Chapter 15 : Circles - Rd Sharma Solutions for Class 9 Maths CBSE

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Chapter 15 - Circles Excercise Ex. 15.1

Question 1

Fill in the blanks:

(i) All points lying inside/outside a circle are called ...... points/ ... points.

(ii) Circles having the same centre and different radii are called ... circles.

(iii) A point whose distance from the centre of a circle is greater than its radius lies in ... of the circle.

(iv) A continuous piece of a circle is ... of the circle.

(v) The longest chord of a circle is a ... of the circle.

(vi) An arc is a ... when its ends are the ends of a diameter.

(vii) Segment of a circle is the region between an arc and ... of the circle.

(viii) A circle divides the plane, on which it lies, in .... parts.

Solution 1

(i) interior/exterior

(ii) concentric

(iii) the exterior

(iv) arc

(v) diameter

(vi) semi-circle

(vii) centre

(viii) three

Question 2

Write the truth value (T/F) of the following with suitable reasons:

(i) A circle is a plane figure.

(ii) Line segment joining the centre to any point on the circle is a radius of the circle.

(iii) If a circle is divided into three equal arcs each is a major arc.

(iv) A circle has only finite number of equal chords.

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

(vi) Sector is the region between the chord and its corresponding arc.

(vii) The degree measure of an arc is the complement of the central angle containing the arc.

(viii) The degree measure of a semi-circle is 180o.

Solution 2

(i) T

(ii) T

(iii) T

(iv) F

(v) T

(vi) T

(vii) F

(viii) T

Chapter 15 - Circles Excercise Ex. 15.2

Question 1

Solution 1



Question 2

Solution 2



Question 3

Solution 3



Question 4

Give a method to find the centre of a given circle.

Solution 4



Steps of construction:

(1) Take three point A, B and C on the given circle.

(2) Join AB and BC.

(3) Draw the perpendicular bisectors of chord AB and BC which interesect each other at O.

(4) Point O will be the required circle because we know that the perpendicular bisector of a chord always passes through the centre.

Question 5

Solution 5



Question 6

Solution 6



Question 7

Solution 7



Question 8

Solution 8



Question 9

Solution 9

Question 10

Solution 10

Question 11

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 form the centre?

Solution 11
                                            
 
Distance of smaller chord AB from centre of circle = 4 cm.
OM = 4 cm

In OMB
In OND
OD=OB=5cm             (radii of same circle)
 
So, distance of bigger chord from centre is 3 cm.
Question 12

Solution 12



Question 13

Solution 13



Question 14

Prove that two different circles cannot intersect each other at more than two points.

Solution 14

Suppose two different circles can intersect each other at three points then they will pass through the three common points but we know that there is one and only one circle with passes through three non-collinear points, which contradicts our supposition.

Hence, two different circles cannot intersect each other at more than two points.

Question 15
Two chords AB and CD of lengths 5 cm and 11cm 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.
Solution 15
Draw OM  AB and ON  CD. Join OB and OD
 
                     
                     (Perpendicular from centre bisects the chord)
Let ON be x, so OM will be 6 - x
In MOB
In NOD
 
We have OB = OD             (radii of same circle)
So, from equation (1) and (2)
 
From equation (2)
 
So, radius of circle is found to be  cm.

Chapter 15 - Circles Excercise Ex. 15.3

Question 1

Three girls Ishita, Isha and Nisha are playing a game by standing on a circle of radius 20 m drawn in a park. Ishita throws a ball to Isha, Isha to Nisha, Nisha to Ishita. If the distance between Ishita and Isha and between Isha and Nisha is 24 m each, what is the distance between Ishita and Nisha?

Solution 1

Question 2

A circular park of radius 40 m is situated in a colony. Three boys Ankur, Amit and Anand are sitting at equal distance on its boundary each having a toy telephone in his hands to talk to each other. Find the length of the string of each phone.

Solution 2

Chapter 15 - Circles Excercise Ex. 15.4

Question 1

In fig., O is the centre of the circle. If APB = 50°, find AOB and OAB.

 

Solution 1

Question 2

In fig., O is the centre of the circle. Find BAC.

 

 

Solution 2

Question 3



Solution 3

Question 4

Solution 4

Question 5

Solution 5

Question 6

Solution 6

Question 7

Solution 7

Question 8

Solution 8

Question 9

If O is the centre of the circle. Find the value of x in the following figure:

Solution 9

Question 10

Solution 10

Question 11

If O is the centre of the circle. Find the value of x in the following figure:

Solution 11

Question 12

If O is the centre of the circle. Find the value of x in the following figure:

Solution 12

Question 13

If O is the centre of the circle. Find the value of x in the following figure:

Solution 13

Question 14

Solution 14

Question 15

Solution 15



Question 16

In fig., O is the centre of the circle, BO is  the bisector of ABC. Show that AB = AC.

 

 

 

Solution 16

Question 17

In fig., O and O' are centres of two circles intersecting at B and C. ABD is straight line, find x.



Solution 17

Question 18

In fig., if ACB = 40°, DPB = 120°, find CBD.

 

 

Solution 18

Question 19

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 19



Question 20

In fig., it is given given that O is the centre of the circle and AOC = 150°. Find ABC.

 

 

Solution 20

Question 21

In fig., O is the centre of the circle, prove that x = y + z.

 

 

Solution 21

Question 22

in fig., O is the centre of a circle and PQ is a diameter. If ROS = 40°, find. RTS.

 

 

Solution 22

Chapter 15 - Circles Excercise Ex. 15.5

Question 1

In fig., ΔABC is an equilateral triangle. Find mBEC.

 

 

Solution 1

Question 2

In fig., ΔPQR is an isosceles triangle with PQ = PR and mPQR = 35°. find mQSR and mQTR.

 

Solution 2

Question 3

In fig., O is the centre of the circle. If BOD = 160°, find the values of x and y.

Solution 3

Question 4

In fig., ABCD is a cyclic qudrilateral. If BCD = 100° and ABD = 70°, find ADB.

 

Solution 4

Question 5

If ABCD is a cyclic quadrilateral in which AD  BC. Prove that B = C.

 

Solution 5

Question 6

In fig., O is the centre of the circle. find CBD.

 

Solution 6

Question 7

In fig., AB and CD are diameters of a circle with centre O. If OBD = 50°, find AOC.

 

 

Solution 7

Question 8

Solution 8



Question 9

Solution 9



Question 10

Solution 10



Question 11

In fig., O is the centre of the circle and DAB = 50. calculate the values of x and y.

 

 

Solution 11

Question 12

In fig., if BAC = 60°, and BCA = 20°, find ADC.

 

 

Solution 12

Question 13

In fig., if ABC is an equilateral triangle. Find BDC and BEC.

 

 

Solution 13

Question 14

In fig., O is the centre of the circle. If CEA = 30°, find the values of x, y and z.

 
Solution 14

Question 15

In fig., BAD = 78°, DCF = x° and DEF = y° find the values of x and y.

 

Solution 15

Question 16

Solution 16

Question 17

In fig., ABCD is cyclic qudrilateral. Find the value of x.

 

 
Solution 17

Question 18

Solution 18



Question 19

Solution 19



Question 20

Solution 20



Question 21

Solution 21



Question 22

Solution 22



Question 23

Solution 23



Question 24

Solution 24



Question 25

In fig., ABCD is cyclic quadrilaterial in which AC an BD are its diagonals. If DBC = 55° and BAC = 45°, find BCD.

 

 

 

 

 

 
Solution 25

Question 26

Solution 26

Question 27

Prove that the centre of the circle circumscribing the cyclic rectangle ABCD is the point of intersection of its diagonals.

Solution 27



Let O be the centre of the circle circumscribing the cyclic rectangle ABCD. Since ABC = 90o and AC is a chord of the circle, so, AC is a diameter of the circle. Similarly, BD is a diameter.

Hence, point of intersection of AC and BD is the centre of the circle.

Question 28

Solution 28



Question 29

Solution 29



Question 30

Solution 30



Question 31

Prove that the line segment joining the mid-point of the hypotenuse of a right triangle to its opposite vertex is half of the hypotenuse.

Solution 31



Chapter 15 - Circles Excercise 15.109

Question 1

If the length of a chord of a circle is 16 cm and is at a distance of 15 cm from the centre of the circle, then the radius of the circle is

(a) 15 cm

(b) 16 cm

(c) 17 an

(d) 34 cm

Solution 1

begin mathsize 12px style AB space equals space 16 space cm
OC space equals 15 space cm
straight C space is space the space mid minus point space of space AB.
AC space equals space BC space equals 16 over 2 equals 8 space cm
Consider space triangle OCA comma space space space space
OC equals space 15 space cm comma space AC equals 8 space cm
rightwards double arrow OA space equals square root of open parentheses 15 close parentheses squared plus open parentheses 8 close parentheses squared end root space equals space square root of 225 space plus 64 end root equals square root of 289
rightwards double arrow OA space equals space 17 space cm
Hence comma space correct space option space is space left parenthesis straight c right parenthesis.
end style

Question 2

begin mathsize 12px style The space radius space of space straight a space circle space is space 6 space cm. space The space perpendicular space distance space from space the space centre space of space the space circle space
to space the space chord space which space is space 8 space cm space in space length comma space is
open parentheses straight a close parentheses space space square root of 5 space cm
open parentheses straight b close parentheses space space 2 square root of 5 space cm
open parentheses straight c close parentheses space space 2 square root of 7 space cm
open parentheses straight d close parentheses space space square root of 7 space cm end style

Solution 2

begin mathsize 12px style AB equals 8 space cm
rightwards double arrow AC equals BC equals 4 space cm
Consider space triangle OCB comma space where space BC space equals space 8 space cm comma space OB space equals space 6 space cm
Now comma space open parentheses OC close parentheses squared space plus space open parentheses BC close parentheses squared space equals space open parentheses OB close parentheses squared
rightwards double arrow left parenthesis OC right parenthesis squared space plus space 4 squared equals 6 squared
rightwards double arrow left parenthesis OC right parenthesis squared space plus space 16 space equals space 36
rightwards double arrow left parenthesis OC right parenthesis squared equals 20
rightwards double arrow OC equals square root of 20 space equals space 2 square root of 5
Hence comma space correct space option space is space open parentheses straight b close parentheses. end style

Chapter 15 - Circles Excercise 15.110

Question 1

begin mathsize 12px style If space straight O space is space the space centre space of space straight a space circle space of space radius space straight r space and space AB space is space straight a space chord space of space the space circle space
at space straight a space distance space straight r divided by 2 space from space straight O comma space then space angle BAO space equals end style

(a) 60°

(b) 45°

(c) 30°

(d) 15°

Solution 1

begin mathsize 12px style Let space angle BAO space equals straight theta
Consider space triangle OAC comma
sin space straight theta space equals space OC over OA equals fraction numerator begin display style bevelled straight r over 2 end style over denominator straight r end fraction equals 1 half equals sin space 30 degree
rightwards double arrow straight theta space equals space 30 degree
Hence comma space correct space option space is space open parentheses straight c close parentheses. end style

Question 2

begin mathsize 12px style ABCD space is space straight a space cyclic space quadrilateral space such space that space angle ADB space equals space 30 degree space and space angle DCA space equals 80 degree space comma space then space angle DAB space equals
left parenthesis straight a right parenthesis space 70 degree
left parenthesis straight b right parenthesis space 100 degree
left parenthesis straight c right parenthesis space 125 degree space
left parenthesis straight d right parenthesis space 150 degree space end style

Solution 2

begin mathsize 12px style ABCD space is space straight a space cyclic space Quadrilateral.
Consider space triangle ABD space and space triangle ABC.
Both space are space on space the space same space base space AB space and space angle ADB space and space angle ACB space are space the space angles
in space the space same space segment space AB.
rightwards double arrow angle ADB equals angle ACB equals 30 degree
rightwards double arrow angle BCD space equals space 80 degree plus 30 degree equals 110 degree
In space straight a space cycle space Quadrilateral comma space sum space of space opposite space angles space is space 180 degree.
rightwards double arrow angle straight A plus angle straight C equals 180 degree
rightwards double arrow angle DAB plus angle BCD equals 180 degree
rightwards double arrow angle DAB equals 180 degree minus 110 degree space equals space 70 degree
Hence comma space correct space option space is space open parentheses straight a close parentheses. end style

Question 3

A chord of length 14 cm is at a distance of 6 cm from the centre of a circle. The length of another chord at a distance of 2 cm from the centre of the circle is

(a) 12 cm

(b)) 14 cm

(c) 16 cm

(d) 18 cm

Solution 3

begin mathsize 12px style Chord space AB space space equals space 14 space cm
AC space equals space BC equals 7 space cm
OC equals 6 space cm
rightwards double arrow OB space equals space square root of 7 squared space plus space 6 squared end root equals square root of 49 plus 36 end root equals square root of 85 cm
Consider space triangle ODE
open parentheses OE close parentheses squared plus open parentheses ED close parentheses squared space equals space left parenthesis OD right parenthesis squared
open parentheses 2 close parentheses squared space plus open parentheses ED close parentheses to the power of 2 space end exponent equals left parenthesis OD right parenthesis squared
open parentheses ED close parentheses to the power of 2 space end exponent equals open parentheses OD close parentheses to the power of 2 space end exponent minus 4
rightwards double arrow left parenthesis ED right parenthesis squared equals 85 minus 4 space space space space left parenthesis OD equals OB equals square root of 85 space cm space minus space radii space of space same space circle right parenthesis
rightwards double arrow ED space equals space square root of 81 space equals space 9 space cm
rightwards double arrow Chord space FD space equals space 9 space cross times space 2 space equals space 18 space cm
Hence comma space correct space option space is space open parentheses straight d close parentheses. end style

Question 4

One chord of a circle is known to be 10 cm. The radius of this circle must be

(a) 5 cm

(b) greater than 5 cm

(c) greater than or equal to 5 cm

(d) less than 5 cm

Solution 4

begin mathsize 12px style The space longest space chord space of space straight a space circle space is space its space diameter.
rightwards double arrow Diameter space greater than space 10 space cm
rightwards double arrow 2 space cross times space Radius space greater than 10 space cm
rightwards double arrow Radius space greater than space 5 space cm
Hence comma space correct space option space is space open parentheses straight b close parentheses.
end style

Question 5

ABC is a triangle with B as right angle, AC = 5 cm and AB = 4 cm. A circle is drawn with A as centre and AC as radius. The length of the chord of this circle passing through C and B is

(a) 3 cm

(b) 4 cm

(c) 5 cm

(d) 6 cm

Solution 5

begin mathsize 12px style right enclose AD space and space AC space are space radii space of space same space circle space and space CD space is space straight a space chord.
Consider space triangle ABC comma
BC squared space equals open parentheses AC close parentheses squared minus open parentheses AB close parentheses squared space equals space 5 squared minus 4 squared space equals space 25 minus 16 space equals space 9
rightwards double arrow BC space equals space 3 space cm
Chord space CD space equals space 2 space cross times BC space equals space 6 space cm
Hence comma space correct space option space is space open parentheses straight d close parentheses. end style

Question 6

begin mathsize 12px style If space AB comma space BC space and space CD space are space equal space chords space of space straight a space circle space with space straight O space as space centre space and space AD space diameter comma space then angle AOB space equals space
left parenthesis straight a right parenthesis space 60 degree
left parenthesis straight b right parenthesis space 90 degree
left parenthesis straight c right parenthesis space 120 degree space
left parenthesis straight d right parenthesis space none space of space these space end style

Solution 6


begin mathsize 12px style Chord space AB space equals space Chord space BC space equals space Chord space CD
rightwards double arrow angle AOB space equals angle BOC equals angle COD space space space space space left parenthesis equal space chords space subtend space equal space angles space at space the space center right parenthesis
Now comma space angle AOB space plus space angle BOC plus angle COD equals 180 degree
rightwards double arrow angle AOB space plus space angle AOB plus angle AOB equals 180 degree
rightwards double arrow 3 angle AOB space equals 180 degree
rightwards double arrow angle AOB space equals space 60 degree
Hence comma space correct space option space is space open parentheses straight a close parentheses. end style

Question 7

begin mathsize 12px style Let space straight C space be space the space mid minus point space of space an space arc space AB space of space straight a space circle space such space that space straight m space AB with overparenthesis on top space equals space 183 degree. space
If space the space space region space bounded space by space the space arc space ACB space and space line space segment space AB space is space denoted space by space straight S comma space
then space the space centre space straight O space of space the space circle space lies space
left parenthesis straight a right parenthesis space in space the space interior space of space straight S space
left parenthesis straight b right parenthesis space in space the space exterior space of space straight S space
left parenthesis straight c right parenthesis space on space the space segment space AB
left parenthesis straight d right parenthesis space on space AB space and space bisects space AB end style

Solution 7

begin mathsize 12px style straight m space AB with overparenthesis on top equals 183 degree
straight O space is space the space centre space of space the space circle space and space AB space is space straight a space chord.
The space Regiion space bounded space by space arc space and space line space segment space AB space is space shaded.
We space can space see comma space apostrophe straight O apostrophe comma space the space centre comma space always space lie space in space the space interior space of space straight S.
Hence comma space correct space option space is space left parenthesis straight a right parenthesis. end style

Question 8

In a circle, the major arc is 3 times the minor arc. The corresponding central angles and the degree measures of two arcs are

(a) 90° and 270°

(b) 90° and 90°

(c) 270° and 90°

(d) 60° and 210°

Solution 8

begin mathsize 12px style fraction numerator AB with overparenthesis on top minor over denominator AB with overparenthesis on top major end fraction equals 1 third equals fraction numerator angle space AB with overparenthesis on top minor over denominator angle space AB with overparenthesis on top major end fraction
Let space space angle AB with overparenthesis on top minor space equals space straight x
rightwards double arrow angle space stack AB space with overparenthesis on top major space equals space 3 straight x
Now space we space know space straight x plus 3 straight x equals 360 degree
rightwards double arrow 4 straight x equals 360 degree
rightwards double arrow straight x equals 90 degree
rightwards double arrow 3 straight x equals 270 degree
Hence comma space correct space option space is space open parentheses straight a close parentheses.

O p t i o n space left parenthesis c right parenthesis space c a n space a l s o space b e space a space p o s s i b i l i t y. end style

Question 9

begin mathsize 12px style If space straight A space and space straight B space are space two space points space on space straight a space circle space such space that space straight m open parentheses AB with overparenthesis on top close parentheses space equals space 260 degree space. space
straight A space possible space value space for space the space angle space subtended space by space arc space BA space at space straight a space point space on space the space circle space is
left parenthesis straight a right parenthesis space 100 degree space
left parenthesis straight b right parenthesis space 75 degree space
left parenthesis straight c right parenthesis space 50 degree space
left parenthesis straight d right parenthesis space 25 degree end style

Solution 9

begin mathsize 12px style straight m open parentheses AB with overparenthesis on top close parentheses equals 260 degree
rightwards double arrow straight m open parentheses BA with overparenthesis on top close parentheses space equals space 100 degree
Now space Let space BA with overparenthesis on top space subtend space an space angle space straight theta space at space straight a space point space straight C space on space circle.
Now comma space we space know space that space angle space subtend space by space an space arc space at space the space centre space
is space double space the space angle space subtended space at space any space point space on space the space circle.
rightwards double arrow 100 degree equals 2 straight theta
rightwards double arrow straight theta equals 50 degree
Hence comma space correct space option space is space open parentheses straight c close parentheses. end style

Question 10

begin mathsize 12px style An space equilateral space triangle space ABC space is space inscribed space in space straight a space circle space with space centre space straight O. space The space measures space of angle BOC space is
open parentheses straight a close parentheses space 30 degree space
open parentheses straight b close parentheses right parenthesis space 60 degree space
open parentheses straight c close parentheses right parenthesis space 90 degree space
open parentheses straight d close parentheses 120 degree space end style

Solution 10

begin mathsize 12px style angle BAC space equals space 60 degree space open parentheses angle space of space equilateral space triangle close parentheses
Arc space BC with overparenthesis on top space makes space angle space angle BAC space at space circle space and space angle BOC space at space centre space of space circle.
rightwards double arrow angle BAC space equals 1 half angle BOC
rightwards double arrow 2 space cross times space angle BAC equals angle BOC
rightwards double arrow 2 cross times 60 degree space equals space angle BOC
rightwards double arrow angle BOC space equals space 120 degree
Hence comma space correct space option space is space open parentheses straight d close parentheses. end style

Question 11

If two diameters of a circle intersect each other at right angles, then quadrilateral formed joining their end points is a

(a) rhombus

(b) rectangle

(c) parallelogram

(d) square

Solution 11

begin mathsize 12px style AB space and space CD space are space diameters space of space straight a space circle space and space diameter space makes space 90 degree space at space any space point space on space circle.
rightwards double arrow angle CAD space equals space angle CBD space equals space angle BCA equals angle ADB space equals space 90 degree
Also comma space diagnols space AB space and space CD space are space perpendicular to space each space other.
Thus comma space ABCD space is space straight a space square.
Hence comma space correct space option space is space open parentheses straight d close parentheses. end style

Question 12

begin mathsize 12px style If space ABC space is space an space arc space of space straight a space circle space and space angle ABC equals 135 degree comma space then space the space ratio space of space are space ABC with overparenthesis on top space to space the space circumference space is
open parentheses straight a close parentheses space space 1 space colon space 4
open parentheses straight b close parentheses space space 3 space colon space 4
open parentheses straight c close parentheses space space 3 space colon space 8
open parentheses straight d close parentheses space space 1 space colon space 2 end style

Solution 12

begin mathsize 12px style ABC space is space an space arc space of space circle.
Take space point space straight D space in space the space alternative space segment space and space join space AD space and space CD.
angle ABC equals 135 degree space space space left parenthesis given right parenthesis
angle ABC plus angle ADC space equals space 180 degree space left parenthesis sum space of space opposite space angles space of space cyclic space quadrilateral space is space 180 degree right parenthesis
rightwards double arrow angle ADC equals 180 degree minus angle ABC space equals 180 degree space minus space 135 degree space equals space 45 degree
Now comma space angle AOC equals 2 space cross times space angle ADC space equals space 2 space cross times space 45 degree space equals space 90 degree
ABC with overparenthesis on top space equals space measure space of space the space central space angle space equals angle AOC space equals space 90 degree
rightwards double arrow Required space ratio space equals space fraction numerator arc space ABC with overparenthesis on top over denominator circumference space end fraction space equals space fraction numerator 90 degree over denominator 360 degree end fraction space equals space 1 fourth space equals space 1 space colon space 4
Hence comma space correct space option space is space left parenthesis straight a right parenthesis.
end style

Question 13

The chord of a circle is equal to its radius. The angle subtended by this chord at the minor of the circle is 

(a) 60

(b) 750

(c) 1200

(d) 1500

 

Solution 13

begin mathsize 12px style angle AOB space equals 60 degree space space left parenthesis Since space triangle AOB space is space equilateral space triangle right parenthesis
Now comma space angle ADB space equals space 30 degree space space space space space space
left parenthesis Since space chord space AB space makes space 60 degree space at space centre comma space same space chord space will space make space half space of space the space angle space
at space circumference space of space angle space made space at space centre right parenthesis
Now space angle ACB space is space angle space made space by space chord space at space minor space arc space of space circle.
ACBD space is space cyclic space Quadrilateral.
rightwards double arrow angle straight C space plus angle straight D space equals space 180 degree
rightwards double arrow angle ACB space plus angle ADB space equals space 180 degree
rightwards double arrow angle ACB space equals space 180 degree space minus space 30 degree space equals 150 degree
Hence comma space correct space option space is space open parentheses straight d close parentheses. end style

Chapter 15 - Circles Excercise 15.111

Question 1

begin mathsize 12px style PQRS space is space straight a space cyclic space quadrilateral space such space that space PR space is space straight a space diameter space of space the space circle. space
If space angle QPR space equals space 67 degree space and space angle SPR space equals space 72 degree comma space then space angle QRS space equals space
open parentheses straight a close parentheses space 41 degree
open parentheses straight b close parentheses space 23 degree
space left parenthesis straight c right parenthesis space 67 degree
space left parenthesis straight d right parenthesis space 18 degree end style

Solution 1

begin mathsize 12px style In space straight a space cyclic space quadrilateral comma space opposite space angles space are space supplementary.
rightwards double arrow angle straight P space plus angle straight R space equals 180 degree
Now comma space angle straight P space equals space 67 degree space plus space 72 degree space equals space 139 degree
Thus comma space angle straight R space equals space 180 degree space minus 139 degree equals 41 degree
straight i. straight e. space angle straight R space equals space angle QRS space equals 41 degree
Hence comma space correct space option space is space left parenthesis straight a right parenthesis. end style

Question 2

begin mathsize 12px style If space straight A comma space straight B comma space straight C space are space three space points space on space straight a space circle space with space centre space straight O space such space that space angle straight A 0 straight B equals 90 degree space and space OC space equals 120 degree comma
then space angle ABC space equals space
open parentheses straight a close parentheses space 60 degree space
open parentheses straight b close parentheses space 75 degree
open parentheses straight c close parentheses space 90 degree space
open parentheses straight d close parentheses space 135 degree end style

Solution 2

begin mathsize 12px style angle AOC space equals space angle AOB space plus space angle BOC space equals space 90 degree plus 120 degree space equals 210 degree
angle COA space equals space 360 degree minus 210 degree space equals space 150 degree
If space arc space COA with overparenthesis on top space makes space 150 degree space at space centre comma space then space it space will space make space half space of space angle space of space the space centre space
at space circumference.
rightwards double arrow angle CBA space or space angle ABC space equals fraction numerator 150 degree over denominator 2 end fraction equals 75 degree
Hence comma space correct space option space is space open parentheses straight b close parentheses. space end style

Question 3

The greatest chord of a circle is called its

(a) radius

(b) secant

(c) diameter 

(d) none of these

Solution 3

The greatest chord of the circle is diameter of the circle.

Hence, correct option is (c).

Question 4

Angle formed in minor segment of a circle is

(a) acute

(b) obtuse

(c) right angle 

(d) none of these

Solution 4

Angle formed in a minor segment is always a obtuse angle.

Hence, correct option is (b).

Question 5

Number of circles that can be drawn through three non-collinear points is

(a) 1

(b) 0

(c) 2

(d) 3

Solution 5

Three non-collinear points make a triangle and there is only one circle that can pass through all three points,

i.e. circumcircle of that triangle.

Hence, correct option is (a).

Question 6

In figure, if chords AB and CD of the circle intersect each other at right angles, then x + y =

(a) 450

(b) 600

(c) 750

(d) 900

Solution 6

begin mathsize 12px style angle CAB space equals space angle CDB space equals straight x degree space space space space space space.... left parenthesis Both space are space on space the space same space arc right parenthesis
Consider space triangle ODB comma
angle DOB space equals space 90 degree comma space angle OBD space equals space straight y comma space angle ODB space equals space straight x
In space triangle ODB comma
straight x plus space straight y plus space 90 degree space equals space 180 degree
rightwards double arrow straight x space plus space straight y space equals space 90 degree
Hence comma space correct space option space is space open parentheses straight d close parentheses. end style

Question 7

begin mathsize 12px style In space figure comma space if space angle ABC space equals space 45 degree comma space then space angle AOC space equals
open parentheses straight a close parentheses space 45 degree
open parentheses straight b close parentheses space 60 degree
open parentheses straight c close parentheses space 75 degree
open parentheses straight d close parentheses space 90 degree end style

Solution 7

begin mathsize 12px style angle AOC space is space made space by space arc space AC with overparenthesis on top space at space centre space and space angle ABC space is space made space by space AC with overparenthesis on top space on space circumference space
in space major space segment.
rightwards double arrow angle ABC space equals 1 half space angle AOC
rightwards double arrow angle AOC space equals space 2 space cross times space angle ABC space equals space 2 space cross times space 45 degree space equals space 90 degree
Hence comma space correct space option space is space open parentheses straight d close parentheses. end style

Question 8

In figure, chords AD and BC intersect each other at right angles at a point P. If begin mathsize 11px style angle DAB space equals space 35 degree comma space then space angle ADC space equals space end style

(a) 350

(b) 450

(c) 550

(d) 650

Solution 8

begin mathsize 12px style angle APC space plus angle APB space equals space 180 degree
rightwards double arrow angle APB space equals space 180 degree minus 90 degree equals 90 degree
In space triangle APB comma
angle ABP space equals 180 degree space minus space angle APB space minus space angle BAP space equals space 180 degree space minus space 90 degree minus 35 degree space equals space 55 degree
Now space Arc space AC with overparenthesis on top space makes space angle ABC space and space angle ADC space on space circle.
rightwards double arrow ABC equals angle ADC
rightwards double arrow angle ADC space equals space 55 degree
Hence comma space correct space option space is space open parentheses straight c close parentheses. end style

Question 9

begin mathsize 12px style In space figure comma space straight O space is space the space center space of space the space circle space and space angle BDC space equals space 42 degree. space The space measure space of space angle ACB space is
open parentheses straight a close parentheses space 42 degree
open parentheses straight b close parentheses space 48 degree
open parentheses straight c close parentheses space 58 degree
open parentheses straight c close parentheses space 52 degree end style

Solution 9

begin mathsize 12px style angle ABC space equals space 90 degree space space space space space.... open parentheses Diameter space AC space makes space 90 degree space at space circumference close parentheses
angle CDB equals space angle CAB space space space space space space.... open parentheses angles space on space the space same space arc close parentheses
rightwards double arrow angle CAB space equals space 42 degree
In space triangle ABC comma
angle ACB space equals 180 degree minus 90 degree minus space 42 degree equals 48 degree
Hence comma space correct space option space is space open parentheses straight b close parentheses. end style

Chapter 15 - Circles Excercise 15.112

Question 1

begin mathsize 12px style In space straight a space circle space with space centre space straight O comma space AB space and space CD space are space two space diameters space perpendicular space to space each space other. space
The space length space of space chord space AC space is
open parentheses straight a close parentheses space 2 AB
open parentheses straight b close parentheses space square root of 2
open parentheses straight c close parentheses space 1 half AB
open parentheses straight d close parentheses space fraction numerator 1 over denominator square root of 2 end fraction AB end style

Solution 1

begin mathsize 12px style OC space equals space OA equals straight r space space space space left parenthesis radius right parenthesis
AB equals Diameter space equals space 2 straight r
AC equals square root of open parentheses OA close parentheses squared plus open parentheses OC close parentheses squared end root
space space space space space equals space square root of straight r to the power of space 2 end exponent plus straight r squared end root
space space space space space equals square root of 2 straight r
space space space space space equals space square root of 2 open parentheses AB over 2 close parentheses
rightwards double arrow AC space equals space fraction numerator 1 over denominator square root of 2 end fraction AB
Hence comma space correct space option space is space open parentheses straight d close parentheses. end style

Question 2

begin mathsize 12px style Two space equal space circles space of space radius space straight r space intersect space such space that space each space passes space through space the space centre space
of space the space other. space The space length space of space the space common space chord space of space the space circles space is
open parentheses straight a close parentheses space square root of straight r
open parentheses straight b close parentheses space square root of 2 space straight r space AB
open parentheses straight c close parentheses space square root of 3 straight r
open parentheses straight d close parentheses space fraction numerator square root of 3 over denominator 2 end fraction straight r end style

Solution 2

begin mathsize 12px style Both space the space circles space pass space through space the space center space of space each space other
rightwards double arrow straight O subscript 1 straight O subscript 2 equals straight r
Common space chord space is space AB.
We space know space that space perpendicular space drawn space from space centre space of space circle space to space any space chord space bisects space it.
rightwards double arrow straight P space is space the space midpoint space of space AB
rightwards double arrow PA equals PB
straight O subscript 1 straight A equals straight r space space space space left parenthesis radius space of space circle right parenthesis
Consider space triangle straight O subscript 1 PA
open parentheses straight O subscript 1 straight A close parentheses squared equals AP squared space plus space straight O subscript 1 straight P squared space space
rightwards double arrow straight r squared equals AP squared space plus open parentheses straight r over 2 close parentheses squared space space space space space space space.... open parentheses straight P space is space also space mid minus poinat space of space straight O subscript 1 straight O subscript 2 space space end subscript close parentheses
rightwards double arrow AP squared space equals space straight r squared minus straight r squared over 4 equals fraction numerator 3 straight r squared over denominator 4 end fraction
rightwards double arrow AP space equals space fraction numerator square root of 3 over denominator 2 end fraction straight r
rightwards double arrow Length space of space chord space AB equals 2 AP equals square root of 3 straight r
Hence comma space correct space option space is space open parentheses straight c close parentheses. end style

Question 3

begin mathsize 12px style If space AB space is space straight a space chord space of space straight a space circle comma space straight P space and space straight Q space are space the space two space points space on space the space circle space different space from space straight A space and space straight B comma space
th e n
open parentheses straight a close parentheses space space angle APB equals space angle AQB
open parentheses straight b space close parentheses space angle APB space plus angle AQB space equals 180 degree space or space angle APB space equals space angle AQB space
open parentheses straight c space close parentheses space angle APB space plus space angle AQB space equals space 90 degree
open parentheses straight d close parentheses space angle APB plus space angle AQB equals space 180 degree space end style

Solution 3

begin mathsize 12px style angle APB space and space angle AQB space are space on space the space same space arc.
rightwards double arrow angle APB space equals space angle AQB
But comma space if space AB equals diameter comma space then space angle APB space equals space angle AQB space equals space 90 degree
open parentheses Becasue space diameter space makes space Right space angle space at space any space point space on space circumference space of space circle close parentheses
rightwards double arrow angle APB space plus angle AQB space equals space 180 degree
Hence comma space correct space option space is space open parentheses straight b close parentheses. end style

Question 4

begin mathsize 12px style AB space and space CD space are space two space parallel space chords space of space straight a space circle space with space centre space apostrophe straight O apostrophe space such space that space AB space equals 6 space cm space and space CD space equals space 12 space cm. space
The space chords space are space on space the space same space side space of space the space centre space and space the space distance space between space is space 3 space cm. space
The space radius space of space the space circle space is
open parentheses straight a close parentheses space 6 space cm space
open parentheses straight b close parentheses space 5 square root of 2 space cm
open parentheses straight c close parentheses space 7 space cm
open parentheses straight d close parentheses space 3 square root of 5 space cm end style

Solution 4

begin mathsize 12px style OB space and space OD space are space the space radii space of space straight a space circle.
In space triangle OED comma
straight r squared equals OE squared space plus ED squared equals OE squared space plus space open parentheses 6 close parentheses squared
rightwards double arrow OE space equals space square root of straight r squared space minus space 36 space end root space space space space space space space space space space space space space space space.... open parentheses 1 close parentheses
In space triangle OFB comma
straight r squared space equals OF squared plus BF squared equals OF squared space plus open parentheses 3 close parentheses squared
rightwards double arrow OF space equals space square root of straight r squared space minus space 9 end root space space space space space space space space space space space space.... open parentheses 2 close parentheses
OF space minus space OE space equals space 3 space cm space space space space space space space open parentheses given close parentheses
rightwards double arrow square root of straight r squared minus 9 end root space minus square root of straight r squared minus 36 end root space equals space 3 space
rightwards double arrow square root of straight r squared minus 9 end root equals square root of straight r squared minus 36 end root plus 3 space space space space.... open parentheses 3 close parentheses
Squarring space equation space open parentheses 3 close parentheses comma space we space have
straight r squared space minus 9 space equals space straight r squared space minus space 36 space plus space 9 space plus space 2 cross times 3 space square root of straight r squared minus 36 end root
rightwards double arrow straight r squared space minus space 9 space equals space straight r squared space minus space 27 space plus space 6 space square root of straight r squared minus 36 end root
rightwards double arrow 18 space equals space 6 space square root of straight r squared minus 36 end root
rightwards double arrow 3 space equals space square root of straight r squared minus 36 end root
rightwards double arrow 9 space equals space straight r squared minus 36 space
rightwards double arrow straight r equals square root of 45 space equals space 3 square root of 5 space cm
Hence comma space correct space option space is space open parentheses straight d close parentheses. end style

Question 5

In a circle of radius 17 cm, two parallel chords are drawn on opposite side of a diameter. The distance between the chords is 23 cm. If the length of one chord is 16 cm, then the length of the other is

(a) 34 cm

(b) 15 cm

(c) 23 cm

(d) 30 cm

Solution 5

begin mathsize 12px style PQ space equals space 23 space cm
AB space equals space 16 space cm space
rightwards double arrow BP space equals space AP equals space 8 space cm
straight r space equals space 17 space cm
rightwards double arrow EF space equals space diameter space equals space 2 straight r space equals space 34 space cm
Consider space triangle OPB comma
straight r squared equals OP squared plus BP squared
rightwards double arrow OP squared equals open parentheses 17 close parentheses squared minus open parentheses 8 close parentheses squared space equals 289 space minus space 64 space equals space 225
rightwards double arrow OP space equals space 15 space cm
rightwards double arrow OQ space equals space 23 minus 15 space equals space 8 space cm
Consider space triangle OQD comma
straight r squared space equals space OQ squared plus QD squared
rightwards double arrow QD squared space space equals space straight r squared minus OQ squared equals open parentheses 17 close parentheses squared minus open parentheses 8 close parentheses squared space equals space 225
rightwards double arrow QD space equals space 15 space cm
rightwards double arrow CD equals space 2 cross times space QD space equals space 30 space cm
Hence comma space correct space option space is space open parentheses straight d close parentheses. end style

Question 6

begin mathsize 12px style In space figure comma space straight O space is space the space centre space of space the space circle space such space that space angle AOC space equals space 130 degree comma space then space angle ABC equals
open parentheses straight a close parentheses space 130 degree
open parentheses straight b close parentheses space 115 degree
open parentheses straight c close parentheses space 65 degree
open parentheses straight d close parentheses space 165 degree end style

Solution 6

begin mathsize 12px style angle ADC space equals 1 half angle AOC
open curly brackets angle ADC space and space angle AOC space are space made space by space same space AC with overparenthesis on top space on space centre space and space circumference close curly brackets
rightwards double arrow angle ADC equals 1 half cross times 130 degree space equals space 65 degree
ADCB space is space straight a space cyclic space Quarilateral. space
rightwards double arrow angle straight D plus angle straight B equals 180 degree
rightwards double arrow angle ABC space equals space 180 degree space minus 65 degree space equals space 115 degree
Hence comma space correct space option space is space open parentheses straight b close parentheses. end style

CBSE Class 9 Maths Homework Help

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