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Question-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 centers.

Solution:

Given: AB and CD are two equal chords of congruent circles with centres O and O' respectively.

To Prove: AOB = CO'D

Proof: In Δ OAB and Δ O'CD,
OA = O'C  (∵ Radii of congruent circles)
OB = O'D  (  Radii of congruent circles ) 
AB = CD     ( Given)
Δ OAB Δ O'CD   (  SSS Rule )
AOB = CO'D (  CPCT )

Question-2

Prove that if chords of congruent circles subtend equal angles at their centers, then the chords are equal.

Solution:

Given: AOB and C'OD are the two equal angles subtended by the chords
AB and CD of two congruent circles with centres O and O' respectively.

To Prove: AB = CD.

Proof: In ΔOAB and ΔO'CD,
OA = O'C                   | Radii of congruent circles
OB = O'D                   | Radii of congruent circles
AOB = CO'D         | Given
ΔOAB = ΔO'CD          | SAS Rule
AB = CD.                   | CPCT

Question-3

Draw different pairs of circles. How many points does each pair have in common? What is the maximum number of common points?

Solution:
Each pair has atmost two common points.

The maximum number of common points is two.

Question-4

Suppose you are given a circle. Give a construction to find its centre.

Solution:
Steps of Construction:
(i) Take any three points P, Q and R on the circle.

(ii) Join PQ and QR.

(iii) Draw the perpendicular bisectors of PQ and QR. Let these intersect at O. Then, O is the centre of the circle. (i.e., Locus of a point equidistant from the line segment)

Question-5

If two circles intersect at two points, prove that their centres lie on the perpendicular bisector of the common chord.

Solution:
Given:  Two circles with centres O and P intersecting at A and B

To Prove: OP is the perpendicular bisector of AB.

Construction: Join OA, OB, PA and PB. Let OP intersect AB at M.

Proof: In Δ OAP and Δ OBP,

OA = OB                | Radii of a circle

PA = PB                 | Radii of a circle
OP=OP                  | Common
Δ OAP Δ OBP    | SSS Rule
AOP = BOP   | CPCT
AOM = BOM      …..(1)
In Δ AOM and Δ BOM,

OA = OB               | Radii of a circle

AOM = BOM     | From(1)

OM = OM               | Common

Δ AOM Δ BOM   | SAS Rule
AM = BM    ...(2) | CPCT
and AMO = BMO    ...(3)  | CPCT

But AMO + BMO = 180°   | Linear Pair Axiom

AMO = BMO = 90°    ...(4)

OM, i.e. OP is the perpendicular bisector of AB.   | From (2) and (4)

Question-6

Two circles of radii 5 cm and 3 cm intersect at two points and the distance between their centers is 4 cm. Find the length of the common chord.

Solution:


We know that if two circles intersect each other at two points, then the line joining their centres is the perpendicular bisector of their common chord.

Length of the common chord
= PQ = 2O’P
                                          = 2 × 3 cm = 6 cm.

Question-7

If two equal chords of a circle intersect within the circle, prove that the segments of one chord is equal to corresponding segments of the other chord.

Solution:
Given: A circle with centre O. Its two equal chords AB and CD intersect at E.
To prove: AE = DE and CE = BE
Construction: Draw OM  AB and ON  CD. Join OE.     
  
                                        
Proof: In ΔOME and ΔONE,
OM = ON     (Equal chords of a circle are equidistant from the centre)
OE = OE      (Common )
ΔOME  ΔONE          ( R.H.S.)
 ME = NE                  (CPCT )
AM + ME = DN + NE (∵ AM = DN = = )
AE = DE
AB AE = CD – DE     (Given AB = CD )
BE = CE.

Question-8

If two equal chords of a circle intersect within the circle, prove that the line joining the point of intersection to the center makes equal angles with the chords.

Solution:
                                                                                                              
Given: Two equal chords AB and CD of a circle with centre O intersect within the circle. Their point of intersection is E.
To Prove: OEA = OED
Construction: Join OA and OD      
                                                                           
Proof: In Δ OEA and Δ OED
OE = OE               |Common
OA = OD              | Radius of a circle
AE = DE               | Proved in  
D OEA D OED     | SSS Rule
∠ OEA = ∠ OED    | CPCT

Question-9

If a line interesects two concentric cirles (circles with the same centre ) with centre O at A,B,C and D prove that AB=CD.
 

 

Solution:

Given: A line intersects two concentric circles (circles with the same centre) with centre O at A, B, C and D.
To Prove: AB = CD

Construction: Draw OM BC.
Proof: The perpendicular drawn from the centre of a circle to a chord bisects the chord.
AM = DM                       ….(1)
and BM = CM                     ...(2)
Subtracting (2) from (1), we get
AM – BM = DM  CM
AB = CD. 

Question-10

Three girls Reshma, Salma and Mandip are playing a game by standing on a circle of radius 5m drawn in a park. Reshma throws a ball to Salma, Salma to Mandip and Mandip to Reshma. If the distance between Reshma and Salma and between Salma and Mandip is 6m each, what is the distance between Reshma and Mandip?


Solution:








Let R, S, M are the three girls Reshma, Salma and Mandip respectively are playing a game by standing on a circle with centre 'O'

In Δ NOR and Δ NOM,
ON = ON                 (Common)
NOR = NOM       (Since equal chords of a circle subtend equal angles at the centre)
OR = OM                 (Radii of a circle)

Δ NOR = Δ NOM    (SAS Rule)

ONR = ONM and    (CPCT)

NR = NM                       
(CPCT)  
But ONR + ONM = 180°       (Linear Pair Axiom)
∴ ∠ ONR = ONM = 90°
ON is the perpendicular bisector of RM.
Draw bisector SN of RSM to intersect the chord RM in N.
In Δ RSN and Δ MSN,
RS = MS (= 6 cm each)

SN = SN               (Common)

RSN = MSN     (By construction) 
RSN Δ MSN   (SAS Rule)

RNS = MNS  (CPCT)

and RN = MN          (CPCT)

But RNS + MNS = 180°      (Linear Pair Axiom)

RNS = MNS = 90°

SN is the perpendicular bisector of RM and therefore passes through O when produced.

Let ON = x m
Then SN = (5
– x)m
In right triangle ONR,
x 2 + RN2 = 52            ...(1) ( By Pythagoras Theorem)
In right triangle SNR,
(5 – x)2 + RN2 = 62     ...(2)   ( By Pythagoras Theorem) 
From (1),
RN2 = 52 – x2
From (2),
RN2 = 62 – (5 – x)2
Equating the two values of RN2, we get
52 –  x2 = 62  –  (5 – x)2
25 – x2 = 36 – (25 – 10 x  x2) 25 – x2 = 36  25 + 10x – x2  


25 – x2 = 11 + 10x – x2 25 – 11 = 10x 
14 = 10x   10x  = 14
Substituting x = 1.4 in (1), we get
(1.4)2 + RN2 = 52 RN2 = 25 – 1.96
RN2 = 52 – (1.4)2 RN =
⇒ RN = 4.8  
RM = 2 RN = 2 × 4.8 m = 9.6 m
Hence the distance between Reshma and Mandip is 9.6 m.

Question-11

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 boy has a toy telephone in his hand to talk with each other. Find the length of the string of each phone.

Solution:


Given: 

i.e., Chord AB = Chord BC = Chord CA

⇒ ΔABC is equilateral

Centroid G of ΔABC coincides with its circumstance

AG = BG = CG = 20m

Since G divides AD in the ratio 2 : 1

GD = 10m

In ΔBDG, we have

BG2 = BD2 + GD2

202 = BD2 + 102

BD2 = 10m

BC = 2 BD

BC = 20m.

 The length of the string of each phone =  20m.

Question-12

In the given figure, A, B and C are three points on a circle with center O such that BOC = 300 and AOB= 60°. If D is a point on the circle other than the arc ABC, find ADC.

Solution:
  ADC = AOC
           =( AOB + BOC)

           = (60° + 30°)
           = (90°)
  ADC = 45°.

Question-13

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.

Solution:
                                                                                                                      
Given: In a circle with centre 'O' and its radius OA and OB are equal to the chord AB.
i.e., OA = OB = AB

To find: Angle subtended by the chord at Minor and major arc, i.e., to find the value of  ACB and  ADB
Let C and D be the points on the major and minor arcs  of the circle respectively,
join C and D with A and B
Now OA = OB = AB       (Given )
Δ OAB is equilateral.
 
ÐAOB = 60
°   
ACB = AOB        ( The angle subtended by an arc at the centre is double the angle subtended by it at any
                                      point on the remaining part of the circle) 
          = (60
° 
ACB = 30
°  
Now,  ADBC is a cyclic quadrilateral.
∴ ∠ADB + ACB = 180
° (Opposite angles of a cyclic quadrilateral are supplementary)
ADB+ 30
° =180°
ADB = 180
° - 30° 
ADB = 150
°.

Question-14


In figure, PQR = 1000, where P, Q and R are points on a circle with center O. Find OPR.

Solution:

Take a point S in the major arc. Join PS and RS.
PQRS is a cyclic quadrilateral.

PQR + PSR = 180°  
( The sum of either pair of opposite angles of a cyclic quadrilateral is 180°)
   Þ 100° + PSR = 180°

PSR = 180°-100°

PSR = 80°               ...(1)
Now, POR = 2 PSR  ( The angle subtended by an arc at the centre is double the angle
                                        subtended by it at any point on the remaining part of the circle)
                  = 2 × 80°= 1600 ...(2)            (Using (1))
In Δ OPR,
OP = OR                       (Radii of a circle )
OPR = ORP     ...(3) (Angles opposite to equal sides of a triangle are equal)
In ΔOPR,
OPR + ORP + POR = 180°    (Sum of all the angles of a triangle is 180°)
OPR + OPR + 160° = 180°   (Using (2) and (3))
2 OPR+160° = 180°
2 OPR = 180° - 160° = 20°
OPR = 10°.

Question-15

In figure, ABC= 690, РACB=310, find BDC.

Solution:
In Δ ABC,
BAC + ABC + ACB = 180°    | Sum of all the angles of a triangle is 180°

BAC+ 69°+ 31°= 180°

BAC +100° =180°
BAC =180° -100° = 80°       ...(1)
Now, BDC = BAC               | Angles in the same segment of a circle are equal
         BDC = 80°.                  | Using(1)

Question-16

In Figure, A, B, C and D are Four points on a circle. AC and BD intersect at a point E such that CEB = 1300 and ECD= 200.
Find BAC.

Solution:
CED + BEC = 180°              | Linear Pair Axiom

CED + 130° =180°

CED = 180° - 130° = 50°          ….(1)

ECD = 20°                               ….(2)

In
ΔCED,
CED + ECD + CDE = 180°       | Sum of all the angles of a triangle is 180°

50° + 20° + CDE = 180°

70° + CDE =180°
CDE =180° -70°
CDE =110°
Now, BAC = CDE                  | Angles in the same segment of a circle are equal
          BAC =110°.                    | Using (3)                

Question-17

ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If DBC = 700, BAC is 300, find BCD. Further, if AB= BC, find ECD.

Solution:
Given ABCD is a cyclic quadrilateral, in DCDB and DBAC   
CDB = BAC             |Angles in the same segment of a circle are equal
           = 30°                    ...(1)
DBC = 70°                    ...(2)
In Δ BCD,


BCD + DBC + CDB = 180°       |Sum of all the angles of a triangle is 180°

BCD + 70° + 30° = 180°           | Using (1) and (2)

BCD+100° =180°
BCD =180° - 100°
BCD = 80°                   ….(3)

In Δ ABC,

AB = BC

BCA = BAC         |Angles opposite to equal sides of a triangle are equal ...(4)
BAC = 30°    (given)

Now, BCD = 80°     | From (3)
BCA + ECD = 80°

30° + ECD = 80°

ECD = 80° - 30°
ECD = 50°.

Question-18

If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.

Solution:
                                                                                        
In Δ OAB and Δ OCD,
OA = OC              | Radii of a circle
OB = OD              | Radii of a circle
AOB = COD    | Vertically Opposite Angles
 D OAB D OCD    | SAS Rule
AB = CD               | CPCT
 arc AB=arc CD         ..(1)
Similarly, we can show that
arc AD = arc CB         ….(2)
Adding (1) and (2), we get
Arc AB + Arc AD = Arc CD + Arc CB
Arc BAD = Arc BCD
BD divides the circle into two equal parts (each a semicircle)

A = 90°, C = 90° | Angle of a semi-circle is 90°

Similarly, we can show that
B = 90°, D = 90°
A = B = C = D = 90°
ABCD is a rectangle.

Question-19

If the non-parallel sides of a trapezium are equal, prove that it is cyclic.

Solution:
Given: ABCD is a trapezium whose two non-parallel sides AD and BC are equal.

To Prove: Trapezium ABCD is cyclic

Construction: Draw BE || AD
Proof: AB || DE       | Given
AD || BE                | By construction
Quadrilateral ABED is a parallelogram
                                                                                                
  BAD = BED             ...(1) | Opposite angles of a parallelogram
and AD = BE ...(2)                     | Opposite sides of a parallelogram
But AD = BC ...(3)                     | Given
From (2) and (3),
BE = BC
BEC = BCE           ...(4)  |Angles opposite to equal sides
BEC + BED =180°          | Linear Pair Axiom

BCE + BAD =180°          | From (4) and (1)
Trapezium ABCD is cyclic      |
∵ If a pair of opposite angles of a quadrilateral is 180°, then the quadrilateral is cyclic.

Question-20

 


Solution:
Given: Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively.
To Prove: ACP = QCD.
Proof: ACP = ABP    ...(1) | Angles in the same segment of a circle are equal
QCD = QBD            ...(2) | Angles in the same segment of a circle are equal
ABP = QBD            ...(3) | Vertically Opposite Angles
From (1), (2) and (3),
ACP = QCD.

Question-21

If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.

Solution:
Given: Circles are described with sides AB and AC of a triangle ABC as diameters. They intersect in a point D.
To Prove: D lies on the third side BC of Δ ABC.

 Construction: Join AD.
Proof: Circle described on AB as diameter intersects BC in D.
ADB = 90°                     | Angle in a semi-circle
But ADB + ADC = 180°    | Linear Pair Axiom
ADC = 90°.
Hence, the circle described on AC as diameter must pass through D.

Thus, the two circles intersect in D.

Now, ADB + ADC = 180°.

Points B, D, C are collinear.

D lies on BC.

Question-22

ABC and ADC are two right triangles with common hypotenuse AC. Prove that
CAD = CBD.

Solution:
Given: ABC and ADC are two right triangles with common hypotenuse AC.

To Prove: CAD = CBD.


                                                                                  
 

Proof: AC is the common hypotenuse and ABC and ADC are two right triangles.
ABC = 90° = ADC
Both the triangles are in the same semi-circle.
Points A, B, D and C are concyclic.

DC is a chord

CAD = CBD.         ( Angles in the same segment are equal)

Question-23

Prove that a cyclic parallelogram is a rectangle.

Solution:
Given: ABCD is a cyclic parallelogram
                          
To Prove: ABCD is a rectangle
 

Proof: ABCD is a cyclic quadrilateral
1 + 2=180° ...(1)    (  opposite angles of a cyclic quadrilateral are supplementary)
ABCD is a parallelogram
∠ 1 = 2         ….(2)    (  Opposite angles of a parallelogram)
From equations (1) and (2),
1 = 2 = 90°
Parallelogram ABCD is a rectangle.




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