If the length of each side of an equilateral triangle is increased by 2 unit

A) \(\sqrt{3}\) units

B) \(3\) units

C) \(3\sqrt{3}\) units

D) \(3\sqrt{2}\) units

Correct Answer:

Description for Correct answer:
Let each side of the triangle be a units

\( \Large \Rightarrow \frac{\sqrt{3}}{4}((a+2)^{2}-a^{2})=3+\sqrt{3} \)

\( \Large \frac{1}{4}(a^{2}+4+4a-a^{2})=1+\sqrt{3} \)

\( \Large \frac{1}{4}(4+4a)=1+\sqrt{3} \)

\( \Large 1+a=1+\sqrt{3} \)

\( \Large a=\sqrt{3} units \)

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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 = \(\left( 2\times \frac { 22 }{ 7 } \times \frac { 35 }{ 2 }  \right) cm  \) = 110 cm
∴ Area of circle = πr2  =  \(\left( \frac { 22 }{ 7 } \times \frac { 35 }{ 2 } \times \frac { 35 }{ 2 }  \right)   \) cm2
= 962.5 cm2

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 = \(\left( 2\times \frac { 22 }{ 7 } \times 9.8 \right)      \) = 61.6 cm

Question 4:
Let radius of circle be r
Then, diameter = 2 r
circumference – Diameter = 16.8

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

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
⇒ \(\left( 2\times \frac { 22 }{ 7 } \times r \right)      \) 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

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\(r_1^2  \)  = \(\left( \frac { 22 }{ 7 } \times 23\times 23 \right)  \) cm2
= 1662.5
Area of inner circle = π\(r_2^2  \)  = \(\left( \frac { 22 }{ 7 } \times 12\times 12 \right)  \) 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

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
=  \(\left( 2\times \frac { 22 }{ 7 } \times 84 \right)m      \)  = 528 m
Rate of fencing = Rs. 5 per meter
Total cost of fencing = Rs. (528 × 5) = Rs. 2640
Area of circular ring = πR2  – πr2

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  = \(\left( \frac { 22 }{ 7 } \times 27\times 27 \right)  \) 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:

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 = \(\left( \frac { 22 }{ 7 } \times 5\times 5 \right)  \) = 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 = \(\left( 2\times \frac { 22 }{ 7 } \times 42 \right)      \) = 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

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 = \(\left( 2\times \frac { 22 }{ 7 } \times 30 \right)      \) = \(\frac { 1320 }{ 7 }  \) 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 = \(\left( 2\times \frac { 22 }{ 7 } \times 70 \right)      \) = 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

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
= \(\left( \frac { 22 }{ 7 } \times 10.5\times 10.5 \right)  \)
= 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:

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 = \(\left( \frac { 1 }{ 2 } \times 16\times 16 \right)  \) 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.

If you have any doubts, please comment below. A Plus Topper try to provide online math tutoring for you.

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