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.
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