R S AGGARWAL AND V AGGARWAL Solutions for Class 10 Maths Chapter 16 - Areas of Circle, Sector and Segment
Chapter 18 - Areas of Circle, Sector and Segment MCQ
Chapter 18 - Areas of Circle, Sector and Segment FA
Chapter 18 - Areas of Circle, Sector and Segment Ex. 16A
Circumference of circle = 2r = 39.6 cm
Area of square =
Perimeter of square = 4 side = 4 22 = 88 cm
Circumference of circle = Perimeter of square
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 =
=
Let the radii of circles be x cm and (7 - x) cm
Circumference of the circles are 26 cm and 18 cm
Area of outer circle =
= 1662.5
Area of ring = Outer area - inner area
= (1662.5 - 452.5)
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 =
Let r m and R m be the radii of inner circle and outer boundaries respectively.
Then, 2r = 352 and 2
R = 396
Width of the track = (R - r) m
Area the track =
Length of the arc
Length of arc =
Area of the sector =
Length of arc =
Circumference of circle = 2 r
Area of circle =
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
Let OA = , OB =
And AB = 10 cm
Area of AOB =
Area of minor segment = (area of sector OACBO) - (area of OAB)
=
Area of sector OACBO
Area of minor segment ACBA
Area of major segment BADB
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
Let the major arc be x cm long
Then, length of the minor arc =
Circumference =
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
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
Area which the horse can graze = Area of the quadrant of radius 21 m
Area ungrazed =
Each angle of equilateral triangle is 60
Area that the horse cannot graze is 36.68 m2
Ungrazed area
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
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
Let the radius of circle be r cm
Then diagonal of square = diameter of circle = 2r cm
Area of the circle =
Let the radius of circle be r cm
Let each side of the triangle be a cm
And height be h cm
Radius of the wheel = 42 cm
Circumference of wheel =2r =
Distance travelled = 19.8 km = 1980000 cm
Number of revolutions =
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 =
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
Radius of the wheel
Circumference of the wheel = 2r =
Distance covered in 140 revolution
Distance covered in one hour =
Distance covered by a wheel in 1minute
Circumference of a wheel =
Number of revolution in 1 min =
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 =
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
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
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
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
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°]
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
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
In PQR,
P = 90, PQ = 24 cm, PR = 7 cm
Area of semicircle
Area of PQR =
Shaded area = 245.31 - 84 = 161.31
In ABC,
A = 90°, AB = 6cm, BC = 10 cm
Area of ABC =
Let r be the radius of circle of centre O
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)
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
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
Length of the pendulum = radius of sector = r cm
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°)
= 56o and let radius is r cm
Area of sector =
Hence radius= 6cm
Area of the sector of circle =
Radius = 10.5 cm
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
Length of arc of circle = 44 cm
Radius of circle = 17.5 cm
Area of sector =
Shaded area = (area of quadrant) - (area of DAOD)
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
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
Area of flower bed = (area of quadrant OPQ)
-(area of the quadrant ORS)
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
Other Chapters for CBSE Class 10 Maths
Chapter 1- Real Numbers Chapter 2- Polynomials Chapter 3- Linear equations in two variables Chapter 4- Quadratic Equations Chapter 5- Arithmetic Progressions Chapter 6- Co-ordinate Geometry Chapter 7- Triangles Chapter 8- Circles Chapter 9- Constructions Chapter 10- Trigonometric Ratios Chapter 11- T-Ratios of Some Particular Angles Chapter 12- Trigonometric Ratios of Complementary Angles Chapter 13- Trigonometric Identities Chapter 14- Height and Distance Chapter 15- Perimeter and Areas of Plane Figures Chapter 17- Volume and Surface Areas of Solids Chapter 18- Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive Chapter 19- ProbabilityR S AGGARWAL AND V AGGARWAL Solutions for CBSE Class 10 Subjects
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