rd sharma solutions class 9 circles ex 16.1 rd sharma ......having a toy telephone in his hands to...
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RD Sharma Solutions Class 9 Circles Ex 16.1
RD Sharma Solutions Class 9 Chapter 16 Ex 16.1Q1) Fill in the blanks:
(i) A ll points lying inside/outside a circle are called_____ points/_______ points.
(ii) Circle having the same centre and different radii are called_____circles.
(iii) A point whose distance from the center o f a circle is greater than its radius lies in _________o f the circle.
(iv) A continuous piece o f a circle is _______ o f the circle.
(v) The longest chord o f a circle is a ____________o f the circle.
(vi) An arc is a __________when its ends are the ends o f a diameter.
(vii) Segment o f a circle is a region between an arc and_______o f the circle.
(viii) A circle divides the plane, on which it lies, in ______ parts.
Solution:
(i) Interior/Exterior
(ii) Concentric
(iii) The Exterior
(iv) Arc
(v) Diameter
(vi) Semi circle
(vii) Center
(viii) Three
Q2) Write the truth value (T/F) o f the following with suitable reasons:
(i) A circle is a plane figure.
(ii) Line segment joining the center to any point on the circle is a radius o f the circle,
(iii) I f a circle is divided into three equal arcs each is a major arc.
(iv) A circle has only finite number o f equal chords.
(v) A chord o f a circle, which is twice as long as its radius is the diameter o f the circle.
(vi) Sector is the region between the chord and its corresponding arc.
(vii) The degree measure o f an arc is the complement o f the central angle containing the arc.
(viii) The degree measure o f a semi-circle is 180°.
Solution:
0)T
00 T
(iii) T
('v) F
(v) T
(vi) T
(vii) F
(viii) T
RD Sharma Solutions Class 9 Circles Ex 16.2
RD Sharma Solutions Class 9 Chapter 16 Ex 16.2
Q l) The radius o f a circle is 8 cm and the length o f one o f its chords is 12 cm. Find the distance o f the chord from the centre.
Solution:
Given that,
Radius of circle (OA) = 8cm
Chord (AB) = 12cm
Draw O C ± A B
We know that
The perpendicular from centre to chord bisects the chord
AC = BC = f = 6cm
Now in A OCA, by Pythagoras theorem
AC2 + OC2 = OA2
=>62 + OC2 = 82
=>36 + OC2 = 64
=>0C2 = 64-36
=> OC2 = 28
=>0C = \ /2 8
=>0C = 5.291 cm
Q2) Find the length o f a chord which is a t a distance o f 5 cm from the centre o f a circle o f radius 10 cm.
Solution:
Distance (OC) = 5cm
Radius of the circle (OA) = 10cm
In A OCA, by Pythagoras theorem
OC2 + AC2 = OA2
=>52 + AC2 = 102
=>25 + AC2 = 100
=>AC2 = 1 0 0 - 2 5
=> AC2 = 75
=>AC = \ /7 5
=>AC = 8.66cm
We know that, the perpendicular from the centre to chord bisects the chord
Therefore, AC = BC = 8.66cm
Then the chord AB = 8.66 + 8.66
= 17.32cm
Q3) Find the length o f a chord which is a t a distance o f 4 cm from the centre o f a circle o f radius 6 cm.
Solution:
Radius of the circle (OA) = 6cm
Distance (OC) = 4cm
In A OCA, by Pythagoras theorem
AC2 + OC2 = OA2
=>AC2 + 42 = 62
=>AC2 = 3 6 - 1 6
=> AC2 = 20
=>AC = V 2 0
=>AC = 4.47cm
We know that the perpendicular distance from centre to chord bisects the chord.
AC = BC = 4.47cm
Then AB = 4.47 + 4.47
= 8.94cm
Q4) Two chords AB, CD o f lengths 5 cm, 11 cm respectively o f a circle are parallel. I f the distance between AB and CD is 3 cm, find the radius o f
the circle.
Solution:
Construction: Draw OP -L CD
Chord AB = 5cm
Chord CD = 11 cm
Distance PQ = 3cm
Let OP = x cm
And OC = OA = r cm
We know that the perpendicular from centre to chord bisects it.
C P = P D = f e r n
And AQ = BQ = \cm
In A O C P , by Pythagoras theorem
O C 2 = O P 2 + C P 2
^ r 2 = x2 + ( f )2 ........(i)
In AOQA, by Pythagoras theorem
OA2 = OQ2 + AQ2
^ r 2 = (* + 3)2 + ( f ) 2....... (»)
Compare equation (i) and (ii)
=M* + 3)a + ( f )a = * a + ( £ ) a
=>x2 + 6x + 9 + f = x 2 + !f- ^ x 2 - x 2 + 6x = ^ - ^ - - 94 4
=>■ 6x = 15
Q5) Give a m ethod to find the centre o f a given circle.
Solution:
(1) Take three points A, B and C on the given circle.
(2) Join AB and BC.
(3) Draw the perpendicular bisectors of the chord AB and BC which intersect each other at 0.
(4) Point 0 will give the required circle because we know that, the Perpendicular bisectors of chord always pass through the centre.
Q6) Prove tha t the line jo in ing the m id-point o f a chord to the centre o f the circle passes through the m id-point o f the corresponding m inor arc.
Solution:
C is the mid-point of chord AB.
To prove: D is the mid-point of arc AB.
Proof:
In A OAC and A OBC
OA = OB [Radius of circle]
OC = OC [Common]
AC = BC [C is the mid-point of AB]
Then AO AC = A OBC [By SSS condition]
ZAOC = ZBOC
=> mAD ^ mBD
=> AD = BD
Hence, D is the mid-point of arc AB.
Q7) Prove tha t a diam eter o f a circle which bisects a chord o f the circle also bisects the angle subtended by the chord a t the centre o f the circle.
Solution:
PQ is a diameter of circle which bisects the chord AB at C.
To Prove: PQ bisects ZAOB
Proof:
In ZAOC and ZBOC
OA = OB [Radius of circle]
OC = OC [Common]
AC = BC [Given]
Then AAOC = ABOC [By SSS condition]
ZAOC = ZBOC [C. P. C. T]
Hence PQ bisects ZAOB.
Q8) Prove tha t two different circles cannot intersect each other a t m ore than two points.
Solution:
Suppose two circles intersect in three points A, B, C.
Then A, B, C are non-collinear so a unique circle passes through these three points. This is contradiction to the face that two given circles are
passing through A, B, C. Hence, two circles cannot intersect each other at more than two points.
Q9) A line segm ent AB is o f length 5 cm. Draw a circle o f radius 4 cm passing through A and B. Can you draw a circle o f radius 2 cm passing through A and B? Give reason in support o f your answer.
Solution:
(1) Draw a line segment AB of 5cm.
(2) Draw the perpendicular bisectors of AB.
(3) With centre A and radius of 4cm, draw an arc which intersects the perpendicular bisector at point 0. The point 0 will be the required centre.
(4) Join OA.
(5) With centre 0 and radius OA, draw a circle.
No, we cannot draw a circle of radius 2cm passing through A and B because when we draw an arc of radius 2cm with centre A, the arc will not
intersect the perpendicular bisector and we will not find the centre.
Q10) An equilateral triangle o f side 9 cm is inscribed in a circle. Find the radius o f the circle.
Solution:
Let ABC be an equilateral triangle of side 9cm and let AD is one of its median.
Let G be the centroid of A ABC. Then AG : GD = 2 :1
We know that in an equilateral triangle, centroid coincides with the circum centre.
Therefore, G is the centre of the circumference with circum radius GA.
Also G is the centre and GD is perpendicular to BC.
Therefore, In right triangle ADB, we have
AB2 = AD2 + DB2
=> 92 = AD2 + DB2
=>■ AD = ^ 8 1 - f - = -^-cm
Radius = AG = j AD = 3- /̂3cm
Q11) Given an arc o f a circle, com plete the circle.
Solution:
(1) Take three points A, B and C on the given arc.
(2) Join AB and BC.
(3) Draw the perpendicular bisectors of chords AB and BC which intersect each other at point 0. Then 0 will be the required centre of the
required circle.
(4) Join 0A.
(5) With centre 0 and radius OA, complete the circle.
Q12) Draw different pairs o f circles. How many points does each pa ir have in common? What is the maximum number o f common points?
Solution:
Each pair of circles have 0,1 or 2 points in common.
The maximum number of points in common is "21.
Q13) Suppose you are given a circle. Give a construction to find its centre.
Solution:
(1) Take three points A, B and C on the given circle.
(2) Join AB and BC.
(3) Draw the perpendicular bisectors of chord AB and BC which intersect each other at 0.
(4) Point 0 will be the required centre of the circle because we know that the perpendicular bisector of the chord always passes through the
centre.
Q14) Two chords AB and CD o f lengths 5 cm and 77 cm respectively o f a circle are parallel to each other and are opposite side o f its centre. I f
the distance between AB and CD is 6 cm, find the radius o f the circle.
Solution:
Join OB and 0D.
BM = 4 ^ = j [Perpendicular from the centre bisects the chord]
N D = ™ = %
Let ON be x, so OM will be 6 - x.
A MOB
0M 2 +M B2 = 0B2
(6 - a;)2 + ( f )2 = OB2
36 + a;2 - 12a; + f = OB2 ........(z)
In A NOD
0N2 + ND2 = 0D2
x2 + {^-)2 = OD2
x2 + I f =O D 2 ........(ii)
We have OB = OD. [Radii of same circle]
So, from equation (i) and (ii).
36 + x2 — 12a; + f - = a;2 +
=> 12a: = 36 + f4 4
_ 144+ 25-121 _ 48 _ -^2 4 4
X = 1
From equation (ii)
( l )2 + ( ^ ) = 0£>2
OD2 = 1 + ^ = 1254 4
Q15) The lengths o f two parallel chords o f a circle are 6 cm and 8 cm. I f the sm aller chord is a t a distance o f 4 cm from the centre, what is the
distance o f the other chord from the centre?
Solution:
Distance of smaller chord AB from centre of circle = 4cm, OM = 4cm
In A OMB
OM2 + MB2 = OB2
42 + 92 = OB2
16 + 9 = OB2
OB=^/2E
OB = 5cm
In A OND
OD = OB = 5cm [Radii of same circle]
ON2 + ND2 = OD2 ON2 + 42 = 52
ON2 = 2 5 - 1 6 ON 9 ON = 3cm
So, the distance of bigger chord from the circle is 3cm.
MB = 4p = | = 3cm
ND = | = 4cm
RD Sharma Solutions Class 9 Circles Ex 16.3
RD Sharma Solutions Class 9 Chapter 16 Ex 16.3
Q1) Three girls ishita, Ishe end Nishe ere playing a gam e by standing on a circle o f radius 20m drawn in a park. Ishita throws a b all to Isha, Isha
to Nisha and Nisha to Ishita. I f the distance between Ishita and Isha and between Isha and Nisha is 24m each, what is the distance between
Ishita and Nisha.
Solution:
Let R, S and M be the position of Ishita, Isha and Nisha respectively.
OR = OS = OM = 20 cm [Radii of circle]
In A OAR,
OA2 + AR2 = OR2
OA2 + 122 = 202
OA2 = 400 - 144 = 256m2
OA = 16m
We know that, in an isosceles triangle altitude divides the base.
So in A RSM, ARCS = 90° and RC = CM
Area of A ORS = \ x OA x RS
=> \ x RC x OS = \ x 16 x 24
=> RC x 20 = 16 x 24
=> RC = 19.2
=> RM = 2(19.2) = 38.4m
So, the distance between Ishita and Nisha is 38.4m.
Q2) A circular park o f radius 4 0 m is situated in a colony. Three boys Ankur, A m it and Anand are sitting a t equal distance on its boundary each
having a toy telephone in his hands to talk to each other. Find the length o f the string o f each phone.
Solution:
So, ABC is an equilateral triangle
OA (radius) = 40m
Medians of equilateral triangle pass through the circum centre (0) of the equilateral triangle ABC.
We also know that median intersect each other at the ratio 2 :1 .
As AD is the median of equilateral triangle ABC, we can write:
OA _ 2 OD 1
4 OM _ 2 ^ OD 1
=> OD = 20m
Therefore, AD = OA + OD = (40 + 20) m
= 60 m
In A ADC
By using Pythagoras theorem
AC2 = AD2 + DC2
AC2 = 602 + (4 £ )2
AC2 = 3600 + ^
=> \AC2 = 36004
=> AC2 = 4800
=> AC = 40v^3m
So, length of string of each phone will be 40-\/3m..
RD Sharma Solutions Class 9 Circles Ex 16.4
RD Sharma Solutions Class 9 Chapter 16 Ex 16.4
Q1) In figure 16.120,0 Is tite centre o f the circle. If/.A P B = 50°, f in d /A O B and / O A B .
P
Solution:
ZAPB = 50°
By degree measure theorem
ZAOB = 2ZAPB=> ZAPB = 2 x 50° = 100°since OA = OB [Radius o f circle]Then ZOAB = ZOBA [Angles opposite to equal sides]Let ZOAB = xIn A OAB, by angle sum property ZOAB + ZOBA + ZAOB = 180°=>x + x + 100° = 180°
=>2x = 180° - 100°
=>2x = 80°
=>x = 40°
ZOAB = ZOBA = 40°
Q2) In figure 16.121, i t is given that O is the centre of the circle and ZAOC = 150°. Find ZABC.
ZAOC = 150°ZAOC + reflex ZAOC = 360° [Complex angle]
=> 150° + reflex ZAOC = 360°=> reflex ZAOC = 360° - 150°=> reflex ZAOC = 210°=> 2ZABC = 210° [By degree measure theorem]
=> ZABC = = 105°
Q3) In figure 16.22,0 is the centre o f the circle. Find ZB AC.
We have ZAOB = 80°
And ZAOC = 110°
Therefore, ZAOB + ZAOC + ZBOC = 360° [Complete angle]
=> 80° + 100° + ZBOC = 360°=> ZBOC = 360° - 80° - 110°=> ZBOC = 170°By degree measure theorem
ZBOC = 2 ABAC =► 170° = 2 ABAC => ZBAC = M = 85°
Q4) IfO is the centre o f tire circle, find die value o fx in each o f die following figures.
(0
p
Solution:
ZAOC = 135°ZAOC + ZBOC = 180° [Linear pair o f angles\
=> 135° + ZBOC = 180°ZBOC = 180° - 135°
=> ZBOC = 45°
By degree measure theorem ZBOC = 2 ZCPB => 45° = 2x
00
c
Solution:
We have ZABC = 40°ZACB = 90° [Angle in semicircle\In AABC, by angle sum property ZCAB + ZACB + ZABC = 180°=> ZCAB + 90° + 40° = 180°=> ZCAB = 180° - 90° - 40°=> ZCAB = 50°Now,ZCDB = ZCAB [Angle is same in segment] => x = 50°
p
Solution:
We have ZAOC = 120°By degree measure theorem.ZAOC = 2ZAPC =► 120° = 2 ZAPC => ZAPC = ^ = 60°
ZAPC + ZABC = 180° [Opposite angles o f cyclic quadrilaterals] => 60° + ZABC = 180°=> ZABC = 180° - 60°=>• ZABC = 120°
ZABC + ZDBC = 180° [Linear pair o f angles]=> 120 + x = 180°^ x = 180° - 120° = 60°(hr)
Solution:
W e have ZCBD = 65°
ZABC + ZCBD = 180° [Linear pair o f angles]=> ZABC = 65° = 180° =* ZABC = 180° - 65° = 115°
reflex ZAOC = 2ZABC [By degree measure theorem]=> x = 2 x 115°=* x = 230°(v)
c
Solution:
We have ZOAB = 35°Then, ZOBA = ZOAB = 35° [Angles opposite to equal radii] In AAOB, by angle sum property => AAOB + ZOAB + ZOBA = 180°
ZAOB + 35° + 35° = 180°=> ZAOB = 180° - 35° - 35° = 110°
ZAOB + reflexZAOB = 360° [Complex angle]=> 110° + reflexZAOB = 360°=> reflexZAOB = 360° - 110° = 250°By degree measure theorem reflexZAOB = 2ZACB => 250° = 2x ^ x = *!f = 125°
(vl)
Solution:
We have ZAOB = 60°By degree measure theorem reflex ZAOB = 2ZACB => 60° = 2ZACB=$■ ZACB = = 30° [Angles opposite to equal radii[
=>x = 30°.(vii)
l )
Solution:
We haveZB AC = 50° and ZDBC = 70°
ZBDC = ZBAC = 50° [Angle in same segment] In ABDC, by angle sum property ZBDC + ZBCD + ZDBC = 180°=> 50° + x + 70° = 180°=> x = 180° - 50° - 70° = 60°(viii)
Solution:
W e have,ZDBO = 40° and ZDBC = 90° [Angle in a semi circle] => ZDBO + ZOBC = 90°=* 40° + ZOBC = 90°=* ZOBC = 90° - 40° = 50°By degree measure theorem ZAOC = 2ZOBC => x = 2 x 50° = 100°0x)
Solution:
In ADAB , by angle sum property ZADB + ZDAB + ZABD = 180°
32° + ZDAB + 50° = 180°=> ZDAB = 180° - 32° - 50°=> ZDAB = 98°Now,ZOAB + ZDCB = 180° [Opposite angles o f cyclic quadrilateral]=> 98° + a = 180°= U = 180° - 98° = 82°
00
Solution:
We have,ZBAC = 35°ZBDC = ZBAC = 35° [Angle in same segment\In A BCD, by angle sum property ZBDC + ZBCD + ZDBC = 180°
35° + x + 65° = 180°=► x = 180° - 35° - 65° = 80°(a)
We have,ZABD = 40°ZACD = ZABD = 40° [Angle in same segment] In APCD, by angle sum property ZPCD + ZCPO + ZPDC = 180°=> 40° + 110° + x = 180°
x = 180° - 150°=> x = 30°(xi'O
Given that,ZBAC = 52°Then ZBDC = ZBAC = 52° [Angle in same segment]Since OD = OCThen ZODC = ZOCD [Opposite angle to equal radii]=> x = 52°Q5) 0 is the circumference o f the triangle ABC and Odis perpendicular on BC. Prove that ZBOD = ZA.
Solution:
Given 0 is the circum centre o f triangle ABC and OD Z BC
To prove ZBOD = 2ZA
Proof:
InA OBD and A OCDZODB = ZODC [Each 90°]OB = OC [Radius o f circle[OD = OD [Common]
T h e n A OBD = A OCD [By RHS Condition],
ZBOD = ZCOD .......(i) [PCT\.By degree measure theorem ZBOC = 2ZBAC=>■ 2ZBOD = 2ZBAC [By using (*)]=> ZBOD = ZB AC
Q6) In figure 16.135,0 is the centre o f the circle, BO is the bisector o f Z ABC. Show thatAB =AC.
Given, BO is the bisector of ZABC
To prove AB = BC
Proof:
Since, BO is the bisector of ZABC.
Then ,ZABO = ZC B O .......(i)
Since, OB = OA [Radius of circle]
Then, ZABO = ZD AB ....... ( i i) [opposite angles to equal sides]
Since OB = OC [Radius of circle]
Then, ZOAB = ZOCB .......(Hi) [opposite angles to equal sides]
Compare equations (i), (ii) and (iii)
ZOAB = ZOCB ....... (iv)
In AOAB and AOCB
ZOAB = ZOCB [From (iv)]
ZOBA = ZOBC [Given]
OB = OB [Common]
Then,AOAB = A OCB [By AAS condition]
AB = BC [CPCT]
Q7) In figure 16.136,0 is the centre o f the circle, then prove that Zx = Z y+ Zz.
We have,
Z3 = Z4 [Angles in same segment].’. Zx = 2Z3 [By degree measure theorem]=>• Zx = Z3 + Z3 Zx = Z3 + Z 4.......(i) [Z3 = angle4]But Zy = Z3 + Z1 [By exterior angle property]=> Z3 = Zy — Z 1 .......(ii)from (i) and (ii)Zx = Zy — Z1 + Z4 => Zx = Zy + Z4 — Z1=> Zx = Zy + Zz + Z1 — Z1 [By exterior angle property]=> Zx — Z y+ ZzQ8) In figure 16.137,0 and O'are centers o f two circles intersecting a t B and C. ACD is a straight fine, findx.
By degree measure theorem ZAOB = 2ZACB=> 130° = 2ZACB => ZACB = ^ = 65°
ZACB + ZBCD = 180° [Linera pair o f angles]=> 65° + ZBCD = 180°=> ZBCD = 180° - 65° = 115°By degree measure theorem reflexZBOD = 2 ZBCD => reflexZBOD = 2 x 115° = 230°Now, reflexZBOD + ZBO'D = 360° [Complex angle]=> 230° + x = 360°=> x = 360° - 230°
x = 130°Q9) In figure 16.138,0 is the centre o f a circle and PQ is a diameter. IfZ R O S = 40°, find Z R T S ..
Solution:
Since PQ is diameter
Then,
ZPRQ = 90° [Angle in semi circle]ZPRQ + ZTRQ — 180° [Linear pair o f angle]
90° + ZTRQ = 180°ZTRQ = 180° - 90° = 90°.
By degree measure theorem
ZROS = 2ZRQS => 40° = 2ZRQS => ZRQS = *£ = 20°
In A RQT, by Angle sum property
ZRQT + ZQRT + ZRTS = 180°=> 20° + 90° + ZRTS = 180°=> ZRTS = 180° - 20° - 90° = 70°Q10) In figure 16.139, i f ZACB = 40°, ZDPB = 120°, find ZCBD.
We have,
ZACB = 40°; ZDPB = 120°
ZAPB = ZDCB = 40° [Angle in same segment]In APOB, by angle sum property ZPDB + ZPBD + ZBPD = 180°=> 40° + ZPBD + 120° = 180°=> ZPBD = 180° - 40° - 120°=> ZPBD = 20°
ZCBD = 20°Q11)A chord o f a circle Is equal to die radius o f the circle. Find die angle subtended by the chord a t a point on the m inor arc and also at a point on the major arc.
Solution:
We have,
Radius OA = Chord AB
=>0A = OB = AB
Then triangle OAB is an equilateral triangle.
ZAOB = 60° [one angle o f equilateral triangle]
By degree measure theorem
ZAOB = 2 ZAPB =► 60° = 2Z4RB => ZAPB = ^ = 30°
Now, ZAPB + ZAQB = 180° [opposite angles o f cyclic quadrilateral]=► 30° + ZAQB = 180°=>• ZAQB = 180° - 30° = 150°.Therefore, Angle by chord AB at minor arc = 150°
Angle by chord AB at major arc = 30°