R S AGGARWAL AND V AGGARWAL Solutions for Class 10 Maths Chapter 16 - Areas of Circle, Sector and Segment

Page / Exercise

Chapter 18 - Areas of Circle, Sector and Segment MCQ

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

Solution 20

Solution 21

Solution 22

Solution 23

Solution 24

Solution 25

Solution 26

Chapter 18 - Areas of Circle, Sector and Segment FA

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

Solution 18

Solution 19

 

Solution 20

Chapter 18 - Areas of Circle, Sector and Segment Ex. 16A

Solution 1

Circumference of circle = 2r = 39.6 cm

Solution 2

Solution 3

Solution 4

Area of square =

Perimeter of square = 4 side = 4 22 = 88 cm

Circumference of circle = Perimeter of square

Solution 5

Area of equilateral =

Perimeter of equilateral triangle = 3a = (3 22) cm

                                              = 66 cm

Circumference of circle = Perimeter of circle

2r = 66 r =

Area of circle =

                   =

Solution 6

Solution 7

Let the radii of circles be x cm and (7 - x) cm

Circumference of the circles are 26 cm and 18 cm

Solution 8

Area of outer circle =

                            = 1662.5

Area of ring = Outer area - inner area

                 = (1662.5 - 452.5)

Solution 9(i)

Inner radius of the circular park = 17 m

Width of the path = 8 m

Outer radius of the circular park = (17 + 8)m = 25 m

Area of path =

                

Solution 10

Let r m and R m be the radii of inner circle and outer boundaries respectively.

Then, 2r = 352 and 2R = 396

Width of the track = (R - r) m

                        

Area the track =

                   

Solution 11

Length of the arc

                     

Length of arc =

Area of the sector =

                                     

Solution 12

Solution 13

Length of arc =

 

 

Circumference of circle = 2 r

Area of circle =

                  

Solution 14

Solution 15

OAB is equilateral.

So, AOB = 60

Length of arc BDA = (2 12 - arc ACB) cm

                         = (24 - 4) cm = (20) cm

                         = (20 3.14) cm = 62.8 cm

Area of the minor segment ACBA

                 

Solution 16

Let OA = , OB =

And AB = 10 cm

Area of AOB =

                  

Area of minor segment = (area of sector OACBO) - (area of OAB)

                                 =

Solution 17

Area of sector OACBO

Area of minor segment ACBA

                     

Area of major segment BADB

                    

Solution 18

Let AB be the chord of circle of centre O and radius = 30 cm such that AOB = 60°

Area of the sector OACBO

Area of OAB =

Area of the minor segment ACBA

= (area of the sector OACBO) - (area of the OAB)

=(471 - 389.25) = 81.75

Area of the major segment BADB

= (area of circle) - (area of the minor segment)

= [(3.14 × 30 × 30) - 81.75)] = 2744.25

Solution 19

Let the major arc be x cm long

Then, length of the minor arc =

Circumference =

Solution 20

In 2 days, the short hand will complete 4 rounds

Distance travelled by its tip in 2 days

=4(circumference of the circle with r = 4 cm)

= (4 × 2 × 4) cm = 32 cm

In 2 days, the long hand will complete 48 rounds

length moved by its tip

= 48(circumference of the circle with r = 6cm)

= (48 × 2 × 6) cm = 576 cm

Sum of the lengths moved

= (32 + 576) = 608 cm

= (608 × 3.14) cm = 1909.12 cm

Solution 21

Solution 22

Area of plot which cow can graze  when r = 16 m is

                                                       

                                                        = 804.5 m2

Area of plot which cow can graze when radius is increased to 23 m

                                                                    

Additional ground = Area covered by increased rope - old area

                        = (1662.57 - 804.5) = 858

Solution 23

Area which the horse can graze = Area of the quadrant of radius 21 m

                                            

Area ungrazed =

                    

Solution 24

 

Each angle of equilateral triangle is 60

Area that the horse cannot graze is 36.68 m2

Solution 25

Ungrazed area

                   

Solution 26

OP = OR = OQ = r

Let OQ and PR intersect at S

We know the diagonals of a rhombus bisect each other at right angle.

Therefore we have

Solution 27

Diameter of the inscribed circle = Side of the square = 10 cm

Radius of the inscribed circle = 5 cm

Diameter of the circumscribed circle

= Diagonal of the square

Radius of circumscribed circle =

(i)Area of inscribed circle =

(ii)Area of the circumscribed circle

Solution 28

Let the radius of circle be r cm

Then diagonal of square = diameter of circle = 2r cm

Area of the circle =

Solution 29

Let the radius of circle be r cm

Let each side of the triangle be a cm

And height be h cm

Solution 30

Radius of the wheel = 42 cm

Circumference of wheel =2r =

Distance travelled = 19.8 km = 1980000 cm

Number of revolutions =

Solution 31

Radius of wheel = 2.1 m

Circumference of wheel =

Distance covered in one revolution = 13.2 m

Distance covered in 75 revolutions = (13.2 75) m = 990 m

=

Distance a covered in 1 minute =

Distance covered in 1 hour =

Solution 32

Distance covered by the wheel in 1 revolution

The circumference of the wheel = 198 cm

Let the diameter of the wheel be d cm

Hence diameter of the wheel is 63 cm

Solution 33

Radius of the wheel

Circumference of the wheel = 2r =

                                       

Distance covered in 140 revolution

                                      

Distance covered in one hour =

Solution 34

Distance covered by a wheel in 1minute

Circumference of a wheel =

Number of revolution in 1 min =

Solution 35

Radius of the front wheel = 40 cm =

Circumference of the front wheel=

Distance moved by it in 800 revolution

                        

Circumference of rear wheel = (2 1)m = (2) m

Required number of revolutions =

Solution 36

Each side of the square is 14 cm

 

Then, area of square = (14 × 14)

                               = 196

Thus, radius of each circle 7 cm

Required area = area of square ABCD

                             -4 (area of sector with r = 7 cm,  = 90°)

                   

Area of the shaded region = 42

Solution 37

Let A, B, C, D be the centres of these circles

Join AB, BC, CD and DA

Side of square = 10 cm

Area of square ABCD

                   

Area of each sector =

                            = 19.625

Required area = [area of sq. ABCD - 4(area of each sector)]

                     = (100 - 4 19.625)

                     = (100 - 78.5) = 21.5

Solution 38

Required area = [area of square - areas of quadrants of circles]

Let the side = 2a unit and radius = a units

Area of square = (side side) = (2a 2a) sq. units

                                          

Solution 39

Let A, B, C be the centres of these circles. Joint AB, BC, CA

Required area=(area of ABC with each side a = 12 cm)

                            -3(area of sector with r = 6, = 60°)

      

      

The area enclosed = 5.76 cm2

Solution 40

Let A, B, C be the centers of these circles. Join AB, BC, CA

Required area= (area of ABC with each side 2)

                               -3[area of sector with r = a cm, = 60°]

                     

Solution 41

Solution 42

Solution 43

Solution 44

Solution 45

Solution 46

Solution 47

Solution 48

  

Solution 49

Solution 50

Solution 51

 

 

Solution 52

Area of equilateral triangle ABC = 49

 

Let a be its side

Area of sector BDF =

Area of sector BDF = Area of sector CDE = Area of sector AEF

Sum of area of all the sectors

Shaded area = Area of ABC - sum of area of all sectors

                     

Solution 53

Solution 54

Solution 55

ABCDEF is a hexagon

AOB = 60, Radius = 35 cm

Area of sector AOB

Area of AOB =

                  

Area of segment APB = (641.083 = 530.425)= 110.658

Area of design (shaded area) = 6 110.658= 663.948

                                                            = 663.95

Solution 56

In PQR, P = 90, PQ = 24 cm, PR = 7 cm

Area of semicircle

                   

Area of PQR =

Shaded area = 245.31 - 84 = 161.31

Solution 57

In ABC, A = 90°, AB = 6cm, BC = 10 cm

Area of  ABC =

Let r be the radius of circle of centre O

Solution 58

       

PS = 12 cm

PQ = QR = RS = 4 cm, QS = 8 cm

Perimeter = arc PTS + arc PBQ + arc QES

Area of shaded region = (area of the semicircle PBQ)

+ (area of semicircle PTS)-(Area of semicircle QES)

Solution 59

Length of the inner curved portion

= (400 - 2 90) m

= 220 m

Let the radius of each inner curved part be r

Inner radius = 35 m, outer radius = (35 + 14) = 49 m

Area of the track = (area of 2 rectangles each 90 m 14 m)

+ (area of circular ring with R = 49 m, r = 35 m

Length of outer boundary of the track

Chapter 18 - Areas of Circle, Sector and Segment Ex. 16B

Solution 1

Solution 2

Solution 3

Solution 4

Solution 5

Since square circumscribes a circle of radius a cm, we have

Side of the square = 2 radius of circle = 2a cm

Then, Perimeter of the square = (4 2a) = 8a cm

Solution 6

Solution 7

Solution 8

Solution 9

Solution 10

Solution 11

Solution 12

Solution 13

Solution 14

Solution 15

Solution 16

Solution 17

 

Solution 18

Solution 19

Length of the pendulum = radius of sector = r cm

Solution 20

Angle described by the minute hand in 60 minutes = 360°

Angle described by minute hand in 20 minutes

                                               

Required area swept by the minute hand in 20 minutes

                                          =Area of the sector(with r = 15 cm and  = 120°)

                                                      

Solution 21

= 56o and let radius is r cm

Area of sector =

Hence radius= 6cm

Solution 22

Area of the sector of circle =

Radius = 10.5 cm

Solution 23

Let sector of circle is OAB

Perimeter of a sector of circle =31 cm

OA + OB + length of arc AB = 31 cm

 

 6.5 + 6.5 + arc AB = 31 cm

  arc AB = 31 - 13

            = 18 cm

Solution 24

Length of arc of circle = 44 cm

Radius of circle = 17.5 cm

Area of sector =

                    

Solution 25

  

Solution 26

Solution 27

Solution 28

Solution 29

Solution 30

Solution 31

Solution 32

Solution 33

Solution 34

Shaded area = (area of quadrant) - (area of DAOD)

Solution 35

Solution 36

Solution 37

Solution 38

Solution 39

Solution 40

Solution 41

Solution 42

Side of the square ABCD = 14 cm

Area of square ABCD = 14 14 = 196

Radius of each circle =

Area of the circles = 4 area of one circle

Area of shaded region = Area of square - area of 4 circles

= 196 - 154 = 42

Solution 43

Solution 44

Solution 45

Solution 46

Solution 47

Area of rectangle = (120 × 90)

                          = 10800

Area of circular lawn = [Area of rectangle - Area of park excluding circular lawn]

                              = [10800 - 2950] = 7850

 

Area of circular lawn = 7850

Hence, radius of the circular lawn = 50 m

Solution 48

Area of flower bed = (area of quadrant OPQ)

-(area of the quadrant ORS)

Solution 49

Diameter of bigger circle = AC = 54 cm

Radius of bigger circle =

                              

Diameter AB of smaller circle

                             

Radius of smaller circle =

Area of bigger circle =

                             = 2291. 14

Area of smaller circle =

                              = 1521. 11

Area of shaded region = area of bigger circle - area of smaller circle

                               

Solution 50

 

Solution 51