ = 20°, then double the angle at the circumference]
the measures of ∠POR and ∠PQR are  BAC = 80°.
respectively : Q.25. In the figure, if ∠CAB = 40° and AC

= BC, then ∠ADB equal to :

(a) 90°, 45° (b) 105°, 45° (a) 40° (b) 60° (c) 80° (d) 100°
Sol. (c) AC = BC  CBA = CAB = 40°
(c) 110°, 55° (d) 100°, 50°  ACB = 180° – (40° + 40°) = 100°
ACB + ADB = 180° [Opposite angles
Sol. (a) OP = OQ of a cyclic quadrilateral are supplementary]
 ADB = 180° – 100° = 80°.
 OQP = OPQ = 25° [Radii of same Q.26. In the given figure, ∠DAB = 70° and
∠ABD = 40°, then ∠ACB is equal to : [2010]
circle]
(a) 40° (b) 70° (c) 110° (d) 30°
Similarly, OQR = 20°

 PQR = 25° + 20° = 45°

Also, POR = 2 PQR [Angle at the

centre is double the angle at the circumference]

 POR = 2 × 45° = 90°.

17

Sol. (b) ADB = 180° – (70° + 40°) 1
= 180° – 110° = 70° and ACB Sol. (d) ACB = 2 AOB [Angle at the
= ADB = 70°. centre is double the angle at the circumference]

[Angles in the same segment are equal]  ACB = 45°
Q.27. In the given
figure, if ∠AOC = 130°, O OA = OB  OAB = OBA
is the centre of the circle 180  90
then ∠ABC is : [2010]  OAB = OBA = 2 = 45°
(a) 65° (b) 115°
Now, CAB = 180° – (45° + 30°) = 105°
(c) 130° (d) 50°
 CAO = CAB – OAB
11
= 105° – 45° = 60°.
Sol. (b) AEC = 2 AOC = 2 × 130° = 65°
Now, ABC = 180° – 65° = 115°. Q.31. In the figure,
Q.28. In the given
figure, the value of BC is a diameter of the
(∠BCD + ∠DEB) is :
circle and ∠BAO = 60°.
[2010]
Then ∠ADC is equal to :
(a) 270° (b) 180°
(c) 360° (d) 90° (a) 30° (b) 45°
PRAKASHABNROTH-
Sol. (b) Opposite angles of a cyclic (c) 60° (d) 120°
Sol. (c) OA = OB
Q.29. In the given  OBA = OAB
figure, ∠C = 40° ∠CEB =
105°, then the value of x is : = 60°
[2010]
[Angles in the same segment are equal]
(a) 50° (b) 35°  ADC = 60°.
(c) 20° (d) 40° Q.32. In the given
Sol. (b) D = C = 40°
figure, O is the centre of
[Angles in the same segment] a circle and ∠BOA =
and AED = 105° 90°, ∠COA = 110°. Find
the measure of ∠BAC.
[Vertically opposite angles]
x = 180° – (105 + 40°) = 180° – 145° = 35° [2011 (T-II)]
Q.30. In the figure, ∠AOB = 90° and
∠ABC = 30°, then ∠CAO is equal to : [V. Imp.]L Sol. We have,

(a) 30° (b) 45° (c) 90° (d) 60°YAS ∠BOA = 90° and ∠AOC = 110°
∴ ∠BOC = 360° – (∠BOA + ∠AOC)
GO ⇒ ∠BOC = 360° – (90° + 110°) = 160°
R
11
Now, ∠BAC = 2 ∠BOC = 2 × 160° = 80°

Q.33. In the figure,

E ABC is an equilateral

triangle. Find ∠BDC and

∠BEC. [2010]

Sol. BAC = 60°
[ ABC is equilateral]

BAC = BDC

[Angle in the same segment are equal]

18

 BDC = 60° In s ABD and ACE, we have

Now, BAC + BEC = 180° [Sum of AB = AC [Given]

opposite angles of a cyclic quadrilateral is 180°] ABD = ACE [Angles in the same segment]

 BEC = 180° – 60° = 120°. BD = CE [Given]

Q.34. In the given figure, two circles centred So, by SAS congruence criterion, we have

at C1 and C2 are intersecting at P and Q. If PR ABD = ACE  AD = AE
and PS are diameters, show that RQS is a
Q.37. In the figure, if AOB is a diameter and

straight line. [2010] ∠ADC = 120° find ∠CAB. [V. Imp.]

Sol. We are

given two circles

with centres C1 and
C2 which intersect
each other at P and

Q. PR is a diameter BROTHERS
PSRHKNA
of circle C1 and PS is a diameter of circle C2. Sol. Join AD and AC.
We need to prove that R, Q and S are

collinear. ADC + ABC

 PQR = 90° …[Angle in a semicircle] = 180° [Opposite angles

and PQS = 90° …[Angle in a semicircle] of a cyclic quadrilateral]

 PQR + PQS = 90° + 90° = 180°  ABC = 60°

 RQS is a straight line. ACB = 90°

Q.35. In the given figure, two circles with [Angle in a semicircle

centre O and O intersect at two points A and B. is 90°]

If AD and AC are diameters to circles then prove  CAB = 90° – ABC = 90° – 60° = 30°.

that B lies on the line segment DC. [2010] Q.38. In the given figure, O is the centre of

Sol. Join AB. the circle, BD = DC and DBC = 30°. Find the

ABD = 90° measure of BAC. [2011 (T-II)]

…[Angle in a

semicircle]

ABC = 90°

…[Angle in a

semicircle]

So, ABD + ABC = 90° + 90° = 180°

Therefore, DBC is a line. That is B lies on
G O YA L
the line segment DC. Sol. BD = DC  DCB = DBC = 30°
BDC = 180° – (30° + 30°) = 120°
Q.36. In the given figure, D is a point on the BDC + BAC = 180° [Opposite angles
of a cyclic quadrilateral are supplementary]
circumcircle of ABC  BAC = 180° – 120° = 60°.
Q.39. In the given figure,
in which AB = AC. If
ABCD is a cyclic quadri-
CD is produced to lateral, if BCD = 120° and
ABD = 50°, then find ADB.
point E such that
[2010, 2011 (T-II)]
BD = CE, prove that

AD = AE. [2010]

Sol. We have,

AB = AC and

CE = BD

19

Sol. We have, BCD + BAD = 180° Q.43. In the given figure, if O is the centre of
[Sum of the opp. angles of a cyclic quad.]
 BAD = 180° – 120° = 60° the circle and AOC = 110° and AB is produced
In ABD, ADB = 180° – (50° + 60°)
to D then find AEC and ABC. [2010]
[Angle sum property of a triangle]

= 180° – 110° = 70°.
Q.40. In the given figure, O is the centre of
the circle and ABD = 35°. Find the value of x.

[2010]

GOYAL BROTHERS PRAKASHAN Sol. Since arc ABC makes AOC = 110° at

the centre and AEC at a point E on the

circumference.

Sol. In ABD, ABD = 35° given and 11

BAD = 90° [Angle inscribed in a semicircle]  AEC = 2 AOC = 2 × 110° = 55°
Again arc CEA makes angle reflex AOC

 ADB = 180° – (90° + 35°) = (360° – 110°) = 250° at the centre and ABC

[Angle sum property in a triangle] at a point B on the circumference.

 ADB = 180° – 125° = 55°  ABC = 1 reflex BOD = 1 × 250°
2 2
Now, x = ACB = ADB = 55°
= 125°.
[Angles in the same segment]
Q.44. In the given figure, if
Q.41. In the given figure, O is the centre of
O is the centre of the circle,
the circle if ABO = 45° and ACO = 35°, then
AOC = 50° and COB =
find BOC. [2010]
30°. Find the measure of

Sol. We have, BOC = 30°

and AOC = 50°

AOB = BOC + AOC

= 30° + 50° = 80°

Sol. OA = OB  OAB = OBA = 45° We know that angle subtended by an arc at
OC = OA  OAC = OCA = 35°
BAC = OAB + OAC = 45° + 35° = 80° the centre of a circle is double the angle
Now, BOC = 2BAC = 2 × 80° = 160°
subtended by the same arc on the remaining part
Q.42. In the given figure,
of the circle.
AC is the diameter of the  2ADB = AOB
circle. If ACB = 55°, then
find the value of x. [2010] 11

Sol. In ABC,  ADB = 2 AOB = 2 × 80° = 40°.
ABC = 90° Q.45. In the given figure, if AB = AC, BEC

[Angle in a semicircle] = 100°, then find the values of x and y. [2010]

and ACB = 55° [Given]
 BAC = 180° – (90° + 55°) = 35°
Now, x = BDC = BAC = 35°

[Angles in the same segment]

20

Sol.Since ABEC is a cyclic quadrilateral. and ECD = 180° – 150° = 30°
 BAC + BEC = 180°
 BAC = 180° – 100° = 80° [Linear pair]
Now, BAC = BDC = y = 80°
 DEC = 180° – (110° + 30°)
[Angles in the same segment]
and since ABC is an isosceles triangle. = 180° – 140° = 40°
 ABC = ACB = x
 2x + 80° = 180° Q.48. In the given figure, ABCD is a cyclic

[Angle sum property of a triangle] quadrilateral in which AB is produced to F and
 2x = 180° – 80° = 100°  x = 50°
Hence, x = 50° and y = 80°. BE || DC. If FBE = 20° and DAB = 95°, then
Q.46. In the given figure, AB is diameter of
the circle with centre O and CD || AB. If CAB find ADC. [2010, 2011 (T-II)]
= 25°, then find the measure of CAD. [2010]
GOYAL BROTHERS PRAKASHANSol. Since ABCD is a cyclic quadrilateral.
Sol. In ABC, C = 90°, because angle in a
semicircle.  BAD + BCD = 180°

 ABC = 180° – (90° + 25°) = 65°  95° + BCD = 180°  BCD = 85°
and ADC = 180° – 65° = 115° [Opposite
angles of a cyclic quad. are supplementary] Since, BE || DC therefore, BCD = CBE
Since, CD || AB, therefore,
BAC = ACD = 25° [Alternate angles]  CBE = 85° [Alternate angles]
Now, in ACD, CAD
= 180° – (115° + 25)° = 180° – 140° = 40° Now, BCF = CBE + EBF
Q.47. In the given figure, ABCD is a cyclic
quadrilateral with opposite sides AD and BC = 85° + 20° = 105°
produced to meet at the point E. If DAB = 30°
and ABC = 110°, then find DEC. [2010] Now, since exterior angle formed by

producing a side of a cyclic quadrilateral, is

equal to the interior opposite angle.

 ADC = BCF = 105°.

Q.49. In the given figure, O is the centre of

the circle. If D = 130°, then find BAC.

[2011 (T-II)]

Sol. Since the opposite angles of a cyclic Sol. Since ABCD is a cyclic quadrilateral.
 ADC + ABC = 180°
 BCD = 180° – 30° = 150°  130° + ABC = 180°
and ADC = 180° – 110° = 70°  ABC = 50°
Now, EDC = 180° – 70° = 110° Since ACB is the angle in a semi-circle.
 ACB = 90°
[Linear pair] Now, in ABC, we have
BAC + ACB + ABC = 180°
21  BAC + 90° + 50° = 180°
 BAC = 40°

Q.50. In the given figure, AB is the diameter Sol. Using AB as
of the circle with centre O. If BAD = 70° and diameter, draw a circle
DBC = 30°. Determine ABD and CDB. which passes through A,
D, B and C.
[2011 (T-II)]
BAC = BDC
[Angles in the same

segment are equal]

Sol. Since ABCD is a cyclic quadrilateral. Q.53. Prove that angle bisector of any angle
of a triangle and the perpendicular bisector of
 BCD + BAD = 180° the opposite side if intersect, they will intersect
on the circumcircle of the triangle. [HOTS]
 BCD + 70° = 180°
Sol.
 BCD = 110°

In BCD, we have,

CBD + BCD + BDC = 180°

 30° + 110° + BDC = 180°

 BDC = 40°

Since ADB is the angle in a semicircle.

 ADB = 90°
GOYAL
PBRRAOTKAHSERHSANIn ABD, we have

ABD + ADB + BAD = 180°

 ABD + 90° + 70° = 180° ABC is the given triangle and O is the centre

 ABD = 20° of its circumcircle. Then the perpendicular

Hence, ABD = 20° and BDC = 40° bisector of BC passes through O. It cuts the

Q.51. If a line is drawn parallel to the base circle at P.
BOC = A
of an isosceles triangle to intersect its equal … (i)

sides, prove that the quadrilateral so formed is [Angle at the centre

cyclic. [HOTS] is twice the angle at the circumference]

Sol. AB = AC  ABC = ACB …(i) OB = OC [Radii of the same circle]

ADE = ABC and OD  BC

[Corresponding  BOD = COD = 1 BOC
angles] 2

 ADE = ACB  BOD = COD = A [From (i)]
[From (i)]
Now, CP makes A at the centre O. So, it

 ADE + EDB A
= ACB + EDB
will make 2 at A.

 ACB + EDB = 180° Or, CAP = A
2
[ ADE and EDB form a linear pair]
 AP is the bisector of A.
 BCED is cyclic.
[ Sum of a pair of opposite angles is 180°] Q.54. In the given figure, two circles

Q.52. On a common hypotenuse AB, two intersect each other at C and D. If ADE and BCF
right triangles ACB and ADB are situated on
opposite sides. Prove that ∠BAC = ∠BDC. are straight lines intersecting circles at A, B, F

[HOTS] and E. Prove that AB || EF. [2010]

22

Side BD of the cyclic

produced to A.

ADE = ACB ...(ii)

[ext. ADE = int. opp. C]

From (i) and (ii), we get

Sol. In order to prove that AB || EF, it is But, these are corres-

sufficient to prove that ponding angles.

1 + 3 = 180° Hence, DE || BC.

Since, ADCB is a cyclic quadrilateral. Q.57. In the given figure, PQ and RS are two

parallel chords of a circle. When produced RP
GOYAL BROTHERS PRAKASHAN1 + 2 = 180° …(i)

Now, DEFC is a cyclic quadrilateral and in a and SQ meet at O. Prove that OP = OQ.

cyclic quadrilateral an exterior angle is equal to [2010]

the opposite interior angle. Sol. We have, PQ || RS

 2 = 3 …(ii)  OPQ = ORS and

From equations (i) and (ii), we get OQP = OSR

1 + 3 = 180° [Corresponding angles]

Hence, AB || EF. But, PQRS is a cyclic

Q.55. AOB is a diameter of the circle and C, quadrilateral.
D, E are any three points on the semicircle. Find
the value of ACD + BED. [2011 (T-II)]  OPQ = OSR and

OQP = ORS

 OPQ = OQP

Thus, in OPQ, we have

OPQ = OQP

 OP = OQ. Proved.

Q.58. Prove that the circle drawn on any one

of the equal sides of an isosceles triangle as

diameter bisects the base of the triangle.

[2010, 2011 (T-II)]

Sol. Join BC. Sol. Given : A ABC in which AB = AC and

Then, ACB = 90° [angle in a semicircle] a circle is drawn by taking AB as diameter which

Now, DCBE is a cyclic quadrilateral. intersects the side BC of triangle at D.

 BCD + DEB = 180° To prove : BD = DC

ACB + BCD + DEB = 90° + 180° Construction : Join AD

[ ACB = 90°] Proof : Since angle in

 ACB + DEB = 270° a semicircle is a right

[ ACB + BCD = ACD] angle. Therefore,

Q.56. In an isosceles ABC with AB = AC, ADB = 90°

a circle passing through B and C intersects the ADB + AD

sides AB and AC at D and E respectively. Prove = 180°

that DE || BC. [2011 (T-II)]  90° + ADC = 180°

Sol. AB = AC  ADC = 90°

 ABC = ACB .....(i) Now, in ABD and ACD, we have

AB = AC [Given]

23

ADB = ADC [Each equal to 90°] BCE = 90° and CBE = CBD = 30°
 BCE + CBE + CEB = 180°
and, AD = AD [Common]  90° + 30° + CEB = 180°
 CEB = 60°  AEB = 60°.
 ABD  ACD [By RHS congruence] Q.61. In the given figure, P is the centre of
the circle. Prove that :
 BD = DC. XPZ = 2(XZY + YXZ) [2011 (T-II)]

Q.59. If O is the centre of a circle as shown Sol. Since arc XY subtends XPY at the
centre and XZY at a point Z in the remaining
in given figure, then prove that x + y = z. part of the circle.

[2011 (T-II)] XPY = 2XZY ...(i)
Similarly, arc YZ subtends YPZ at the
Sol. In ACF, side CF is centre and YXZ at a point Y in the remaining
part of the circle.
produced to B.  YPZ = 2YXZ ...(ii)
Adding (i) and (ii), we get
 y = 1 + 3 ...(i) XPY + YPZ = 2XZY + 2YXZ
GOYAL BROTHERS PRAKASHAN  XPZ = 2(XZY + YXZ)
[ext.  = sum of int. Q.62. If O is the circumcentre of a ABC
and OD  BC, prove that BOD = A
opp. angles]
[2011 (T-II)]
In AED, side ED is

produced to B.

 1 + x = 4 ...(ii)

From (i) and (ii), we have

1 + x + y = 1 + 3 + 4

 x + y = 3 + 4 = 23

[ 4 = 3, angles in the same segment]

= z [ AOB = 2ACB]

Hence, x + y = z.

Q.60. In the given figure, AB is diameter of

circle and CD is a chord equal to the radius of

circle. AC and BD when extended intersect at a

point E. Prove that AEB = 60°. [2011 (T-II)]

Sol. Join OC, OD and BC. Sol. Join OB and OC.

In triangle OCD, we have In OBD and OCD, we have

OC = OD = CD [Each equal to radius] OB = OC
 OCD is equilateral.
 COD = 60°

1 [Each equal to the radius of circumcircle]
2
Now, CBD = COD  CBD = 30° ODB = ODC [Each equal to 90°]

Since ACB is angle in a semi-circle. and, OD = OD [Common]

 ACB = 90°  OBD  OCD

BCE = 180° – ACB = 180° – 90° = 90°  BOD = COD

Thus, in BCE, we have  BOC = 2BOD = 2COD

Now, arc BC subtends BOC at the centre

24

and BAC = A at a point in the remaining part Reflex AOB = 2ACB, when AB is a
of the circle. major arc.

 BOC = 2A Construction : Join AB and CO. Produce
2BOD = 2A [ BOC = 2BOD]
 BOD = A. CO to a point D outside the circle.
Q.63. Prove that the angle subtended by an
We know that when one side of a
arc at the centre is double the angle subtended
by it at any point on the remaining part of the triangle is produced then the exterior angle so

circle. [2010] formed is equal to the sum of the interior

Sol. Given : A circle C(O, r) in which arc opposite angles.
 AOD = OAC + OCA
AB subtends AOB at the BOD = OBC + OCB
But, OAC = OCA [ OC = OA = r].
centre and ACB at any and OBC = OCB [ OC = OB = r].
 AOD = 2OCA and BOD = 2OCB
point C on the remaining Now, AOD + BOD = 2OCA + 2OCB
 AOB = 2(OCA + OCB)
part of the circle.  AOB = 2ACB.
GOYAL
PRABKRAOSTHHEARNSTo prove : AOB =

2ACB, when AB is a
minor arc or a semicircle.

PRACTICE EXERCISE 10B

1 Mark Questions (d) 40° 4. In the figure, O is
the centre of the circle and
1. In the given figure, ∠PQR = 100°.
if O is the centre of the
circle and A is a point on Then the reflex ∠POR
the circle such that CBA is :
= 40° and AD  BC, then
the value of x is (a) 280° (b) 200°

(a) 50° (b) 90° (c) 45° (c) 260° (d) none of these
2. In the given figure, if 5. In the given figure,
O is the centre of the circle E is any point in the inte-
and A is a point on the circle rior of the circle with cen-
such that BOA = 120°, tre O. Chord AB = Chord
then the value of x is AC. If ∠OBE = 20°,then the
value of x is :
[2011 (T-II)]
(a) 40° (b) 45°
(a) 120° (b) 60°
(c) 90° (d) 30° (c) 50° (d) 70°

3. In the given figure, O 6. In the figure, ∠ABC
is the centre of the circle. If = 79°, ∠ACB = 41°, then
AOB = 160°, then ACB ∠BDC is :
is [2011 (T-II)]
(a) 41° (b) 79°
(a) 160° (b) 200°
(c) 80° (d) 100° (c) 60° (d) 50°

25

2 Marks Questions 12. In the given figure,
A, B, C and D are four points
7. In the given figure, on the circle. AC and BD
intersect at a point E such
ABC = 45°, prove that that BEC = 130° and
ECD = 20°. Find BAC.
OA  OC. [2011 (T-II)]
3 Marks Questions
8. In the given figure, ABCD is a cyclic quad- [2011 (T-II)]

rilateral and ABC = 85°. Find the measure of

9. In the figure, O is theG 13. ABC is an isosceles triangle in which AB
centre of the circle and BACO = AC. A circle passing through B and C intersects
= 60°. Find the value of x.YA AB and AC at D and E respectively. Prove that
L BC || DE.
[2011 (T-II)]PRAKBARSOHTAHNERS
14. O is the circumcentre of the triangle ABC
10. O is the centre of the and D is the mid-point of the base BC. Prove that
∠BOD = ∠A.
circle as shown in figure. Find
15. A quadrilateral ABCD is inscribed in a
CBD. [2011 (T-II)] circle such that AB is a diameter and ∠ADC = 130°.
Find ∠BAC.

16. If two sides of a cyclic quadrilateral are
parallel, prove that remaining two sides are equal
and both diagonals are equal.

4 Marks Questions

17. In the figure, O is
the centre of the circle. If
BD = OD and CD  AB,
find ∠CAB. [HOTS]

11. ABCD is a cyclic 18. Prove that the angles in a segment greater
quadrilateral and AB = AC if than a semi-circle is less than a right angle.
ACB = 70°, find BDC.
[HOTS]
[2011 (T-II)]

TEXTBOOK’S EXERCISE 10.6 (OPTIONAL)

Q.1. Prove that the line of centres of two To prove : OAO = OBO

intersecting circles subtends equal angles at the Construction : Join AO, BO, AO and BO.

two points of intersection. [2011 (T-II)] Proof : In AOO and BOO, we have

Sol. Given : Two AO = BO [Radii of the same circle]
intersecting circles, in which
OO is the line of centres and AO = BO [Radii of the same circle]
A and B are two points of intersection.
OO = OO [Common]

 AOO  BOO [SSS axiom]

26

 OAO= OBO [CPCT] Substituting x =1 in (i), we get

Hence, the line of centres of two intersecting r2 = (6 – x)2 +  5 2  r2 = (6 – 1)2 +  5 2
 2   2 
circles subtends equal angles at the two points of

intersection. Proved.  5 2 25
 2  4
Q.2. Two chords AB and CD of lengths 5 cm  r2 = (5)2 + = 25 +

and 11 cm respectively of a circle are parallel to 125 55
4 2
each other and are on opposite sides of its  r2 = r=

centre. If the distance between AB and CD is 55

6 cm, find the radius of the circle. [2010] Hence, radius r = 2 cm.
Q.3. The lengths of two parallel chords of a
GOYAL BROTHERS PRAKASHAN
circle are 6 cm and 8 cm. If the smaller chord is

at distance 4 cm from the centre, what is the

distance of the other chord from the centre?

Sol. Let O be the centre of the circle and let [2010]

its radius be r cm.

Draw OM  AB and OL  CD.

15

Then, AM = 2 AB = 2 cm

and, CL = 1 11 cm Sol. Let PQ and RS be two parallel chords
2
2 CD =

Since, AB || CD, it follows that the points O, of a circle with centre O.

L, M are collinear and therefore, LM = 6 cm. We have, PQ = 8 cm and RS = 6 cm.

Let OL = x cm. Then OM = (6 – x) cm Draw perpendicular bisector OL of RS

Join OA and OC. Then OA = OC = r cm. which meets PQ in M. Since, PQ || RS,

Now, from right-angled OMA and OLC, therefore, OM is also perpendicular bisector of

we have PQ.

OA2 = OM2 + AM2 and OC2 = OL2 + CL2 1

[By Pythagoras Theorem] Also, OL = 4 cm and RL = 2 RS
 RL = 3 cm
 5 2
 r2 = (6 – x)2 +  2  ...(i) and 1  PM

 11 2 and PM = 2 PQ = 4 cm
 2 
 r2 = x2 + ... (ii) In ORL, we have

 5 2  11 2 OR2 = RL2 + OL2 [Pythagoras theorem]
 2   2 
 (6 – x)2 + = x2 +  OR2 = 32 + 42 = 9 + 16

 OR2 = 25  OR = 25

[From (i) and (ii)]  OR = 5 cm

 36 + x2 – 12x + 25 = x2 + 121  OR = OP [Radii of the circle]
4 4
 OP = 5 cm
121 25
 – 12x = 4 – 4 – 36 Now, in OPM

96 OM2 = OP2 – PM2 [Pythagoras theorem]
4
 – 12x = – 36  – 12x = 24 – 36  OM2 = 52 – 42 = 25 – 16 = 9

 – 12x = – 12  x = 1 OM = 9 = 3 cm

27

Hence, the distance of the other chord from z

the centre is 3 cm.  OEB = 90° + 2 ... (v)

Q.4. Let the vertex of an angle ABC be y

located outside a circle and let the sides of the Also, OED = ODE = 90° – 2 ... (vi)
from (iv), (v) and (vi), we have
angle intersect equal chords AD and CE with the

circle. Prove that  ABC is equal to half the z  y 
2  2 
difference of the angles subtended by the chords BDE = BED = 90° + – 90 

AC and DE at the centre. [HOTS]

  BDE = BED = yz
2
GOYAL BROTHERS PRAKASHAN
  BDE = BED = y + z ... (vii)
... (viii)
  BDE = 180° – (y + z)

  ABC = 180° – (y + z)

Now, y z  360  y  2z  y
2 2
Sol. Given : Two equal chords AD and CE
= 180° – (y + z) ... (ix)
of a circle with centre O. When meet at B when
From (viii) and (ix), we have
produced.
xy
1 ABC = 2 Proved.

To Prove : ABC = 2 (AOC – DOE) Q.5. Prove that the circle drawn with any
Proof : Let AOC = x, DOE = y, AOD = z
side of a rhombus as diameter, passes through
EOC = z
the point of intersection of its diagonals.
[Equal chords subtends equal angles at the

centre]

x + y + 2z = 360° .. (i) [Angle at a point]
OA = OD  OAD = ODA

 In OAD, we have
OAD + ODA + z = 180°

2OAD = 180° – z [ OAD = OBA]

 OAD = 90° – z ... (ii) Sol. Given : A rhombus ABCD whose
2
diagonals intersect each other at O.

Similarly, OCE = 90° – z ... (iii) To prove : A circle with AB as diameter
2
passes through O.

 ODB = OAD +ODA Proof : AOB = 90° [Diagonals of a

[Exterior angle property] rhombus bisect each other at 90°]

 OEB = 90° – z +z [From (ii)]  AOB is a right triangle right angled at O.
2
 AB is the hypotenuse of right AOB.

z  If we draw a circle with AB as diameter,

 ODB = 90° + 2 ... (iv) then it will pass through O because angle in a
Also, OEB = OCE + COE
semicircle is 90° and AOB = 90° Proved.
[Exterior angle property]
Q.6. ABCD is a parallelogram. The circle

 OEB = 90° – z + z [From (iii)] through A, B and C intersect CD (produced if
2
necessary) at E. Prove that AE = AD.

[2011 (T-II)]

28

Sol. In order to prove that AE = AD it is OA = OC [Given]
sufficient to prove that AED = ADE.
OB = OD [Given]

AOB = COD [Vertically opposite angles]

 AOB  COD [SAS congruence]

 ABO = CDO and BAO = DCO
[CPCT]

  AB || DC ... (i)

Since ABCE is a cyclic quadrilateral. Similarly, we can prove BC || AD ... (ii)

GOYAL BROTHERS PRAKASHAN AED + ABC = 180°…(i) Hence, ABCD is a parallelogram.

Now, CDE is a straight line But ABCD is a cyclic parallelogram.

 ADE + ADC = 180° …(ii)  ABCD is a rectangle.

But, ADC and ABC are opposite angles [Proved in Q.12 of textbooks exercise 10.5]
of a parallelogram.
 ABC = 90° and BCD = 90°

 ADC = ABC  AC is a diameter and BD is a diameter.
[Angle in a semicircle is 90°] Proved.

 ABC + ADE = 180° …(ii) Q.8. Bisectors of angles A, B and C of a

From equations (i) and (ii), we get triangle ABC intersect its circumcircle at D, E

AED + ABC = ADE + ABC and F respectively. Prove that the angles of the
11
 AED = ADE triangle DEF are 90° – A, 90° – B and
2 2
Thus, in AED, we have 1
90° – 2 C. [HOTS]
Sol. Given : ABC and its circumcircle.
 AE = AD
AD, BE, CF are bisectors of A, B, C

Q.7. AC and BD are chords of a circle which respectively.
bisect each other. Prove that (i) AC and BD are
diameters, (ii) ABCD is rectangle.

Sol. Given : A circle with chords AB and Construction : Join DE, EF and FD.
CD which bisect each other at O.
Proof : We know that angles in the same
To Prove : (i) AC and BD are diameters
segment are equal.
(ii) ABCD is a rectangle.
 5 = C and 6 = B ..(i)
Proof : In OAB and OCD, we have 2 2

29

1 = A and 2 = C ...(ii) But the circles are congruent.
2 2
arc ADB = arc AEB APB = AQB

A B [Equal arcs subtend equal angles]
2 2
4 = and 3 = ...(iii)  BP = BQ

From (i), we have [Sides opposite to equal angles are equal]

C B Proved.
2 2
5 + 6 = + Q.10. In any triangle ABC, if the angle

C B bisector of A and perpendicular bisector of BC
2 2
 D = + ...(iv) intersect, prove that they intersect on the

[ 5 + 6 = D] circumcircle of the triangle ABC. [HOTS]
GOYAL BROTHERS PRAKASHAN
But A + B + C = 180°

 B + C = 180° – A

 B + C = 90° – A
2 2 2

 (iv) becomes, D = 90° – A .
2

Similarly, from (ii) and (iii), we can prove Sol. Let angle bisector of A intersect

that E = 90° – B and F = 90° – C circumcircle of ABC at D.
2 2
Join DC and DB.

Proved. BCD = BAD

Q.9. Two congruent circles intersect each [Angles in the same segment]

other at points A and B. Through A any line 1

segment PAQ is drawn so that P, Q lie on the two  BCD = BAD = 2  A …(i)

circles. Prove that BP = BQ. [2011 (T-II)] [AD is bisector of A]

1

Similarly, DBC = DAC = 2  A ... (ii)
From (i) and (ii) DBC = BCD

  BD = DC

[Sides opposite to equal angles are equal]

Sol. Given : Two congruent circles which  D lies on the perpendicular bisector
intersect at A and B. PAB is a line through A.
of BC.
To Prove : BP = BQ.
Construction : Join AB. Hence, angle bisector of A and
Proof : AB is a common chord of both the
circles. perpendicular bisector of BC intersect on the

circumcircle of ABC. Proved.

B. FORMATIVE ASSESSMENT

Activity-1

Objective : To verify that the angle subtended by an arc at the centre of a circle is twice the angle
subtended by the same arc at any other point on the remaining part of the circle, using the
method of paper cutting, pasting and folding.

Materials Required : White sheets of paper, tracing paper, a pair of scissors, gluestick, colour pencils,
geometry box, etc.

30

Procedure :
1. On a white sheet of paper, draw a circle of any convenient
radius with centre O. Mark two points A and B on the boundary

of the circle to get arc AB. Colour the minor arc AB green.

GOYAL BROTHERS PRAKASHAN Figure-1

2. Take any point P on the remaining part
of the circle. Join OA, OB, PA and PB.

Figure-2

3. Make two replicas of APB using tracing
paper. Shade the angles using different

colours.

Figure-3

4. Paste the two replicas of APB adjacent Figure-4
to each other on AOB as shown in the
figure.

31

GOYAL BROTHERS PRAKASHANObservations :
1. In figure 2, AOB is the angle subtended by arc AB at the centre and APB is the angle
subtended by arc AB on the remaining part of the circle.
2. In figure 3, each angle is a replica of APB.
3. In figure 4, we see that the two replicas of APB completely cover the angle AOB.
So, AOB = 2APB.

Conclusion : From the above activity, it is verified that the angle subtended by an arc at the centre of a
circle is twice the angle subtended by the same arc at any other point on the remaining
part of the circle.

Do Yourself : Verify the above property by taking three circles of different radii.
Activity-2

Objective : To verify that the angles in the same segment of a circle are equal, using the method of
paper cutting, pasting and folding.

Materials Required : White sheets of paper, tracing paper, a pair of scissors, gluestick, colour pencils,
geometry box, etc.

Procedure :
1. On a white sheet of paper, draw a circle of any
convenient radius. Draw a chord AB of the
circle.

Figure-1

2. Take any three points P, Q and R on the major arc AB of the circle. Join A to P, B to P, A to Q, B
to Q, A to R and B to R.

Figure-2

32

GOYAL BROTHERS PRAKASHAN3. On a tracing paper, trace each of the angles APB, AQB and ARB. Shade the traced copies using
different colours.

Figure-3

4. Place the three cut outs one over the other such that the vertices P, Q and R coincide and PA, QA
and RA fall along the same direction.

Figure-4

Observations :
1. In figure 2, APB, AQB and ARB are the angles in the same major segment AB.
2. In figure 4, we see that APB, AQB and ARB coincide.
So, APB = AQB = ARB

Conclusion : From the above activity, it is verified that the angles in the same segment of a circle are
equal.

Do Yourself : Verify the above property by taking three circles of different radii.
Activity-3

Objective : To verify using the method of paper cuting, pasting and folding that
(a) the angle in a semi circle is a right angle
(b) the angle in a major segment is acute
(c) the angle in a minor segment is obtuse.

Materials Required : White sheets of paper, tracing paper, cut out of a right angle, colour pencils, a
pair of scissors, gluestick, geometry box, etc.

33

GOYAL BROTHERS PRAKASHANProcedure : (a) To verify that the angle in a semicircle is a right angle :
1. On a white sheet of paper, draw a circle of any convenient radius with centre O. Draw its
diameter AB as shown.

Figure-1

2. Take any point P on the semicircle. Join A to P
and B to P.

Figure-2

3. Make two replicas of APB on tracing paper. Shade the replicas using different colours.

Figure-3

4. On a white sheet of paper, draw a straight line XY. Paste the replicas obtained in figure 3 on
XY and adjacent to each other such that AP and BP coincide as shown in the figure.

Figure-4

34

(b) To verify that the angle in a major segment is acute :

1. On a white sheet of paper, draw a circle of any
convenient radius with centre O. Draw a chord
AB which does not pass through O.

Figure 5

2. Take any point P on the major segment. Join P to A and P to B.
GOYAL BROTHERS PRAKASHAN
3. Trace APB on a tracing paper. Figure-6

Figure-7

4. Paste the traced copy of APB on the cut out of a right angled triangle XYZ, right-angled at Y
such that PA falls along YZ.

Figure-8

35

(c) To verify that the angle in a minor segment is obtuse :

1. On a white sheet of paper, draw a circle of any convenient
radius with centre O. Draw any chord AB which does not pass
through O.

2. Take any point P on the minor segment. Join P to A and P to B.

GOYAL BROTHERS PRAKASHAN Figure-9

Figure-10

3. Trace APB on a tracing paper.

Figure-11

4. Paste the traced copy of APB on the cut out of a right-angled triangle XYZ, right angled at Y,
such that PA falls along YZ.

Figure-12

Observations :
1. In figure 2, APB is a semicircle. So,APB is an angle in a semicircle.
2. In figure 4, we see that PB and PA fall along XY.

36

GOYAL BROTHERS PRAKASHANOr APB + APB = a straight angle = 180°
2APB = 180°
APB = 90°
Hence, angle in a semicircle is a right angle.
3. In figure 7, APB is an angle formed in the major segment of a circle.
4. In figure 8, we see that the side PB of APB lies to the right of XY of XYZ,
ie, APB is less than a right angle, or PB is acute.
Hence, the angle in a major segment is acute.
5. In figure 11, APB is an angle formed in the minor segment of a circle.
6. In figure 12, we see that the side PB of PAB lies to the left of XY of XYZ
ie, APB is greater than XYZ or APB is obtuse.
Hence, the angle in a minor segment is obtuse.
Conclusion : From the above activity, it is verified that :
(a) the angle in a semicircle is a right angle.
(b) the angle in a major segment is acute.
(c) the angle in a minor segment is obtuse.

Activity-4
Objective : To verify using the method of paper cutting, pasting and folding that

(a) the sum of either pair of opposite angles of a cyclic quadrilateral is 180°
(b) in a cyclic quadrilateral the exterior angle is equal to the interior opposite angle.
Materials Required : White sheets of paper, tracing paper, colour pencils, a pair of scissors, gluestick,

geometry box, etc.
Procedure :
(a) 1. On a white sheet of paper, draw a circle of any convenient radius. Mark four points P, Q, R, S on

the circumference of the circle. Join P to Q, Q to R, R to S and S to P.

Figure-1

37

2. Colour the quadrilateral PQRS as shown in the figure and cut it into four parts such that each
part contains one angle, ie, P, Q, R and S.

GOYAL BROTHERS PRAKASHAN
Figure-2

3. On a white sheet of paper, paste P and R adjacent to each other. Similarly, paste Q and S
adjacent to each other.

Figure-3

(b) 1. Repeat step 1 of part (a).
2. Extend PQ to PT to form an exterior angle RQT. Shade RQT.

Figure-4 Figure-5

3. Trace PSR on a tracing paper and colour it.

38

4. Paste the traced copy of PSR on RQT such that S falls at Q and SP falls along QT.

GOYAL BROTHERS PRAKASHAN
Figure-6

Observations :
1. In figure 2, P, Q, R and S are the four angles of the cyclic quadrilateral PQRS.
2. In figure 3(a), we see that R and P form a straight angle and in figure 3(b), Q and S form
a straight angle.
So, P + R = 180° and Q + S = 180°.
Hence, the sum of either pair of opposite angles of a cyclic quadrilateral is 180°.
3. In figure 5, PSR is the angle opposite to the exterior angle RQT.
4. In figure 6, we see that PSR completely covers TQR.
Hence, in a cyclic quadrilateral the exterior angle is equal to the interior opposite angle.

Conclusion : From the above activity, it is verified that
(a) the sum of either pair of opposite angles of a cyclic quadrilateral is 180°.
(b) in a cyclic quadrilateral, the exterior angle is equal to the interior opposite angle.