**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 =

Circumference of circle = 2πr = = 110 cm

∴ Area of circle = πr^{2} = cm^{2}

= 962.5 cm^{2}

**More Resources**

**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 = πr^{2} = 301.84

Circumference of circle = 2πr = = 61.6 cm

**Question 4:**

Let radius of circle be r

Then, diameter = 2 r

circumference – Diameter = 16.8

Circumference of circle = 2πr = 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 = = 22 cm

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

Circumference of circle = Perimeter of square

**Question 7:**

Area of equilateral = = 121√3

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

= 66 cm

Circumference of circle = Perimeter of circle

2πr = 66

⇒ cm = 66

⇒ r = 10.5 cm

Area of circle = πr^{2} = 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 = π = cm^{2}

= 1662.5

Area of inner circle = π = 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

= = 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 =

= cm = 27 cm

Diameter AB of smaller circle = AC – BC = 54-10 = 44 cm

Radius of smaller circle = cm = 22 cm

Area of bigger circle = πR^{2} = cm^{2}

= 2291. 14 cm^{2}

Area of smaller circle = πr^{2} = 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 = = 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 = = 264 cm

Distance travelled = 19.8 km = 1980000 cm

Number of revolutions = = 7500

**Question 25:**

Radius of wheel = 2.1 m

Circumference of wheel = 2πr = = 13.2 m

Distance covered in one revolution = 13.2 m

Distance covered in 75 revolutions = (13.2 × 75) m = 990 m

= km

Distance a covered in 1 minute = km

Distance covered in 1 hour = 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 = = 30 cm

Circumference of the wheel = 2πr = = cm

Distance covered in 140 revolution

Distance covered in one hour = = 15.84 km

**Question 28:**

Distance covered by a wheel in 1minute

Circumference of a wheel = 2πr = = 440 cm

Number of revolution in 1 min = = 275

**Question 29:**

Area of quadrant = 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 = = 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}

=

= 804.5 m^{2}

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

=

= 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 = = 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} = = 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} = = 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 × = 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 = × 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

= × 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 = cm^{2}

= 128 cm^{2}

Area of segment DEB

Area of segment DFB = cm^{2}

Total area of segments = 2 × cm^{2} = 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 × ) cm = 55 cm

Area of the sector =

= ( × 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 = x cm

Circumference =

**Question 76:**

Radius of the front wheel = 40 cm = 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 =

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 **

- Real Numbers
- Polynomials
- Linear equations in two variables
- Triangles
- Trigonometric Ratios
- T-Ratios of Some Particular Angles
- Trigonometric Identities
- Trigonometric Ratios of Complementary Angles
- Mean, Median, Mode of Grouped Data, of Grouped Data Ex.9A Ex.9B Ex.9C and 9D Ex.9E
- Quadratic Equations Ex.10A Ex.10B Ex.10C Ex.10D
- Arithmetic Progressions
- Circles
- Constructions
- Height and Distance
- Probability
- Coordinate Geometry
- Perimeter and Areas of Plane Figures
- Areas of Circle, Sector and Segment
- Volume and Surface Areas of Solids