ROUTERA


Chapter 16 Area of Circle, Sector and Segment

Class 10th R. S. Aggarwal Maths Solution
CBSE Class 10 Maths
R. S. Aggarwal Solution


Area of Circle, Sector and Segment Exercise 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 =

                

(ii)

Diameter of park = 7 m

Radius r1 = 3.5 m

Width of park = 0.7 m

Bigger radius = r2 = 0.7 + 3.4 = 4.2 m

Now,

Area of path = Area of the bigger circle - Area of the smaller circle

equals πr subscript 2 squared minus πr subscript 1 squared
equals 22 over 7 open parentheses 4.2 squared minus 3.5 squared close parentheses
equals 16.94 space straight m squared
Also comma space cost space of space expenditure space equals space rate space cross times space area space equals space 110 space cross times space 16.94
equals 1863.4
Hence comma space cost space for space cementing space the space path space is space Rs. space 1863.4

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

We know that the area of a minor segment of angle θo in a circle of radius r is given by

  

So, we have

Area of a circle = pr2 = 3.14 × 10 × 10 = 314 cm2

Therefore, area of major segment

= Area of a circle - Area of minor segment

= (314 - 9) cm2

= 305 cm2

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 the circle of centre O and radius = 30 cm such that AOB = 60°

Area of the sector OACBO

Area of OAB 

=

=225 × 1.73 = 389.7 cm2

Area of the minor segment ACBA

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

=(471 - 389.7) = 81.3 

Area of the major segment BADB

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

= [(3.14 × 30 × 30) - 81.3)] = 2744.7

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 an equilateral triangle is 60

=62.352 - 25.67

=36.69 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

Radius of the wheel = r = 35 cm

Let the wheel of a motorcycle makes 'n' revolutions per minute to keep a speed of 66 km/hr.

Then, distance covered by the wheel in one revolution

= Circumference of the wheel

⇒ Distance covered by the wheel in 'n' revolutions = (220 × n) cm

⇒ Distance covered by the wheel in one minute = (220 × n) cm

Given, speed of the motorcycle = 66 km/hr

⇒ Wheel covers 66 km in one hour.

⇒ Distance covered by the wheel in one minute   

Hence, the wheel makes 500 revolutions per minute.

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

equals fraction numerator square root of 3 over denominator 4 end fraction cross times 12 squared minus fraction numerator 3.14 cross times 6 squared cross times 60 over denominator 360 end fraction
equals 36 square root of 3 minus 18.84
equals 43.44 space cm squared
therefore Area space of space shaded space region space equals space Area space of space shaded space circular space part space plus space Area space of space shaded space triangular space region
equals 94.2 space plus space 43.44
equals 137.64 space cm squared


Solution 44

Solution 45

Solution 46

Solution 47

Solution 48

  

Solution 49

Solution 50

Solution 51

 

=6594 m2

cost of levelling per m2 = Rs. 20

Hence, the cost of levelling 6594 m2 = Rs. 20 × 6594 = Rs. 1,31,880

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

Solution 60

Area of rectangle ABCD = 21 cm × 14 cm = 294 cm2

Diameter of a semicircle = BC = 14 cm

⇒ Radius of a semicircle = 7 cm

Then, area of a semicircle   cm2

Therefore, area of shaded region

= Area of rectangle ABCD - Area of semicircle

= (294 - 77) cm2

= 217 cm2

Now, perimeter of the shaded region

Solution 61

Area of the region ABDC = Area of sector AOB - Area of sector COD

Area of the circular ring   

= (5544 - 1386) cm2

= 4158 cm2

Therefore, area of the shaded region

= Area of the circular ring - Area of the region ABDC

= (4158 - 693) cm2

= 3465 cm2

Solution 62

Area of each semicircle of diameter 3 cm

Area of the circle of diameter 4.5 cm   

Area of the semicircle of radius 4.5 cm   

Now, area of the shaded region

= Area of the semicircle of radius 4.5 cm - Area of the circle - 2(area of semicircle of diameter 3 cm) + Area of semicircle of diameter 3 cm

= Area of the semicircle of radius 4.5 cm - Area of the circle -Area of semicircle of diameter 3 cm

= (31.82 - 15.92 - 3.54) cm222

= 12.36 cm2

Solution 63

Area of a square = (28 cm)2 = 784 cm2

Radius of each circle =   

Then, area of each circle =   

Now, area of each quadrant of circle =   

Therefore, area of the shaded region

= Area of a square + Area of two circles - Area of two quadrants

= [784 + 2(616) - 2(154)] cm2

= [784 + 1232 - 308] cm2

= 1708 cm2

Solution 64

  

 

Let ∠A = q1, ∠B = q2, and ∠C = q3.

Now, area of three sectors with central angles q1, q2 and q3 and each with radius 5 cm

 

Sides of triangle ABC are as follows:

a = AB = 14 cm

b = BC = 48

c = CA = 50 cm

Then,

(s - a) = (56 - 14) = 42 cm

(s - b) = (56 - 48) = 8 cm

(s - c) = (56 - 50) = 6 cm

Therefore, area of triangle ABC

 

Hence, area of the shaded region

= Area of triangle ABC - Area of three sectors

= 336 - 39.25

= 296.75 cm2

Solution 65

Let the diameters of the given concentric circles be x, 2x and 3x respectively.

  

So, we have

Area of R1 = Area of circle with radius x units

Area of R2 = Area of circle with radius 2x units - Area of circle with radius x units  

Area of R3 = Area of circle with radius 3x units - Area of circle with radius 2x units  

Therefore, ratio of the areas of three regions

Solution 66

Let the radius of a circular playground = r

Now, area of a circular playground = 22176 cm2

Then, circumference of a circular playground   

Cost of fencing per metre ground = Rs. 50

So, cost of fencing 5.28 m ground = Rs. 50 × 5.28 = Rs. 264

Area of Circle, Sector and Segment Exercise 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

Area of Circle, Sector and Segment Exercise 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

Area of Circle, Sector and Segment Exercise Test Yourself

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