Class 10 Class 10 Maths RS Aggarwal Solutions Class 10

RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment

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RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment is a part of RS Aggarwal Solutions Class 10. Here we have given RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment.

Exercise 18

  

 

Question 1:
Radius = \frac { Diameter }{ 2 } =\frac { 35 }{ 2 } cm  
Circumference of circle = 2πr = \left( 2\times \frac { 22 }{ 7 } \times \frac { 35 }{ 2 }  \right) cm   = 110 cm
∴ Area of circle = πr2  =  \left( \frac { 22 }{ 7 } \times \frac { 35 }{ 2 } \times \frac { 35 }{ 2 }  \right)   cm2
= 962.5 cm2

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Question 2:
Circumference of circle = 2πr = 39.6 cm
rs-aggarwal-class-10-solutions-areas-of-circle-sector-and-segment-18-q2

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Question 3:
Area of circle = πr2  =  301.84
rs-aggarwal-class-10-solutions-areas-of-circle-sector-and-segment-18-q3
Circumference of circle = 2πr = \left( 2\times \frac { 22 }{ 7 } \times 9.8 \right)       = 61.6 cm

Question 4:
Let radius of circle be r
Then, diameter = 2 r
circumference – Diameter = 16.8
rs-aggarwal-class-10-solutions-areas-of-circle-sector-and-segment-18-q4
Circumference of circle = 2πr = \left( 2\times \frac { 22 }{ 7 } \times 3.92 \right)       cm = 24.64 cm

Question 5:
Let the radius of circle be r cm
Then, circumference – radius = 37 cm
rs-aggarwal-class-10-solutions-areas-of-circle-sector-and-segment-18-q5

Question 6:
Area of square = (side)2 = 484 cm2
⇒ side = \sqrt { 484 } cm   = 22 cm
Perimeter of square = 4 × side = 4 × 22 = 88 cm
Circumference of circle = Perimeter of square
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment ex 18

Question 7:
Area of equilateral = \frac { \sqrt { 3 }  }{ 4 } { a }^{ 2 }   = 121√3
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment ex 18
Perimeter of equilateral triangle = 3a = (3 × 22) cm
= 66 cm
Circumference of circle = Perimeter of circle
2πr = 66
\left( 2\times \frac { 22 }{ 7 } \times r \right)       cm = 66
⇒ r = 10.5 cm
Area of circle = πr2  = \left( \frac { 22 }{ 7 } \times 10.5\times 10.5 \right)   cm2
= 346.5 cm2

Question 8:
Let the radius of park be r meter
rs-aggarwal-class-10-solutions-areas-of-circle-sector-and-segment-18-q8

Question 9:
Let the radii of circles be x cm and (7 – x) cm
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment ex 18
Circumference of the circles are 26 cm and 18 cm

Question 10:
Area of first circle = πr2 = 962.5 cm2
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment ex 18
Area of second circle = πR2 = 1386 cm2
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment ex 18
Width of ring R – r = (21 – 17.5) cm = 3.5 cm

Question 11:
Area of outer circle = πr_1^2    = \left( \frac { 22 }{ 7 } \times 23\times 23 \right)   cm2
= 1662.5
Area of inner circle = πr_2^2    = \left( \frac { 22 }{ 7 } \times 12\times 12 \right)   cm2
= 452.2 cm2
Area of ring = Outer area – inner area
= (1662.5 – 452.5) cm2 = 1210 cm2

Question 12:
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 = π[(25)2-(17)2] = cm2
RS Aggarwal Class 10 Solutions Chapter 18 Areas of Circle, Sector and Segment ex 18
Area = 1056 m2

Question 13:
Let the inner and outer radii of the circular tacks be r meter and R meter respectively. Then
Inner circumference = 440 meter
RS Aggarwal Class 10 Solutions Chapter 18 Areas of Circle, Sector and Segment ex 18
Since the track is 14 m wide every where.
Therefore,
Outer radius R = r + 14m = (70 + 14) m = 84 m
Outer circumference = 2πR
=  \left( 2\times \frac { 22 }{ 7 } \times 84 \right)m        = 528 m
Rate of fencing = Rs. 5 per meter
Total cost of fencing = Rs. (528 × 5) = Rs. 2640
Area of circular ring = πR2  – πr2
RS Aggarwal Class 10 Solutions Chapter 18 Areas of Circle, Sector and Segment ex 18
Cost of levelling = Rs 0.25 per m2
Cost of levelling the track = Rs(6776 × 0.25) = Rs. 1694

Question 14:
Let r m and R m be the radii of inner circle and outer boundaries respectively.
Then, 2r = 352 and 2R = 396
RS Aggarwal Class 10 Solutions Chapter 18 Areas of Circle, Sector and Segment ex 18
Width of the track = (R – r) m
RS Aggarwal Class 10 Solutions Chapter 18 Areas of Circle, Sector and Segment ex 18
Area the track = π(R2  – r) = π (R+r)(R-r)
RS Aggarwal Class 10 Solutions Chapter 18 Areas of Circle, Sector and Segment ex 18

Question 15:
Area of rectangle = (120 × 90)
= 10800 m2
Area of circular lawn = [Area of rectangle – Area of park excluding circular lawn]
= [10800 – 2950] m2 = 7850 m2
Area of circular lawn = 7850 m2
⇒  πr2 = 7850 m2
RS Aggarwal Class 10 Solutions Chapter 18 Areas of Circle, Sector and Segment ex 18
Hence, radius of the circular lawn = 50 m

Question 16:
RS Aggarwal Class 10 Solutions Chapter 18 Areas of Circle, Sector and Segment ex 18
Area of the shaded region = (area of circle with OA as diameter) + (area of semicircle ∆DBC) – (area of ∆BCD)
Area of circle with OA as diameter = πr2
rs-aggarwal-class-10-solutions-areas-of-circle-sector-and-segment-18-q16-2
OB = 7 cm, CD = AB = 14 cm
Area of semicircle ∆DBC = rs-aggarwal-class-10-solutions-areas-of-circle-sector-and-segment-18-q16-3
= 72
Solution of RS Aggarwal Class 10 Chapter 18 Areas of Circle, Sector and Segment ex 18

Question 17:
Solution of RS Aggarwal Class 10 Chapter 18 Areas of Circle, Sector and Segment ex 18
Diameter of bigger circle = AC = 54 cm
Radius of bigger circle = \frac { AC }{ 2 }  
=  (\frac { 54 }{ 2 })   cm = 27 cm
Diameter AB of smaller circle = AC – BC = 54-10 = 44 cm
Radius of smaller circle = \frac { 44 }{ 2 }   cm = 22 cm
Area of bigger circle = πR2  = \left( \frac { 22 }{ 7 } \times 27\times 27 \right)   cm2
= 2291. 14 cm2
Area of smaller circle = πr2  = \left( \frac { 22 }{ 7 } \times 22\times 22 \right)   cm2
= 1521. 11 cm2
Area of shaded region = area of bigger circle – area of smaller circle
=  (2291. 14 – 1521. 11) cm2  = 770 cm2

Question 18:
Solution of RS Aggarwal Class 10 Chapter 18 Areas of Circle, Sector and Segment ex 18
PS = 12 cm
PQ = QR = RS = 4 cm, QS = 8 cm
Perimeter = arc PTS + arc PBQ + arc QES
Solution of RS Aggarwal Class 10 Chapter 18 Areas of Circle, Sector and Segment ex 18
Area of shaded region = (area of the semicircle PBQ) + (area of semicircle PTS)-(Area of semicircle QES)
Solution of RS Aggarwal Class 10 Chapter 18 Areas of Circle, Sector and Segment ex 18

Question 19:
Solution of RS Aggarwal Class 10 Chapter 18 Areas of Circle, Sector and Segment ex 18
Length of the inner curved portion
= (400 – 2 × 90) m
= 220 m
Let the radius of each inner curved part be r
Solution of RS Aggarwal Class 10 Chapter 18 Areas of Circle, Sector and Segment ex 18
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
RS Aggarwal Solutions Class 10 2017 Chapter 18 Areas of Circle, Sector and Segment ex 18
Length of outer boundary of the track
RS Aggarwal Solutions Class 10 2017 Chapter 18 Areas of Circle, Sector and Segment ex 18

Question 20:
RS Aggarwal Solutions Class 10 2017 Chapter 18 Areas of Circle, Sector and Segment ex 18
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
RS Aggarwal Solutions Class 10 2017 Chapter 18 Areas of Circle, Sector and Segment ex 18

Question 21:
Diameter of the inscribed circle = Side of the square = 10 cm
Radius of the inscribed circle = 5 cm
RS Aggarwal Solutions Class 10 2017 Chapter 18 Areas of Circle, Sector and Segment ex 18
Diameter of the circumscribed circle
= Diagonal of the square
= (√2×10) cm
Radius of circumscribed circle = 5√2 cm
(i) Area of inscribed circle = \left( \frac { 22 }{ 7 } \times 5\times 5 \right)   = 78.57 cm2
(ii) Area of the circumscribed circle RS Aggarwal Solutions Class 10 2017 Chapter 18 Areas of Circle, Sector and Segment ex 18

Question 22:
Let the radius of circle be r cm
RS Aggarwal Solutions Class 10 2017 Chapter 18 Areas of Circle, Sector and Segment ex 18
Then diagonal of square = diameter of circle = 2r cm
Area of the circle = πr2 cm2
RS Aggarwal Solutions Class 10 2017 Chapter 18 Areas of Circle, Sector and Segment ex 18

Question 23:
Let the radius of circle be r cm
RS Aggarwal Solutions Class 10 2017 Chapter 18 Areas of Circle, Sector and Segment ex 18
Let each side of the triangle be a cm
And height be h cm
RS Aggarwal Solutions Class 10 2017 Chapter 18 Areas of Circle, Sector and Segment ex 18

Question 24:
Radius of the wheel = 42 cm
Circumference of wheel = 2πr = \left( 2\times \frac { 22 }{ 7 } \times 42 \right)       = 264 cm
Distance travelled = 19.8 km = 1980000 cm
Number of revolutions = \frac { 1980000 }{ 264 }   = 7500

Question 25:
Radius of wheel = 2.1 m
Circumference of wheel = 2πr = \left( 2\times \frac { 22 }{ 7 } \times 2.1 \right)       = 13.2 m
Distance covered in one revolution = 13.2 m
Distance covered in 75 revolutions = (13.2 × 75) m = 990 m
\frac { 990 }{ 1000 }   km
Distance a covered in 1 minute = \frac { 99 }{ 100 }   km
Distance covered in 1 hour = \frac { 99 }{ 100 } \times 60 km = 59.4 km

Question 26:
Distance covered by the wheel in 1 revolution
RS Aggarwal Solutions Class 10 2017 Chapter 18 Areas of Circle, Sector and Segment ex 18
The circumference of the wheel = 198 cm
Let the diameter of the wheel be d cm
Class 10 RS Aggarwal Solutions Chapter 18 Areas of Circle, Sector and Segment ex 18
Hence diameter of the wheel is 63 cm

Question 27:
Radius of the wheel = r = \frac { 60 }{ 2 }   = 30 cm
Circumference of the wheel = 2πr = \left( 2\times \frac { 22 }{ 7 } \times 30 \right)      \frac { 1320 }{ 7 }   cm
Distance covered in 140 revolution
Class 10 RS Aggarwal Solutions Chapter 18 Areas of Circle, Sector and Segment ex 18
Distance covered in one hour = \frac { 264 }{ 1000 } \times 60 = 15.84 km

Question 28:
Distance covered by a wheel in 1minute
Class 10 RS Aggarwal Solutions Chapter 18 Areas of Circle, Sector and Segment ex 18
Circumference of a wheel = 2πr = \left( 2\times \frac { 22 }{ 7 } \times 70 \right)       = 440 cm
Number of revolution in 1 min = \frac { 121000 }{ 440 }   = 275

Question 29:
Area of quadrant = \frac { 1 }{ 4 }   πr2
Circumference of circle = 2πr = 22
Class 10 RS Aggarwal Solutions Chapter 18 Areas of Circle, Sector and Segment ex 18

Question 30:
Class 10 RS Aggarwal Solutions Chapter 18 Areas of Circle, Sector and Segment ex 18
Area which the horse can graze = Area of the quadrant of radius 21 m
Class 10 RS Aggarwal Solutions Chapter 18 Areas of Circle, Sector and Segment ex 18
Area ungrazed = [(70×52) – 346.5] m2
= 3293.5 m2

Question 31:
Each angle of equilateral triangle is 60°
RS Aggarwal Class 10 Maths Book Solutions pdf Chapter 18 Areas of Circle, Sector and Segment ex 18
RS Aggarwal Class 10 Maths Book Solutions pdf Chapter 18 Areas of Circle, Sector and Segment ex 18
Area that the horse cannot graze is 36.68 m2

Question 32:
RS Aggarwal Class 10 Maths Book Solutions pdf Chapter 18 Areas of Circle, Sector and Segment ex 18
Each side of the square is 14 cm
Then, area of square = (14 × 14) cm2
= 196 cm2
Thus, radius of each circle 7 cm
Required area = area of square ABCD – 4 (area of sector with r = 7 cm, θ= 90°)
RS Aggarwal Class 10 Maths Book Solutions pdf Chapter 18 Areas of Circle, Sector and Segment ex 18
Area of the shaded region = 42 cm2

Question 33:
RS Aggarwal Class 10 Maths Book Solutions pdf Chapter 18 Areas of Circle, Sector and Segment ex 18
Area of square = (4 × 4) cm2
= 16 cm2
Area of four quadrant corners
RS Aggarwal Class 10 Book pdf download Chapter 18 Areas of Circle, Sector and Segment ex 18
Radius of inner circle = 2/2 = 1 cm
Area of circle at the center = πr2 = (3.14 × 1 × 1) cm2
= 3.14 cm2
Area of shaded region = [area of square – area of four corner quadrants – area of circle at the centre]
= [16 – 3.14 – 3.14] cm= 9.72 cm2

Question 34:
RS Aggarwal Class 10 Book pdf download Chapter 18 Areas of Circle, Sector and Segment ex 18
Area of rectangle = (20 × 15) m= 300 m2
Area of 4 corners as quadrants of circle
RS Aggarwal Class 10 Book pdf download Chapter 18 Areas of Circle, Sector and Segment ex 18
Area of remaining part = (area of rectangle – area of four quadrants of circles)
= (300 – 38.5) m2 = 261.5 m2

Question 35:
RS Aggarwal Class 10 Book pdf download Chapter 18 Areas of Circle, Sector and Segment ex 18
Ungrazed area
RS Aggarwal Class 10 Book pdf download Chapter 18 Areas of Circle, Sector and Segment ex 18

Question 36:
RS Aggarwal Class 10 Book pdf download Chapter 18 Areas of Circle, Sector and Segment ex 18
Shaded area = (area of quadrant) – (area of DAOD)
RS Aggarwal Class 10 Book pdf download Chapter 18 Areas of Circle, Sector and Segment ex 18

Question 37:
RS Aggarwal Class 10 Book pdf download Chapter 18 Areas of Circle, Sector and Segment ex 18
Area of flower bed = (area of quadrant OPQ) – (area of the quadrant ORS)
Class 10 Maths RS Aggarwal Solutions Chapter 18 Areas of Circle, Sector and Segment ex 18

Question 38:
Class 10 Maths RS Aggarwal Solutions Chapter 18 Areas of Circle, Sector and Segment ex 18
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, θ = 60°)
Class 10 Maths RS Aggarwal Solutions Chapter 18 Areas of Circle, Sector and Segment ex 18
The area enclosed = 5.76 cm2

Question 39:
Class 10 Maths RS Aggarwal Solutions Chapter 18 Areas of Circle, Sector and Segment ex 18
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, θ = 60°]
rs-aggarwal-class-10-solutions-areas-of-circle-sector-and-segment-18-q39-2

Question 40:
Class 10 Maths RS Aggarwal Solutions Chapter 18 Areas of Circle, Sector and Segment ex 18
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
= (10 × 10) cm2
= 100 cm2
Area of each sector = Maths RS Aggarwal Class 10 Solutions Chapter 18 Areas of Circle, Sector and Segment ex 18
= 19.625 cm2
Required area = [area of sq. ABCD – 4(area of each sector)]
= (100 – 4 × 19.625) cm2
= (100 – 78.5) = 21.5 cm2

Question 41:
Maths RS Aggarwal Class 10 Solutions Chapter 18 Areas of Circle, Sector and Segment ex 18
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 = 4a2 sq.units
Maths RS Aggarwal Class 10 Solutions Chapter 18 Areas of Circle, Sector and Segment ex 18

Question 42:
Maths RS Aggarwal Class 10 Solutions Chapter 18 Areas of Circle, Sector and Segment ex 18
Let the side of square = a m
Area of square = (a × a) cm  = a2m2
Maths RS Aggarwal Class 10 Solutions Chapter 18 Areas of Circle, Sector and Segment ex 18
Side of square = 40 m
Therefore, radius of semi circle = 20 m
Area of semi circle = Maths RS Aggarwal Class 10 Solutions Chapter 18 Areas of Circle, Sector and Segment ex 18
= 628 m2
Area of four semi circles = (4 × 628) m= 2512 m2
Cost of turfing the plot of of area 1 m2 = Rs. 1.25
Cost of turfing the plot of area 2512 m= Rs. (1.25 × 2512)
= Rs. 3140

Question 43:
Maths RS Aggarwal Class 10 Solutions Chapter 18 Areas of Circle, Sector and Segment ex 18
Area of rectangular lawn in the middle
= (50 × 35) = 1750 m2
Radius of semi circles = \frac { 35 }{ 2 }   = 17.5 m
RS Aggarwal Solutions Class 10 2018 Chapter 18 Areas of Circle, Sector and Segment ex 18
Area of lawn = (area of rectangle + area of semi circle)
= (1750 + 962.5) m2 = 2712.5 m2

Question 44:
Area of plot which cow can graze when r = 16 m is πr2
\left( \frac { 22 }{ 7 } \times 10.5\times 10.5 \right)  
= 804.5 m2
Area of plot which cow can graze when radius is increased to 23 m
\left( \frac { 22 }{ 7 } \times 10.5\times 10.5 \right)  
= 1662.57 m2
Additional ground = Area covered by increased rope – old area
= (1662.57 – 804.5)m2 = 858 m2

Question 45:
RS Aggarwal Solutions Class 10 2018 Chapter 18 Areas of Circle, Sector and Segment ex 18
Given: ABC is right angled at A with AB = 6 cm and AC = 8 cm
RS Aggarwal Solutions Class 10 2018 Chapter 18 Areas of Circle, Sector and Segment ex 18
Let us join OA, OB and OC
ar(∆AOC) + ar(∆OAB) + ar(∆BOC) = ar(∆ABC)
RS Aggarwal Solutions Class 10 2018 Chapter 18 Areas of Circle, Sector and Segment ex 18

Question 46:
RS Aggarwal Solutions Class 10 2018 Chapter 18 Areas of Circle, Sector and Segment ex 18

Question 47:
RS Aggarwal Solutions Class 10 2018 Chapter 18 Areas of Circle, Sector and Segment ex 18
Area of region ABCDEFA = area of square ABDE + area of semi circle BCD – area of ∆AFE
RS Aggarwal Solutions Class 10 2018 Chapter 18 Areas of Circle, Sector and Segment ex 18

Question 48:
RS Aggarwal Solutions Class 10 2018 Chapter 18 Areas of Circle, Sector and Segment ex 18
Side of the square ABCD = 14 cm
Area of square ABCD = 14 × 14 = 196 cm2
Radius of each circle = \frac { 14 }{ 4 }   = 3.5 cm
Area of the circles = 4 × area of one circle
RS Aggarwal Solutions Class 10 2018 Chapter 18 Areas of Circle, Sector and Segment ex 18
Area of shaded region = Area of square – area of 4 circles
= 196 – 154 = 42 cm2

Question 49:
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment ex 18
Diameter AC = 2.8 + 1.4
= 4.2 cm
Radius r1 = \frac { 4.2 }{ 2 }   = 2.1 cm
Length of semi-circle ADC = πr=  π × 2.1 = 2.1 π cm
Diameter AB = 2.8 cm
Radius r2  =  1.4 cm
Length of semi- circle AEB = πr2 =  π × 1.4 = 1.4 π cm
Diameter BC = 1.4 cm
Radius r3 = \frac { 1.4 }{ 2 }   = 0.7 cm
Length of semi – circle BFC = π × 0.7 = 0.7 π  cm
Perimeter of shaded region = 2.1 + 1.4 + 0.7 = 4.2 π cm
= 4.2 × \frac { 22 }{ 7 }   = 13.2 cm

Question 50:
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment ex 18
Area of shaded region = Area of ∆ABC + Area of semi-circle APB + Area of semi circle AQC – Area of semicircle BAC
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment ex 18
Further in ∆ABC, ∠A = 90°
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment ex 18
Adding (1), (2), (3) and subtracting (4)
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment ex 18

Question 51:
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment ex 18
In ∆PQR, ∠P = 90°, PQ = 24 cm, PR = 7 cm
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment ex 18
Area of semicircle
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment ex 18
Area of ∆PQR = \frac { 1 }{ 2 }    × 7 × 24 cm2 = 84 cm2
Shaded area = 245.31 – 84 = 161.31 cm2

Question 52:
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment ex 18
ABCDEF is a hexagon.
∠AOB = 60°, Radius = 35 cm
Area of sector AOB
RS Aggarwal Class 10 Solutions Chapter 18 Areas of Circle, Sector and Segment ex 18
Area of ∆AOB = RS Aggarwal Class 10 Solutions Chapter 18 Areas of Circle, Sector and Segment ex 18
= 530.425 cm2
Area of segment APB = (641.083 – 530.425) cm= 110.658 cm2
Area of design (shaded area) = 6 × 110.658 cm2 = 663.948 cm2
= 663.95 cm2

Question 53:
RS Aggarwal Class 10 Solutions Chapter 18 Areas of Circle, Sector and Segment ex 18
In ∆ABC, ∠A = 90°, AB = 6cm, BC = 10 cm
RS Aggarwal Class 10 Solutions Chapter 18 Areas of Circle, Sector and Segment ex 18
Area of ∆ABC = RS Aggarwal Class 10 Solutions Chapter 18 Areas of Circle, Sector and Segment ex 18
Let r be the radius of circle of centre O
RS Aggarwal Class 10 Solutions Chapter 18 Areas of Circle, Sector and Segment ex 18

Question 54:
Area of equilateral triangle ABC = 49√3 cm2
RS Aggarwal Class 10 Solutions Chapter 18 Areas of Circle, Sector and Segment ex 18
Let a be its side
RS Aggarwal Class 10 Solutions Chapter 18 Areas of Circle, Sector and Segment ex 18
Area of sector BDF = RS Aggarwal Class 10 Solutions Chapter 18 Areas of Circle, Sector and Segment ex 18
RS Aggarwal Class 10 Solutions Chapter 18 Areas of Circle, Sector and Segment ex 18
Area of sector BDF = Area of sector CDE = Area of sector AEF
Sum of area of all the sectors
\frac { 77 }{ 3 }   × 3 cm2 = 77 cm2
Shaded area = Area of ∆ABC – sum of area of all sectors
= 49√3 – 77 = (84.77 – 77.00) cm2
= 77.7 cm2

Question 55:
RS Aggarwal Class 10 Solutions Chapter 18 Areas of Circle, Sector and Segment ex 18
In ∆ABC, ∠B = 90°, AB = 48 cm, BC = 14 cm
RS Aggarwal Class 10 Solutions Chapter 18 Areas of Circle, Sector and Segment ex 18
Area of semi-circle APC
RS Aggarwal Class 10 Solutions Chapter 18 Areas of Circle, Sector and Segment ex 18
Area of quadrant BDC with radius 14 cm
rs-aggarwal-class-10-solutions-areas-of-circle-sector-and-segment-18-q55-4
Shaded area = Area of ∆ABC + Area of semi-circle APC – Area of quadrant BDC
= ( 336+982.14-154 ) cm2
= ( 1318.14-154 ) cm= 1164.14 cm2

Question 56:
Solution of RS Aggarwal Class 10 Chapter 18 Areas of Circle, Sector and Segment ex 18
Radius of quadrant ABED = 16 cm
Its area = Solution of RS Aggarwal Class 10 Chapter 18 Areas of Circle, Sector and Segment ex 18
Area of ∆ABD = \left( \frac { 1 }{ 2 } \times 16\times 16 \right)   cm2
= 128 cm2
Area of segment DEB
Solution of RS Aggarwal Class 10 Chapter 18 Areas of Circle, Sector and Segment ex 18
Area of segment DFB = \frac { 512 }{ 7 }   cm2
Total area of segments = 2 × \frac { 512 }{ 7 }   cm2 =  \frac { 1024 }{ 7 }   cm2
Shaded area = Area of square ABCD – Total area of segments
Solution of RS Aggarwal Class 10 Chapter 18 Areas of Circle, Sector and Segment ex 18

Question 57:
Solution of RS Aggarwal Class 10 Chapter 18 Areas of Circle, Sector and Segment ex 18
Radius of circular table cover = 70 cm
Area of the circular cover = rs-aggarwal-class-10-solutions-areas-of-circle-sector-and-segment-18-q57-2
Solution of RS Aggarwal Class 10 Chapter 18 Areas of Circle, Sector and Segment ex 18
Shaded area = Area of circle – Area of ∆ABC
= (15400 – 6365.1)

Question 58:
Area of the sector of circle = RS Aggarwal Solutions Class 10 2017 Chapter 18 Areas of Circle, Sector and Segment ex 18
r = 14 cm and θ = 45°
RS Aggarwal Solutions Class 10 2017 Chapter 18 Areas of Circle, Sector and Segment ex 18

Question 59:
Length of the arc RS Aggarwal Solutions Class 10 2017 Chapter 18 Areas of Circle, Sector and Segment ex 18
RS Aggarwal Solutions Class 10 2017 Chapter 18 Areas of Circle, Sector and Segment ex 18
Length of arc = ( 17.5 × \frac { 22 }{ 7 }   ) cm = 55 cm
Area of the sector = rs-aggarwal-class-10-solutions-areas-of-circle-sector-and-segment-18-q59-3
= ( \frac { 22 }{ 7 }   × 183.75 ) cm2 = 577.5 cm2

Question 60:
Length of arc of circle = 44 cm
Radius of circle = 17.5 cm
Area of sector = RS Aggarwal Solutions Class 10 2017 Chapter 18 Areas of Circle, Sector and Segment ex 18
= ( 22 × 17.5) cm2 = 385 cm2

Question 61:
Let sector of circle is OAB
Perimeter of a sector of circle =31 cm
OA + OB + length of arc AB = 31 cm
RS Aggarwal Solutions Class 10 2017 Chapter 18 Areas of Circle, Sector and Segment ex 18
6.5 + 6.5 + arc AB = 31 cm
arc AB = 31 – 13
= 18 cm
Class 10 RS Aggarwal Solutions Chapter 18 Areas of Circle, Sector and Segment ex 18

Question 62:
Area of the sector of circle = Class 10 RS Aggarwal Solutions Chapter 18 Areas of Circle, Sector and Segment ex 18
Radius = 10.5 cm
Class 10 RS Aggarwal Solutions Chapter 18 Areas of Circle, Sector and Segment ex 18

Question 63:
Length of the pendulum = radius of sector = r cm
Class 10 RS Aggarwal Solutions Chapter 18 Areas of Circle, Sector and Segment ex 18

Question 64:
Length of arc = Class 10 RS Aggarwal Solutions Chapter 18 Areas of Circle, Sector and Segment ex 18
Class 10 RS Aggarwal Solutions Chapter 18 Areas of Circle, Sector and Segment ex 18

Circumference of circle = 2πr
Class 10 RS Aggarwal Solutions Chapter 18 Areas of Circle, Sector and Segment ex 18
Area of circle =
RS Aggarwal Class 10 Maths Book Solutions pdf Chapter 18 Areas of Circle, Sector and Segment ex 18
= 962.5 cm2

Question 65:
Circumference of circle = 2πr
RS Aggarwal Class 10 Maths Book Solutions pdf Chapter 18 Areas of Circle, Sector and Segment ex 18

Question 66:
Angle described by the minute hand in 60 minutes θ = 360°
Angle described by minute hand in 20 minutes
RS Aggarwal Class 10 Maths Book Solutions pdf Chapter 18 Areas of Circle, Sector and Segment ex 18
Required area swept by the minute hand in 20 minutes
= Area of the sector(with r = 15 cm and θ = 120°)
RS Aggarwal Class 10 Maths Book Solutions pdf Chapter 18 Areas of Circle, Sector and Segment ex 18

Question 67:
θ = 56° and let radius is r cm
Area of sector = RS Aggarwal Class 10 Maths Book Solutions pdf Chapter 18 Areas of Circle, Sector and Segment ex 18
RS Aggarwal Class 10 Maths Book Solutions pdf Chapter 18 Areas of Circle, Sector and Segment ex 18
Hence radius = 6cm

Question 68:
RS Aggarwal Class 10 Maths Book Solutions pdf Chapter 18 Areas of Circle, Sector and Segment ex 18

Question 69:
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

Question 70:
∆OAB is equilateral.
So, ∠AOB = 60°
RS Aggarwal Class 10 Maths Book Solutions pdf Chapter 18 Areas of Circle, Sector and Segment ex 18
Length of arc BDA = (2π × 12 – arc ACB) cm
= (24π – 4π) cm = (20π) cm
= (20 × 3.14) cm = 62.8 cm
Area of the minor segment ACBA
rs-aggarwal-class-10-solutions-areas-of-circle-sector-and-segment-18-q70-2

Question 71:
Let AB be the chord of circle of centre O and radius = 6 cm such that ∠AOB = 90°
RS Aggarwal Class 10 Book pdf download Chapter 18 Areas of Circle, Sector and Segment ex 18
Area of sector = OACBO
RS Aggarwal Class 10 Book pdf download Chapter 18 Areas of Circle, Sector and Segment ex 18
Area of ∆AOB = RS Aggarwal Class 10 Book pdf download Chapter 18 Areas of Circle, Sector and Segment ex 18
Area of minor segment ACBA
= (area of sector OACBO) – (area of ∆OAB)
= (28.29 – 18) cm2 = 10.29 cm2
Area of major segment BDAB
RS Aggarwal Class 10 Book pdf download Chapter 18 Areas of Circle, Sector and Segment ex 18

Question 72:
Let OA = 5√2 cm , OB = 5√2 cm
And AB = 10 cm
RS Aggarwal Class 10 Book pdf download Chapter 18 Areas of Circle, Sector and Segment ex 18
Area of ∆AOB = Class 10 Maths RS Aggarwal Solutions Chapter 18 Areas of Circle, Sector and Segment ex 18
= 25 cm2
Area of minor segment = (area of sector OACBO) – (area of ∆OAB)
= ( 39.25 – 25 ) cm2 = 14.25 cm2
Class 10 Maths RS Aggarwal Solutions Chapter 18 Areas of Circle, Sector and Segment ex 18

Question 73:
Area of sector OACBO
Class 10 Maths RS Aggarwal Solutions Chapter 18 Areas of Circle, Sector and Segment ex 18
Area of minor segment ACBA
Class 10 Maths RS Aggarwal Solutions Chapter 18 Areas of Circle, Sector and Segment ex 18
Area of major segment BADB
Class 10 Maths RS Aggarwal Solutions Chapter 18 Areas of Circle, Sector and Segment ex 18

Question 74:
Let AB be the chord of circle of centre O and radius = 30 cm such that AOB = 60°
Class 10 Maths RS Aggarwal Solutions Chapter 18 Areas of Circle, Sector and Segment ex 18
Area of the sector OACBO
Maths RS Aggarwal Class 10 Solutions Chapter 18 Areas of Circle, Sector and Segment ex 18
Area of ∆OAB = Maths RS Aggarwal Class 10 Solutions Chapter 18 Areas of Circle, Sector and Segment ex 18
Maths RS Aggarwal Class 10 Solutions Chapter 18 Areas of Circle, Sector and Segment ex 18
Area of the minor segment ACBA
= (area of the sector OACBO) – (area of the ∆OAB)
=(471 – 389.25) cm2 = 81.75 cm2
Area of the major segment BADB
= (area of circle) – (area of the minor segment)
= [(3.14 × 30 × 30) – 81.75)] cm2 = 2744.25 cm2

Question 75:
Let the major arc be x cm long
Then, length of the minor arc = \frac { 1 }{ 5 }   x cm
Circumference = Maths RS Aggarwal Class 10 Solutions Chapter 18 Areas of Circle, Sector and Segment ex 18
Maths RS Aggarwal Class 10 Solutions Chapter 18 Areas of Circle, Sector and Segment ex 18

Question 76:
Radius of the front wheel = 40 cm = \frac { 2 }{ 5 }   m
Circumference of the front wheel = RS Aggarwal Solutions Class 10 2018 Chapter 18 Areas of Circle, Sector and Segment ex 18
Distance moved by it in 800 revolution
RS Aggarwal Solutions Class 10 2018 Chapter 18 Areas of Circle, Sector and Segment ex 18
Circumference of rear wheel = (2π × 1)m = (2π) m
Required number of revolutions = RS Aggarwal Solutions Class 10 2018 Chapter 18 Areas of Circle, Sector and Segment ex 18

We hope the given Solutions of RS Aggarwal Solutions Class 10 Chapter 18 are helpful to complete your math homework. If you have any doubts, please comment below. We will try to provide online math tutoring for you.


RS Aggarwal Solutions Class 10

  1. Real Numbers
  2. Polynomials
  3. Linear equations in two variables
  4. Triangles
  5. Trigonometric Ratios
  6. T-Ratios of Some Particular Angles
  7. Trigonometric Identities
  8. Trigonometric Ratios of Complementary Angles
  9. Mean, Median, Mode of Grouped Data, of Grouped Data Ex.9A Ex.9B Ex.9C and 9D Ex.9E
  10. Quadratic Equations Ex.10A Ex.10B Ex.10C Ex.10D
  11. Arithmetic Progressions
  12. Circles
  13. Constructions
  14. Height and Distance
  15. Probability
  16. Coordinate Geometry
  17. Perimeter and Areas of Plane Figures
  18. Areas of Circle, Sector and Segment
  19. Volume and Surface Areas of Solids

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