Circle, Geometrical Constructions Statistics & Probability

CIRCLES

CHORD PROPERTIES OF A CIRCLE



CIRCLE

A circle is the locus of a point which moves in a place in such a way that its distance from a fixed point remains constant. The fixed point is called the centre and the constant distance is called the radius of the circle.

The given figure consists of a circle with centre O and radius equal to r units.

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

TERMS AND FACTS RELATED TO CIRCLES.



Radius : A line segment joining the centre and a point on the circle is called its radius, generally denoted by r In

the adjoining figure, OA, OB and OC are the radii of a circle.



Circumference : The perimeter of a circle is called its circumference. Circumference = 2r



Position of a Point With Respect To a Circle

Let us consider a circle with centre O and radius r. (i) inside the circle, if OP < r:

(ii) on the circle, if OP = r.

(iii) outside the circle, if OP > r.

In the adjoining figure of a circle with centre O and radius r. (i) The points A, O, B lie inside the circle.

(ii) The points P, Q, R lie on the circle; (iii) The points X, Y, Z lie outside the circle.



Interior and Exterior of a Circle.

The region consisting of all those points which lie inside a circle, is called the interior of the circle. The region consisting of all those points which lie outside a circle, is called the exterior of the circle.



Circular Region or Circular Disc

The region consisting of all those points which are either on the circle or lie inside the side, is called the circular region.



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

Chord : A line segment joining any two points on a circle is called a chord of the circle.



Diameter : A chord of the circle passing through the centre of a circle is called its diameter.

In the figure. AOB is a diameter of a circler with centre O.

Diameter = 2 × Radius

Properties : (i) Diameter is the largest chord of a circle. (ii) All diameters of a circle are equal in length.



Secant : A line which intersects a circle in two distinct points in called a secant of the circle.

In the adjoining figure, the line  cuts the circle in two points C and D. So,  is a secant of the circle.

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

Tangent : A line that intersects the circle is exactly one point is called a tangent of the circle.

The point at which the tangent intersects the circle is called its point of contact. In the adjoining figure, SPT is a tangent at the point P of the circle with centre O. Clearly, P is the point of contact of the tangent with the circle.

Facts About Tangents :

(i) No tangent can be drawn to a circle through a point inside the circle: (ii) One and only one tangent can be drawn to a circle at a point on the circle. (iii) Two tangents can be drawn to a circle from a point outside it.

In the adjoining figure, PT₁and PT₂are the tangents to the circle from point P.

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

Touching Circles : Two circles are said to touch each other if and only if they have one and only one point in

common. Two circles may touch externally (Fig. (i) o internally [Fig (ii)].

The common point is called the point of contact, and the line joining their centre is called the line of centre. A line touching the two circles is called common tangent.

Thus, in the above figure, P is the point of contact, AB is the line of centers are PT is a common tangent.



Direct Common Tangents : A common tangent to two circles is called a direct common tangent if both the



Transverse Common Tangents : A common tangent to two circles is called a transverse common tangent if the

circles lie on its opposite sides.

In the adjoining figure, PQ and RS are two transverse common tangents.

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

Arc : A continuous piece of a circle is called on arc of the circle.

Let P and Q be any two points on a circle with centre O. Then, clearly the whole circle has been divided into two pieces.

namely arc PAQ and arc QBP, to be denoted by PAQ and QBP respectively. We any denote them by PQ and QP respectively.



Minor and Major Arc: An arc less than one-half of the whole arc of a circle is called a minor arc, and an arc

greater than one-half of the whole are of a circle is called a major arc of the circle. Thus, in the above figure, PQ is a minor arc, while QP is a major arc.



Central Angle : An angle subtended by an arc at the centre of a circle is called its central angle. In the given

figure, central angle of PQ =



POQ.



Degree Measure of An Arc :

Let PQ be an arc of a circle with centre O.

If



POQ =  , we say that the degree measure of PQ is0  and we write, m(PQ)0 0. If m(PQ)= , then0 m(PQ) = (360 - )0 0_{. Degree measure of a circle is 360}0



Congruent Arcs : Two arcs AB and CD are said to be congruent, if they have same degree measure.

 



CD m(AB) m(CD)

AB



AOB =



COD.



Semi-Circle : A diameter divides a circle into two equal arcs. Each of these two arcs is called a semi-circle. The

degree measure of a semi-circle is 1800_{. In the given figure of a circle with centre O, ABC as well as ADC is} semi-circle.



Congruent Circles : Two circles of equal radii are said to be congruent.



Concentric Circles : Circles having same centre but different radii are called concentric circles.



Concyclic Points : The points, which lie on the circumference of the same circle, are called concyclic points. In

the adjoining figure, points A, B, C and D lie on the same circle and hence, they are concyclic.

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

Segment : A segment is a part of circular region bounded by an arc and a chord, including the arc and the

chord. The segment containing the minor arc is called a minor segment, while the other one is a major segment. The centre of the circle lies in the major segment.

Alternate Segments of a Circle : The minor and major segments of a circle are called alternate segments of

each other.

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

Sector of a Circle : The part of the plane region enclosed by an arc of a circle and its two bounding radii is

called a sector of the circle.

Thus, the region OABO is the sector of a circle with centre O.

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

Quadrant : One-fourth of a circular disc if called a quadrant.



Cyclic Quadrilateral : If all the four vertices of a quadrilateral lie on a circle, then such a quadrilateral is called a

cyclic quadrilateral.

Theorem 1. The straight line drawn from the centre of a circle to bisect a chord, which is not a diameter, is perpendicular to the chord.

Given : AB is a chord, other than the diameter of a circle

with centre O and OL bisects AB.

To prove : OL  AB.

Construction : Join OA and OB.

Proof :

Theorem 2. (Converse of Theorem 1): The perpendicular to a chord from the centre of a circle bisects the chord.

Given : AB is a chord of a circle with centre O and OL  AB. To prove : LA = LB.

Construction : Join OA and OB. Proof :

STATEMENT

REASON

1. InOLA and OLB, we have (i) OA = OB

(ii) AL = BL (iii) OL = OL OLA  OLB



 

OLA =



OLB ....(I)

2.



OLA +



OLB = 1800 _....(II)

3.



OLA =



OLB = 900

Radii of the same circle Given, OL bisects AB Common

SSS-Axiom of congruence

Corresponding part of congruents are congruent. ALB is a straight line

From (I) and (II).

Hence, OL



AB

STATEMENT

REASON

InOLA and OLB, we have

(i) OA = OB

(ii)



OLA =



OLB (iii) OL = OL

 OLA  OLB



LA = LB

Radii of the same circle

Each equal to 900_{, since OL}  AB Common

RHS axiom of congruency c.p.c.t.

Theorem 3. One and only one circle can be drawn, passing through three non-collinear points. Given : Three non-collinear points A, B, C.

To prove : One and only one circle can be drawn, passing through A, B and C. Construction : Join AB and BC. Draw the perpendicular bisector of AB

and BC, meeting at a point O.

Proof :

STATEMENT

REASON

1. O lines on the perpendicular bisector of AB



OA = OB ...(i)

2. O lies on the perpendicular bisector of BC



OB = OC ...(ii)

3. OA = OB = OC



O is equidistant from A, B and C



Any circle drawn with centre O and radius OA will pass through B and C also.

4. O is the only point equidistant from A, B and C

Each point on perpendicular bisector of AB is equidistant From A and B.

Each point on perpendicular bisector of BC is equidistant From B and C

From (i) and (ii)

Perpendicular bisector of AB and BC cut each other at point O only.

Hence, one and only circle can be drawn to pass through three non-collinear points, A B and C.

Theorem 4. Equal chords of circle are equidistant from the centre. Given : A circle with centre O in which chord AB = chord CD;

OL  AB and OM  CD. To prove : OL = OM. Proof :

STATEMENT

REASON

1. AL AB 2 1  …..(i) 2. CM CD 2 1  …..(ii) 3. Now, AB = CD



2 1 AB = 2 1 CD



AL = CM ....(iii)

4. InOLA and OMC, we have

(i) OA = OC

(ii) AL = CM (iii)



OLA =



OMC

 OLA  OMC

Perpendicular from centre bisects the chord. Perpendicular from centre bisects the chord. Given

Halves of equals are equal. From (i) and (ii).

Radii of the same circle. From (iii).

Each equal to 900_{, as OL}  AB and OM  CD. RHS-axiom of congruency ofs.



OL = OM

Hence, the chords AB and CD are equidistant from the centre O.

Theorem 5. (Converse of Theorem 4) : Chords of a circle that are equidistant from the centre of the circle, are equal.

Given : AB and CD are two chords of a circle with centre O.

OL  AB, OM  CD and OL = OM.

To prove : AB = CD

Construction : Join OA and OC.

Proof :

STATEMENT

REASON

InOLA and OMC we have

(i) OL = OM

(ii) OA = OC (iii)



OLA =



OMC

 OLA =  OMC



AL = CM



AB CD 2 1 2 1 _



AB = CD Given,

Radii of the same circle.

Each equal to 900_{, as OL  AB and OM  CD.} RHS-axiom of congruency of



s.

c.pc.t.

Perpendicular from centre bisect the chord Doubles of equal are equal.

Hence, chord AB = chord CD

Ex.1 A chord of length 30 cm is drawn in a circle of radius 17 cm. Find its distance from the centre o the circle.

Sol. Let AB be a chord of a circle with centre O and radius = 17 cm such that AB = 30 cm. From O, draw OL  AB. Join OA.

Since perpendicular to the chord from the centre bisects the chord, we have

AL = AB 30 cm 15cm. 2 1 2 1         

OA = Radius of the circle = 17 cm. From right-anglesOLA, we have

2

2 _AL

OA

. cm ) ( ) (172 152  289225 648 

Ex.2 In the figure, the diameter CD of a circle with centre O is perpendicular to the chord AB. If AB = 12 cm and CE = 3 cm, find the radius of the circle.

Sol. Join OA.

Since perpendicular from the centre to a chord, bisects the chord, we have AE = EB = 6 cm.

Let r be the radius of the circle. The, OA = OC = r cm. OE (OC - CE) = (r - 3) cm.

From right anglesOEA, we have

OA2_{= AE}2_{+ OE}2 _{[By Pythagoras Theorem]}



r2_{= (6)}2_{+ (r - 3)}2 _{[OA = r cm, AE = 6 cm & OE = (r - 3) cm]}



r2_{= 36 + r}2_{- 6r + 9}



6r = 45



r = 7.5.

Hence, the radius of the circle is 7.5 cm.

Ex.3 AB and CD are two parallel chords of a circle of length 24 cm and 10 cm respectively and lie on the same side of its centre O. If the distance between the chords is 7 cm, find the radius of the circle.

Sol. Draw OL  AB and OM  CD.

Since AB || CD, it follows that O, L, M are collinear. Now, OL  AB and OM  CD



AL AB 24 cm 12cm 2 1 2 1 _          and CM = CD 10 cm 5cm. 2 1 2 1         

Also, distance between the chords = LM = 7 cm. Join OA and OC.

Let OL = x cm. Then, OM = (x + 7) cm. Let the radius of the circle be r cm. Then, OA = OC = r cm.

Now, from right-anglesOLA and OMC, we have OA2_{= OL}2_{+ AL}2_{and OC}2_{= OM}2_{+ CM}2



r2_{= x}2_{+ (12)}2_{.... (i) and r}2_{= (x + 7)}2_{+ (5)}2 _...(ii)



x2_{+ (12)}2_{= (x + 7)}2_{+ 5}2 _{[From (i) and (ii)]}



x2_{+ 144 = x}2_{+ 14x + 74}



14x = 70



x = 5

Substituting = 5 in (i), we get :



r = 169 = 13 cm.

Hence, the radius of the circle is 13 cm.

Ex.4 AB and CD are two parallel chords of a circle such that AB = 10 cm and CD = 24 cm. If the chords are on opposite sides of the centre and the distance between them is 17 cm, find the radius of the circle.

Sol. Let O be the centre of the circle and let its radius be r cm. Draw OL  AB and OM  CD Then AL = 2 1 AB = 5 cm and CD = 2 1 CD = 12 cm.

Since AB || CD, it follows that the points O, L, M are collinear and therefore, LM = 17 cm.

Let OL = x xm. Then, OM = (17 - x) cm. Join OA and OC. Then, OA = OC = r cm.

Now, from right-anglesOLA and OMC, we have

OA2_{= OL}2_{+ AL}2_{and OC}2_{= OM}2_{+ CM}2 _{[By Pythagoras Theorem]}



r2_{= x}2_{+ 5}2_{....(i) and r}2_{= (17 - x)}2_{+ (12)}2_...(ii)



x2_{+ 5}2_{= (17 - x)}2_{+ (12)}2 _{[From (i) and (ii)]}



x2_{+ 25 = x}2_{- 34x + 433}



34x = 408



x= 12.

Substituting x = 12 in (i), we get : r2_{= (12)}2_{+ 5}2_{= (144 + 25) = 169}



r = 169 = 13 c,

Hence, the radius of the circle is 13 cm.

Ex.5 In the adjoining figure, AB and AC are two equal chords of a circle with centre O. Show that O lies on the

bisector of



BAC.

Sol. Given : AB and AC are equal chords of circle with centre O and O has been joined with A.

To prove :



BAO =



CAO. Construction : Join OB and OC.

Proof :

REASON

STATEMENT

1. InOAB and OAC, we have

(i) AB = AC Given

(ii) OB = OC Radii of the same circle.

(iii) OA = OA Common.





BAO =



CAO c.p.c.t.

Ex.6 In the adjoining figure, AB and AC are two equal chords of circle of radius 5 cm. IF AB = AC = 6 cm, find

the length of chord BC.

Sol. Let O be the centre of the circle, Join OB.

Let AD be the bisector of



BAC, meeting BC and D. Ins BAD and CAD, we have

AB = AC [Given]



BAD =



CAD [By contraction]

AD = AD [Common]

BAD  CAD [SAS-axiom of congruency]





BDA =



CDA [c.p.c.t.]

But,



BDA = +



CDA = 1800 [BDC is a straight line] 



BDA =



CDA = 900

This shows that AD in the perpendicular bisector of BC and so, it when produced passes through O.  OB = OA = 5 cm.

Let OD = x cm and BD = y cm. The, AD = (OA - OD) = (5 - x)cm.

From right anglesOBD and ADB, we get: OB2_{= OD}2_{+ BD}2_{and AB}2_{= AD}2_{+ BC}2



52_{= x}2_{+ y}2_{and 6}2_{= (5 - x)}2_{+ y}2



y2_{= 25 - x}2 _{...(i) and y}2_{= 36 - (5 - x)}2_...(ii) From (i) and (ii), we get:

25 = x2_{= 36 - (5 - x)}2



10x = 14 i.e. x = 1.4. Substituting x = 1.4 in (i) we get :

y2_{= 25 - (1.4)}2_{= 23.04}



_{y = 23.04 = 4.8 cm.}  BD = 4.8 cm.

 BC = 2 × BC = (2 × 4.8)cm = 9.6 cm Hence, the length of chord BC = 9.6 cm.

Ex.7 If two equal chords of a circle intersect, then prove that their segments are equal. Sol. Given : Two equal chords AB and CD of a circle with centre O,

intersect at a point P.

To prove : BP = CP and AP = DP.

Construction : Draw OL  AB, OM  CD. Join OP. Proof :

STATEMENT

REASON

1. AL = BL = 2 1 AB 2. DM = CD = 2 1 CD 3. AL = DM ...(i) BL = CM ...(ii)

4. In rights OLP and OMP, we have:

(i) OL = OM

(ii)



OLP =



OMP (iii) OP = OP  PLP  OMP



LP = MP ....(iii) 5. AL + LP = CM + MP



AP = DP 6. BL - LP = CM - MP



BP = CP

Perpendicular from centre bisects the chord. Perpendicular from centre bisects the chord. From (i) and (ii).

Equal chords are equidistant from the centre, Each equal to 900

RHS-axiom of congruence. c.p.c.t

From (i) and (iii), adding the corresponding sides.

From (ii) and (iii)

Hence, BP = CP and AP = DP

ANGLE PROPERTIES OF A CIRCLE

Theorem-6. The angle subtended by an arc of a circle at the centre is double the angle subtended by its an any point on the remaining part of the circle.

Given : A circle with centre O and an arc AB subtends



AOB at the centre and



ACB at any point C on the remaining part of the circle.

To prove :



AOB = 2



ACB.

Remark : In the above theorem, fig (i) refers to minor arc while fig. (iii) refers to major arc. Theorem-7 : Angles in the same segment of a circle are equal.

Given : A circle with centre O and two angles



ACB and



ADB in the same segment of the circle.

To prove :



ACN =



ADE/

Construction : Join OA and OB.

STATEMENT

REASON

1. InAOC, we have OA = OC





OAC =



OCA ....(i)

2.



AOD =



OAC +



OCA =



OCA +



OCA = 2



OCA ....(ii)

3. Similarly, we have



BOD = 2



OCB ....(iii)

4. In figure (i), we have :



AOD +



BOD = 2



OCA + 2



OCB = 2(



OCA +



OCB) = 2



ACB 



AOB = 2



ACB.

In Figure (iii), we have :



AOD +



BOD = 2



OCA + 2



OCB. =2



ACB

In figure (ii), we have.



BOD -



AOD = 2



OCB - 2



OCA = 2(



OCB -



OCA) = 2



ACB 



AOB = 2



ACB.

Radii of the same circle.

Angle opposite to equal sides of a are equal. Ext. angle of a = Sum of its into opp.



s. Using (i)

Adding the corresponding sides of (ii) and (iii).

Adding the corresponding sides of (ii) and (iii).

Subtracting the corresponding sides of (iii) and (ii).

Hence,

AOB = 2ACB.

(ii

)

(i)

Proof :

STATEMENT

REASON

In fig (i) :

1. Arc AB subtends



AOB at the centre and



ACB at a point C of the remaining part of the circle.





AOB - 2



ACB ...(i)

2. Arc AB subtends



AOB at the centre and



ADB at a point D on the remaining part of the circle.





AOB = 2



ADB ....(ii)

3. 2



ACB = 2



ADB 



ACB =



ADB

4. Similarly, in fig (ii)



ACB =



ADB 2 1

 reflex



AOB 



ACB =



ADB.

Angle at the centre is double the angle at any point on remaining part of the circle.

Same as above From (i) and (ii)

Hence, the angels in the same segment of a circle are equal.

Theorem-8. The angle in a semi-circle is a right angle.

Given : A semi-circle ACB of a circle with centre O.

To prove :



ACB = 900

STATEMENT

REASON

1. Arc AB subtends



AOB at the centre and



ACB at a point C on the remaining part of the circle.





AOB =2



ACB

 

ACB = 2 1

_

AOB ...(i) 2.



AOB = 1800 _....(ii) 3.



ACB =        0 180 2 1 = 900

Angle at the centre is double the angle at any point on remaining part of the circle.

AOB is a straight line.

Hence, the angle in a semi circle is a right angle.

Theorem-9. (Converse of Theorem 8) : if an arc of a circle subtends a right angle at any point on the remaining part of the circle, then the arc is a semi-circle.

Given : A circle with centre O and an arc AB subtending



ACB at a point C on the remaining part of the circle such that



ACB 900_.

To prove : Arc AB is semi-circle. Construction : Join OA and OB Proof :

STATEMENT

REASON

1. Arc AB subtends



AOB at the centre and



ACB at a point C on the remaining part of the circle.





AOB = 2



ACB ...(i)

2.



ACB = 900

3. 



AOB = (2 × 900_{) = 180}0



AOB is a straight line



AOB is a diameter



Arc AB is a semi-circle.

Angle at the centre is double the angle at a point of the remaining part of the circle.

Given

From (i) and (ii)

Chord AB passes through the centre O.

Hence, arc AB is a semi-circle

Theorem-10. The opposite angles of a quadrilateral inscribed in a circle are supplementary. OR

The sum of the opposite angles of a cyclic quadrilateral is 1800

Given : A quadrilateral ABCD inscribed in a circle with centre O.

To prove :



ADC +



ABC = 1800_and



_{BAD +}



_{BCD = 180}0

Construction : Join OA and OC. Proof :

STATEMENT

REASON

1. Arc ABC subtends



AOC at the centre

and



ADC at a point D on the remaining part of the circle.





AOC = 2



ADC

 

ADC = 2 1

_

AOC ...(i)

2. Major arc CDA subtends reflex



AOC at the centre and



ABC at a point B on the remaining part of the circle.  reflex



AOC = 2



AOC

 

ABC = 2 1

reflex



AOC ...(ii)

3. From (i) and (ii), we get



ADC +



ABC = 2 1

_

AOC + 2 1 reflex



AOC = 2 1

(



AOC + reflex



AOC)

=       ₃₆₀0 2 1 = 1800 



ADC +



ABC = 1800 4. Similarly



BAD +



BCD = 1800

Angle at the centre is double the angle at any point on remaining part of the circle.

Same as above

(



AOC + reflex



AOC)

=sum of the angles around a point O = 3600

Hence, the opposite angles of a cyclic quadrilateral are supplementary.

Theorem-11. (Converse of Theorem 10) : If a pair of opposite angles of a quadrilateral are supplementary, then the quadrilateral is cyclic.

Given : A quadrilateral ABCD in which



B +



D = 1800

To prove : ABCD is a cyclic quadrilateral.

Construction : if possible, let ABCD be not a cyclic

quadrilateral . Draw a circle passing through three non-collinear points A, B, C. Suppose this circle meets CD or CD produced at D’, as shown in Fig. (i) and Fig. (ii) respectively. Join D’A.

Proof :

STATEMENT

REASON

1.



B +



D = 1800 2.



B +



D = 1800 3.



B +



D =



B +



D

 

D =



D’

4. But, this is not possible.

Given.

ABCD is a cyclic quadrilateral and so its opposite



s are supplementary.

From (i) and (ii)

An exterior angle of a



is never equal to its into opp.

(ii

)

(i)

 Our supposition is wrong. angle.

Hence, ABCD is a cyclic quadrilateral.

Theorem-12. The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.

Given : A cyclic quadrilateral whose side AB is produced to a point E.

To prove :



CBE =



ADC

Proof :

STATEMENT

REASON

1.



ABC +



ADC = 1800

2.



ABC +



CBE = 1800

3.



ABC +



ADC =



ABC +



CBE

 

ADC =



CBE

ABCD is a cyclic quadrilateral and so the sum of its opp.



s is 1800

ABE is a straight line. From (i) and (ii).



ABC is common to both sides. Hence, the exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.

Ex.8 In the given circle with diameter AB, find the value of x. Sol.



ADB = 900 _{[Angle in a semi-circle]}



ABD =



ACD = 300 _[



_{s in the same segment]} InADB, we have :



BAD +



ADB +



ABD = 1800_{(Sum of the}



_{s of a} is 1800₎



x + 900_{+ 30}0_{= 180}0



x = (1800_{- 120}0_{) = 60}0 Hence, x = 600

Ex.9 If O is the centre of the circle, find the value of x in each of the following figure, giving reasons.

Sol. We have:

(i)



BDC, =



BAC = 350 _[



_{s in the same segment].}



DBC +



BCD +



BDC = 1800 _{[Sum of the}



_{s of a} is 1800_]



700_{+ x}0_{+ 35}0_{= 180}0



x0_{= (180}0_{- 105}0_{) = 75}0 Hence x = 750_.

(ii)



BCD = 900_{. [Angle is semi-circle]} InBCD, we have

(iii

)

(iv

)

(ii

)

(i)



BCD +



CDB +



DBC = 1800 _{[Sum of the}



_{s of a} is 1800_]



900₊



_{CDB + 55}0_{= 180}0





CDB = (1800_{- 145}0_{) = 35}0

 x =



BAC =



CDB = 350 _[



_{s in the same segment]} Hence, x = 350_.

(iii) Reflex



AOC = (360 - 110)0_{= 250}0

Major arc CA subtends reflex



AOC at the centre and



ABC at a point B on the remaining part of the circle. 



ABC = 2 1 reflex



AOC = 2 1 × 2500_{= 125}0 Hence, x = 1250_.

(iv) Since ABD is a straight line, we have



ABC +



CBD = 1800





ABC + 700_{= 180}0





ABC = (1800_{- 70)}0_{= 110}0_. Reflex



AOC = 2



ABC = (2 × 1100_{) = 220}0

[Major arc CA subtends reflex



AOC at the centre and



ABC at a point B on remaining part of the circle.]

 x = 2200

Ex.10 In the adjoining figure, AB is a diameter of a circle with centre O and CD || BA.

If



BAC = 200_{, find the value of}

(i)



BOC (ii)



DOC (iii)



DAC (IV)



ADC.

Sol. (i) Arc BC subtends



BOC at the centre and



BAC at the circumference.





BOC = 2



BAC = (2 × 200_{) = 40}0 _{[Angle at the centre is double the angle at circumference]} (ii)



OCD =



BOC = 400 _{[Alt. into}



_{s as CD || BA]}

Now, OC = OD [Radii of the same circle]





ODC =



OCD = 400 InOCD, we have



DOC +



OCD = 40 InOCD, we have



DOC +



OCD +



ODC = 1800 _{[Sum of the}



_{s of a is 180}0_]





DOC + 400_{+ 40}0_{= 180}0

(iii)



DAC = 2 1

_

DOC =        0 100 2 1

= 500 _{[Angle at the centre is double the}



_{at circumference]} (iv)



ACD =



CAB = 200 _{[Alt. Int}



_{s, as CD || BA\}

InACD, we have



ADC +



ACD +



DAC = 1800 _{[Sum of the}



_{s of a} is 1800_]





ADC + 200_{+ 50}0_{= 180}0





ADC = (1800_{- 70}0_{) = 110}0

Hence,



BOC = 400_,



_{DOC = 100}0_,



_{DAC = 50}0_and



_{ACD = 110}0_.

Ex.11 Two circles with centers O and O’ intersect at points A and B. AC and AD are their diameters respectively. Prove that the points C, B and Dare collinear.

Sol. Given: Two circle with centre O and O’ intersect at points

A and B; AC is a diameter of circle with centre of O and AD is a diameter of circle with centre O’.

To Prove : The points C, B, D are collinear.

Construction : Join AB, CB and BD. Proof :

STATEMENT

REASON

1.



CBA = 900 2.



DBA = 900 3.



CBA +



DBA = 1800



CBD is a straight line



C, B, D are collinear Angle in a semi-circle Angle in a semi-circle. Adding 1 and 2

Hence, the points C, B and D are collinear

Ex.12 Prove that every cyclic parallelogram is a rectangle.

Sol. Given : ABCD is a cyclic parallelogram. To prove : ABCD is a rectangle.

Proof :

1.



ABC =



ADC

2.



ABC +



ADC = 1800

3.



ABC =



ADC = 900  ABCD is a rectangle

Opposite



s of a ||gm are equal.

Sum of opposite



s of a cyclic quad. is 180_. From (i) and (ii).

A || gm one of whose



s is 900_{is a rectangle.} Hence, every cyclic parallelogram is a rectangle.

Ex.13 Prove that an isosceles trapezium is always cyclic.

Sol. Given : A trapezium ABCD in which AB || DC and AD = BC.

To prove : ABCD is a cyclic trapezium.

Construction : Draw DE  AB and CF  AB. Proof :

STATEMENT

REASON

1. InDEA and CFB, we have

(i) AD = BC

(ii)



DEA =



CFB = 900 (iii) DE = CF

 DEA  CFB





A =



B ....(i)

and



ADE =



BCF ...(ii)

2.



ADE =



BCF





ADE + 900₌



_{BCF + 90}0





ADE +



CDE =



BCF +



DCF





D =



C ...(iii) 3. 



A =



B + and



C =



D 4.



A +



B +



C +



D = 3600



2 (



B +



D) = 3600

 

B +



D = 1800



Sum of a pair of opp.



s of quad. ABCD is 1800



ABCD is a cyclic trapezium,

Given

DE  AB and CF  AB

Distance between parallel lines remains constant. RHS axiom of congruency.

c.p.c.t. c.p.c.t. From (ii)



ADE +



CDE =



D,



BCF +



DCF =



C. From (i) and (iii)

Sum of the



s of quadrilateral is 3600 Using (3)

Hence, ABCD is a cyclic trapezium.

Ex.14 Prove that the quadrilateral formed by angle bisectors of a cyclic quadrilateral ABCD is also cyclic. Sol. Given : A cyclic quadrilateral ABCD in which AP, BP, CR and DR

are the bisector of



A,



B,



C and



D respectively , forming a quadrilateral PQRS.

To prove : PQRS is a cyclic quadrilateral Proof :

STATEMENT

REASON

1.



APB +



PA B +



PBA = 1800





APB + 2 1

_

A + 2 1

_

B = 1800_(i) 2.



CRD +



RCD +



RDC = 1800

 

CRD + 2 1

_

C + 2 1

_

D = 1800_....(ii) 3.



APB +



CRD + 2 1 (



A +



B +



C +



D) = 3600





APB +



CRD +



× 3600_{= 360}0





APB +



CRD = 1800



Sum of a pair of opposite



s of quad, PQRS is 1800



PQRS is a cyclic quadrilateral.

Sum of the



s ofPAB is 1800



PAB = 2 1

_

A and



PBA = 2 1

_

B. Sum of the



s ofRCD is 1800



RCD = 2 1

_

C and



RDC = 2 1

_

D.

Adding (i) and (ii)

Hence, PQRS is a cyclic quadrilateral.

ARC PROPERTIES OF A CIRCLE

Theorem-13. In equal (or in the same circle), if two arcs subtend equal angles at the centre, they are equal. Given : Two equal circles C₁and C₂with O and O.

as their centre respectively. AB subtends



OB and CD subtends



CO’D such that



AOB =



CO’D

To prove AB = CD Proof :

STATEMENT

REASON

1. Place circle C₁on circle C₂such that O falls on O’ and OA falls along O’C.

2. Then, A falls on C and OB falls along O’D.

3. Clearly, B falls on D.

 AB completely coincides with CD.

OA = O’C (Radii of equal circles).



AOB’ =



CO’D (Given) OB = O’D (Radii of equal circles)

A falls on C, B falls on D and AB falls along CD, as circles

are equal.

Hence, AB=CD

Theorem-14. In equal circles (or in the same circle), if two arcs are equal, they subtend equal angles at the centre.

Given : Two equal circle C₁and C₂with O and O’ as their respective centre such that AB = CD

To prove:



AOB -



CO’D

Proof :

STATEMENT

REASON

1. Place circle C₁on circle C₁such that A falls on C, AO falls along CO’ and AB = CD

2. Then, O falls on O’ and B falls on D.  OB falls on O’D

3. Sector AOB completely coincides with sector CO’D.

AO = CO’ (Radii of equal circles.) and AB = CD (Given)





AOB =



CO’D.

Theorem-15. In equal circles (or in the same circle), if two chords are equal, they cut off equal arcs.

Given : Two equal circle C₁and C₂with centers O And, chord AB = chord CD

To prove : AB CD

Proof :

STATEMENT

REASON

Case - I When AB and CD are Minor Arcs

1. InOAB and O’CD, we have (i) OA = O’C (ii) OB = O’D (iii) AB = CD 2.  OAB  O’CD

 

AOB =CO’D



AB = CD ...(i)

Case -II When AB and CD are Major Arcs

In this case, BA and DC are Minor Arcs.  AB = CD



BA = DC



BA = DC



AB = CD

Case-III. When AB and CD are diameters

In this case, AB and CD are semi-circles.

 AB = CD

Radii of equal circles Radii of equal circles. Given,

SSS-axiom of congruency. c.p.c.t.

In equal circles, two arcs subtending equal



s at the centre, are equal

Chord AB = chord BA, chord CD = chord DC Result being true for Minor Arcs

Equal arcs subtracted from equal circles give equal arcs.

Si mi-circles of equal circles are equal.

Hence, chord AB = chord CD

 AB = CD

Theorem-16. In equal circles (or in the same circle), if two arcs are equal, then their chords are equal.

Given : Two equal circles C₁and C₂with centers O and, O’ respectively and AB = CD

To prove : Chord AB = Chord CD.

STATEMENT

REASON

Case-I, When AB and CD are Minor Arcs

1. AB = CD

 

AOB =



CO’D ...(i)

2. InOAB and O’CD, we have (i) OA = O’C

(ii) OB = O’D (iii)



AOB =



CO’D  OAB  O’CD



Chord AB = chord CD ....(ii)

Case-II, When AB and CD are Major Arcs

In this case BA and DC are Minor Arcs. Now, AB = CD



BA = DC



BA = DC



B = CD

Case-III. When AB and CD are semi-circles

In this case, AB and CD are diameters.

 AB = CD

Equal arcs of equal circles subtend equal angles at the centre.

Radii of equal circles. Radii of equal circles. From (i)

SAS-axiom of congruency. c.p.c.t.

Chord BA = chord AB, chord DC = chord CD.

Diameters of equal circles are equal.

Hence, in all the case AB = CD

 chord AB = chord CD.

Ex.15 In the given diagram, O is the centre of the circle and chord AB = chord BC. (i) What is the relation between arc AB and arc BC ?

(ii) What is the relation between



AOB and



BOC ? Sol. (i) Since equal chords is a circle cut equal arcs, so

chord AB = chord BC



arc AB = arc BC.

(ii) Equal arcs in a circle make equal angles at the centre.  chord AB = chord BC



arc AB = arc BC

 

AOB =



BOC.

Ex.16 In the adjoining figure A, D, B, C are four points on the circumference of a circle with centre O. Arc AB = 2 Arc BC and



AOB = 1080_{, find :}

(i)



ACB (ii)



CAB (iii)



ADB Sol. Let



BOC = x0_{. Them}



_{AOB = 2x}0

Now 2x0_{= 108}



_{x = 54.}





BOC = 540_and



_{AOB = 108}0 (i)



ACB = 2 1

_

AOB = 2 1

× 1080_{= 54}0 _{[Angle at the centre is double the}



_{at a point on the} circumference] (ii)



CAB = 2 1

_

CPB = 2 1 ×



BOC = 2 1 × 540_{= 27}0

(iii)



ADB = 2 1 reflex



AOB = 2 1 (3600_{- 108}0₍₃₆₀0_{- 108}0_{) =} 2 1 × 2520_{= 126}0

Ex.17 The adjoining figure shows a pentagon inscribed in a circle with centre O. Given AB = BC = CD and



ABC = 1300

Find (i)



AEB (ii)



AED (iii)



COD. Sol. (i) ABCE is a cyclic quadrilateral.





ABC +



AEC = 1800



1300₊



_{AEC = 180}0

 

AEC = (1800_{- 130}0_{) = 50}0

 

AEB +



BEC = 500



2



AEB = 500 [



_{BEC =}



_{AEB, since equal chords subtend equal angles at a point on the} circumference]

 

AEB = 250 (ii) CD = AB

 

CED =



AEB = 250 _{[Equal chords subtend equal}



_{s at a point on the circumference]} 



CED = 250





AED =



AEC +



CED = (500_{+ 25}0_{) = 75}0

(iii)



COD = 2



CED [ Angle at the centre is double the



at a point on the circumference] = (2 × 250_{) = 50}0_.





COD = 500

Ex.18 In the given figure, O is the centre of the circle. Chord AB is parallel to chord CD and CB is a diameter. Prove that : arc AC = arc BD.

Sol. Given : Chord AB || chord CD and COB is a diameter. To prove : Arc AC = Arc BD.

Construction : Join OA and OD. Proof :

STATEMENT

REASON

1.



AOC =2



ABC





ABC = 2 1

_

AOC ...(i) 2.



BOD = 2



BCD

 

BCD = 2 1

_

BOD ...(ii) 3. But



ABC =



BCD

Angle at the centre is double the angle at a point

on the circumference.

Same as above

Alternate



s, as AB || CD From (i) and (ii)



2 1

_

AOC 2 1

_

BOD





AOC =



BOD



arc AC = arc BD

In a circle, the arcs subtending equal



s at the centre are equal

Ex.19 If a pair of opposite sides of a cyclic quadrilateral are equal, prove that its diagonals are equal. Sol. Given : A cyclic quadrilateral ABCD in which AB = DC.

To Prove : Diagonal AC = Diagonal BD. Proof :

STATEMENT

REASON

1.



BAC =



BDC ...(i)

2.



CAD =



ADB ....(ii)

3.



BAC +



CAD =



BDC +



ADB





BAD =



ADC



BD = AC



s in the same segment of a circle.

Equal arcs CD and AB subtend equal



s at the circumference

Adding (i) and (ii)

Equal



s on the same circle cut off equal chords.

Hence, diagonal AC = diagonal BD.

EXERCISE

CIRCLE

SUBEJCTIVE TYPE QUESTIONS.

(A) CHORD PROPERTIES OF A CIRCLE

1. A chord of length 16 cm is drawn is a circle of radius 10 cm. Calculate the distance of the chord from the centre of the circle.

2. A circle of radius 2.5 cm has a chord of length 4.8 cm. Find the distance of the chord from the centre of the circle.

3. The radius of a circle is 40 cm and the length of perpendicular drawn from its centre to chord is 24 cm. Find the length of the chord.

4. A chord of length 48 cm is drawn at a distance of 7 cm from the centre of a circle. Calculate the radius of the circle.

5. A chord of length 16 cm is at a distance of 15 cm from the centre of the circle. Find the length of the chord of the same which is at a distance of 8 cm from the centre.

6. Two parallel chords of lengths 30 cm and 16 cm are drawn on the opposite sides of the centre of circle of radius 17 cm. Find the distance between the chords.

7. Two parallel chords of lengths 80 cm and 18 cm drawn on the same side of the centre of circle of radius 41 cm. Find the distance between the chords.

8. Two parallel chords AB and CD are 3.9 cm apart and lie on the opposite sides of the centre of a circle. If AB = 1.4 cm and CD = 4 cm, find the radius of the circle.

9. AB and CD are two parallel chords of lengths 8 cm and 6 cm respectively. If they are 1 cm apart and lie on the same side of the centre of the circle, find the radius of the circle.

10. PQR is an isosceles triangle inscribed in a circle. If PQ = PR = 25 cm and QR = 14 cm, calculate the radius of the circle to the nearest cm.

11. An isoscelesABC is inscribed in a circle. IF AB = AC = 12 5 cm and BC = 24 cm, find the radius of the circle.

12. An equilateral triangle of side 9 cm is inscribed in a circle. Find the radius of the circle.

13. If a line  intersects two concentric circles at the points, A, B, C and D, as shown in the figure, prove that AB = CD.

14. The radii of two concentric circles are 17 cm and 10 cm. A line segment PQRS cuts the larger circle at P and S and the smaller circle at Q and R. If QR = 12 cm, find the length PQ.

15. Two circles of radii 17 cm and 25 cm intersect each other at two points A and B. If the length of common chord AB of the circle be 30 cm, find the distance between the centers of the circles.

16. In the adjoining figure, BC is a diameter of a circle with centre O. If AB and CD are two chords such that AB || CD. prove that AB = CD.

17. The adjoining figure shows a circle with centre O in which a diameter AB bisects the chord PQ at point R. If PR = RQ = 8 cm and RB = 4 cm, find the radius of the circle.

18. In the adjoining figure, AB is a chord of a circle with centre O and BC is a diameter. If OD  AB, show that CA = 2 OD and CA || OD.

19. In the adjoining figure, P is a point of intersection of two circles with centre C and D. If the straight line APB is parallel to CD, prove that AB = 2 CD.

20. If a diameter of a circle bisects each of the two chords of a circle, then prove that the chords are parallel.

21. If two chords of a circle are equally inclined to the diameter through their point of intersection, prove that the chords are equal.

22. Show that equal chords of a circle subtend equal angles at the centre of the circle.

23. In the given figure, equal chords AB and CD of a circle with centre O, cut at right angles at P. If L and M are mid-points of AB and CD respectively, prove that OLPM is a square.

24. Prove that the perpendicular bisector of a chord of a circle always passes through the centre.

25. AB and CD are two parallel chords of a circle and line  is the perpendicular bisector of AB. Show that  is the perpendicular bisector of CD also.

26. Prove that the diameter of a circle perpendicular to one of the two parallel chords of a circle is perpendicular to the other and bisects it.

27. Prove that a diameter of a circle, which bisects a chord of the circle also bisects the angels subtended by the chord at the center of the circle.

28. In the given figure, L and M are mid-points of two equal chords AB and CD of a circle with centre O. Prove that

29. In the given figure, AB and AC are equal chords of a circle with centre O and OP  AB, OQ  AC. Prove that PB = QC.

30. In an equilateral triangle, prove that the centroid and the circum centre of the triangle coincide.

(B)

ANGLE PROPERTIES OF A CIRCLE

1. In the given figure, O is the centre of the circle



OAB = 300_and



_{OCB = 40}0_{. Calculate}



_AOC.

2. In the given figure, O is the centre of the circle and



AOC = 1300_{. Find}



_ABC.

3. In the given figure, O is the centre of the circle and



AOB = 1100_{. Calculate (i)}



_{ACO (ii)}



_CAO.

4. In the given figure, AB || D and



BAD = 1000_{. Calculate : (i)}



_{BCD (ii)}



_{ADC (iii)}



_ABC.

5. In the given figure,



ACB = 520_and



_{BDC = 43}0_{Calculate (i)}



_{ADB (ii)}



_{BAC (iii)}



_ABC.

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



AOB = 1400 _and



_{OAC = 50}0_{, find} (i)



ABC

(ii)



BCO (iii)



OAB

7. In the given figure,



BAD = 700_,



_{ABD = 56}0_and



_{ADC = 72}0_{. Calculate (i)}



_{DBC (ii) BCD (iii)}



_BCA

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



ADC = 1400_{, find}



_BAC.

9. In the given figure, O is the centre of the circle andABC is equilateral. Find (i)



BDC (ii)



BEC.

10. In the given figure, O is the centre of the circle and



AOC = 1600_{. Prove that 3}



_{y - 2}



_{x = 140}0

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



CBD = 250_and



_{APB = 120}0_{, find}



_ADB.

12. (i) In the given figure, AOB is a diameter of the circle O and



AOC = 1000_{, find}



_BDC.

13. In the figure, AB is parallel to DC,



BCE = 800_and



_{BAC = 25}0_{. Find : (i)}



_{CAD (ii)}



_{CBD (iii)}



_ADC.

14. In the given figure, O is the centre of the circle and



OBC = 500_{Calculate (i)}



_{ADC (ii)}



_AOC.

15. In the given figure, ABCD is a cyclic quadrilateral in which



CAD = 250_,



_{ABC = 50}0_and



_{ACB = 35}0_{. Calculate} (i)



CBD (ii)



DAB (iii)



ADB

16. In the adjoining figure,



BAD = 650_,



_{ABD = 70}0_and



_{BDC = 45}0_{. Find (i)}



_{BCD (ii)}



_{ADB Hence, show that} AC is a diameter.

17. In the given figure, AB is a diameter of a circle with centre O and chord ED is parallel to AB and



EAB = 650 Calculate (i)



EBA (ii)



BED (iii)



BCD

18. In the given figure, ABCD is a cyclic quadrilateral whose side CD has been produced to E. If BA = BC and



BAC = 460_{, find}



_ADE.

19. In the given figure, O is the centre of a circle and ABE is a straight line. If



CBE = 550_{, find :} (i)



ADC (ii)



ABC (iii) the value of x.

20. In the given figure AB and CD are two parallel chords of a circle. If BDE and ACE are straight lines, intersecting at E, prove thatAEB is isosceles.

21. In the given figure, chords AB and CD of a circle are produced to meet at O. Prove that ODB and OAC are similar. If BO = 3 cm, DO = 6 cm and CD = 2 cm, find AB.

22. In the given figure, O is the centre of the circle, IfAOD = 1400_and



_{CAB = 50}0_{, calculate :} (i)



EDB (ii)



EBD

23. In the given figure, AB is diameter of a circle with centre O. If ADE and CBE are straight lines, meeting at E such that



BAD = 350_and



_{BED = 25}0_{, find : (i)}



_{DCB (ii)}



_{DBC (iii)}



_BDC

25. In the given figure, the straight lines AB and CD pass through the centre O of the circle. if



AOD = 750 _and



OCE = 400_{, find (i)}



_{CDE (ii)}



_OBE.

26. In the given figure, the two circles intersect at P and Q. If



A = 800_and



_{D = 84}0_{calculate :} (i)



QBC (ii)



BCP

27. In the adjoining figure, AB = AC = CD,



ADC = 350_{. Calculate : (i)}



_{ABC (ii)}



_BEC

28. In the adjoining figure, two circles intersect at A and B. The centre of the smaller circle is O and lies on the circumference of the larger circle. If PAC and PBD are straight lines and



APB = 750_{, find}

(i)



AOB (ii)



ACB (iii)



ADB.

29. The exterior angles B and C inABC are bisected to meet at a point P. Prove that



BPC = 900 2 A

 . Is ABPC a

cyclic quadrilateral ?

30. In the given figure, is the incentre ofABC. AT produced meets the circum circle of ABC at D:



ABC = 550_and

(C) ARC Properties of Circles

1. In the given figure, arc AC and arc BD are two equal arcs of a circle. Prove that chord AB and chord CD are parallel.

2. Prove that the angle subtended at the centre of a circle, is bisected by the radius through the mid-pint of the arc.

3. In the given figure, P is the mid-point of arc APB and M is the midpoint of chord AB of a circle with centre O. Prove that :

(i) PM  AB

(ii) PM produced will pass through the centre O; (iii) PM produced will bisect the major arc AB.

4. Prove that in a cyclic trapezium, the non-parallel sides are equal.

5. P is a point on a circle with centre O. if P is equidistant from the two radii OA and OB, prove that arc AP = arc BP.

6. In the given figure, two chords AC and BD of a circle intersect at E. If arc AB = arc CD, prove that: BE = EC and AE = ED.

7. In the given figure, two chords AB and CD of a circle intersect at a point P. If AB = CD, prove that : arc AD = are CB.

8. If two sides of a cyclic quadrilateral are parallel, prove that : (i) its other two sides are equal, (ii) its diagonals are equal.

9. In the given figure, AB BC and CD are equal chords of a circle with centre O and AD is a diameter. If DEF = 1100_{find (i)}AEF _(ii)



_FAB.

10. In the given figure, ABCDE is a pentagon inscribed in a circle. If AB = BC = CD,



BCD = 1100 _and



_{BAE =} 1200_{, find : (i)}



_{ABC (ii)}



_{CDE (iii)}



_{AED (iv)}



_EAD

11. In the given figure,ABC is an isosceles triangle inscribed in a circle with centre O. If AB = AC, prove that : AP = bisects



BPC.

12. In the given figure, AB is a side a regular 6-sided polygon and AC is a side a regular 8-sided polygon inscribed in a circle with centre O. Find : (i)



AOB (ii)



ACB



ABC.

(D) OBJECTIVE TYPE QUESTIONS :

1. O is the centre of the circle, If chord AD = chord CD, then x = (A) 700

(B) 500 (C) 350 (D) 450

2. O is the centre of the circle. If 1 2, then (A) x > z

(B) x < y (C) x + y = 2z (D) None of these

3. O is the centre of the circle having radius 5 cm. OM  on chord AB. If OM = 4 cm, then the length o the chord AB =

(A) 6 cm (B) 5 cm (C) 8 cm

4. O is the centre of the circle, AB is a chord of the circle. OM  AB. If AB = 20 cm, OM =  cm, then radius of the circle is -(A) 15 cm (B) 12 cm (C) 10 cm (D) 11 cm

5. O is the centre of the circle with radius 5 cm. Chords AB and CD are parallel. AB = 6 cm and CD = 8cm. If PQ is distance between AB and CD then PQ =

(A) 10 cm (B) 8 cm (C) 7 cm (D) 7 2 cm

6. O is the centre of the circle. AB and CD are two chords of the circle. OM  AB and ON  CD. If OM = ON = 3 cm and AM = BN = 4.5 cm, then CD =

(A) 8 cm (B) 9 cm (C) 10 cm

(D) None of these

7. If BADADC, then (A) AB = CD (B) AB  CD (C) AD = BC (D) AD  BC 8. If ABCD, then (A)



1 =



2 (B)



2 =



3 (C)



3 =



4 (D) None of these

9. O is the centre of the circle having radius 5 cm. AB and AC are two chords such that AB = AC = 6 cm. If OA meets BC at M then OM =

(A) 3.6 cm (B) 1.4 cm (C) 2 cm (D) 3 cm

10. O is the centre of the circle. BC is a diameter of the circle. OD  AB (chord). If OD = 4 cm, BD = 5 cm, then CD =

(A) 13 cm (B) 71 cm

(C) 89 cm (D) None of these

ANSWER KEY

(A) CHORD PROPERTIES OF A CIRCLE

1. 6 cm,2. 07 cm 3. 64 cm 4. 25 cm 5. 30 cm 6. 23 cm 7. 31 cm

8. 2.5 cm 9. 5cm 10. 13 cm 11. 15 cm 12. 3 3 14. 9 cm 15. 28 cm 17. 10 cm

(B) ANGLE PROPERTIES OF A CIRCLE

1.



AOC = 1400 _2.



_{ABC = 115}0 _{3. (i) 55}0_{(ii) 55}0 _{4. (i)}



_{BCD = 80}0_(ii)



_{ADC = 80}0_(ii)



_{ABC = 100}0

5. (i)



ADB = 520_(ii)



_{BAC = 43}0_(iii)



_{ABC = 85}0

6. (i)



ABC = 400_(ii)



_{BCO = 60}0_(iii)



_{OAB = 20}0_(iv)



_{BCA = 110}0

7. (i)



BDC = 1800_(ii)



_{BCD = 60}0_(iii)



_{BCA = 54}0

8.



BAC = 500 _{9. (i)}



_{BDC = 60}0_(iii)



_{BEC = 120}0 _{10. (i)}



_{BAD = 62.5}0_(ii)



_{BCD = 117}50

11.



ADB = 950_12.



_{BDC = 40}0_(ii)



_{OCB = 60}0 _{13. (i)}



_{CAD = 55}0_(ii)



_{CBD = 55}0_(iii)



_{ADC - 100}0

14. (i)



ADC = 1300_(ii)



_{AOC = 100}0 _15.



_{CBD = 25}0_(ii)



_{DAB = 70}0_(iii)



_{ADB = 35}0

17. (i)



EBA = 250_(ii)



_{BED = 25}0_(iii)



_{BCD = 155}0 _{18. 88}0

19. (i)



ADC = 550_(ii)



_{ABC = 125}0_{(iii) x = 250 21. AB = 13 cm}

22. (i)



EDB = 500_(ii)



_{EBD = 110}0 _{23. (i)}



_{DCB = 35}0_(ii)



_{DBC = 115}0 _(iii)



_{DBC = 30}0

24. (i) Yes (ii) No. 25. (i)



CDE = 500_(ii)



_{OBE = 25}0 _{26. (i)}



_{QBC = 100}0_(ii)



_{BCP = 96}0

27. (i)



ABC = 400_(ii)



_{BEC = 40}0 _{28. (i)}



_{AOB = 150}0_(ii)



_{ACB = 30}0_(iii)



_{ADB = 30}0

29. No 30. (i)



BCD = 250_(ii)



_{CBD = 35}0_(iii)



_{DCI = 55}0_(iv)



_{BIC = 120}0

(C) ARC Properties of Circles

9. (i)



AEF = 200_(ii)



_{FAB = 130}0 _{10. (i)}



_{ABC = 110}0_(ii)



_{CDE = 95}0_(iii)



_{AED = 105}0_(iv)



_{EAD = 50}0

12. (i)



AOB = 600_(ii)



_{ACB = 30}0 _(iii)



_{ABC = 22}0₃₀

(D) Objective Type Questions :

GEOMETRICAL CONSTRCUTIONS



INTRODUCTION

In the chapter “Lines and Angles” and “Triangles” we have proved many theorem and properties by using diagrams in which angles and sides of triangles were drawn in approximate measurement. The diagrams were drawn to have the idea of the situations according to the given conditions. In this chapter, we shall construct some angles and triangles in precise measurement by using only two geometrical instruments. These two instruments’ are, a graduated ruler and a compass.



CONSTRCUTION OF PERPENDICULAR BISECTOR OF A LINE SEGMENT

We want to construct the perpendicular bisector of the given line segment AB.

Steps of Construction:

1. Taking A and B as centers and radius AB 2 1

 , draw, arcs on both sides of the line segment AB.

2. The arcs are drawn in such as way that on both sides of AB, we get intersection points P and Q.

3. Join PQ. PQ intersects AB at M. Here, PMQ is the required perpendicular bisector of AB.

Justification. In figure, Join AP, AQ, BT and BQ.

Here, AP = BP = BQ = AQ (Each = radius of the arc)



APBQ is a rhombus.



Diagonals AB and PQ are right bisectors of each other. Hence, PQ is perpendicular bisector of the chord AB.



CONSTRUCTION OF BISECTOR OF A GIVEN ANGLE

We want to construct the bisector of given angle ABC.

Steps of Construction:

1. Taking B as centre, we draw an arc of circle which meets BC at P and BA at Q.

2. Now, taking P and Q as centre and radius PQ 2 1

 draw two arcs so that they intersect at a point M.

3. Join BM.

Here, they ray BM is the required bisector of



ABC.

Justification. In figure, Join PM and QM.

InBPM and BQM, we have

BP = BQ (Radius of the first arc) PM = QM (Radius of the second arc)

BM = BM (Common)



BPM  BQM

By CPCT, we have



ABC. 

Construction of 60

0

_angle.

Step of Construction :

1. Taking A centre and radius = r (say), we draw an arc of a circle. The arc intersects AB and P.

2. Now, taking P as centre and same radius r, we again draw an arc of a circle which intersects the previous arc at Q.

3. Join AQ and produce this as ray AC.

Here,



CAB = 600

Justification :

Joint PQ.

We know that AP = AQ = PQ = r.



PAQ is equilateral





PAQ = 600 

Construction of 30

0

_Angle

We will construct 300_{and at the initial point A of the given ray AB.}

Step of construction :

1. Taking A as centre and radius = r (say), we draw an arc of a circle. The arc intersect AB at P.

2. Now, taking P as centre and same radius r, we again draw an arc of a circle which intersects the previous arc at Q.

3. Join AQ and produce the ray AC. Here



BAC = 600

4. Now, taking P and Q as centers and some radius r > r, we draw arcs which intersect at R.

5. Join AR and produce the ray AD along AR.

6. Here, AD is bisector of



BAC = 600 Therefore, we have



BAD = 300 

Construction of 45

0

_Angle

We will construct 450_{angle at the initial point A of the given ray AB.}

Steps of construction :

1. First of all, we construct



BAC = 600

2. We find AD bisector of



BAC. Here;



BAD =



DAC = 300

3. Now, we find AE bisector of



CAD. Here,



DAE =



CAE = 150

4.



BAE =



BAD +



DAE = 300_{+ 15}0_{= 45}0 

Construction of 90

0

_Angle

Steps of Construction :

1. We construct



CAB =



CAD = 600

2. Now, we find AE bisector of



CAD.

3.



BAE =



BAC +



CAE =



BAC + 2 1

_

CAD = 600₊ 2 1 × 600

Ex.1 Take a line segment AB = 6.3 cm. Find the right bisector (perpendicular bisector) of AB. Sol. Steps of Construction:

1. We take AB = 6.3 cm.

2. Taking A and B as centers and radius > 2 1

AB. we draw arcs on both sides of B.

3. Two arcs intersect at P on one side and

other two arcs intersect at Q on the second side of AB.

4. We join PQ ; it meets AB at M.

Now PMQ is the required right bisector of AB.

Ex.2 Construct an angle of 750_{at the initial point of a given ray. Justify the construction.}

Sol. We have



BAC = 600 and



CAD = 600_, Then,



CAE =

2 1

× 600_{= 30}0 Further, AF bisect



CAE, Then



CAF = 2 1

_

CAE = 2 1 × 300_{= 15}0

Hence,



BAF =



BAC +



CAF = 600_{+ 15}0_{= 75}0_.



CONSTRCUTION OF TRAINGLES

In this section, we shall construct some triangles with given data by using a graduated ruler and a compass. 

To construct a triangle, given its base, a base angle and sum of other two sides.

We have to construct



ABC. When base BC,



B and AB + AC is given. Here, AB + AC > BC.

Steps of Construction :

1. Draw the given base BC.

2. Construct the base angle



CBX =



B as given.

3. Cut the line segment BD = AB + AC along BX.

4. Join CD.

5. Draw perpendicular bisector PQ of CD. PQ meets CD at L and BC at A.

6. Join AC.

Here,ABC is the required triangle.

Justification.

InABC, AL is perpendicular bisector of CD.



AD = AC.

Now, inABC we have

AB + AC = AB + AD = BC

Note :

If sum of two sides is less than the given base, then triangle is not possible.

Case (a) :

We shall constructABC when base BC and



B are given. Also, we are given AB - AC. Here, AB > AC.