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

## Areas of Circle, Sector and Segment Exercise Ex. 16A

### Solution 1

Circumference of circle = 2r = 39.6 cm

### 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 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 13

Length of arc =

Circumference of circle = 2 r

Area of circle =

### 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

### 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 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

(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

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 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 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

## Areas of Circle, Sector and Segment Exercise Ex. 16B

### 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 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 =

### Solution 22

Area of the sector of circle =

### 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 42

Side of the square ABCD = 14 cm

Area of square ABCD = 14 14 = 196

Area of the circles = 4 area of one circle

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

= 196 - 154 = 42

### 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)

### Solution 49

Diameter of bigger circle = AC = 54 cm

Diameter AB 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