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Question Bank In Mathematics Class IX (Term II)
10
CIRCLES
A. SUMMATIVE ASSESSMENT
10.1 CIRCLES AND ITS RELATED
TERMS : A REVIEW
through the centre of the circle. In the given
figure, AOB is the diameter of the circle. A
diameter is the longest chord of a circle.
Diameter = 2 × radius
6. A piece of a circle between two points is
called an arc. Look at the pieces of the circle
between two points P and Q in the given figure.
You find that there are two pieces, one longer
and the other smaller. The longer one is called
the major arc PQ and the shorter one is called
the minor arc PQ.
S
PR
AK
AS
HA
N
1. The collection of all the points in a plane,
which are at a fixed distance from a fixed point
in the plane, is called a circle.
2. The fixed point is called the centre of the
circle and the fixed distance is called the radius
of the circle.
TH
ER
In the given figure, O is the centre and the
length OP is the radius of the circle.
3. A circle divides the plane on which it lies
into three parts. They are : (i) inside the circle,
which is also called the interior of the circle;
(ii) the circle and (iii) outside the circle, which is
also called the exterior of the circle. The circle
and its interior make up the circular region.
G
O
YA
L
BR
O
7. The length of the complete circle is called
its circumference. The
region between a chord and
either of its arcs is called a
segment of the circular
region or simply a segment
of the circle. You will find
that there are two types of
segments also, which are the major segment and
the minor segment.
8. The region between an arc and the two
radii, joining the centre to the end points of the
arc is called a sector. Like
segments, you find that the
minor arc corresponds to
the minor sector and the
major arc corresponds to
the major sector.
4. A chord of a circle
is a line segment joining
any two points on the
circle. In the given figure
PQ, RS and AOB are the
chords of a circle.
5. A diameter is a chord of a circle passing
1
TEXTBOOK’S EXERCISE 10.1
(iii) If a circle is divided into three equal
arcs, each is a major arc.
(iv) A chord of a circle, which is twice as
long as its radius, is a diameter of the circle.
HA
N
(v) Sector is the region between the chord
and its corresponding arc.
(vi) A circle is a plane figure.
AS
Sol. (i) True, because all points on the circle
are equidistant from its centre.
AK
(ii) False, because there are infinitely many
points on the circle.
(iii) False, because for each arc, the
remaining arc will have greater length.
(iv) True, because of definition of diameter.
ER
S
Q.2. Write True or False: Give reasons for
your answers.
(i) Line segment joining the centre to any
point on the circle is a radius of the circle.
(ii) A circle has only finite number of equal
chords.
PR
Q.1. Fill in the blanks :
(i) The centre of a circle lies in
___________ of the circle. (exterior/interior)
(ii) A point, whose distance from the centre
of a circle is greater than its radius lies in
__________ of the circle. (exterior/interior)
(iii) The longest chord of a circle is a
__________ of the circle.
(iv) An arc is a __________ when its ends
are the ends of a diameter.
(v) Segment of a circle is the region between
an arc and __________ of the circle.
(vi)A circle divides the plane, on which it
lies, in __________ parts.
Sol. (i) interior (ii) exterior (iii) diameter
(iv) semicircle (v) the chord (vi) three
(v) False by virtue of its definition.
(vi) True as it is a part of a plane.
TH
10.2 ANGLE SUBTENDED BY A CHORD
AT A POINT
1. Equal chords of a circle subtend equal
angles at the centre.
BR
O
2. If the angles subtended by the chords of a
circle at the centre are equal, then the chords are
equal.
TEXTBOOK’S EXERCISE 10.2
Proof : In triangles AOB and COD,
AB = CD
[Given]
YA
L
Q.1. Recall that two circles are congruent if
they have the same radii. Prove that equal
chords of congruent circles subtend equal angles
at their centres
[V. Imp]
O

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]
G
Sol.
AO = CO
 [Radii of congruent circles]
BO = DO
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
AO = CO

BO = DO
Given : Two congruent circles with centres O
and O. AB and CD are chords of circles with
centre O and O respectively such that AOB =
COD
To Prove : AB = CD
Proof : In triangles AOB and COD,
 AOB = COD
 AOB  COD
 AB = CD
10.3 PERPENDICULAR FROM THE
CENTRE TO A CHORD
[Given]
[SAS axiom]
[CPCT] Proved.
2. The line drawn through the centre of a circle
to bisect a chord is perpendicular to the chord.
3. There is one and only one circle passing
through three given non-collinear points.
AS
TEXTBOOK’S EXERCISE 10.3
HA
N
1. The perpendicular from the centre of a
circle to a chord bisects the chord.
PR
AK
Sol. Given : AB is the common chord of two
intersecting circles (O, r) and (O, r).
To Prove : Centres of both circles lie on the
perpendicular bisector of chord AB, i.e., AB is
bisected at right angle by OO.
Construction : Join AO, BO, AO and BO.
O
TH
ER
S
Q.1. Draw different pairs of circles. How
many points does each pair have in common?
What is the maximum number of common points?
Sol.
[Radii of congruent circle]
BR
Proof : In AOO and BOO
AO = OB [Radii of the circle (O, r)]
AO = BO [Radii of the circle (O, r)]
OO = OO [Common]
 AOO BOO [SSS congruency]
 AOO= BOO [CPCT]
Now in AOC and BOC,
AOC = BOC [AOO = BOO]
AO = BO [Radii of the circle (O, r)]
OC = OC [Common]
 AOC  BOC [SAS congruency]
AC = BC and ACO = BCO ...(i) [CPCT]
ACO + BCO = 180° ..(ii) [Linear pair]
ACO = BCO = 90° [From (i) and (ii)]
Hence, OO lie on the perpendicular bisector
of AB. Proved.
Maximum number of common points = 2
G
O
YA
L
Q.2. Suppose you are given a circle. Give a
construction to find its centre.
Sol. Steps of Construction :
1. Take arc PQ of the given circle.
2. Take a point R on the arc PQ and draw
chords PR and RQ.
3. Draw perpendicular
bisectors of PR and RQ.
These perpendicular bisectors
intersect at point O.
Hence, point O is the
centre of the given circle.
Q.3. If two circles intersect at two points,
prove that their centres lie on the perpendicular
bisector of the common chord.
[2011 (T-II)]
3
10.4 EQUAL CHORDS AND THEIR
DISTANCES FROM THE CENTRE
circles) are equidistant from the centre (or centres).
2. Chords equidistant from the centre of a
circle are equal in length.
1. Equal chords of a circle (or of congruent
TEXTBOOK’S EXERCISE 10.4
Now, AB = CD

1
1
AB = CD
2
2
AM = CN ...(ii) [Perpendicular from
HA
centre bisects the chord]
(ii), we get
= EN + CN
= CE
...(iii)
= CD
...(iv)
= CD – AE
[From (iii)]
= CD – CE Proved.
PR
AK
AS
Adding (i) and
EM + AM

AE
Now,
AB
 AB – AE

BE
Q.3. If two equal chords of a circle intersect
within the circle, prove that the line joining the
point of intersection to the centre makes equal
angles with the chords.
[2011 (T-II)]
ER
S
Sol. In AOO,
AO2 = 52 = 25
AO2 = 32 = 9
OO2 = 42 = 16
AO2 + OO2
= 9 + 16 = 25 = AO2
 AOO= 90° [By converse of
Pythagoras theorem]
Similarly, BOO = 90°.
 AOB= 90° + 90° = 180°
AOB is a straight line, whose mid-point
is O.
 AB = (3 + 3) cm = 6 cm

[Given]
N
Q.1. Two circles of radii 5 cm and 3 cm
intersect at two points and the distance between
their centres is 4 cm. Find the length of the
common chord.
[2011 (T-II)]
TH
Sol. Given : AB and
CD are two equal chords
of a circle which meet at
E within the circle and a
line PQ joining the point
of intersection to the
centre.
To Prove : AEQ = DEQ
Construction : Draw OL  AB and
OM  CD.
Proof : In OLE and OME, we have
G
O
YA
L
BR
O
Q.2. If two equal chords of a circle intersect
within the circle, prove that the segments of one
chord are equal to corresponding segments of
the other chord.
[V. Imp.]
Sol. Given : AB and CD are two equal
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
ONE, OM = ON
[Equal chords
are equidistant]
OE = OE [Common]
OME = ONE
[Each equal to 90°]
 OME  ONE [RHS axiom]
 EM = EN...(i)
[CPCT]
OL = OM [Equal chords are equidistant]
OE = OE
[Common]
OLE = OME
[Each = 90°]
 OLE  OME [RHS congruence]
 LEO = MEO [CPCT]
Q.4. If a line intersects two concentric
circles (circles with the same centre) with centre
O at A, B, C and D, prove that AB = CD (See
fig.)
[2011 (T-II)]
4
Sol. Given : A line
AD intersects two
concentric circles at A,
B, C and D, where O is
the centre of these
circles.
To prove : AB = CD
Construction : Draw OM  AD.
Proof : AD is the chord of larger circle.
AM = DM ..(i) [OM bisects the chord]
 KR =
12  2 24
= 4.8 m  RM = 2KR

5
5
 RM = 2 × 4.8 = 9.6 m
Hence, distance between Reshma and
Mandip is 9.6 m.
HA
N
Q.6. A circular park of radius 20 m is
situated in a colony. Three boys Ankur, Syed and
David are sitting at equal distance on its
boundary each having a toy telephone in his
hands to talk each other. Find the length of the
string of each phone.
[HOTS]
AS
BC is the chord of
smaller circle
 BM = CM ...(ii)
[OM bisects the chord]
Subtracting (ii) from
(i), we get
AM – BM = DM – CM  AB = CD Proved.
PR
AK
Sol. Let Ankur, Syed and David be
represented by A, S and D respectively.
ER
S
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
Salma and between Salma and Mandip is 6 m
each, what is the distance between Reshma and
Mandip?
[HOTS]
Sol. Let Reshma, Salma and Mandip be
represented by R, S and M respectively.
Draw OL  RS, OL2 = OR2 – RL2
OL2 = 52 – 32 [RL = 3 m, because OL  RS]
= 25 – 9 = 16
TH
Let PD = SP = SQ = QA = AR = RD = x
In OPD, OP2 = 400 – x2
BR
O
 OP =
400  x2
YA
L
 AP = 2 400  x 2  400  x 2
[ centroid divides the
median in the ratio 2 : 1]
= 3 400  x 2
Now, in APD, PD2 = AD2 – AP2
O
OL = 16 = 4 m
Now, area of
triangle ORS

G
 x2 = (2x)2 – 3 400  x 2

2
 x2 = 4x2 – 9(400 – x2)
 x2 = 4x2 – 3600 + 9x2  12x2 = 3600
1
= × KR × OS
2
1
= × KR × 5
2
 x2 =
3600
= 300  x = 10 3
12
Now, SD = 2x = 2 × 10 3 = 20 3
 ASD is an equilateral triangle.
1
Also, area of ORS =
× RS × OL
2
1
=
× 6 × 4 = 12 m2
2
1

× KR × 5 = 12
2
 SD = AS = AD = 20 3
Hence, length of the string of each phone is
20 3 m.
5
OTHER IMPORTANT QUESTIONS
Q.5. There are three non-collinear points.
The number of circles passing through them
are :
[2010]
(a) 2
(b) 1
(c) 3
(d) 4
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 =
13 cm and AB = 24 cm,
then OM equals : [2010]
(a) 3 cm (b) 4 cm
AK
AS
HA
N
Q.1. In the given figure, O is the centre of
the circle. If OA = 5 cm and OC = 3 cm, then the
length of AB is :
[2011 (T-II)]
Sol. (c) AC = AO2 – OC2 = 25 – 9 cm
= 4 cm
PR
(a) 4 cm (b) 6 cm (c) 8 cm (d) 15 cm
O
TH
ER
S
AB = 2 × AC = 2 × 4 cm = 8 cm.
Q.2. Three chords AB, CD and EF of a circle
are respectively 3 cm, 3.5 cm and 3.8 cm away
from the centre. Then which of the following
relations is correct ?
[HOTS]
(a) AB > CD > EF (b) AB < CD < EF
(c) AB = CD = EF (d) none of these
Sol. (a) We know that longer the chord,
shorter is its distance from the centre.
Q.3. In a circle, chord AB of length 6 cm is
at a distance of 4 cm from the centre O. The
length of another chord CD which is also 4 cm
away from the centre is :
(a) 6 cm (b) 4 cm (c) 8 cm (d) 3 cm
Sol. (a) Chords equidistant from the centre
are equal.
Q.4. In the figure,
chord AB is greater than
chord CD. OL and OM
are the perpendiculars
from the centre O on these
two chords as shown in
the figure. The correct
releation between OL and OM is :
[HOTS]
(a) OL = OM
(b) OL < OM
(c) OL > OM
(d) none of these
Sol. (b) Longer the chord, shorter is its
distance from the centre.
(c) 5 cm (d) 4.7 cm
Sol. (c) In AMO, M = 90°
 OA2 = AM2 + OM2
 OM =
OA 2  AM 2
 OM =
(13)2  (12)2 = 5 cm
G
O
YA
L
BR
Q.7. In the given figure,
in a circle with centre O, a
chord AB is drawn and C is
its mid-point ACO will be :
[2010]
(a) more than 90° (b) less than 90°
(c) 90°
(d) none of these
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
at the centre of the circle. If chord AB = 8 cm,
then chord CD is :
[2011 (T-II)]
(a) 4 cm (b) 8 cm (c) 16 cm (d) 12 cm
Sol. (b) Equal chords subtend equal angles at
the centre.
6
Q.9. In the given
figure, a circle with
centre O is shown,
where ON > OM. Then
which of the following
relations
is
true
between the chord AB
and chord CD ?
[HOTS]
(a) AB = CD
(b) AB > CD
(c) AB < CD
(d) none of these
Sol. (b) Longer the chord, shorter is its
distance from the centre.
(c) 5 cm (d)
RS
R
PAKH
SAN
-E
H
T
O
R
Sol. AB = 2 AO 2 – OC 2
= 2 132 – 52 cm
Sol. (a) OC = AO2 – AC2
= 2 × 12 cm = 24 cm.
Q.14. Two concentric circles with centre O
have A, B, C and D as
points of intersection
with a line l as shown
in the figure. If AD =
12 cm and BC = 8 cm,
find the length of AB
and CD.
[2011 (T-II)]
Sol. Since OM  BC
= 25 – 16 cm = 3 cm
YA
L
Since, OD = OA = 5 cm
 CD = OD – OC = (5 – 3) cm = 2 cm.
Q.11. In the given figure,
O is the centre of the circle
of radius 5 cm. OP ⊥ AB,
OQ ⊥ CD, AB | | CD, AB =
8 cm and CD = 6 cm. The
length of PQ is : [2011 (T-II)]
(a) 8 cm
(b) 1 cm
(c) 6 cm
(d) none of these
1
BC = 4 cm
2
Similarly, OM  AD
 BM = CM =
O
G
5 3
cm
4
Sol. (c) We have, OA = 5 cm = OB
[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.
(a) 2 cm (b) 3 cm (c) 4 cm (d) 5 cm
Sol. (b) OQ =
5
5 3
cm (b)
cm
2
2
B
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)
OC 2 – CQ 2
1
AD = 6 cm
2
Now, AB = AM – BM = (6 – 4) cm = 2 cm
Also, CD = DM – CM = (6 – 4) cm = 2 cm
Hence, AB = CD = 2 cm
Q.15. Two circles of radii 10 cm and 8 cm
intersect and the length of the common chord is
12 cm. Find the distance between their centres.
 AM = DM =
= 25  9 cm = 4 cm
OP = OA 2 – AP 2 = 25 – 16 cm = 3 cm
 PQ = OQ – OP = 1 cm.
Q.12. In the given figure, ∠AOB chord AB
subtends angle equal to 60° at the centre of the
circle. If OA = 5 cm, then length of AB (in cm)
is :
[2010]
[2011 (T-II)]
7
 AB2 = AC2 + BC2 [Pythagoras theorem]
Q.18. Two chords of a circle of lengths
10 cm and 8 cm are at the distances 8 cm and
3.5 cm respectively from the centre. Check
whether the above statement is true or not.
Sol. False, because larger the chord, shorter
is its distance from the centre.
Sol. Let O and O
be the centres of the
circles of radii 10 cm
and 8 cm respectively
and let PQ be their
common chord.
We have,
OP = 10 cm, OP = 8 cm and PQ = 12 cm
1
 PL =
PQ = 6 cm
2
In right OLP, we have OP2 = OL2 + LP2
 OL =
R
S
OP 2  LP 2 = 102  62
= 64 cm = 8 cm
In right OLP,
we have OP2 = OL2 + LP2
Q.19. If the perpendicular bisector of a
chord AB of a circle PXAQBY intersects the
circle at P and Q, prove that arc PXA  arcc
PYB.
[HOTS]
Sol.
In PAO and PBO,
AO = BO [Given]
POA = POB
= 90°
[Given]
PO = PO
[Common]
 PAO  PBO [SAS]
[CPCT]
 PA = PB
 arc PXA = arc PYB
Proved.
 arc PXA  arc PYB
 OL =
B
R
O
TH
E
OP 2  LP 2  82  62  28 cm
= 5.29 cm
 OO = OL + LO = (8 + 5.29) cm
= 13.29 cm.
Q.16. Two congruent circles with centres O
and O intersect at two points A and B. Check
whether ∠AOB = ∠AO B or not.
[V. Imp.]
Sol. OA = OB = OA = OB
G
PR O Y
AK A L
AS
HA
N
Q.20. Show that two circles cannot intersect
at more than two points.
[Imp.]
Sol.Let us assume that two circles intersect
at three points say A, B and C. Then clearly, A,
B and C are not collinear. But, through three
non-collinear points we can draw one and only
one circle. Therefore, we cannot have two circles
passing A, B and C. Or two circles cannot
intersect at more than two points.
Q.21. AB and AC are two chords of a circle
of radius r such that AB = 2AC. If p and q are
the distances of AB and AC from the centre,
prove that 4q2 = p2 + 3r2.
[2011 (T-II)]
Sol. Draw OD  AB,
OE  AC and join AO.
Let AC = 2x, then
AE = CE = x
So, AB = 4x and
AD = BD = 2x.
In  AOB and  AOB, we have
AO = AO
OB = OB
AB = AB
[Common]
 AOB  AO B [SSS congruency axiom]
[CPCT]
 AOB = AO B
Q.17. AOB is a diameter of a circle and C is
a point on the circle.
Check whether AC2 + BC2
= AB 2 is true or
not. [Imp.]
Sol. True. We know
that ACB = 90°
[Angle made in semi-circle]
8
In AOD,
AO2 = AD2 + OD2 [By Pythagoras theorem]
 AD2 = AO2 – OD2
… (i)
 4x2 = r2 – p2
In AOE, we have
AO2 = OE2 + AE2  AE2 = AO2 – OE2
 x 2 = r 2 – q2
… (ii)
 4x2 = 4r2 – 4q2
From (i) and (ii), we have
4r2 – 4q2 = r2 – p2  4q2 = p2 + 3r2.
Proved.
OM = ON
[From (i)]
OME = ONE
[each equal to 90°]
OE = OE
[common]
 OME  ONE [by SAS congruence]
 ME = NE
[CPCT]
Thus, in quad. OMEN, we have
OM = ON, ME = NE and
OME = ONE = 90°
Hence, OMEN is a square. Proved.
Q.23. In the given
figure, OD is perpendicular to the chord AB of a
circle whose centre is O. If
BC is a diameter, show
that CA = 2OD.
Q.22. In the given
figure, equal chords AB
and CD of a circle cut
at right angles at E. If
M and N are the midpoints of AB and CD
respectively. Prove that
OMEN is a square.
[2011 (T-II)]
Sol. Join OE.
Since the line joining the centre of a circle to
the mid-point of a chord is perpendicular to the
chord, we have OM  AB and ON  CD
 OMB = 90° and OND = 90°
 OME = 90° and ONE = 90°
Also, equal chords of a circle are equidistant
from the centre.
 OM = ON ..... (i)
Now, in OME and ONE, we have
[2011 (T-II)]
YA
L
B
PR R O
AK T
H
AS E R
HA S
N
Sol. Since OD  AB and the perpendicular
drawn from the centre to a chord bisects the
chord.
 D is the mid-point of AB
Also, O being the centre, is the mid-point of
BC.
Thus, in ABC, D and O are mid-points of
AB and BC respectively.
1
CA
2
[ segment joining the mid-points of two
sides of a triangle is half of the third side.]
 CA = 2OD. Proved.
 OD || AC and OD =
PRACTICE EXERCISE 10A
O
1 Mark Questions
G
1. In the given figure, O
is the centre and AB = BC. If
BOC = 80°, then AOB is
(a) 80°
(b) 70°
(c) 85°
(d) 90°
(a) 10 cm
(b) 8 cm
(c) 12 cm
(d) 16 cm
3. In the given figure, O is the centre of the
circle, OB = 5 cm and AB = 8 cm. The distance of
AB from the centre is
(a) 4 cm
(b) 3 cm
2. In the given figure, O is
the centre of the circle, OA =
10 cm and OC = 6 cm. The
length of AB is
(c) 89 cm
(d) 6 cm
9
10. In a circle of radius 5 cm, there are two
parallel chords of length 6 cm and 4 cm. Find the
distance between them, when they are on (i) opposite sides of the centre (ii) same side of the centre.
11. Two concentric circles are of radii 7 cm,
4 cm. A line PQRS cuts one circle at P, S and other
at Q, R. If QR = 6 cm, find the length of PS.
12. If in the figure, AN
= NB = l and ND = h,
prove that the diameter of
2 Marks Questions
4. Calculate the length of a chord which is at a
distance 5 cm from the centre of a circle whose
radius is 13 cm.
5. Find the distance from the centre to a chord
70 cm in length in a circle whose diameter is 74 cm.
6. Two points P and Q are 9 cm apart. A circle
of radius 5.1 cm passes through P and Q. Calculate
the distance of its centre from the chord PQ.
7. Two circles of radii 26 cm and 25 cm intersect at two points which are 48 cm apart. Find the
distance between their centres.
8. Two parallel chords of a circle whose diameter is 13 cm are respectively 5 cm and 12 cm.
Find the distance between them if they lie on opposite sides of the centre.
the circle is
l2  h2
.
h
[HOTS]
3/4 Marks Questions
13. AB and CD are two equal chords of a circle.
M and N are their mid-points respectively. Prove
that MN makes equal angles with AB and CD.
9. PQ is a variable chord of a circle of radius
7.5 cm. If PQ = 9 cm, find the radius of the circle
which is the locus of the mid-point of PQ.
14. A chord AB of a circle (O, r) is produced
to P so that BP = 2 AB. Prove that
OP2 = OA2 + 6AB2.
[HOTS]
10.5 ANGLE SUBTENDED BY AN ARC OF
A CIRCLE AND CYCLIC QUADRILATERAL
1. The angle subtended by an arc at the
centre is double the angle subtended by it at any
point on the remaining part of the circle.
2. Angles in the same segment of a circle are
equal.
3. If a line segment joining two points
subtends equal angles at two other points lying
on the same side of the line containing the line
segment, the four points lie on a circle (i.e., they
are concyclic).
4. The sum of either pair of opposite angles
of a cyclic quadrilateral is 180°.
5. If the sum of a pair of opposite angles of
a quadrilateral is 180°, the quadrilateral is cyclic.
G
P R OY
A K AL
AS BR
H A OT
N H
ER
S
[HOTS]
TEXTBOOK’S EXERCISE 10.5
Q.1. In the figure, A, B and C are three
points on a circle with centre
O such that  BOC = 30°
and  AOB = 60°. If D is a
point on the circle other than
the arc ABC, find  ADC.
the centre of a circle is double the angle
subtended by the same arc on the remaining part
of the circle.
 2ADC = AOC
1
1
 ADC = AOC = × 90°
2
2
 ADC = 45°.
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.]
[2010]
Sol. We have, BOC = 30° and AOB = 60°
AOC = AOB + BOC = 60° + 30° = 90°
We know that angle subtended by an arc at
10
 69° + 31° + BAC = 180°
 BAC = 180° – 100° = 80°
Also, BAC = BDC
[Angles in the same segment]
 BDC = 80°
Q.5. In the figrue, A, B,
C and D are four points on
a circle. AC and BD
intersect at a point E such
that BEC = 130° and
ECD = 20°. Find BAC.
Sol. We have, OA = OB = AB
Therefore, OAB is a equilateral triangle.
 AOB = 60°
We know that angle
subtended by an arc at the
centre of a circle is double
the angle subtended by the
same arc on the remaining
part of the circle.
    AOB = 2ACB
1
1
AOB = × 60°
2
2
    ACB = 30°
1
reflex AOB
2
1
1
= (360° – 60°) = × 300° = 150°
2
2
Also, ADB =
[2011 (T-II)]
Sol. BEC + DEC = 180° [Linear pair]
 130° + DEC = 180°
 DEC= 180° – 130° = 50°
Now, in DEC,
 DEC + DCE + CDE = 180°
[Angle sum property of a triangle]
 50° + 20° + CDE = 180°
 CDE = 180° – 70° = 110°
Also, CDE = BAC
[Angles in same segment]
 BAC = 110°
BR
PR O
AK TH
A S ER
HA S
N
    ACB =
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,
PQR = 100°, where P, Q
and R are points on a circle
with centre O. Find OPR.
Q.6. ABCD is a cyclic
quadrilateral whose diagonals intersect at a point
E. If DBC = 70°,
BAC = 30°, find BCD.
Further, if AB = BC, find
ECD.
[V. Imp.]
Sol. CAD = DBC = 70° [Angles in the
same segment]
Therefore, DAB = CAD + BAC
= 70° + 30° = 100°
But, DAB + BCD = 180°
[Opposite
angles of a cyclic quadrilateral]
So, BCD = 180° – 100° = 80°
Now, we have AB = BC
Therefore, BCA = 30° [Opposite angles
of an isosceles triangle]
Again, DAB + BCD = 180° [Opposite
angles of a cyclic quadrilateral]
 100° + BCA + ECD = 180°
[ BCD = BCA + ECD]
[2011 (T-II)]
G
O
Y
A
L
Sol. Reflex angle
POR = 2PQR
= 2 × 100° = 200°
Now, angle POR = 360° – 200° = 160°
Also, PO = OR
[Radii of a circle]
 OPR = ORP
[Opposite angles of
isosceles triangle]
In OPR, POR = 160°
 OPR = ORP = 10°
[Angle sum property of a triangle]
Q.4. In the figure, ABC = 69°, ACB = 31°,
find  BDC. [2011 (T-II)]
Sol. In ABC, we have
ABC + ACB + BAC
= 180°
[Angle sum property
of a triangle]
11
 100° + 30° + ECD = 180°
 130° + ECD = 180°
 ECD = 180° – 130° = 50°
Hence, BCD = 80° and ECD = 50°.
Q.7. If diagonals of a cyclic quadrilateral
are diameters of the circle through the vertices
of the quadrilateral, prove that it is a rectangle.
Sol. Given : A trapezium ABCD in which
AB || CD and AD = BC.
To Prove : ABCD is a cyclic trapezium.
Construction : Draw DE  AB and CF  AB.
Proof : In DEA and CFB, we have
AD = BC [Given]
DEA = CFB = 90°
[DE  AB and CF  AB]
DE = CF
[Distance between parallel
lines remains constant]
 DEA  CFB
[RHS axiom]
 A = B
...(i)
[CPCT]
and, ADE = BCF ...(ii)
[CPCT]
Since, ADE = BCF
[From (ii)]
 ADE + 90° = BCF + 90°
 ADE + CDE = BCF + DCF
 D = C
...(iii)
[ADE + CDE = D,
BCF + DCF = C]
 A = B and C = D
…(iv)
[From (i) and (iii)]
A + B + C + D = 360°
[Sum of the angles
of a quadrilateral is 360°]
 2(B + D) = 360° [Using (iv)]
 B + D = 180°
 Sum of a pair of opposite angles of
quadrilateral ABCD is 180°.
 ABCD is a cyclic trapezium Proved.
BR
O
TH
ER
S
PR
AK
AS
HA
N
[2010]
Sol. Given : ABCD is a
cyclic quadrilateral, whose
diagonals AC and BD are
diameter of the circle passing
through A, B, C and D.
To Prove : ABCD is a
rectangle.
Proof : In AOD and COB
AO = CO [Radii of a circle]
OD = OB [Radii of a circle]
AOD = COB
[Vertically
opposite angles]
 AOD  COB [SAS axiom]
 OAD = OCB [CPCT]
But these are alternate interior angles made
by the transversal AC, intersecting AD and BC.
 AD || BC
Similarly, AB || CD.
Hence, quadrilateral ABCD is a parallelogram.
Also, ABC = ADC ..(i)
[Opposite angles of a ||gm are equal]
And, ABC + ADC = 180° ...(ii)
[Sum of opposite angles of a
cyclic quadrilateral is 180°]
ABC = ADC = 90° [From (i) and (ii)]
 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)]
G
O
YA
L
Q.9. 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 (see Fig.). [2011 (T-II)]
Prove that ACP = QCD.
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
PR
AK
AS
HA
N
Proof : Let O be the
mid-point of AC.
Then OA = OB = OC
= OD
Mid point of the hypotenuse of a right triangle is
equidistant from its vertices
with O as centre and radius equal to OA, draw
a circle to pass through A, B, C and D.
We know that angles in the same segment of
a circle are equal.
Since, CAD and CBD are angles of the
same segment.
Therefore, CAD = CBD. Proved.
Q.12. Prove that a cyclic parallelogram is a
rectangle.
[2010]
Or
Prove that a parallelogram inscribed in a
circle is a rectangle. [2011 (T-II)]
Sol. Given : ABCD is a
cyclic parallelogram.
To prove : ABCD is a
rectangle.
Proof :
ABC = ADC ...(i)
[Opposite angles of a ||gm are equal]
But, ABC + ADC = 180°
... (ii)
[Sum of opposite angles of a cyclic
quadrilateral is 180°]
 ABC = ADC = 90° [From (i) and (ii)]
 ABCD is a rectangle
[A ||gm one of whose angles is
90° is a rectangle]
Hence, a cyclic parallelogram is a rectangle.
Proved.
G
O
YA
L
BR
O
TH
ER
S
To Prove : ACP = QCD.
Proof : ACP = ABP ...(i) [Angles in
the same segment]
QCD = QBD
..(ii) [Angles in
the same segment]
But, ABP = QBD
..(iii) [Vertically
opposite angles]
By (i), (ii) and (iii), we get
ACP = QCD.
Proved.
Q.10. If circles are drawn taking two sides
of a triangle as diameters, prove that the point
of intersection of these circles lie on the third
side.
[2011 (T-II)]
Sol. Given : Sides AB
and AC of a triangle ABC
are diameters of two circles
which intersect at D.
To Prove : D lies on BC.
Proof : Join AD
ADB = 90°
...(i)
[Angle in a semicircle]
Also, ADC = 90° ...(ii)
Adding (i) and (ii), we get
ADB + ADC = 90° + 90°
 ADB + ADC = 180°
 BDC is a straight line.
 D lies on BC
Hence, point of intersection of circles lie on
the third side BC.
Proved.
Q.11. ABC and ADC are two right triangles
with common hypotenuse AC. Prove that
CAD = CBD.
[2011 (T-II)]
Sol. Given : ABC and ADC are two right
triangles with common hypotenuse AC.
To Prove : CAD = CBD
OTHER IMPORTANT QUESTIONS
Q.1. In the figure, O
is the centre of the circle
with AB as diameter. If
AOC = 40°, the value
of x is equal to : [Imp.]
(a) 50°
(b) 60°
(c) 70°
(d) 80°
Sol. (c) OA = OC  OAC = OCA
 Now, OAC + OCA + AOC = 180°
2x + 40° = 180° x =
13
180  40
= 70°
2
1
× 70°
2
= 35° [Angle at the
centre is double the angle
at the circumference]
y = x = 35°
[Angles in the same segment are equal]
Q.7. In the given
figure, if POQ is a
diameter of the circle
and PR = QR, then
RPQ is equal to :
TH
-
Sol. (a) x =
[2011 (T-II)]
N
HA
BR
O
(a) 30°
(b) 60°
(c) 90°
(d) 45°
Sol. (d) PRQ = 90°
[ angle in a semicircle is a right angle]
 PR = PQ
 P = Q = 45°
Q.8. ABCE is a cyclic
quadrilateral. 'O' is the
centre of the circle and
 AOC = 150°, then
CBD is : [2011 (T-II)]
(a) 225° (b) 128°
(c) 150° (d) 75°
Sol.(d) Since the angle subtended by an arc
at the centre of a circle is twice the angle
sustended at a point on the remaining part of the
circumference, we have
1
1
AEC =  AOC =
× 150° = 75°
2
2
Now, ABCE is a cyclic quadrilateral whose
side AB is produce to D
 CBD = AEC = 75°
[ ext.  of cyclic quad. = int. opp. ]
Q.9. 'O' is the centre
of the circle QPS = 65°;
 PRS = 33°,  PSQ is
equal to : [2011 (T-II)]
AK
AS
Q.2. For what value
of x in the figure, points
A, B, C and D are
concyclic ? [2011 (T-II)]
(a) 9°
(b) 10°
(c) 11°
(d) 12°
Sol. (b) Since, opposite angles of a cyclic
quadrilateral are supplementary.
 81° + x + 89° = 180°
 x = 180° – 170° = 10°.
Q.3. In the given
figure, O is the centre
of the circle. If CAB
= 40° and  CBA =
110°, the value of x is :
(a) 50°
(b) 80°
(c) 55°
(d) 60°
Sol. (d) ACB = 180° – (110° + 40°) = 30°
AOB = 2ACB [Angle at the centre is
twice the angle at the circumference]
 x = 2 × 30° = 60°.
Q.4. Angle inscribed in a semicircle is :
PR
[2010]
E
G
O
R YA
S L
(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
the circle. If QPR is 50°,
then QOR is : [2010]
(a) 130°
(b) 40°
(c) 100°
(d) 50°
Sol. (c) QOR = 2QPR
[The angle subtended by an arc at the
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°
(c) 70° – x
(b) 70° + x
(d) 140°
14
(a) 90°
(b) 82°
(c) 102°
(d) 42°
AK
AS
HA
N
BR
O
TH
-
(c) 140° (d) 110°
Sol. (c) OA = OB
OBA = OAB = 30°
OC = OB  OBC = OCB = 40°
ABC = OBA + OBC = 30° + 40°
= 70°
Now, AOC = 2ABC = 2 × 70° = 140°
Q.14. ABCD is a cyclic
quadrilateral as shown in
the figure. The value of
(x + y) is : [2011 (T-II)]
(a) 200° (b) 100°
(c) 180° (d) 160°
Sol. (d) x + 90° = 180°
 x = 90°
y + 110° = 180°  y = 70°
 x + y = 90° + 70° = 160°
Q.15. Arc ABC subtends an angle of 130° at
the centre O of the circle. AB is extended to P.
Then CBP equals :
[2010]
PR
Sol. (b) R = Q = 33°
[Angles in the
same segment are equal]
Now, in PQS, P + Q + S = 180°
 PSQ = 82°
Q.10. In the given
figure, AB is a diameter of
the circle. CD || AB and
BAD = 40°, then ACD
is :
[2011 (T-II)]
(a) 40°
(b) 90°
(c) 130° (d) 140°
Sol. (c) ADB = 90° [Angle in a semicircle
is a right angle]
BAD + ADB + ABD = 180°
 ABD = 180° – (40° + 90°) = 50°
ACD = 130° [opposite angles of a cyclic
quad. are supplementary]
Q.11. In the given
figure the values of x and y
is :
[2011 (T-II)]
(a) 20°, 30°
(b) 36°, 60°
(c) 15°, 30°
(d) 25°, 30°
Sol. (b) 2x + 3x = 180°  x = 36°
y + 2y = 180°  y = 60° [opp. angles of a
cyclic quad. are supplementary]
Q.12. In the given
figure, if AOB is the
diameter of the circle and
B = 35°, then x is equal
to :
[2011 (T-II)]
(a) 90°
(b) 55°
(c) 75°
(d) 45°
Sol. (b) BCA = 90°
[Angle in a semicircle is a right angle]
Now, BCA + CBA + CAB = 180°
 x = 180° – (90° + 35°) = 55°
Q.13. In the given
figure, O is the centre of
the circle. If OAB = 30°
and OCB = 40°, then
measure of AOC is :
G
E OY
R AL
S
(a) 60°
(b) 65° (c) 70° (d) 130°
Sol. (b) Take a point E on the remaining part
of the circumference. Join EA and EC.
Since the angle subtended by an arc at the
centre of a circle is twice the angle subtended at
a point on the remaining part of the
circumference, we have
AEC =
Now, ABCE is a cyclic quadrilateral whose
side AB is produced to P.
 CBP = AEC = 65°
[ ext.  of a cyclic quad. = int. opp. ]
Hence, CBP = 65°.
Q.16. In the figure, O is the centre of the
circle and ∠AOB = 80°. The value of x is :
[2011 (T-II)]
(a) 70°
1
1
AOC =
× 130° = 65°
2
2
(b) 220°
[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°
N
1
Sol. (b) x = AOB [Angle at the centre
2
1
× 80° = 40°.
2
(a) 20°
(b) 40° (c) 60° (d) 10°
Sol. (b) AOC = 2ABC [Angle at the centre
is twice the angle at the circumference]
 AOC = 40°.
Q.20. In the given figure, a circle is centred
at O. The value of x is :
[2010]
AS

HA
is double the angle at the circumference]
AK
Q.17. In the figure, O is the centre of the
circle. If ∠OAB = 40°, then ∠ACB is equal to :
TH
ER
S
PR
[Imp.]
1
BR
O
(a) 50°
(b) 40° (c) 60° (d) 70°
Sol. (a) OA = OB  OAB = OBA = 40°
 AOB = 180° – (40° + 40°) = 100°
(a) 55°
(b) 70° (c) 110° (d) 125°
Sol. (c) OA = OC  OAC = OCA = 20°
OC = OB  OCB = OBC = 35°
BCA = OCA + OCB = 20° + 35° = 55°
Now, x° = AOB = 2BCA = 2 × 55° = 110°
 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]
YA
L
ACB =
AOB [Angle at the centre is
2
double the angle at the circumference]
 ACB = 50°.
G
O
Q.18. In the figure , if ∠DAB = 60°, ∠ABD
= 50°, then ∠ACB is equal to :
(a) 60°
(b) 50° (c) 70° (d) 80°
Sol. (c) ADB = 180° – (60° + 50°) = 70°
16
Q.24. In the figure, O is the centre of the
circle with ∠AOB = 85° and ∠AOC = 115°.
Then ∠BAC is :
[Imp.]
HA
N
(a) 22°
(b) 33° (c) 44° (d) 68°
Sol. (a) We have, ABC = 90°
( angle in a semicircle is a right angle)
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.]
AS
(a) 115° (b) 85° (c) 80° (d) 100°
Sol. (c) BOC = 360° – (85° + 115°) = 160°
1
BOC [Angle at the centre is
2
PR
double the angle at the circumference]
 BAC = 80°.
Q.25. In the figure, if ∠CAB = 40° and AC
= BC, then ∠ADB equal to :
BR
O
TH
ER
S
(a) 65°
(b) 70° (c) 74° (d) 76°
Sol. (a) PSR = 180° – (73° + 42°) = 65°
PSR = PQR
[Angles in the same segment are equal]
 PQR = 65°.
Q.23. In the figure, O is the cnetre of the
circle. If ∠OPQ = 25° and ∠ORQ = 20°, then
the measures of ∠ POR and ∠ PQR are
respectively :
AK
BAC =
O
YA
L
(a) 40°
(b) 60° (c) 80° (d) 100°
Sol. (c) AC = BC  CBA = CAB = 40°
 ACB = 180° – (40° + 40°) = 100°
ACB + ADB = 180° [Opposite angles
of a cyclic quadrilateral are supplementary]
 ADB = 180° – 100° = 80°.
Q.26. In the given figure, ∠DAB = 70° and
∠ABD = 40°, then ∠ACB is equal to : [2010]
G
(a) 90°, 45°
(b) 105°, 45°
(c) 110°, 55°
(d) 100°, 50°
Sol. (a) OP = OQ
 OQP = OPQ = 25° [Radii of same
circle]
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°.
(a) 40°
17
(b) 70°
(c) 110° (d) 30°
Sol. (b) ADB = 180° – (70° + 40°)
= 180° – 110° = 70° and ACB
= ADB = 70°.
[Angles in the same segment are equal]
1
Sol. (d) ACB =
AOB [Angle at the
2
centre is double the angle at the circumference]
 ACB = 45°
OA = OB  OAB = OBA
Q.27. In the given
figure, if ∠AOC = 130°, O
is the centre of the circle
then ∠ABC is :
[2010]
(a) 65°
(b) 115°
(c) 130°
(d) 50°
(∠BCD + ∠DEB) is :
[2010]
(a) 270°
(c) 360°
Now, CAB = 180° – (45° + 30°) = 105°
 CAO = CAB – OAB
= 105° – 45° = 60°.
Q.31. In the figure,
BC is a diameter of the
circle and ∠BAO = 60°.
Then ∠ADC is equal to :
(a) 30°
(b) 45°
(c) 60°
(d) 120°
1
1
AOC = × 130° = 65°
2
2
Now, ABC = 180° – 65° = 115°.
Q.28. In the given
figure, the value of
(b) 180°
(d) 90°
Sol. (b) Opposite angles of a cyclic
quadrilateral are supplementary.
Q.29. In the given
figure, ∠C = 40° ∠CEB =
105°, then the value of x is :
[2010]
Sol. (c) OA = OB
 OBA = OAB
= 60°
ABC = ADC
[Angles in the same segment are equal]
 ADC = 60°.
Q.32. In the given
figure, O is the centre of
a circle and ∠ BOA =
90°, ∠COA = 110°. Find
the measure of ∠BAC.
[2011 (T-II)]
Sol. We have,
∠BOA = 90° and ∠AOC = 110°
∴ ∠BOC = 360° – (∠BOA + ∠AOC)
⇒ ∠BOC = 360° – (90° + 110°) = 160°
1
1
Now, ∠BAC =
∠BOC =
× 160° = 80°
2
2
O
Y
R AL
S
(a) 50°
(b) 35°
(c) 20°
(d) 40°
Sol. (b) D = C = 40°
[Angles in the same segment]
and AED = 105°
[Vertically opposite angles]
x = 180° – (105 + 40°) = 180° – 145° = 35°
180  90
= 45°
2
PR
BR
AK
O
TH
AS
HA
N
Sol. (b) AEC =
 OAB = OBA =
G
Q.30. In the figure, ∠AOB = 90° and
∠ABC = 30°, then ∠CAO is equal to : [V. Imp.]
E
Q.33. In the figure,
ABC is an equilateral
triangle. Find ∠BDC and
∠BEC.
[2010]
Sol. BAC = 60°
[ ABC is equilateral]
BAC = BDC
(a) 30°
(b) 45°
(c) 90°
(d) 60°
[Angle in the same segment are equal]
18
In s ABD and ACE, we have
AB = AC
[Given]
ABD = ACE [Angles in the same segment]
BD = CE
[Given]
So, by SAS congruence criterion, we have
ABD = ACE  AD = AE
Q.37. In the figure, if AOB is a diameter and
∠ADC = 120° find ∠CAB.
[V. Imp.]
R
H
A
K
N
PS
 BDC = 60°
Now, BAC + BEC = 180° [Sum of
opposite angles of a cyclic quadrilateral is 180°]
 BEC = 180° – 60° = 120°.
Q.34. In the given figure, two circles centred
at C1 and C2 are intersecting at P and Q. If PR
and PS are diameters, show that RQS is a
straight line.
[2010]
Sol. We are
given two circles
with centres C1 and
C2 which intersect
each other at P and
Q. PR is a diameter
of circle C1 and PS is a diameter of circle C2.
We need to prove that R, Q and S are
collinear.
 PQR = 90° …[Angle in a semicircle]
and PQS = 90° …[Angle in a semicircle]
 PQR + PQS = 90° + 90° = 180°
 RQS is a straight line.
Q.35. In the given figure, two circles with
centre O and O intersect at two points A and B.
If AD and AC are diameters to circles then prove
that B lies on the line segment DC.
[2010]
Sol. Join AB.
ABD = 90°
…[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
the line segment DC.
Q.36. In the given figure, D is a point on the
circumcircle of ABC
in which AB = AC. If
CD is produced to
point E such that
BD = CE, prove that
AD = AE.
[2010]
Sol. We have,
AB = AC and
CE = BD
L
BR
O
TH
E
RS
Sol. Join AD and AC.
ADC + ABC
= 180° [Opposite angles
of a cyclic quadrilateral]
 ABC = 60°
ACB = 90°
[Angle in a semicircle
is 90°]
 CAB = 90° – ABC = 90° – 60° = 30°.
Q.38. In the given figure, O is the centre of
the circle, BD = DC and DBC = 30°. Find the
measure of BAC.
[2011 (T-II)]
G
O
YA
Sol. BD = DC  DCB = DBC = 30°
BDC = 180° – (30° + 30°) = 120°
BDC + BAC = 180° [Opposite angles
of a cyclic quadrilateral are supplementary]
 BAC = 180° – 120° = 60°.
Q.39. In the given figure,
ABCD is a cyclic quadrilateral, if BCD = 120° and
ABD = 50°, then find ADB.
[2010, 2011 (T-II)]
19
Sol. We have, BCD + BAD = 180°
[Sum of the opp. angles of a cyclic quad.]
 BAD = 180° – 120° = 60°
In ABD, ADB = 180° – (50° + 60°)
[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.
Q.43. In the given figure, if O is the centre of
the circle and AOC = 110° and AB is produced
to D then find AEC and ABC.
[2010]
N
[2010]
 AEC =
1
1
AOC =
× 110° = 55°
2
2
PR
AK
Again arc CEA makes angle reflex AOC
= (360° – 110°) = 250° at the centre and ABC
at a point B on the circumference.
 ABC =
1
1
reflex BOD =
× 250°
2
2
= 125°.
Q.44. In the given figure, if
O is the centre of the circle,
AOC = 50° and COB =
30°. Find the measure of
ADB.
[2010]
Sol. We have, BOC = 30°
and AOC = 50°
AOB = BOC + AOC
= 30° + 50° = 80°
We know that angle subtended by an arc at
the centre of a circle is double the angle
subtended by the same arc on the remaining part
of the circle.
 2ADB = AOB
BR
O
TH
ER
S
Sol. In ABD, ABD = 35° given and
BAD = 90° [Angle inscribed in a semicircle]
 ADB = 180° – (90° + 35°)
[Angle sum property in a triangle]
 ADB = 180° – 125° = 55°
Now, x = ACB = ADB = 55°
[Angles in the same segment]
Q.41. In the given figure, O is the centre of
the circle if ABO = 45° and ACO = 35°, then
find BOC.
[2010]
AS
HA
Sol. Since arc ABC makes AOC = 110° at
the centre and AEC at a point E on the
circumference.
G
O
YA
L
Sol. OA = OB  OAB = OBA = 45°
OC = OA  OAC = OCA = 35°
BAC = OAB + OAC = 45° + 35° = 80°
Now, BOC = 2BAC = 2 × 80° = 160°
Q.42. In the given figure,
AC is the diameter of the
circle. If ACB = 55°, then
find the value of x. [2010]
Sol. In ABC,
ABC = 90°
[Angle in a semicircle]
and ACB = 55° [Given]
 BAC = 180° – (90° + 55°) = 35°
Now, x = BDC = BAC = 35°
[Angles in the same segment]
 ADB =
1
1
AOB =
× 80° = 40°.
2
2
Q.45. In the given figure, if AB = AC, BEC
= 100°, then find the values of x and y. [2010]
20
and ECD = 180° – 150° = 30°
[Linear pair]
 DEC = 180° – (110° + 30°)
= 180° – 140° = 40°
Q.48. In the given figure, ABCD is a cyclic
quadrilateral in which AB is produced to F and
BE || DC. If FBE = 20° and DAB = 95°, then
find ADC.
[2010, 2011 (T-II)]
AS
HA
N
Sol.Since ABEC is a cyclic quadrilateral.
 BAC + BEC = 180°
 BAC = 180° – 100° = 80°
Now, BAC = BDC = y = 80°
[Angles in the same segment]
and since ABC is an isosceles triangle.
 ABC = ACB = x
 2x + 80° = 180°
[Angle sum property of a triangle]
 2x = 180° – 80° = 100°  x = 50°
Hence, x = 50° and y = 80°.
Q.46. In the given figure, AB is diameter of
the circle with centre O and CD || AB. If CAB
= 25°, then find the measure of CAD. [2010]
O
TH
ER
S
Sol. In ABC, C = 90°, because angle in a
semicircle.
 ABC = 180° – (90° + 25°) = 65°
and ADC = 180° – 65° = 115° [Opposite
angles of a cyclic quad. are supplementary]
Since, CD || AB, therefore,
BAC = ACD = 25° [Alternate angles]
Now, in ACD, CAD
= 180° – (115° + 25)° = 180° – 140° = 40°
Q.47. In the given figure, ABCD is a cyclic
quadrilateral with opposite sides AD and BC
produced to meet at the point E. If DAB = 30°
and ABC = 110°, then find DEC.
[2010]
PR
AK
Sol. Since ABCD is a cyclic quadrilateral.
 BAD + BCD = 180°
 95° + BCD = 180°  BCD = 85°
Since, BE || DC therefore, BCD = CBE
 CBE = 85°
[Alternate angles]
Now, BCF = CBE + EBF
= 85° + 20° = 105°
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.
G
O
YA
L
BR
[2011 (T-II)]
Sol. Since ABCD is a cyclic quadrilateral.
 ADC + ABC = 180°
 130° + ABC = 180°
 ABC = 50°
Since ACB is the angle in a semi-circle.
 ACB = 90°
Now, in ABC, we have
BAC + ACB + ABC = 180°
 BAC + 90° + 50° = 180°
 BAC = 40°
Sol. Since the opposite angles of a cyclic
quadrilateral are supplementary.
 BCD = 180° – 30° = 150°
and ADC = 180° – 110° = 70°
Now, EDC = 180° – 70° = 110°
[Linear pair]
21
Q.50. In the given figure, AB is the diameter
of the circle with centre O. If BAD = 70° and
DBC = 30°. Determine ABD and CDB.
Sol. Using AB as
diameter, draw a circle
which passes through A,
D, B and C.
BAC = BDC
[Angles in the same
segment are equal]
[2011 (T-II)]
Q.53. Prove that angle bisector of any angle
of a triangle and the perpendicular bisector of
the opposite side if intersect, they will intersect
on the circumcircle of the triangle.
[HOTS]
Sol.
PR TH
A K ER
AS S
HA
N
Sol. Since ABCD is a cyclic quadrilateral.
 BCD + BAD = 180°
 BCD + 70° = 180°
 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°
In ABD, we have
ABD + ADB + BAD = 180°
 ABD + 90° + 70° = 180°
 ABD = 20°
Hence, ABD = 20° and BDC = 40°
Q.51. If a line is drawn parallel to the base
of an isosceles triangle to intersect its equal
sides, prove that the quadrilateral so formed is
cyclic.
[HOTS]
Sol. AB = AC  ABC = ACB …(i)
ADE = ABC
[Corresponding
angles]
 ADE = ACB
[From (i)]
 ADE + EDB
= ACB + EDB
 ACB + EDB = 180°
[ ADE and EDB form a linear pair]
 BCED is cyclic.
[ Sum of a pair of opposite angles is 180°]
Q.52. On a common hypotenuse AB, two
right triangles ACB and ADB are situated on
opposite sides. Prove that ∠BAC = ∠BDC.
YA
L
BR
O
ABC is the given triangle and O is the centre
of its circumcircle. Then the perpendicular
bisector of BC passes through O. It cuts the
circle at P.
BOC = A
… (i)
[Angle at the centre
is twice the angle at the circumference]
OB = OC
[Radii of the same circle]
and OD  BC
 BOD = COD =
1
BOC
2
G
O
 BOD = COD = A [From (i)]
Now, CP makes A at the centre O. So, it
will make
A
at A.
2
Or, CAP =
A
2
 AP is the bisector of A.
Q.54. In the given figure, two circles
intersect each other at C and D. If ADE and BCF
are straight lines intersecting circles at A, B, F
and E. Prove that AB || EF.
[2010]
[HOTS]
22
Side BD of the cyclic
quadrilateral BCED is
produced to A.
ADE = ACB ...(ii)
[ext. ADE = int. opp. C]
From (i) and (ii), we get
ABC = ADE.
But, these are corresponding angles.
Hence, DE || BC.
Q.57. In the given figure, PQ and RS are two
parallel chords of a circle. When produced RP
and SQ meet at O. Prove that OP = OQ.
HA
N
Sol. In order to prove that AB || EF, it is
sufficient to prove that
1 + 3 = 180°
Since, ADCB is a cyclic quadrilateral.
1 + 2 = 180°
…(i)
Now, DEFC is a cyclic quadrilateral and in a
cyclic quadrilateral an exterior angle is equal to
the opposite interior angle.
 2 = 3
…(ii)
From equations (i) and (ii), we get
1 + 3 = 180°
Hence, AB || EF.
AS
AK
PR
BR
O
TH
ER
S
Q.55. AOB is a diameter of the circle and C,
D, E are any three points on the semicircle. Find
the value of ACD + BED. [2011 (T-II)]
G
O
YA
L
Sol. Join BC.
Then, ACB = 90° [angle in a semicircle]
Now, DCBE is a cyclic quadrilateral.
 BCD + DEB = 180°
ACB + BCD + DEB = 90° + 180°
[ ACB = 90°]
 ACB + DEB = 270°
[ ACB + BCD = ACD]
Q.56. In an isosceles ABC with AB = AC,
a circle passing through B and C intersects the
sides AB and AC at D and E respectively. Prove
that DE || BC.
[2011 (T-II)]
Sol. AB = AC
 ABC = ACB
.....(i)
23
[2010]
Sol. We have, PQ || RS
 OPQ = ORS and
OQP = OSR
[Corresponding angles]
But, PQRS is a cyclic
quadrilateral.
 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. Given : A ABC in which AB = AC and
a circle is drawn by taking AB as diameter which
intersects the side BC of triangle at D.
To prove : BD = DC
Construction : Join AD
Proof : Since angle in
a semicircle is a right
angle. Therefore,
ADB = 90°
ADB + AD
= 180°
 90° + ADC = 180°
 ADC = 90°
Now, in ABD and ACD, we have
AB = AC
[Given]
ADB = ADC
[Each equal to 90°]
and, AD = AD
[Common]
 ABD  ACD [By RHS congruence]
 BD = DC.
Q.59. If O is the centre of a circle as shown
in given figure, then prove that x + y = z.
BCE = 90° and CBE = CBD = 30°
 BCE + CBE + CEB = 180°
 90° + 30° + CEB = 180°
 CEB = 60°  AEB = 60°.
Q.61. In the given figure, P is the centre of
the circle. Prove that :
XPZ = 2(XZY + YXZ)
[2011 (T-II)]
HA
AS
PR
AK
Sol. Since arc XY subtends XPY at the
centre and XZY at a point Z in the remaining
part of the circle.
XPY = 2XZY
...(i)
Similarly, arc YZ subtends YPZ at the
centre and YXZ at a point Y in the remaining
part of the circle.
 YPZ = 2YXZ
...(ii)
Adding (i) and (ii), we get
XPY + YPZ = 2XZY + 2YXZ
 XPZ = 2(XZY + YXZ)
Q.62. If O is the circumcentre of a ABC
and OD  BC, prove that BOD = A
O
TH
ER
S
Sol. In ACF, side CF is
produced to B.
 y = 1 + 3 ...(i)
[ext.  = sum of int.
opp. angles]
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)]
N
[2011 (T-II)]
YA
L
BR
[2011 (T-II)]
O
Sol. Join OC, OD and BC.
In triangle OCD, we have
OC = OD = CD
[Each equal to radius]
 OCD is equilateral.
 COD = 60°
G
Sol. Join OB and OC.
In OBD and OCD, we have
OB = OC
[Each equal to the radius of circumcircle]
ODB = ODC
[Each equal to 90°]
and, OD = OD
[Common]
 OBD  OCD
 BOD = COD
 BOC = 2BOD = 2COD
Now, arc BC subtends BOC at the centre
1
COD  CBD = 30°
2
Since ACB is angle in a semi-circle.
 ACB = 90°
BCE = 180° – ACB = 180° – 90° = 90°
Thus, in BCE, we have
Now, CBD =
24
and BAC = A at a point in the remaining part
of the circle.
 BOC = 2A
2BOD = 2A [ BOC = 2BOD]
 BOD = A.
Q.63. Prove that the angle subtended by an
arc at the centre is double the angle subtended
by it at any point on the remaining part of the
circle.
[2010]
Sol. Given : A circle C(O, r) in which arc
AB subtends AOB at the
centre and ACB at any
point C on the remaining
part of the circle.
To prove : AOB =
O
 is a
2ACB, when AB
minor arc or a semicircle.
AK TH
A S ER
HA S
N
 is a
Reflex AOB = 2ACB, when AB
major arc.
Construction : Join AB and CO. Produce
CO to a point D outside the circle.
We know that when one side of a
triangle is produced then the exterior angle so
formed is equal to the sum of the interior
opposite angles.
 AOD = OAC + OCA
BOD = OBC + OCB
But, OAC = OCA [ OC = OA = r].
and OBC = OCB [ OC = OB = r].
 AOD = 2OCA and BOD = 2OCB
Now, AOD + BOD = 2OCA + 2OCB
 AOB = 2(OCA + OCB)
 AOB = 2ACB.
PR
4. In the figure, O is
the centre of the circle and
∠PQR = 100°.
Then the reflex ∠POR
is :
(a) 280° (b) 200°
(c) 260° (d) none of these
5. In the given figure,
E is any point in the interior of the circle with centre O. Chord AB = Chord
AC. If ∠OBE = 20°,then the
value of x is :
(a) 40°
(b) 45°
(c) 50°
(d) 70°
6. In the figure, ∠ABC
= 79°, ∠ACB = 41°, then
∠BDC is :
(a) 41°
(b) 79°
(c) 60°
(d) 50°
(d) 40°
O
YA
L
1 Mark Questions
1. In the given figure,
if O is the centre of the
circle and A is a point on
the circle such that CBA
= 40° and AD  BC, then
the value of x is
(a) 50°
(b) 90° (c) 45°
2. In the given figure, if
O is the centre of the circle
and A is a point on the circle
such that BOA = 120°,
then the value of x is
BR
PRACTICE EXERCISE 10B
G
(a) 120°
(c) 90°
[2011 (T-II)]
(b) 60°
(d) 30°
3. In the given figure, O
is the centre of the circle. If
AOB = 160°, then ACB
is
[2011 (T-II)]
(a) 160° (b) 200°
(c) 80°
(d) 100°
25
2 Marks Questions
7. In the given figure,
ABC = 45°, prove that
OA  OC.
[2011 (T-II)]
12. In the given figure,
A, B, C and D are four points
on the circle. AC and BD
intersect at a point E such
that BEC = 130° and
ECD = 20°. Find BAC.
8. In the given figure, ABCD is a cyclic quadrilateral and ABC = 85°. Find the measure of
ADE.
[2011 (T-II)]
[2011 (T-II)]
13. ABC is an isosceles triangle in which AB
= AC. A circle passing through B and C intersects
AB and AC at D and E respectively. Prove that
BC || DE.
14. O is the circumcentre of the triangle ABC
and D is the mid-point of the base BC. Prove that
∠BOD = ∠A.
15. A quadrilateral ABCD is inscribed in a
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]
L
10. O is the centre of the
circle as shown in figure. Find
CBD.
[2011 (T-II)]
3 Marks Questions
PR BR
AK O
AS THE
HA RS
N
9. In the figure, O is the
centre of the circle and BAC
= 60°. Find the value of x.
[2011 (T-II)]
18. Prove that the angles in a segment greater
than a semi-circle is less than a right angle.
[2011 (T-II)]
[HOTS]
G
O
Y
A
11. ABCD is a cyclic
quadrilateral and AB = AC if
ACB = 70°, find BDC.
TEXTBOOK’S EXERCISE 10.6 (OPTIONAL)
To prove : OAO = OBO
Construction : Join AO, BO, AO and BO.
Proof : In AOO and BOO, we have
AO = BO
[Radii of the same circle]
AO = BO
[Radii of the same circle]
OO = OO
[Common]
 AOO  BOO [SSS axiom]
Q.1. Prove that the line of centres of two
intersecting circles subtends equal angles at the
two points of intersection.
[2011 (T-II)]
Sol. Given : Two
intersecting circles, in which
OO is the line of centres and
A and B are two points of intersection.
26
 OAO= OBO [CPCT]
Hence, the line of centres of two intersecting
circles subtends equal angles at the two points of
intersection.
Proved.
Q.2. Two chords AB and CD of lengths 5 cm
and 11 cm respectively of a circle are parallel to
each other and are on opposite sides of its
centre. If the distance between AB and CD is
6 cm, find the radius of the circle.
[2010]
Substituting x =1 in (i), we get
5
2
5
2
r2 = (6 – x)2 +    r2 = (6 – 1)2 +  
2
2
5
2
25
 r2 = (5)2 +   = 25 +
4
2
 r2 =
125
5 5
r=
4
2
5 5
cm.
2
N
Hence, radius r =
AS
HA
Q.3. The lengths of two parallel chords of a
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?
PR
1
5
AB =
cm
2
2
1
11
and, CL = CD =
cm
2
2
ER
S
Then, AM =
YA
L
BR
O
TH
Since, AB || CD, it follows that the points O,
L, M are collinear and therefore, LM = 6 cm.
Let OL = x cm. Then OM = (6 – x) cm
Join OA and OC. Then OA = OC = r cm.
Now, from right-angled OMA and OLC,
we have
OA2 = OM2 + AM2 and OC2 = OL2 + CL2
[By Pythagoras Theorem]
5
 r2 = (6 – x)2 +  
2
 11 
O
G
1
2
...(i)
5
Sol. Let PQ and RS be two parallel chords
of a circle with centre O.
We have, PQ = 8 cm and RS = 6 cm.
Draw perpendicular bisector OL of RS
which meets PQ in M. Since, PQ || RS,
therefore, OM is also perpendicular bisector of
PQ.
Also, OL = 4 cm and RL = RS
2
 RL = 3 cm
and
and PM =
2
 r2 = x2 +  
2
 11 
1
PQ  PM = 4 cm
2
In ORL, we have
OR2 = RL2 + OL2 [Pythagoras theorem]
 OR2 = 32 + 42 = 9 + 16
... (ii)
2
[2010]
AK
Sol. Let O be the centre of the circle and let
its radius be r cm.
Draw OM  AB and OL  CD.
2
 (6 – x)2 +   = x2 +  
2
2
[From (i) and (ii)]
 OR2 = 25  OR = 25
 OR = 5 cm
 OR = OP
[Radii of the circle]
 OP = 5 cm
Now, in OPM
OM2 = OP2 – PM2 [Pythagoras theorem]
 OM2 = 52 – 42 = 25 – 16 = 9
25
121
= x2 +
4
4
121
25
 – 12x =
–
– 36
4
4
96
 – 12x =
– 36  – 12x = 24 – 36
4
 36 + x2 – 12x +
 – 12x = – 12  x = 1
OM =
27
9 = 3 cm
Hence, the distance of the other chord from
the centre is 3 cm.
Q.4. Let the vertex of an angle ABC be
located outside a circle and let the sides of the
angle intersect equal chords AD and CE with the
circle. Prove that  ABC is equal to half the
difference of the angles subtended by the chords
AC and DE at the centre.
[HOTS]
 OEB = 90° +
z
2
... (v)
Also, OED = ODE = 90° –
BDE = BED = 90° +
y

z
–  90  
2
2


N
yz
2
HA
  BDE = BED = y + z
  BDE = 180° – (y + z)
  ABC = 180° – (y + z)
Sol. Given : Two equal chords AD and CE
of a circle with centre O. When meet at B when
produced.
AK
= 180° – (y + z)
From (viii) and (ix), we have
ABC =
PR
... (vii)
... (viii)
yz
360   y  2 z  y

2
2
AS
Now,
xy
2
... (ix)
Proved.
Q.5. Prove that the circle drawn with any
side of a rhombus as diameter, passes through
the point of intersection of its diagonals.
z
... (ii)
2
YA
L
 OAD = 90° –
BR
O
TH
ER
S
Proof : Let AOC = x, DOE = y, AOD = z
EOC = z
[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]
... (vi)
from (iv), (v) and (vi), we have
  BDE = BED =
1
To Prove : ABC =
(AOC – DOE)
2
y
2
Similarly, OCE = 90° –
Sol. Given : A rhombus ABCD whose
diagonals intersect each other at O.
To prove : A circle with AB as diameter
passes through O.
Proof : AOB = 90°
[Diagonals of a
rhombus bisect each other at 90°]
 AOB is a right triangle right angled at O.
 AB is the hypotenuse of right AOB.
 If we draw a circle with AB as diameter,
then it will pass through O because angle in a
semicircle is 90° and AOB = 90° Proved.
Q.6. ABCD is a parallelogram. The circle
through A, B and C intersect CD (produced if
necessary) at E. Prove that AE = AD.
z
... (iii)
2
G
O
 ODB = OAD +ODA
[Exterior angle property]
 OEB = 90° –
z
+z
2
 ODB = 90° +
z
2
[From (ii)]
... (iv)
Also, OEB = OCE + COE
[Exterior angle property]
 OEB = 90° –
z
2
+ z [From (iii)]
[2011 (T-II)]
28
Sol. In order to prove that AE = AD it is
sufficient to prove that AED = ADE.
OA = OC
[Given]
OB = OD
[Given]
AOB = COD [Vertically opposite angles]
 AOB  COD
[SAS congruence]
 ABO = CDO and BAO = DCO
[CPCT]
  AB || DC
N
Similarly, we can prove BC || AD ... (ii)
HA
Hence, ABCD is a parallelogram.
But ABCD is a cyclic parallelogram.
AS
 ABCD is a rectangle.
[Proved in Q.12 of textbooks exercise 10.5]
AK
 ABC = 90° and BCD = 90°
PR
 AC is a diameter and BD is a diameter.
[Angle in a semicircle is 90°] Proved.
Q.8. Bisectors of angles A, B and C of a
triangle ABC intersect its circumcircle at D, E
and F respectively. Prove that the angles of the
TH
ER
S
Since ABCE is a cyclic quadrilateral.
 AED + ABC = 180°
…(i)
Now, CDE is a straight line
 ADE + ADC = 180°
…(ii)
But, ADC and ABC are opposite angles
of a parallelogram.
 ADC = ABC
 ABC + ADE = ADC + ADE
 ABC + ADE = 180°
…(ii)
From equations (i) and (ii), we get
AED + ABC = ADE + ABC
 AED = ADE
Thus, in AED, we have
ADE = AED
 AE = AD
... (i)
triangle DEF are 90° –
90° –
1
C.
2
1
1
A, 90° –
B and
2
2
[HOTS]
O
Sol. Given : ABC and its circumcircle.
AD, BE, CF are bisectors of A, B, C
respectively.
G
O
YA
L
BR
Q.7. AC and BD are chords of a circle which
bisect each other. Prove that (i) AC and BD are
diameters, (ii) ABCD is rectangle.
Sol. Given : A circle with chords AB and
CD which bisect each other at O.
To Prove : (i) AC and BD are diameters
Construction : Join DE, EF and FD.
Proof : We know that angles in the same
segment are equal.
(ii) ABCD is a rectangle.
 5 =
Proof : In OAB and OCD, we have
29
B
C
and 6 =
2
2
..(i)
1 =
A
C
and 2 =
2
2
...(ii)
4 =
B
A
and 3 =
2
2
...(iii)
But the circles are congruent.
arc ADB = arc AEB APB = AQB
[Equal arcs subtend equal angles]
 BP = BQ
[Sides opposite to equal angles are equal]
Proved.
Q.10. In any triangle ABC, if the angle
bisector of A and perpendicular bisector of BC
intersect, prove that they intersect on the
circumcircle of the triangle ABC.
[HOTS]
From (i), we have
B
C
+
2
2

B
C
 D =
+
2
2
5 + 6 =
...(iv)
 (iv) becomes, D = 90° –
HA
AS
B
C
A
+
= 90° –
2
2
2
A
.
2
AK

N
[ 5 + 6 = D]
But A + B + C = 180°
 B + C = 180° – A
Sol. Let angle bisector of A intersect
circumcircle of ABC at D.
Join DC and DB.
BCD = BAD
[Angles in the same segment]
Similarly, from (ii) and (iii), we can prove
B
C
and F = 90° –
2
2
TH
ER
S
Proved.
Q.9. Two congruent circles intersect each
other at points A and B. Through A any line
segment PAQ is drawn so that P, Q lie on the two
circles. Prove that BP = BQ.
[2011 (T-II)]
PR
that E = 90° –
O
Similarly, DBC = DAC =
1
 A ... (ii)
2
BR
From (i) and (ii) DBC = BCD
  BD = DC
[Sides opposite to equal angles are equal]
 D lies on the perpendicular bisector
of BC.
Hence, angle bisector of A and
perpendicular bisector of BC intersect on the
circumcircle of ABC.
Proved.
O
YA
L
Sol. Given : Two congruent circles which
intersect at A and B. PAB is a line through A.
To Prove : BP = BQ.
Construction : Join AB.
Proof : AB is a common chord of both the
circles.
G
1
 BCD = BAD =  A
…(i)
2
[AD is bisector of A]
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
N
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.
AS
HA
Figure-1
AK
2. Take any point P on the remaining part
of the circle. Join OA, OB, PA and PB.
ER
S
PR
Figure-2
BR
O
TH
3. Make two replicas of APB using tracing
paper. Shade the angles using different
colours.
G
O
YA
L
Figure-3
4. Paste the two replicas of APB adjacent
to each other on AOB as shown in the
figure.
Figure-4
31
Observations :
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.
N
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.
HA
Do Yourself : Verify the above property by taking three circles of different radii.
AS
Activity-2
AK
Objective : To verify that the angles in the same segment of a circle are equal, using the method of
paper cutting, pasting and folding.
S
PR
Materials Required : White sheets of paper, tracing paper, a pair of scissors, gluestick, colour pencils,
geometry box, etc.
ER
Procedure :
BR
O
TH
1. On a white sheet of paper, draw a circle of any
convenient radius. Draw a chord AB of the
circle.
Figure-1
G
O
YA
L
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
HA
N
3. On a tracing paper, trace each of the angles APB, AQB and ARB. Shade the traced copies using
different colours.
Figure-3
Figure-4
O
TH
ER
S
PR
AK
AS
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.
Observations :
BR
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.
YA
L
So, APB = AQB = ARB
O
Conclusion : From the above activity, it is verified that the angles in the same segment of a circle are
equal.
G
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
Procedure : (a) To verify that the angle in a semicircle is a right angle :
HA
N
1. On a white sheet of paper, draw a circle of any convenient radius with centre O. Draw its
diameter AB as shown.
ER
S
PR
2. Take any point P on the semicircle. Join A to P
and B to P.
AK
AS
Figure-1
Figure-2
YA
L
BR
O
TH
3. Make two replicas of APB on tracing paper. Shade the replicas using different colours.
Figure-3
G
O
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.
AS
HA
N
Figure 5
PR
AK
2. Take any point P on the major segment. Join P to A and P to B.
Figure-6
Figure-7
YA
L
BR
O
TH
ER
S
3. Trace APB on a tracing paper.
G
O
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.
PR
AK
AS
HA
N
Figure-9
O
TH
3. Trace APB on a tracing paper.
ER
S
Figure-10
Figure-11
G
O
YA
L
BR
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
Or 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.
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Hence, the angle in a major segment is acute.
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5. In figure 11, APB is an angle formed in the minor segment of a circle.
ie, APB is greater than XYZ or APB is obtuse.
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Conclusion : From the above activity, it is verified that :
(a) the angle in a semicircle is a right angle.
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Hence, the angle in a minor segment is obtuse.
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6. In figure 12, we see that the side PB of PAB lies to the left of XY of XYZ
(b) the angle in a major segment is acute.
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(c) the angle in a minor segment is obtuse.
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Activity-4
Objective : To verify using the method of paper cutting, pasting and folding that
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(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.
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Materials Required : White sheets of paper, tracing paper, colour pencils, a pair of scissors, gluestick,
geometry box, etc.
Procedure :
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(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
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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.
Figure-2
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3. On a white sheet of paper, paste P and R adjacent to each other. Similarly, paste Q and S
adjacent to each other.
(b) 1. Repeat step 1 of part (a).
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Figure-3
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2. Extend PQ to PT to form an exterior angle RQT. Shade RQT.
Figure-4
3. Trace PSR on a tracing paper and colour it.
Figure-5
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4. Paste the traced copy of PSR on RQT such that S falls at Q and SP falls along QT.
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Figure-6
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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.
ANSWERS
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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.
6. 2.4 cm
7. 17 cm
8. 8.5 cm
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Practice Exercise 10A
1. (a) 2. (d) 3. (b) 4. 24 cm 5. 12 cm
9. 6 cm 10. 8.58 cm, 11. 13 cm
Practice Exercise 10B
1. (d) 2. (b) 3. (d)
15. 40° 17. 30°
4. (b)
5. (d) 6. (c) 8. 85° 9. 240° 10. 50° 11. 140°
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12. 110°