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.

Textbook | NCERT |

Class | Class 10 |

Subject | Maths |

Chapter | Chapter 18 |

Chapter Name | Areas of Circle, Sector and Segment |

Exercise | 18 |

Category | RS 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 = πr^{2} = (227×352×352) cm^{2}

= 962.5 cm^{2}

**Question 2:**

Circumference of circle = 2πr = 39.6 cm

**Question 3:**

Area of circle = πr^{2} = 301.84

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

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

**Question 6:**

Area of square = (side)^{2} = 484 cm^{2}

⇒ side = 484−−−√cm = 22 cm

Perimeter of square = 4 × side = 4 × 22 = 88 cm

Circumference of circle = Perimeter of square

**Question 7:**

Area of equilateral = 3√4a2 = 121√3

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 = πr^{2} = (227×10.5×10.5) cm^{2}

= 346.5 cm^{2}

**Question 8:**

Let the radius of park be r meter

**Question 9:**

Let the radii of circles be x cm and (7 – x) cm

Circumference of the circles are 26 cm and 18 cm

**Question 10:**

Area of first circle = πr^{2} = 962.5 cm^{2}

Area of second circle = πR^{2} = 1386 cm^{2}

Width of ring R – r = (21 – 17.5) cm = 3.5 cm

**Question 11:**

Area of outer circle = πr21 = (227×23×23) cm^{2}

= 1662.5

Area of inner circle = πr22 = (227×12×12) cm^{2}

= 452.2 cm^{2}

Area of ring = Outer area – inner area

= (1662.5 – 452.5) cm^{2} = 1210 cm^{2}

**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}] = cm^{2}

Area = 1056 m^{2}

**Question 13:**

Let the inner and outer radii of the circular tacks be r meter and R meter respectively. Then

Inner circumference = 440 meter

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 = πR^{2} – πr^{2}

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

Width of the track = (R – r) m

Area the track = π(R^{2} – r^{2 }) = π (R+r)(R-r)

**Question 15:**

Area of rectangle = (120 × 90)

= 10800 m^{2}

Area of circular lawn = [Area of rectangle – Area of park excluding circular lawn]

= [10800 – 2950] m^{2} = 7850 m^{2}

Area of circular lawn = 7850 m^{2}

⇒ πr^{2} = 7850 m^{2}

Hence, radius of the circular lawn = 50 m

**Question 16:**

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 = πr^{2}

OB = 7 cm, CD = AB = 14 cm

Area of semicircle ∆DBC =

= 72

**Question 17:**

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 = πR^{2} = (227×27×27) cm^{2}

= 2291. 14 cm^{2}

Area of smaller circle = πr^{2} = (227×22×22) cm^{2}

= 1521. 11 cm^{2}

Area of shaded region = area of bigger circle – area of smaller circle

= (2291. 14 – 1521. 11) cm^{2} = 770 cm^{2}

**Question 18:**

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)

**Question 19:**

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

**Question 20:**

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

**Question 21:**

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

= (√2×10) cm

Radius of circumscribed circle = 5√2 cm

(i) Area of inscribed circle = (227×5×5) = 78.57 cm^{2}

(ii) Area of the circumscribed circle

**Question 22:**

Let the radius of circle be r cm

Then diagonal of square = diameter of circle = 2r cm

Area of the circle = πr^{2} cm^{2}

**Question 23:**

Let the radius of circle be r cm

Let each side of the triangle be a cm

And height be h cm

**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

The circumference of the wheel = 198 cm

Let the diameter of the wheel be d cm

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

Distance covered in one hour = 2641000×60 = 15.84 km

**Question 28:**

Distance covered by a wheel in 1minute

Circumference of a wheel = 2πr = (2×227×70) = 440 cm

Number of revolution in 1 min = 121000440 = 275

**Question 29:**

Area of quadrant = 14 πr^{2}

Circumference of circle = 2πr = 22

**Question 30:**

Area which the horse can graze = Area of the quadrant of radius 21 m

Area ungrazed = [(70×52) – 346.5] m^{2}

= 3293.5 m^{2}

**Question 31:**

Each angle of equilateral triangle is 60°

Area that the horse cannot graze is 36.68 m^{2}

**Question 32:**

Each side of the square is 14 cm

Then, area of square = (14 × 14) cm^{2}

= 196 cm^{2}

Thus, radius of each circle 7 cm

Required area = area of square ABCD – 4 (area of sector with r = 7 cm, θ= 90°)

Area of the shaded region = 42 cm^{2}

**Question 33:**

Area of square = (4 × 4) cm^{2}

= 16 cm^{2}

Area of four quadrant corners

Radius of inner circle = 2/2 = 1 cm

Area of circle at the center = πr^{2} = (3.14 × 1 × 1) cm^{2}

= 3.14 cm^{2}

Area of shaded region = [area of square – area of four corner quadrants – area of circle at the centre]

= [16 – 3.14 – 3.14] cm^{2 }= 9.72 cm^{2}

**Question 34:**

Area of rectangle = (20 × 15) m^{2 }= 300 m^{2}

Area of 4 corners as quadrants of circle

Area of remaining part = (area of rectangle – area of four quadrants of circles)

= (300 – 38.5) m^{2} = 261.5 m^{2}

**Question 35:**

Ungrazed area

**Question 36:**

Shaded area = (area of quadrant) – (area of DAOD)

**Question 37:**

Area of flower bed = (area of quadrant OPQ) – (area of the quadrant ORS)

**Question 38:**

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°)

The area enclosed = 5.76 cm^{2}

**Question 39:**

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°]

**Question 40:**

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) cm^{2}

= 100 cm^{2}

Area of each sector =

= 19.625 cm^{2}

Required area = [area of sq. ABCD – 4(area of each sector)]

= (100 – 4 × 19.625) cm^{2}

= (100 – 78.5) = 21.5 cm^{2}

**Question 41:**

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 = 4a^{2} sq.units

**Question 42:**

Let the side of square = a m

Area of square = (a × a) cm = a^{2}m^{2}

Side of square = 40 m

Therefore, radius of semi circle = 20 m

Area of semi circle =

= 628 m^{2}

Area of four semi circles = (4 × 628) m^{2 }= 2512 m^{2}

Cost of turfing the plot of of area 1 m^{2} = Rs. 1.25

Cost of turfing the plot of area 2512 m^{2 }= Rs. (1.25 × 2512)

= Rs. 3140

**Question 43:**

Area of rectangular lawn in the middle

= (50 × 35) = 1750 m^{2}

Radius of semi circles = 352 = 17.5 m

Area of lawn = (area of rectangle + area of semi circle)

= (1750 + 962.5) m^{2} = 2712.5 m^{2}

**Question 44:**

Area of plot which cow can graze when r = 16 m is πr^{2}

= (227×10.5×10.5)

= 804.5 m^{2}

Area of plot which cow can graze when radius is increased to 23 m

= (227×10.5×10.5)

= 1662.57 m^{2}

Additional ground = Area covered by increased rope – old area

= (1662.57 – 804.5)m^{2} = 858 m^{2}

**Question 45:**

Given: ABC is right angled at A with AB = 6 cm and AC = 8 cm

Let us join OA, OB and OC

ar(∆AOC) + ar(∆OAB) + ar(∆BOC) = ar(∆ABC)

**Question 46:**

**Question 47:**

Area of region ABCDEFA = area of square ABDE + area of semi circle BCD – area of ∆AFE

**Question 48:**

Side of the square ABCD = 14 cm

Area of square ABCD = 14 × 14 = 196 cm^{2}

Radius of each circle = 144 = 3.5 cm

Area of the circles = 4 × area of one circle

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

= 196 – 154 = 42 cm^{2}

**Question 49:**

Diameter AC = 2.8 + 1.4

= 4.2 cm

Radius r_{1} = 4.22 = 2.1 cm

Length of semi-circle ADC = πr_{1 }= π × 2.1 = 2.1 π cm

Diameter AB = 2.8 cm

Radius r_{2} = 1.4 cm

Length of semi- circle AEB = πr_{2}= π × 1.4 = 1.4 π cm

Diameter BC = 1.4 cm

Radius r_{3} = 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:**

Area of shaded region = Area of ∆ABC + Area of semi-circle APB + Area of semi circle AQC – Area of semicircle BAC

Further in ∆ABC, ∠A = 90°

Adding (1), (2), (3) and subtracting (4)

**Question 51:**

In ∆PQR, ∠P = 90°, PQ = 24 cm, PR = 7 cm

Area of semicircle

Area of ∆PQR = 12 × 7 × 24 cm^{2} = 84 cm^{2}

Shaded area = 245.31 – 84 = 161.31 cm^{2}

**Question 52:**

ABCDEF is a hexagon.

∠AOB = 60°, Radius = 35 cm

Area of sector AOB

Area of ∆AOB =

= 530.425 cm^{2}

Area of segment APB = (641.083 – 530.425) cm^{2 }= 110.658 cm^{2}

Area of design (shaded area) = 6 × 110.658 cm^{2} = 663.948 cm^{2}

= 663.95 cm^{2}

**Question 53:**

In ∆ABC, ∠A = 90°, AB = 6cm, BC = 10 cm

Area of ∆ABC =

Let r be the radius of circle of centre O

**Question 54:**

Area of equilateral triangle ABC = 49√3 cm^{2}

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

= 773 × 3 cm^{2} = 77 cm^{2}

Shaded area = Area of ∆ABC – sum of area of all sectors

= 49√3 – 77 = (84.77 – 77.00) cm^{2}

= 77.7 cm^{2}

**Question 55:**

In ∆ABC, ∠B = 90°, AB = 48 cm, BC = 14 cm

Area of semi-circle APC

Area of quadrant BDC with radius 14 cm

Shaded area = Area of ∆ABC + Area of semi-circle APC – Area of quadrant BDC

= ( 336+982.14-154 ) cm^{2}

= ( 1318.14-154 ) cm^{2 }= 1164.14 cm^{2}

**Question 56:**

Radius of quadrant ABED = 16 cm

Its area =

Area of ∆ABD = (12×16×16) cm^{2}

= 128 cm^{2}

Area of segment DEB

Area of segment DFB = 5127 cm^{2}

Total area of segments = 2 × 5127 cm^{2} = 10247 cm^{2}

Shaded area = Area of square ABCD – Total area of segments

**Question 57:**

Radius of circular table cover = 70 cm

Area of the circular cover =

Shaded area = Area of circle – Area of ∆ABC

= (15400 – 6365.1)

**Question 58:**

Area of the sector of circle =

r = 14 cm and θ = 45°

**Question 59:**

Length of the arc

Length of arc = ( 17.5 × 227 ) cm = 55 cm

Area of the sector =

= ( 227 × 183.75 ) cm^{2} = 577.5 cm^{2}

**Question 60:**

Length of arc of circle = 44 cm

Radius of circle = 17.5 cm

Area of sector =

= ( 22 × 17.5) cm^{2} = 385 cm^{2}

**Question 61:**

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

**Question 62:**

Area of the sector of circle =

Radius = 10.5 cm

**Question 63:**

Length of the pendulum = radius of sector = r cm

**Question 64:**

Length of arc =

Circumference of circle = 2πr

Area of circle =

= 962.5 cm^{2}

**Question 65:**

Circumference of circle = 2πr

**Question 66:**

Angle described by the minute hand in 60 minutes θ = 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 θ = 120°)

**Question 67:**

θ = 56° and let radius is r cm

Area of sector =

Hence radius = 6cm

**Question 68:**

**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°

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

**Question 71:**

Let AB be the chord of circle of centre O and radius = 6 cm such that ∠AOB = 90°

Area of sector = OACBO

Area of ∆AOB =

Area of minor segment ACBA

= (area of sector OACBO) – (area of ∆OAB)

= (28.29 – 18) cm^{2} = 10.29 cm^{2}

Area of major segment BDAB

**Question 72:**

Let OA = 5√2 cm , OB = 5√2 cm

And AB = 10 cm

Area of ∆AOB =

= 25 cm^{2}

Area of minor segment = (area of sector OACBO) – (area of ∆OAB)

= ( 39.25 – 25 ) cm^{2} = 14.25 cm^{2}

**Question 73:**

Area of sector OACBO

Area of minor segment ACBA

Area of major segment BADB

**Question 74:**

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) cm^{2} = 81.75 cm^{2}

Area of the major segment BADB

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

= [(3.14 × 30 × 30) – 81.75)] cm^{2} = 2744.25 cm^{2}

**Question 75:**

Let the major arc be x cm long

Then, length of the minor arc = 15 x cm

Circumference =

**Question 76:**

Radius of the front wheel = 40 cm = 25 m

Circumference of the front wheel =

Distance moved by it in 800 revolution

Circumference of rear wheel = (2π × 1)m = (2π) m

Required number of revolutions =

**All Chapter RS Aggarwal Solutions For Class 10 Maths**

—————————————————————————–**All Subject NCERT Exemplar Problems Solutions For Class10**

**All Subject NCERT Solutions For Class 10**

*************************************************

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment on the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share rsaggarwalsolutions.in to your friends.