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

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

These Solutions are 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( 2times 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

More Resources

• RS Aggarwal Solutions Class 10
• RD Sharma Class 10 Solutions
• NCERT Solutions

Question 2:

Circumference of circle = 2πr = 39.6 cm

Read More:

• Parts of a Circle
• Perimeter of A Circle
• Common Chord of Two Intersecting Circles
• Construction of a Circle
• The Area of A Circle
• Properties of Circles
• Sector of A Circle
• The Area of A Segment of A Circle
• The Area of A Sector of A Circle

Question 3:

Area of circle = πr²  =  301.84



Circumference of circle = 2πr = (left( 2times 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



Circumference of circle = 2πr = (left( 2times 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

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

Question 7:

Area of equilateral = (frac { sqrt { 3 }  }{ 4 } { a }^{ 2 }   ) = 121√3



Perimeter of equilateral triangle = 3a = (3 × 22) cm

= 66 cm

Circumference of circle = Perimeter of circle

2πr = 66

⇒ (left( 2times frac { 22 }{ 7 } times r right)      ) cm = 66

⇒ r = 10.5 cm

Area of circle = πr²  = (left( frac { 22 }{ 7 } times 10.5times 10.5 right)  ) cm²

= 346.5 cm²

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 = πr² = 962.5 cm²



Area of second circle = πR² = 1386 cm²



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

Question 11:

Area of outer circle = π(r_1^2  )  = (left( frac { 22 }{ 7 } times 23times 23 right)  ) cm²

= 1662.5

Area of inner circle = π(r_2^2  )  = (left( frac { 22 }{ 7 } times 12times 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



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



Since the track is 14 m wide every where.

Therefore,

Outer radius R = r + 14m = (70 + 14) m = 84 m

Outer circumference = 2πR

=  (left( 2times 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



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(R²  – r² ) = π (R+r)(R-r)

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



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 = πr²



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 = (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 27times 27 right)  ) cm²

= 2291. 14 cm²

Area of smaller circle = πr²  = (left( frac { 22 }{ 7 } times 22times 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:



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 = (left( frac { 22 }{ 7 } times 5times 5 right)  ) = 78.57 cm²

(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 = πr² cm²

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 = (left( 2times 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( 2times 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



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 = (frac { 60 }{ 2 }  ) = 30 cm

Circumference of the wheel = 2πr = (left( 2times frac { 22 }{ 7 } times 30 right)      ) = (frac { 1320 }{ 7 }  ) cm

Distance covered in 140 revolution



Distance covered in one hour = (frac { 264 }{ 1000 } times 60) = 15.84 km

Question 28:

Distance covered by a wheel in 1minute



Circumference of a wheel = 2πr = (left( 2times 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

Question 30:



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



Area ungrazed = [(70×52) – 346.5] m²

= 3293.5 m²

Question 31:

Each angle of equilateral triangle is 60°





Area that the horse cannot graze is 36.68 m²

Question 32:



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°)



Area of the shaded region = 42 cm²

Question 33:



Area of square = (4 × 4) cm²

= 16 cm²

Area of four quadrant corners



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] cm² = 9.72 cm²

Question 34:



Area of rectangle = (20 × 15) m² = 300 m²

Area of 4 corners as quadrants of circle



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

= (300 – 38.5) m² = 261.5 m²

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 cm²

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) cm²

= 100 cm²

Area of each sector =

= 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:



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

Question 42:



Let the side of square = a m

Area of square = (a × a) cm  = a²m²



Side of square = 40 m

Therefore, radius of semi circle = 20 m

Area of semi circle =

= 628 m²

Area of four semi circles = (4 × 628) m² = 2512 m²

Cost of turfing the plot of of area 1 m² = Rs. 1.25

Cost of turfing the plot of area 2512 m² = Rs. (1.25 × 2512)

= Rs. 3140

Question 43:



Area of rectangular lawn in the middle

= (50 × 35) = 1750 m²

Radius of semi circles = (frac { 35 }{ 2 }  ) = 17.5 m



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 10.5times 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 10.5times 10.5 right)  )

= 1662.57 m²

Additional ground = Area covered by increased rope – old area

= (1662.57 – 804.5)m² = 858 m²

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 cm²

Radius of each circle = (frac { 14 }{ 4 }  ) = 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 cm²

Question 49:



Diameter AC = 2.8 + 1.4

= 4.2 cm

Radius r₁ = (frac { 4.2 }{ 2 }  ) = 2.1 cm

Length of semi-circle ADC = πr₁ =  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:



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 = (frac { 1 }{ 2 }  )  × 7 × 24 cm² = 84 cm²

Shaded area = 245.31 – 84 = 161.31 cm²

Question 52:



ABCDEF is a hexagon.

∠AOB = 60°, Radius = 35 cm

Area of sector AOB



Area of ∆AOB =

= 530.425 cm²

Area of segment APB = (641.083 – 530.425) cm² = 110.658 cm²

Area of design (shaded area) = 6 × 110.658 cm² = 663.948 cm²

= 663.95 cm²

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 cm²



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

= (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:



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 ) cm²

= ( 1318.14-154 ) cm² = 1164.14 cm²

Question 56:



Radius of quadrant ABED = 16 cm

Its area =

Area of ∆ABD = (left( frac { 1 }{ 2 } times 16times 16 right)  ) cm²

= 128 cm²

Area of segment DEB



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

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 × (frac { 22 }{ 7 }  ) ) cm = 55 cm

Area of the sector =

= ( (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 =

= ( 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



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 cm²

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) cm² = 10.29 cm²

Area of major segment BDAB

Question 72:

Let OA = 5√2 cm , OB = 5√2 cm

And AB = 10 cm



Area of ∆AOB =

= 25 cm²

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

= ( 39.25 – 25 ) cm² = 14.25 cm²

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) 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 =

Question 76:

Radius of the front wheel = 40 cm = (frac { 2 }{ 5 }  ) 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 =

Hope given RS Aggarwal Solutions Class 10 Chapter 18 Areas of Circle, Sector and Segment are helpful to complete your math homework.

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