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

In this chapter, we provide RS Aggarwal Solutions for Class 10 Chapter 18 Areas of Circle, Sector and Segment Maths for English medium students, Which will very helpful for every student in their exams. Students can download the latest RS Aggarwal Solutions for Class 10 Chapter 18 Areas of Circle, Sector and Segment Maths pdf, free RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment Maths book pdf download. Now you will get step by step solution to each question.

TextbookNCERT
ClassClass 10
SubjectMaths
ChapterChapter 18
Chapter NameAreas of Circle, Sector and Segment
Exercise18
CategoryRS Aggarwal Solutions

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

Question 1:
Radius = Diameter2=352cm
Circumference of circle = 2πr = (2×227×352)cm = 110 cm
∴ Area of circle = πr2  =  (227×352×352) cm2
= 962.5 cm2

Question 2:
Circumference of circle = 2πr = 39.6 cm
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 2.1

Question 3:
Area of circle = πr2  =  301.84
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 3.1
Circumference of circle = 2πr = (2×227×9.8) = 61.6 cm

Question 4:
Let radius of circle be r
Then, diameter = 2 r
circumference – Diameter = 16.8
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 4.1
Circumference of circle = 2πr = (2×227×3.92) cm = 24.64 cm

Question 5:
Let the radius of circle be r cm
Then, circumference – radius = 37 cm
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 5.1

Question 6:
Area of square = (side)2 = 484 cm2
⇒ side = 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 9a 6.1

Question 7:
Area of equilateral = 3√4a2 = 121√3
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 7.1
Perimeter of equilateral triangle = 3a = (3 × 22) cm
= 66 cm
Circumference of circle = Perimeter of circle
2πr = 66
⇒ (2×227×r) cm = 66
⇒ r = 10.5 cm
Area of circle = πr2  = (227×10.5×10.5) cm2
= 346.5 cm2

Question 8:
Let the radius of park be r meter
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 8.1

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 9a 9.1
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 9a 10.1
Area of second circle = πR2 = 1386 cm2
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 10.2
Width of ring R – r = (21 – 17.5) cm = 3.5 cm

RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment a6

Question 11:
Area of outer circle = πr21  = (227×23×23) cm2
= 1662.5
Area of inner circle = πr22  = (227×12×12) 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 Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment Q12
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 Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 13.1
Since the track is 14 m wide every where.
Therefore,
Outer radius R = r + 14m = (70 + 14) m = 84 m
Outer circumference = 2πR
=  (2×227×84)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 Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment Q13
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 Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 14.1
Width of the track = (R – r) m
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 14.2
Area the track = π(R2  – r) = π (R+r)(R-r)
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 14.3

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 Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 15.1
Hence, radius of the circular lawn = 50 m

Question 16:
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 16.1
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 Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 16.2
OB = 7 cm, CD = AB = 14 cm
Area of semicircle ∆DBC = RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 16.3
= 72
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 16.4

Question 17:
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 17.1
Diameter of bigger circle = AC = 54 cm
Radius of bigger circle = AC2
=  (542) cm = 27 cm
Diameter AB of smaller circle = AC – BC = 54-10 = 44 cm
Radius of smaller circle = 442 cm = 22 cm
Area of bigger circle = πR2  = (227×27×27) cm2
= 2291. 14 cm2
Area of smaller circle = πr2  = (227×22×22) 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:
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 18.1
PS = 12 cm
PQ = QR = RS = 4 cm, QS = 8 cm
Perimeter = arc PTS + arc PBQ + arc QES
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 18.2
Area of shaded region = (area of the semicircle PBQ) + (area of semicircle PTS)-(Area of semicircle QES)
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 18.3

Question 19:
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 19.1
Length of the inner curved portion
= (400 – 2 × 90) m
= 220 m
Let the radius of each inner curved part be r
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 19.2
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 Chapter 18 Areas of Circle, Sector and Segment 9a 19.3
Length of outer boundary of the track
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 19.4

Question 20:
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 20.1
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 Chapter 18 Areas of Circle, Sector and Segment 9a 20.2

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 Chapter 18 Areas of Circle, Sector and Segment 9a 21.1
Diameter of the circumscribed circle
= Diagonal of the square
= (√2×10) cm
Radius of circumscribed circle = 5√2 cm
(i) Area of inscribed circle = (227×5×5) = 78.57 cm2
(ii) Area of the circumscribed circle RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 21.2

Question 22:
Let the radius of circle be r cm
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 22.1
Then diagonal of square = diameter of circle = 2r cm
Area of the circle = πr2 cm2
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 22.2

Question 23:
Let the radius of circle be r cm
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 23.1
Let each side of the triangle be a cm
And height be h cm
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 23.2

Question 24:
Radius of the wheel = 42 cm
Circumference of wheel = 2πr = (2×227×42) = 264 cm
Distance travelled = 19.8 km = 1980000 cm
Number of revolutions = 1980000264 = 7500

Question 25:
Radius of wheel = 2.1 m
Circumference of wheel = 2πr = (2×227×2.1) = 13.2 m
Distance covered in one revolution = 13.2 m
Distance covered in 75 revolutions = (13.2 × 75) m = 990 m
= 9901000 km
Distance a covered in 1 minute = 99100 km
Distance covered in 1 hour = 99100×60 km = 59.4 km

Question 26:
Distance covered by the wheel in 1 revolution
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 26.1
The circumference of the wheel = 198 cm
Let the diameter of the wheel be d cm
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 26.2
Hence diameter of the wheel is 63 cm

Question 27:
Radius of the wheel = r = 602 = 30 cm
Circumference of the wheel = 2πr = (2×227×30) = 13207 cm
Distance covered in 140 revolution
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 27.1
Distance covered in one hour = 2641000×60 = 15.84 km

Question 28:
Distance covered by a wheel in 1minute
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 28.1
Circumference of a wheel = 2πr = (2×227×70) = 440 cm
Number of revolution in 1 min = 121000440 = 275

Question 29:
Area of quadrant = 14 πr2
Circumference of circle = 2πr = 22
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 29.1

Question 30:
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 30.1
Area which the horse can graze = Area of the quadrant of radius 21 m
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 30.2
Area ungrazed = [(70×52) – 346.5] m2
= 3293.5 m2

Question 31:
Each angle of equilateral triangle is 60°
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 31.1
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 31.2
Area that the horse cannot graze is 36.68 m2

Question 32:
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 32.1
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 Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 32.2
Area of the shaded region = 42 cm2

Question 33:
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 33.1
Area of square = (4 × 4) cm2
= 16 cm2
Area of four quadrant corners
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 33.2
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 Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 34.1
Area of rectangle = (20 × 15) m= 300 m2
Area of 4 corners as quadrants of circle
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 34.2
Area of remaining part = (area of rectangle – area of four quadrants of circles)
= (300 – 38.5) m2 = 261.5 m2

Question 35:
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 35.1
Ungrazed area
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 35.2

Question 36:
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 36.1
Shaded area = (area of quadrant) – (area of DAOD)
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 36.2

Question 37:
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 37.1
Area of flower bed = (area of quadrant OPQ) – (area of the quadrant ORS)
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 37.2

Question 38:
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 38.1
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°)
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 38.2
The area enclosed = 5.76 cm2

Question 39:
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 39.1
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 Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 39.2

Question 40:
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 40.1
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 = RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 40.2
= 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:
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 41.1
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
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 41.2

Question 42:
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 42.1
Let the side of square = a m
Area of square = (a × a) cm  = a2m2
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 42.2
Side of square = 40 m
Therefore, radius of semi circle = 20 m
Area of semi circle = RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 42.3
= 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:
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 43.1
Area of rectangular lawn in the middle
= (50 × 35) = 1750 m2
Radius of semi circles = 352 = 17.5 m
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 43.2
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
= (227×10.5×10.5)
= 804.5 m2
Area of plot which cow can graze when radius is increased to 23 m
= (227×10.5×10.5)
= 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 Chapter 18 Areas of Circle, Sector and Segment 9a 45.1
Given: ABC is right angled at A with AB = 6 cm and AC = 8 cm
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 45.2
Let us join OA, OB and OC
ar(∆AOC) + ar(∆OAB) + ar(∆BOC) = ar(∆ABC)
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 45.3

Question 46:
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 46.1

Question 47:
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 47.1
Area of region ABCDEFA = area of square ABDE + area of semi circle BCD – area of ∆AFE
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 47.2

Question 48:
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 48.1
Side of the square ABCD = 14 cm
Area of square ABCD = 14 × 14 = 196 cm2
Radius of each circle = 144 = 3.5 cm
Area of the circles = 4 × area of one circle
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 48.2
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 9a 49.1
Diameter AC = 2.8 + 1.4
= 4.2 cm
Radius r1 = 4.22 = 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 = 1.42 = 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 × 227 = 13.2 cm

Question 50:
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 50.1
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 9a 50.2
Further in ∆ABC, ∠A = 90°
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 50.3
Adding (1), (2), (3) and subtracting (4)
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 50.4

Question 51:
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 51.1
In ∆PQR, ∠P = 90°, PQ = 24 cm, PR = 7 cm
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 51.2
Area of semicircle
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 51.3
Area of ∆PQR = 12  × 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 9a 52.1
ABCDEF is a hexagon.
∠AOB = 60°, Radius = 35 cm
Area of sector AOB
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 52.2
Area of ∆AOB = RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 52.3
= 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 Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 53.1
In ∆ABC, ∠A = 90°, AB = 6cm, BC = 10 cm
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 53.2
Area of ∆ABC = RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 53.3
Let r be the radius of circle of centre O
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 53.4

Question 54:
Area of equilateral triangle ABC = 49√3 cm2
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 54.1
Let a be its side
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 54.2
Area of sector BDF = RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 54.3
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 54.4
Area of sector BDF = Area of sector CDE = Area of sector AEF
Sum of area of all the sectors
= 773 × 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 Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 55.1
In ∆ABC, ∠B = 90°, AB = 48 cm, BC = 14 cm
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 55.2
Area of semi-circle APC
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 55.3
Area of quadrant BDC with radius 14 cm
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 55.4
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:
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 56.1
Radius of quadrant ABED = 16 cm
Its area = RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 56.2
Area of ∆ABD = (12×16×16) cm2
= 128 cm2
Area of segment DEB
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 56.3
Area of segment DFB = 5127 cm2
Total area of segments = 2 × 5127 cm2 =  10247 cm2
Shaded area = Area of square ABCD – Total area of segments
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 56.4

Question 57:
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 57.1
Radius of circular table cover = 70 cm
Area of the circular cover = RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 57.2
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 57.3
Shaded area = Area of circle – Area of ∆ABC
= (15400 – 6365.1)

Question 58:
Area of the sector of circle = RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 58.1
r = 14 cm and θ = 45°
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 58.2

Question 59:
Length of the arc RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 59.1
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 59.2
Length of arc = ( 17.5 × 227 ) cm = 55 cm
Area of the sector = RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 59.3
= ( 227 × 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 Chapter 18 Areas of Circle, Sector and Segment 9a 60.1
= ( 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 Chapter 18 Areas of Circle, Sector and Segment 9a 61.1
6.5 + 6.5 + arc AB = 31 cm
arc AB = 31 – 13
= 18 cm
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 61.2

Question 62:
Area of the sector of circle = RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 62.1
Radius = 10.5 cm
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 62.2

Question 63:
Length of the pendulum = radius of sector = r cm
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 63.1

Question 64:
Length of arc = RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 64.1
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 64.2

Circumference of circle = 2πr
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 64.3
Area of circle =
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 64.4
= 962.5 cm2

Question 65:
Circumference of circle = 2πr
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 65.1

Question 66:
Angle described by the minute hand in 60 minutes θ = 360°
Angle described by minute hand in 20 minutes
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 66.1
Required area swept by the minute hand in 20 minutes
= Area of the sector(with r = 15 cm and θ = 120°)
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 66.2

Question 67:
θ = 56° and let radius is r cm
Area of sector = RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 67.1
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 67.2
Hence radius = 6cm

Question 68:
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 68.1

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 Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 70.1
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 Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 70.2

Question 71:
Let AB be the chord of circle of centre O and radius = 6 cm such that ∠AOB = 90°
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 71.1
Area of sector = OACBO
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 71.2
Area of ∆AOB = RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 71.4
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 Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 71.5

Question 72:
Let OA = 5√2 cm , OB = 5√2 cm
And AB = 10 cm
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 72.1
Area of ∆AOB = RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 72.2
= 25 cm2
Area of minor segment = (area of sector OACBO) – (area of ∆OAB)
= ( 39.25 – 25 ) cm2 = 14.25 cm2
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 72.3

Question 73:
Area of sector OACBO
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 73.1
Area of minor segment ACBA
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 73.2
Area of major segment BADB
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 73.3

Question 74:
Let AB be the chord of circle of centre O and radius = 30 cm such that AOB = 60°
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 74.1
Area of the sector OACBO
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 74.2
Area of ∆OAB = RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 74.3
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 74.4
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 = 15 x cm
Circumference = RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 75.1
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 75.2

Question 76:
Radius of the front wheel = 40 cm = 25 m
Circumference of the front wheel = RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 76.1
Distance moved by it in 800 revolution
RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 76.2
Circumference of rear wheel = (2π × 1)m = (2π) m
Required number of revolutions = RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment 9a 76.3

All Chapter RS Aggarwal Solutions For Class 10 Maths

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All Subject NCERT Exemplar Problems Solutions For Class10

All Subject NCERT Solutions For Class 10

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