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CIRCLES

A. SUMMATIVE ASSESSMENT

10

10.1 CIRCLES AND ITS RELATEDTERMS : A REVIEW

1. The collection of all the points in a plane,which are at a fixed distance from a fixed pointin the plane, is called a circle.

2. The fixed point is called the centre of thecircle and the fixed distance is called the radiusof the circle.

In the given figure, O is the centre and thelength OP is the radius of the circle.

3. A circle divides the plane on which it liesinto 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 isalso called the exterior of the circle. The circleand its interior make up the circular region.

4. A chord of a circleis a line segment joiningany two points on thecircle. In the given figurePQ, RS and AOB are thechords of a circle.

5. A diameter is a chord of a circle passing

through the centre of the circle. In the givenfigure, AOB is the diameter of the circle. Adiameter is the longest chord of a circle.

Diameter = 2 × radius6. A piece of a circle between two points is

called an arc. Look at the pieces of the circlebetween two points P and Q in the given figure.You find that there are two pieces, one longerand the other smaller. The longer one is calledthe major arc PQ and the shorter one is calledthe minor arc PQ.

7. The length of the complete circle is calledits circumference. Theregion between a chord andeither of its arcs is called asegment of the circularregion or simply a segmentof the circle. You will findthat there are two types ofsegments also, which are the major segment andthe minor segment.

8. The region between an arc and the tworadii, joining the centre to the end points of thearc is called a sector. Likesegments, you find that theminor arc corresponds tothe minor sector and themajor arc corresponds tothe major sector.

Question Bank In Mathematics Class IX (Term II)

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TEXTBOOK’S EXERCISE 10.1

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 centreof 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 endsare the ends of a diameter.

(v) Segment of a circle is the region betweenan arc and __________ of the circle.

(vi)A circle divides the plane, on which itlies, in __________ parts.

Sol. (i) interior (ii) exterior (iii) diameter(iv) semicircle (v) the chord (vi) three

Q.2. Write True or False: Give reasons foryour answers.

(i) Line segment joining the centre to anypoint on the circle is a radius of the circle.

(ii) A circle has only finite number of equalchords.

(iii) If a circle is divided into three equalarcs, each is a major arc.

(iv) A chord of a circle, which is twice aslong as its radius, is a diameter of the circle.

(v) Sector is the region between the chordand its corresponding arc.

(vi) A circle is a plane figure.

Sol. (i) True, because all points on the circleare equidistant from its centre.

(ii) False, because there are infinitely manypoints on the circle.

(iii) False, because for each arc, theremaining arc will have greater length.

(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 CHORDAT A POINT

1. Equal chords of a circle subtend equalangles at the centre.

2. If the angles subtended by the chords of acircle at the centre are equal, then the chords areequal.

TEXTBOOK’S EXERCISE 10.2

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]

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 = COD

Proof : In triangles AOB and COD,AB = CD [Given]

AO = CO

BO = DO

[Radii of congruent circles]

AOB COD [SSS axiom]

AOB = COD [CPCT] Proved.

Q.2. Prove that if chords of congruent

circles subtend equal angles at their centres,

then the chords are equal. [2010]

Sol.

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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.

Maximum number of common points = 2

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)]

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 =

COD

To Prove : AB = CD

Proof : In triangles AOB and COD,

AO = CO

BO = DO

[Radii of congruent circle]

AOB = COD [Given]

AOB COD [SAS axiom]

AB = CD [CPCT] Proved.

10.3 PERPENDICULAR FROM THECENTRE TO A CHORD

1. The perpendicular from the centre of acircle to a chord bisects the chord.

2. The line drawn through the centre of a circleto bisect a chord is perpendicular to the chord.

3. There is one and only one circle passingthrough three given non-collinear points.

TEXTBOOK’S EXERCISE 10.3

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.

Proof : In AOO and BOO

AO = OB [Radii of the circle (O, r)]

AO= BO [Radii of the circle (O, r)]

OO = OO [Common]

AOOBOO [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 andACO =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.

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TEXTBOOK’S EXERCISE 10.4

10.4 EQUAL CHORDS AND THEIRDISTANCES FROM THE CENTRE

1. Equal chords of a circle (or of congruent

circles) are equidistant from the centre (or centres).

2. Chords equidistant from the centre of acircle are equal in length.

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)]

Sol. In AOO,

AO2 = 52 = 25

AO2 = 32 = 9

OO2 = 42 = 16

AO2 + OO2

= 9 + 16 = 25 = AO2

AOO= 90° [By converse of

Pythagoras theorem]

Similarly, BOO = 90°.

AOB= 90° + 90° = 180°

AOB is a straight line, whose mid-pointis O.

AB = (3 + 3) cm = 6 cm

Q.2. If two equal chords of a circle intersectwithin the circle, prove that the segments of onechord are equal to corresponding segments ofthe other chord. [V. Imp.]

Sol. Given : AB and CD are two equalchords of a circle which meet at E.

To prove : AE = CE and BE = DE

Construction : Draw OM AB andON 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]

Now, AB = CD [Given]

1

2AB =

1

2CD

AM = CN ...(ii) [Perpendicular from

centre bisects the chord]

Adding (i) and (ii), we get

EM + AM = EN + CN

AE = CE ...(iii)

Now, AB = CD ...(iv)

AB – AE = CD – AE [From (iii)]

BE = CD – CE Proved.

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)]

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 = DEQConstruction : Draw OL AB and

OM CD.

Proof : In OLE and OME, we have

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 centreO at A, B, C and D, prove that AB = CD (Seefig.) [2011 (T-II)]

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Sol. Given : A lineAD intersects twoconcentric circles at A,B, C and D, where O isthe centre of thesecircles.

To prove : AB = CDConstruction : Draw OM AD.Proof : AD is the chord of larger circle.AM = DM ..(i) [OM bisects the chord]

BC is the chord ofsmaller circle

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

Subtracting (ii) from(i), we get

AM – BM = DM – CM AB = CD Proved.

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 berepresented by R, S and M respectively.

Draw OL RS, OL2 = OR2 – RL2

OL2 = 52 – 32 [RL = 3 m, because OL RS]

= 25 – 9 = 16

OL = 16 = 4 mNow, area of

triangle ORS

=1

2× KR × OS

=1

2× KR × 5

Also, area of ORS =1

2× RS × OL

=1

2× 6 × 4 = 12 m2

1

2× KR × 5 = 12

KR =12 2 24

5 5

= 4.8 m RM = 2KR

RM = 2 × 4.8 = 9.6 mHence, distance between Reshma and

Mandip is 9.6 m.

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]

Sol. Let Ankur, Syed and David be

represented by A, S and D respectively.

Let PD = SP = SQ = QA = AR = RD = x

In OPD, OP2 = 400 – x2

OP = 2400 x

AP = 2 22 400 400x x

[ centroid divides the

median in the ratio 2 : 1]

= 23 400 x

Now, in APD, PD2 = AD2 – AP2

x2 = (2x)2 – 2

23 400 x

x2 = 4x2 – 9(400 – x2)x2 = 4x2 – 3600 + 9x2 12x2 = 3600

x2 =3600

12= 300 x = 10 3

Now, SD = 2x = 2 × 10 3 = 20 3

ASD is an equilateral triangle.

SD = AS = AD = 20 3Hence, length of the string of each phone is

20 3 m.

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OTHER IMPORTANT QUESTIONS

Q.1. In the given figure, O is the centre ofthe circle. If OA = 5 cm and OC = 3 cm, then thelength of AB is : [2011 (T-II)]

(a) 4 cm (b) 6 cm (c) 8 cm (d) 15 cm

Sol. (c) 2 2AC = AO – OC = 25 – 9 cm

= 4 cm

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 isat a distance of 4 cm from the centre O. Thelength of another chord CD which is also 4 cmaway from the centre is :

(a) 6 cm (b) 4 cm (c) 8 cm (d) 3 cmSol. (a) Chords equidistant from the centre

are equal.

Q.4. In the figure,chord AB is greater thanchord CD. OL and OMare the perpendicularsfrom the centre O on thesetwo chords as shown inthe figure. The correctreleation between OL and OM is : [HOTS]

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

(c) OL > OM (d) none of theseSol. (b) Longer the chord, shorter is its

distance from the centre.

Q.5. There are three non-collinear points.The number of circles passing through themare : [2010]

(a) 2 (b) 1 (c) 3 (d) 4Sol. (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

(c) 5 cm (d) 4.7 cm

Sol. (c) In AMO, M = 90° OA2 = AM2 + OM2

OM = 2 2OA AM

OM = 2 2(13) (12) = 5 cm

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 theseSol. (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 atthe centre.

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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 theseSol. (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) 2 2OC = AO – AC

= 25 – 16 cm = 3 cm

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

Sol. (b) OQ = 2 2OC – CQ

= 25 9 cm = 4 cm

2 2OP = OA – AP = 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]

(a)5

2cm (b)

5 3

2cm

(c) 5 cm (d)5 3

4cm

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.

Sol. 2 2

2 2

AB = 2 AO – OC

= 2 13 – 5 cm

= 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

BM = CM =1

2BC = 4 cm

Similarly, OM AD

AM = DM =1

2AD = 6 cm

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.

[2011 (T-II)]

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Sol. Let O and Obe the centres of thecircles of radii 10 cmand 8 cm respectivelyand let PQ be theircommon chord.

We have,OP = 10 cm, OP = 8 cm and PQ = 12 cm

PL =1

2PQ = 6 cm

In right OLP, we have OP2 = OL2 + LP2

OL = 2 2OP LP = 2 210 6

= 64 cm = 8 cm

In right OLP,

we have OP2 = OL2 + LP2

OL = 2 2 2 2O P LP 8 6 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 = OA = OB

In AOB and AOB, we haveAO = AOOB = OBAB = AB [Common]

AOBAO B [SSS congruency axiom]

AOB = AO B [CPCT]Q.17. AOB is a diameter of a circle and C is

a point on the circle.

Check whether AC2 + BC2

= AB2 is true or

not. [Imp.]

Sol. True. We know

that ACB = 90°

[Angle made in semi-circle]

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.

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]

PA = PB [CPCT]

arc PXA = arc PYB

arc PXA arc PYB Proved.

Q.20. Show that two circles cannot intersectat more than two points. [Imp.]

Sol.Let us assume that two circles intersectat three points say A, B and C. Then clearly, A,B and C are not collinear. But, through threenon-collinear points we can draw one and onlyone circle. Therefore, we cannot have two circlespassing A, B and C. Or two circles cannotintersect at more than two points.

Q.21. AB and AC are two chords of a circleof radius r such that AB = 2AC. If p and q arethe distances of AB and AC from the centre,prove that 4q2 = p2 + 3r2. [2011 (T-II)]

Sol. Draw OD AB,

OEAC and join AO.

Let AC = 2x, then

AE = CE = x

So, AB = 4x and

AD = BD = 2x.

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In AOD,

AO2 = AD2 + OD2 [By Pythagoras theorem]

AD2 = AO2 – OD2

4x2 = r2 – p2 … (i)In AOE, we haveAO2 = OE2 + AE2 AE2 = AO2 – OE2

x2 = r2 – q2

4x2 = 4r2 – 4q2 … (ii)From (i) and (ii), we have

4r2 – 4q2 = r2 – p2 4q2 = p2 + 3r2.Proved.

Q.22. In the givenfigure, equal chords ABand CD of a circle cutat right angles at E. IfM and N are the mid-points of AB and CDrespectively. Prove thatOMEN 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

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 perpendi-cular to the chord AB of acircle whose centre is O. IfBC is a diameter, showthat CA = 2OD.

[2011 (T-II)]

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.

OD || AC and OD =1

2CA

[ segment joining the mid-points of two

sides of a triangle is half of the third side.]

CA = 2OD. Proved.

PRACTICE EXERCISE 10A

1 Mark Questions

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°

2. In the given figure, O is

the centre of the circle, OA =

10 cm and OC = 6 cm. The

length of AB is

(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

(c) 89 cm

(d) 6 cm

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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 circleof radius 5.1 cm passes through P and Q. Calculatethe distance of its centre from the chord PQ.

7. Two circles of radii 26 cm and 25 cm inter-sect at two points which are 48 cm apart. Find thedistance between their centres.

8. Two parallel chords of a circle whose diam-eter is 13 cm are respectively 5 cm and 12 cm.Find the distance between them if they lie on op-posite sides of the centre.

3/4 Marks Questions

9. PQ is a variable chord of a circle of radius7.5 cm. If PQ = 9 cm, find the radius of the circlewhich is the locus of the mid-point of PQ.

10. In a circle of radius 5 cm, there are twoparallel chords of length 6 cm and 4 cm. Find thedistance between them, when they are on (i) oppo-site 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 otherat 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

the circle is2 2l h

h

.

[HOTS]

13. AB and CD are two equal chords of a circle.M and N are their mid-points respectively. Provethat MN makes equal angles with AB and CD.

[HOTS]

14. A chord AB of a circle (O, r) is producedto P so that BP = 2 AB. Prove that

OP2 = OA2 + 6AB2. [HOTS]

10.5 ANGLE SUBTENDED BY AN ARC OFA CIRCLE AND CYCLIC QUADRILATERAL

1. The angle subtended by an arc at thecentre is double the angle subtended by it at anypoint on the remaining part of the circle.

2. Angles in the same segment of a circle areequal.

3. If a line segment joining two points

subtends equal angles at two other points lyingon the same side of the line containing the linesegment, the four points lie on a circle (i.e., theyare concyclic).

4. The sum of either pair of opposite anglesof a cyclic quadrilateral is 180°.

5. If the sum of a pair of opposite angles ofa quadrilateral is 180°, the quadrilateral is cyclic.

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.

[2010]

Sol. We have, BOC = 30° and AOB = 60°

AOC = AOB + BOC = 60° + 30° = 90°

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. 2ADC = AOC

ADC =1

2AOC =

1

2× 90°

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.]

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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 = 2ACB

ACB =1

2AOB =

1

2× 60°

ACB = 30°

Also, ADB =1

2reflex AOB

=1

2(360° – 60°) =

1

2× 300° = 150°

Hence, angle subtended by the chord at apoint on the minor arc is 150° and at a point onthe major arc is 30°

Q.3. In the figure,PQR = 100°, where P, Qand R are points on a circlewith centre O. Find OPR.

[2011 (T-II)]

Sol. Reflex anglePOR = 2PQR

= 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 propertyof a triangle]

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 ona circle. AC and BDintersect at a point E suchthat BEC = 130° andECD = 20°. Find BAC.

[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°

Q.6. ABCD is a cyclicquadrilateral whose dia-gonals intersect at a pointE. If DBC = 70°,BAC = 30°, find BCD.Further, if AB = BC, findECD. [V. Imp.]

Sol. CAD = DBC = 70° [Angles in thesame segment]

Therefore, DAB = CAD + BAC= 70° + 30° = 100°

But, DAB + BCD = 180° [Oppositeangles of a cyclic quadrilateral]

So, BCD = 180° – 100° = 80°Now, we have AB = BCTherefore, 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]

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100° + 30° + ECD = 180° 130° + ECD = 180° ECD = 180° – 130° = 50°

Hence, BCD = 80° and ECD = 50°.

Q.7. If diagonals of a cyclic quadrilateralare diameters of the circle through the verticesof the quadrilateral, prove that it is a rectangle.

[2010]

Sol. Given : ABCD is acyclic quadrilateral, whosediagonals AC and BD arediameter of the circle passingthrough A, B, C and D.

To Prove : ABCD is arectangle.

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 || BCSimilarly, 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 acyclic quadrilateral is 180°]

ABC = ADC = 90° [From (i) and (ii)]

ABCD is a rectangle. [A ||gm one ofwhose angles is 90° is a rectangle] Proved.

Q.8. If the non-parallel sides of a trapeziumare equal, prove that it is cyclic.

[2010, 2011 (T-II)]

Sol. Given : A trapezium ABCD in whichAB || 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.

Q.9. Two circles intersect at two points Band C. Through B, two line segments ABD andPBQ are drawn to intersect the circles at A, Dand P, Q respectively (see Fig.). [2011 (T-II)]

Prove that ACP = QCD.

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

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OTHER IMPORTANT QUESTIONS

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 getACP = QCD. Proved.

Q.10. If circles are drawn taking two sidesof a triangle as diameters, prove that the pointof intersection of these circles lie on the thirdside. [2011 (T-II)]

Sol. Given : Sides ABand AC of a triangle ABCare diameters of two circleswhich intersect at D.

To Prove : D lies on BC.Proof : Join ADADB = 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 BCHence, point of intersection of circles lie on

the third side BC. Proved.

Q.11. ABC and ADC are two right triangleswith common hypotenuse AC. Prove thatCAD = CBD. [2011 (T-II)]

Sol. Given : ABC and ADC are two righttriangles with common hypotenuse AC.

To Prove : CAD = CBD

Proof : Let O be themid-point of AC.

Then OA = OB = OC= OD

Mid point of the hypo-tenuse of a right triangle isequidistant from its verticeswith O as centre and radius equal to OA, drawa circle to pass through A, B, C and D.

We know that angles in the same segment ofa circle are equal.

Since, CAD and CBD are angles of thesame segment.

Therefore, CAD = CBD. Proved.Q.12. Prove that a cyclic parallelogram is a

rectangle. [2010]

OrProve that a parallelogram inscribed in a

circle is a rectangle. [2011 (T-II)]

Sol. Given : ABCD is acyclic parallelogram.

To prove : ABCD is arectangle.

Proof :ABC = ADC ...(i)

[Opposite angles of a ||gm are equal]But, ABC + ADC = 180° ... (ii)

[Sum of opposite angles of a cyclicquadrilateral is 180°]

ABC = ADC = 90° [From (i) and (ii)] ABCD is a rectangle

[A ||gm one of whose angles is90° is a rectangle]

Hence, a cyclic parallelogram is a rectangle.Proved.

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 =180 40

2

= 70°

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Q.2. For what valueof x in the figure, pointsA, B, C and D areconcyclic ? [2011 (T-II)]

(a) 9°(b) 10°(c) 11°(d) 12°

Sol. (b) Since, opposite angles of a cyclicquadrilateral are supplementary.

81° + x + 89° = 180°

x = 180° – 170° = 10°.Q.3. In the given

figure, O is the centreof 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 = 2ACB [Angle at the centre is

twice the angle at the circumference]

x = 2 × 30° = 60°.Q.4. Angle inscribed in a semicircle is :

[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 ofthe circle. If QPR is 50°,then QOR is : [2010]

(a) 130°(b) 40°(c) 100°(d) 50°

Sol. (c) QOR = 2QPR[The angle subtended by an arc at the

centre is double the angle subtended by it atany 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°

Sol. (a) x =1

2× 70°

= 35° [Angle at thecentre is double the angleat the circumference]

y = x = 35°[Angles in the same segment are equal]

Q.7. In the givenfigure, if POQ is adiameter of the circleand PR = QR, thenRPQ is equal to :

[2011 (T-II)]

(a) 30° (b) 60°(c) 90° (d) 45°

Sol. (d) PRQ = 90°[ angle in a semicircle is a right angle]

PR = PQP = Q = 45°Q.8. ABCE is a cyclic

quadrilateral. 'O' is thecentre of the circle andAOC = 150°, thenCBD is : [2011 (T-II)]

(a) 225° (b) 128°(c) 150° (d) 75°

Sol.(d) Since the angle subtended by an arcat the centre of a circle is twice the anglesustended at a point on the remaining part of thecircumference, we have

AEC =1

2AOC =

1

2× 150° = 75°

Now, ABCE is a cyclic quadrilateral whoseside AB is produce to D

CBD = AEC = 75°[ ext. of cyclic quad. = int. opp. ]

Q.9. 'O' is the centreof the circle QPS = 65°;PRS = 33°, PSQ isequal to : [2011 (T-II)]

(a) 90° (b) 82°

(c) 102° (d) 42°

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Sol. (b) R = Q = 33° [Angles in thesame segment are equal]

Now, in PQS, P + Q + S = 180°PSQ = 82°Q.10. In the given

figure, AB is a diameter ofthe circle. CD || AB andBAD = 40°, then ACDis : [2011 (T-II)]

(a) 40° (b) 90°(c) 130° (d) 140°

Sol. (c) ADB = 90° [Angle in a semicircleis 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 yis : [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 thediameter of the circle andB = 35°, then x is equalto : [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 ofthe circle. If OAB = 30°and OCB = 40°, thenmeasure of AOC is :

[2011 (T-II)]

(a) 70° (b) 220°

(c) 140° (d) 110°Sol. (c) OA = OBOBA = OAB = 30°OC = OB OBC = OCB = 40°ABC = OBA + OBC = 30° + 40°

= 70°Now, AOC = 2ABC = 2 × 70° = 140°Q.14. ABCD is a cyclic

quadrilateral as shown inthe 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]

(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 ata point on the remaining part of thecircumference, we have

AEC =1

2AOC =

1

2× 130° = 65°

Now, ABCE is a cyclic quadrilateral whoseside 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 :[Imp.]

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(a) 30° (b) 40° (c) 60° (d) 160°

Sol. (b) x =1

2AOB [Angle at the centre

is double the angle at the circumference]

1

2× 80° = 40°.

Q.17. In the figure, O is the centre of thecircle. If ∠OAB = 40°, then ∠ACB is equal to :

[Imp.]

(a) 50° (b) 40° (c) 60° (d) 70°Sol. (a) OA = OB OAB = OBA = 40°

AOB = 180° – (40° + 40°) = 100°

ACB =1

2AOB [Angle at the centre is

double the angle at the circumference] ACB = 50°.

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°

ADB = ACB [Angles in thesame 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) 20° (b) 40° (c) 60° (d) 10°Sol. (b) AOC = 2ABC [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]

(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 = 2BCA = 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]

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(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.]

(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°, thenthe measures of ∠POR and ∠PQR arerespectively :

(a) 90°, 45° (b) 105°, 45°(c) 110°, 55° (d) 100°, 50°

Sol. (a) OP = OQ

OQP = OPQ = 25° [Radii of samecircle]

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°.

Q.24. In the figure, O is the centre of thecircle with ∠AOB = 85° and ∠AOC = 115°.Then ∠BAC is : [Imp.]

(a) 115° (b) 85° (c) 80° (d) 100°Sol. (c) BOC = 360° – (85° + 115°) = 160°

BAC =1

2BOC [Angle at the centre is

double the angle at the circumference]

BAC = 80°.Q.25. In the figure, if ∠CAB = 40° and AC

= BC, then ∠ADB equal to :

(a) 40° (b) 60° (c) 80° (d) 100°Sol. (c) AC = BC CBA = CAB = 40°

ACB = 180° – (40° + 40°) = 100°ACB + ADB = 180° [Opposite anglesof 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]

(a) 40° (b) 70° (c) 110° (d) 30°

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Sol. (b) ADB = 180° – (70° + 40°)= 180° – 110° = 70° and ACB= ADB = 70°.

[Angles in the same segment are equal]

Q.27. In the givenfigure, if ∠AOC = 130°, Ois the centre of the circlethen ∠ABC is : [2010]

(a) 65° (b) 115°

(c) 130° (d) 50°

Sol. (b)AEC =1

2AOC =

1

2× 130° = 65°

Now, ABC = 180° – 65° = 115°.

Q.28. In the givenfigure, the value of

(∠BCD + ∠DEB) is :[2010]

(a) 270° (b) 180°(c) 360° (d) 90°

Sol. (b) Opposite angles of a cyclicquadrilateral are supplementary.

Q.29. In the givenfigure, ∠C = 40° ∠CEB =105°, then the value of x is :

[2010]

(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°

Q.30. In the figure, ∠AOB = 90° and∠ABC = 30°, then ∠CAO is equal to : [V. Imp.]

(a) 30° (b) 45° (c) 90° (d) 60°

Sol. (d) ACB =1

2AOB [Angle at the

centre is double the angle at the circumference]

ACB = 45°OA = OB OAB = OBA

OAB = OBA =180 90

2

= 45°

Now,CAB = 180° – (45° + 30°) = 105°

CAO = CAB – OAB= 105° – 45° = 60°.

Q.31. In the figure,BC is a diameter of thecircle and ∠BAO = 60°.Then ∠ADC is equal to :

(a) 30° (b) 45°(c) 60° (d) 120°

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 ofa circle and ∠BOA =90°, ∠COA = 110°. Findthe measure of ∠BAC.

[2011 (T-II)]

Sol. We have,∠BOA = 90° and ∠AOC = 110°∴ ∠BOC = 360° – (∠BOA + ∠AOC)⇒ ∠BOC = 360° – (90° + 110°) = 160°

Now, ∠BAC =1

2∠BOC =

1

2× 160° = 80°

Q.33. In the figure,ABC is an equilateraltriangle. Find ∠BDC and∠BEC. [2010]

Sol. BAC = 60°[ ABC is equilateral]

BAC = BDC

[Angle in the same segment are equal]

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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 PRand PS are diameters, show that RQS is astraight line. [2010]

Sol. We aregiven two circleswith centres C1 andC2 which intersecteach other at P andQ. PR is a diameterof circle C1 and PS is a diameter of circle C2.

We need to prove that R, Q and S arecollinear.

PQR = 90° …[Angle in a semicircle]andPQS = 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 provethat 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 ABCin which AB = AC. IfCD is produced topoint E such thatBD = CE, prove thatAD = AE. [2010]

Sol. We have,AB = AC andCE = BD

In s ABD and ACE, we haveAB = AC [Given]ABD = ACE [Angles in the same segment]BD = CE [Given]So, by SAS congruence criterion, we haveABD = ACE AD = AEQ.37. In the figure, if AOB is a diameter and

∠ADC = 120° find ∠CAB. [V. Imp.]

Sol. Join AD and AC.ADC + ABC

= 180° [Opposite anglesof 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 themeasure of BAC. [2011 (T-II)]

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 quadri-lateral, if BCD = 120° andABD = 50°, then find ADB.

[2010, 2011 (T-II)]

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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.[2010]

Sol. In ABD, ABD = 35° given andBAD = 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°, thenfind BOC. [2010]

Sol. OA = OB OAB = OBA = 45°OC = OA OAC = OCA = 35°BAC =OAB +OAC = 45° + 35° = 80°Now, BOC = 2BAC = 2 × 80° = 160°Q.42. In the given figure,

AC is the diameter of thecircle. If ACB = 55°, thenfind 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]

Q.43. In the given figure, if O is the centre ofthe circle and AOC = 110° and AB is producedto D then find AEC and ABC. [2010]

Sol. Since arc ABC makes AOC = 110° atthe centre and AEC at a point E on thecircumference.

AEC =1

2AOC =

1

2× 110° = 55°

Again arc CEA makes angle reflex AOC= (360° – 110°) = 250° at the centre and ABCat a point B on the circumference.

ABC =1

2reflex BOD =

1

2× 250°

= 125°.Q.44. In the given figure, if

O is the centre of the circle,AOC = 50° and COB =30°. Find the measure ofADB. [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 anglesubtended by the same arc on the remaining partof the circle.

2ADB = AOB

ADB =1

2AOB =

1

2× 80° = 40°.

Q.45. In the given figure, if AB = AC, BEC= 100°, then find the values of x and y. [2010]

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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 = x2x + 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]

Sol. In ABC, C = 90°, because angle in asemicircle.

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 BCproduced to meet at the point E. If DAB = 30°and ABC = 110°, then find DEC. [2010]

Sol. Since the opposite angles of a cyclicquadrilateral are supplementary.

BCD = 180° – 30° = 150°and ADC = 180° – 110° = 70°Now, EDC = 180° – 70° = 110°

[Linear pair]

and ECD = 180° – 150° = 30°[Linear pair]

DEC = 180° – (110° + 30°)= 180° – 140° = 40°

Q.48. In the given figure, ABCD is a cyclicquadrilateral in which AB is produced to F andBE || DC. If FBE = 20° and DAB = 95°, thenfind ADC. [2010, 2011 (T-II)]

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, isequal 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 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 haveBAC + ACB + ABC = 180°BAC + 90° + 50° = 180°BAC = 40°

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Q.50. In the given figure, AB is the diameterof the circle with centre O. If BAD = 70° andDBC = 30°. Determine ABD and CDB.

[2011 (T-II)]

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 haveABD + 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 equalsides, prove that the quadrilateral so formed iscyclic. [HOTS]

Sol. AB = AC ABC = ACB …(i)

ADE = ABC

[Correspondingangles]

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, tworight triangles ACB and ADB are situated onopposite sides. Prove that ∠BAC = ∠BDC.

[HOTS]

Sol. Using AB asdiameter, draw a circlewhich passes through A,D, B and C.

BAC = BDC[Angles in the same

segment are equal]

Q.53. Prove that angle bisector of any angleof a triangle and the perpendicular bisector ofthe opposite side if intersect, they will intersecton the circumcircle of the triangle. [HOTS]

Sol.

ABC is the given triangle and O is the centreof its circumcircle. Then the perpendicularbisector of BC passes through O. It cuts thecircle 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

2BOC

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

will makeA

2

at A.

Or, CAP =A

2

AP is the bisector of A.

Q.54. In the given figure, two circlesintersect each other at C and D. If ADE and BCFare straight lines intersecting circles at A, B, Fand E. Prove that AB || EF. [2010]

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Sol. In order to prove that AB || EF, it issufficient 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 tothe opposite interior angle.

2 = 3 …(ii)From equations (i) and (ii), we get1 + 3 = 180°Hence, AB || EF.

Q.55. AOB is a diameter of the circle and C,D, E are any three points on the semicircle. Findthe value of ACD + BED. [2011 (T-II)]

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 thesides AB and AC at D and E respectively. Provethat DE || BC. [2011 (T-II)]

Sol. AB = ACABC = ACB .....(i)

Side BD of the cyclicquadrilateral BCED isproduced to A.

ADE = ACB ...(ii)[ext. ADE = int. opp. C]

From (i) and (ii), we getABC = ADE.But, these are corres-

ponding angles.Hence, DE || BC.Q.57. In the given figure, PQ and RS are two

parallel chords of a circle. When produced RPand SQ meet at O. Prove that OP = OQ.

[2010]

Sol. We have, PQ || RSOPQ = ORS and

OQP = OSR[Corresponding angles]

But, PQRS is a cyclicquadrilateral.

OPQ = OSR andOQP = ORS

OPQ = OQPThus, in OPQ, we have

OPQ = OQPOP = OQ. Proved.Q.58. Prove that the circle drawn on any one

of the equal sides of an isosceles triangle asdiameter 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 whichintersects the side BC of triangle at D.

To prove : BD = DCConstruction : Join ADProof : Since angle in

a semicircle is a rightangle. Therefore,

ADB = 90°ADB +AD

= 180° 90° + ADC = 180° ADC = 90°Now, in ABD and ACD, we haveAB = AC [Given]

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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.[2011 (T-II)]

Sol. In ACF, side CF isproduced 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 have1 + x + y = 1 + 3 + 4x + y = 3 + 4 = 23[ 4 = 3, angles in the same segment]

= z [ AOB = 2ACB]Hence, x + y = z.Q.60. In the given figure, AB is diameter of

circle and CD is a chord equal to the radius ofcircle. AC and BD when extended intersect at apoint E. Prove that AEB = 60°. [2011 (T-II)]

Sol. Join OC, OD and BC.In triangle OCD, we haveOC = OD = CD [Each equal to radius]OCD is equilateral.COD = 60°

Now, CBD =1

2COD CBD = 30°

Since ACB is angle in a semi-circle.ACB = 90°BCE = 180° – ACB = 180° – 90° = 90°Thus, in BCE, we have

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)]

Sol. Since arc XY subtends XPY at thecentre and XZY at a point Z in the remainingpart of the circle.

XPY = 2XZY ...(i)Similarly, arc YZ subtends YPZ at the

centre and YXZ at a point Y in the remainingpart of the circle.

YPZ = 2YXZ ...(ii)Adding (i) and (ii), we getXPY + YPZ = 2XZY + 2YXZXPZ = 2(XZY + YXZ)Q.62. If O is the circumcentre of a ABC

and OD BC, prove that BOD = A[2011 (T-II)]

Sol. Join OB and OC.In OBD and OCD, we haveOB = OC

[Each equal to the radius of circumcircle]ODB = ODC [Each equal to 90°]and, OD = OD [Common]OBD OCDBOD = CODBOC = 2BOD = 2CODNow, arc BC subtends BOC at the centre

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and BAC = A at a point in the remaining partof the circle.

BOC = 2A2BOD = 2A [ BOC = 2BOD]BOD = A.Q.63. Prove that the angle subtended by an

arc at the centre is double the angle subtendedby it at any point on the remaining part of thecircle. [2010]

Sol. Given : A circle C(O, r) in which arcAB subtends AOB at thecentre and ACB at anypoint C on the remainingpart of the circle.

To prove : AOB =

2ACB, when AB is aminor arc or a semicircle.

Reflex AOB = 2ACB, when AB is amajor arc.

Construction : Join AB and CO. ProduceCO to a point D outside the circle.

We know that when one side of atriangle is produced then the exterior angle soformed is equal to the sum of the interioropposite angles.

AOD = OAC + OCABOD = OBC + OCB

But, OAC = OCA [ OC = OA = r].andOBC = OCB [ OC = OB = r]. AOD = 2OCA and BOD = 2OCBNow, AOD + BOD = 2OCA + 2OCB AOB = 2(OCA + OCB) AOB = 2ACB.

PRACTICE EXERCISE 10B

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 thatCBA

= 40° and AD BC, then

the value of x is

(a) 50° (b) 90° (c) 45° (d) 40°

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

[2011 (T-II)]

(a) 120° (b) 60°

(c) 90° (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°

4. In the figure, O is

the centre of the circle and

∠PQR = 100°.

Then the reflex ∠PORis :

(a) 280° (b) 200°

(c) 260° (d) none of these5. In the given figure,

E is any point in the inte-rior of the circle with cen-tre O. Chord AB = ChordAC. If ∠OBE = 20°,then thevalue 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°

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2 Marks Questions

7. In the given figure,

ABC = 45°, prove that

OA OC. [2011 (T-II)]

8. In the given figure, ABCD is a cyclic quad-

rilateral and ABC = 85°. Find the measure of

ADE. [2011 (T-II)]

9. In the figure, O is the

centre of the circle andBAC

= 60°. Find the value of x.

[2011 (T-II)]

10. O is the centre of the

circle as shown in figure. Find

CBD. [2011 (T-II)]

11. ABCD is a cyclic

quadrilateral and AB = AC if

ACB = 70°, find BDC.

[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.

[2011 (T-II)]

3 Marks Questions

13. ABC is an isosceles triangle in which AB= AC. A circle passing through B and C intersectsAB and AC at D and E respectively. Prove thatBC || DE.

14. O is the circumcentre of the triangle ABCand D is the mid-point of the base BC. Prove that∠BOD = ∠A.

15. A quadrilateral ABCD is inscribed in acircle such thatAB is a diameter and ∠ADC = 130°.Find ∠BAC.

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

4 Marks Questions

17. In the figure, O isthe centre of the circle. IfBD = OD and CD AB,find ∠CAB. [HOTS]

18. Prove that the angles in a segment greaterthan a semi-circle is less than a right angle.

[HOTS]

TEXTBOOK’S EXERCISE 10.6 (OPTIONAL)

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.

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]

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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]

Sol. Let O be the centre of the circle and let

its radius be r cm.

Draw OM AB and OL CD.

Then, AM =1

2AB =

5

2cm

and, CL =1

2CD =

11

2cm

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]

r2 = (6 – x)2 +2

5

2

...(i) and

r2 = x2 +2

11

2

... (ii)

(6 – x)2 +2

5

2

= x2 +2

11

2

[From (i) and (ii)]

36 + x2 – 12x +25

4= x2 +

121

4

– 12x =121

4–

25

4– 36

– 12x =96

4– 36 – 12x = 24 – 36

– 12x = – 12 x = 1

Substituting x =1 in (i), we get

r2 = (6 – x)2 +2

5

2

r2 = (6 – 1)2 +2

5

2

r2 = (5)2 +2

5

2

= 25 +25

4

r2 =125

4 r =

5 5

2

Hence, radius r =5 5

2cm.

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?

[2010]

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 =1

2RS

RL = 3 cm

and PM =1

2PQ PM = 4 cm

In ORL, we have

OR2 = RL2 + OL2 [Pythagoras theorem]

OR2 = 32 + 42 = 9 + 16

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

OM = 9 = 3 cm

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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]

Sol. Given : Two equal chords AD and CE

of a circle with centre O. When meet at B when

produced.

To Prove : ABC =1

2(AOC – DOE)

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°

2OAD = 180° – z [ OAD = OBA]

OAD = 90° –2

z... (ii)

Similarly, OCE = 90° –2

z... (iii)

ODB = OAD +ODA[Exterior angle property]

OEB = 90° –2

z+ z [From (ii)]

ODB = 90° +2

z... (iv)

Also, OEB = OCE + COE[Exterior angle property]

OEB = 90° –2

z+ z [From (iii)]

OEB = 90° +2

z... (v)

Also, OED = ODE = 90° –2

y... (vi)

from (iv), (v) and (vi), we have

BDE = BED = 90° +2

z– 90

2

y

BDE = BED =2

y z

BDE = BED = y + z ... (vii)BDE = 180° – (y + z)ABC = 180° – (y + z) ... (viii)

Now,360 2

2 2

y z y z y

= 180° – (y + z) ... (ix)From (viii) and (ix), we have

ABC =2

x yProved.

Q.5. Prove that the circle drawn with any

side of a rhombus as diameter, passes through

the point of intersection of its diagonals.

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.

[2011 (T-II)]

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Sol. In order to prove that AE = AD it is

sufficient to prove that AED = ADE.

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 anglesof 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

Q.7. AC and BD are chords of a circle whichbisect each other. Prove that (i) AC and BD arediameters, (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

(ii) ABCD is a rectangle.

Proof : In OAB and OCD, we have

OA = OC [Given]

OB = OD [Given]

AOB = COD [Vertically opposite angles]

AOB COD [SAS congruence]

ABO = CDO and BAO = DCO

[CPCT]

AB || DC ... (i)

Similarly, we can prove BC || AD ... (ii)

Hence, ABCD is a parallelogram.

But ABCD is a cyclic parallelogram.

ABCD is a rectangle.

[Proved in Q.12 of textbooks exercise 10.5]

ABC = 90° and BCD = 90°

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 atriangle ABC intersect its circumcircle at D, Eand F respectively. Prove that the angles of the

triangle DEF are 90° –1

2AA, 90° –

1

2B and

90° –1

2C. [HOTS]

Sol. Given : ABC and its circumcircle.

AD, BE, CF are bisectors of A, B, C

respectively.

Construction : Join DE, EF and FD.

Proof : We know that angles in the same

segment are equal.

5 =C

2

and 6 =

B

2

..(i)

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1 =A

2

and 2 =

C

2

...(ii)

4 =A

2

and 3 =

B

2

...(iii)

From (i), we have

5 + 6 =C

2

+

B

2

D =C

2

+

B

2

...(iv)

[5 + 6 = D]But A + B + C = 180°B + C = 180° – A

B

2

+

C

2

= 90° –

A

2

(iv) becomes, D = 90° –A

2

.

Similarly, from (ii) and (iii), we can prove

that E = 90° –B

2

and F = 90° –

C

2

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)]

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.

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]

Sol. Let angle bisector of A intersect

circumcircle of ABC at D.

Join DC and DB.

BCD = BAD

[Angles in the same segment]

BCD = BAD =1

2A …(i)

[AD is bisector of A]

Similarly, DBC = DAC =1

2A ... (ii)

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.

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.

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ANFigure-1

Figure-2

Figure-3

Figure-4

Procedure :

1. On a white sheet of paper, draw a circle of any convenientradius with centre O. Mark two points Aand B on the boundary

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

2. Take any point P on the remaining part

of the circle. Join OA, OB, PA and PB.

3. Make two replicas of APB using tracingpaper. Shade the angles using different

colours.

4. Paste the two replicas of APB adjacentto each other on AOB as shown in thefigure.

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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 = 2APB.

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 anyconvenient radius. Draw a chord AB of the

circle.

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, Bto Q, A to R and B to R.

Figure-2

Figure-1

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3. On a tracing paper, trace each of the angles APB, AQB and ARB. Shade the traced copies usingdifferent 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 areequal.

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, apair of scissors, gluestick, geometry box, etc.

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Procedure : (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 itsdiameter AB as shown.

Figure-1

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

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 onXY and adjacent to each other such that AP and BP coincide as shown in the figure.

Figure-4

Figure-2

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

1. On a white sheet of paper, draw a circle of anyconvenient radius with centre O. Draw a chordAB which does not pass through O.

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

3. Trace APB on a tracing paper.

Figure-7

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

Figure-8

Figure 5

Figure-6

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

1. On a white sheet of paper, draw a circle of any convenientradius with centre O. Draw any chord AB which does not passthrough O.

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

Figure-10

3. Trace APB on a tracing paper.

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.

Figure-9

Figure-11

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Or APB + APB = a straight angle = 180°

2APB = 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

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Figure-5

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

Figure-2

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

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

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ANSWERS

Practice Exercise 10A

1. (a) 2. (d) 3. (b) 4. 24 cm 5. 12 cm 6. 2.4 cm 7. 17 cm 8. 8.5 cm

9. 6 cm 10. 8.58 cm, 11. 13 cm

Practice Exercise 10B

1. (d) 2. (b) 3. (d) 4. (b) 5. (d) 6. (c) 8. 85° 9. 240° 10. 50° 11. 140° 12. 110°

15. 40° 17. 30°

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

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 forma 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.