Samacheer Kalvi Class 7 Maths Term 2 Chapter 2 Measurements Solutions
Welcome to NCTB Solutions. Here with this post we are going to help 7th class students by providing Solutions for Samacheer Kalvi Class 7 Maths Term 2 chapter 2 Measurements. Here students can easily find all the solutions for Measurements Exercise 2.2, 2.2, 2.3 and 2.4. Also here our Expert Maths Teacher’s solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get chapter 2 solutions. Here all the solutions are based on Tamil Nadu State Board latest syllabus.
Measurements Exercise 2.1 Solutions :
(1) Find the missing values in the following table for the circles with radius (r), diameter (d) and Circumference (C).
| Sr. No | redius (r) | diameter (d) | Circumference (C) |
| (i) | 15 cm | ||
| (ii) | 1760 cm | ||
| (iii) | 24 cm |
Solution :
(i) radius (r) = 15 cm
diameter (d) = 30 cm
Circumference (C) = 94.28 cm
(ii) radius (r) = 280 cm
diameter (d) = 560 cm
Circumference (C) = 1760 cm
(iii) radius (r) = 12 m
diameter (d) = 24 cm
Circumference (C) = 75.42 m
(2) Diameters of different circles are given below. Find their circumference
(i) d c = 70 m
(ii) d m = 56
(iii) d m = 28 m
Solution :
(i) Given, d = 70 cm
We know that the circumference of circle = πd units.
= 22/7 × 70
= 22 × 10
= 220 cm
(ii) Given, d = 56 m
We know that the circumference of circle = πd units.
= 22/7 × 56
= 22 × 8
= 176 m
(iii) Given, d = 28 mm
We know that the circumference of circle = πd units.
= 22/7 × 28
= 22 × 4
= 88 mm
(3) Find the circumference of the circles whose radii are given below.
(i) 49 cm
(ii) 91 mm
Solution :
(i) Given, r = 49 cm
We know that the circumference of circle = 2πr
= 2 × 22/7 × 49
= 2 × 22 × 7
= 44 × 7
=308 cm
(ii) Given, r = 91 mm
We know that the circumference of circle = 2πr
= 2 × 22/7 × 91
= 44 × 13
= 572 mm
(4) The diameter of a circular well is 4.2 m. What is its circumference?
Solution :
Given, the diameter of a circular well is 4.2 m.
We know that the circumference of circle = πd units.
= 22/7 × 4.2
= 22 × 0.6
= 13.2 m
Therefore, the circumference of well is 13.2 m.
(5) The diameter of the bullock cart wheel is 1.4 m. Find the distance covered by it in 150 rotations?
Solution :
Given, the diameter of the bullock cart wheel is 1.4 m.
Thus, d = 1.4 m
We know that the circumference of circle = πd units.
= 22/7 × 1.4
= 22 × 0.2
= 4.4 m
The distance covered in one rotation = 4.4 m
∴ The distance covered in 150 rotations,
= 150 × 4.4
= 660 m.
(6) A ground is in the form of a circle whose diameter is 350 m. An athlete makes 4 revolutions. Find the distance covered by the athlete.
Solution :
Given, diameter of ground is 350 m.
We know that the circumference of circle = πd units.
= 22/7 × 350
= 22 × 50
= 1100 m
The distance covered in one revolution = 1100 m
∴ The distance covered in 4 revolutions
= 4 × 1100
= 4400 m
(7) A wire of length 1320 cm is made into circular frames of radius 7 cm each. How many frames can be made?
Solution :
Given, radius of circular frame is 7 cm.
We know that the circumference of circle = 2πr units.
= 2 × 22/7 ×7
= 2 × 22
= 44 cm
From wire length 44 cm made frame with 7 cm radius = 1
Thus, from wire length 1320 cm made frame with 7 cm radius
= 1320/44
= 30
Hence, there are 30 circular frames made.
Objective Type Question Solutions :
(9) Formula used to find the circumference of a circle is
(i) 2πr units
(ii) πr2 + 2r units
(iii) πr2 sq. units
(iv) πr3 cu. units
Solution :
Correct Option → (i)
Formula used to find the circumference of a circle is 2πr units.
(10) In the formula, C = 2πr, ‘r’ refers to
(i) circumference
(ii) area
(iii) rotation
(iv) radius
Solution :
Correct Option → (iv)
In the formula, C = 2πr, ‘r’ refers to radius.
(11) If the circumference of a circle is 82p , then the value of ‘r’ is
(i) 41 cm
(ii) 82 cm
(iii) 21 cm
(iv) 20 cm
Solution :
Correct Option → (i)
If the circumference of a circle is 82π, then the value of ‘r’ is 41 cm.
(12) Circumference of a circle is always
(i) three times of its diameter
(ii) more than three times of its diameter
(iii) less than three times of its diameter
(iv) three times of its radius
Solution :
Correct Option → (ii)
Circumference of a circle is always more than three times of its diameter.
Measurements Exercise 2.2 Solutions :
(1) Find the area of the dining table whose diameter is 105 cm.
Solution :
Given, diameter of the dining table = 105 cm
Thus, radius = 105/2 = 52.5 cm
We know, area of the circle = πr²
= 22/7 × 52.5 × 52.5
= 22 × 7.5×52.5
= 8662.5 sq.cm
Therefore, the area of the dining table 8662.5 sq.cm.
(3) A sprinkler placed at the centre of a flower garden sprays water covering a circular area. If the area watered is 1386 cm2, find its radius and diameter.
Solution :
Given, the area watered = 1386 cm²
We know, area of the circle = πr²
1386 = 22/7 × r²
r² = 1386 × 7/22
r² = 441
Taking square root on both sides,
r = 21 cm
Now, d = 2r = 2 × 21
= 42 cm
Hence, radius and diameter is 21 cm and 42 cm respectively.
(4) The circumference of a circular park is 352 m. Find the area of the park.
Solution :
Given, the circumference of a circular park is 352 m.
We know, the circumference of a circle = 2πr
352 = 2 × 22/7 × r
352 × 7/2 × 22 = r
r = 16 × 7/2
r = 8 × 7
r = 56
Thus the radius of circular park is 56 m
Now, area of the circle = πr²
= 22/7 × 56 × 56
= 22 × 8 × 56
= 9856 sq.m
Thus, the area of the park is 9856 sq.m.
(5) In a grass land, a sheep is tethered by a rope of length 4.9 m. Find the maximum area that the sheep can graze.
Solution :
Given, In a grass land, a sheep is tethered by a rope of length 4.9 m.
Thus, r = 4.9 m
We know, area of the circle = πr²
= 22/7 × 4.9 × 4.9
= 22 × 0.7 × 4.9
= 75.46 sq.m
Hence, the maximum area that the sheep can graze is 75.46 sq.m.
(6) Find the length of the rope by which a bull must be tethered in order that it may be able to graze an area of 2464 m2
Solution :
Given, graze an area of 2464 m²
That is, area of circle = 2464 sq.m
We know, area of the circle = πr²
2464 = 22/7 × r²
r² = 2464 × 7/22
r² = 17248/22
r² = 784
Taking square root on both sides,
r = 28 m
Hence, the length of rope is 28 m.
(7) Lalitha wants to buy a round carpet of radius is 63 cm for her hall. Find the area that will be covered by the carpet.
Solution :
Given, r = 63.
We know, area of the circle = πr²
= 22/7 × 63 × 63
= 22 × 9 × 63
= 12,474 sq.cm
Hence, carpet covered by the carpet is 12,474 sq.cm.
(8) Thenmozhi wants to level her circular flower garden whose diameter is 49 m at the rate of ₹150 per m2. Find the cost of levelling.
Solution :
First, r = 49/2 = 24.5 m
Now, area of circle = πr²
= 22/7 × 24.5 × 24.5
= 22 × 3.5 × 24.5
= 1886.5 sq.m
Since, for 1 sq.m = ₹150
Thus, for 1886.5 sq.m
= ₹(150 × 1886.5)
= ₹2,82,975
Hence, the cost of levelling on circular flower garden is ₹2,82,975
(9) The floor of the circular swimming pool whose radius is 7 m has to be cemented at the rate of ₹18 per m2. Find the total cost of cementing the floor.
Solution :
Given, floor of the circular swimming pool whose radius is 7 m
Thus, r = 7m
Now, area of circle = πr²
= 22/7 × 7 × 7
= 22 × 7
= 154 sq.m
Since, cost of cemented for 1 sq.m = ₹18
Therefore, cost of cemented for 154 sq.m
= ₹(18×154)
= ₹2772
Objective Type Question Solutions :
(10) The formula used to find the area of the circle is __ sq. units
(i) 4πr2
(ii) πr2
(iii) 2πr2
(iv) πr2 + 2r
Solution :
Correct Option → (ii)
The formula used to find the area of the circle is πr² sq. units.
(11) The ratio of the area of a circle to the area of its semicircle is
(i) 2 : 1
(ii) 1 : 2
(iii) 4 : 1
(iv) 1 : 4
Solution :
Correct Option → (i)
The ratio of the area of a circle to the area of its semicircle is 2 : 1.
(12) Area of a circle of radius n units is
(i) 2πrp sq. units
(ii) πm2 sq. units
(iii) πr2 sq. units
(iv) πn2 sq. units
Solution :
Correct Option → (iv)
Area of a circle of radius n units is πn² sq. units.
Measurements Exercise 2.3 Solutions :
(1) Find the area of a circular pathway whose outer radius is 32 cm and inner radius is 18 cm.
Solution :
The radius of the outer circle, R = 32 cm
The radius of the inner circle, r = 18 cm
The area of the circular path = π (R² – r²)
= 22/7 × (32 × 32 – 18 × 18)
= 22/7 × (1024 – 324)
= 22/7 × 700
= 22 × 100
= 2200 sq.cm
Therefore, the area of a circular pathway is 2200 sq.cm
(2) There is a circular lawn of radius 28 m. A path of 7 m width is laid around the lawn. What will be the area of the path?
Solution :
The radius of the inner circle, r = 28 m
The radius of the outer circle, R = 28 + 7 = 35 m
The area of the circular path = π (R² – r²)
= 22/7 × (35 × 35 – 28×28)
= 22/7 × (1225 – 784)
= 22/7 × 441
= 22 × 63
= 1386 sq.m
Therefore, the area of the path is 1386 sq.m.
(3) A circular carpet whose radius is 106 cm is laid on a circular hall of radius 120 cm. Find the area of the hall uncovered by the carpet.
Solution :
Given, r = 106 cm and R = 120 cm
Therefore, area uncovered by carpet = area of circular hall – area of carpet
= π (R² – r²)
= 22/7 × ( 120×120 – 106×106)
= 22/7 × ( 14400 – 11236)
= 22/7 × 3164
= 22 × 452
= 9944 sq.cm
Therefore, the area of the hall uncovered by the carpet is 9944 sq.cm
(5) The figure shown is the aerial view of the pathway. Find the area of the pathway.
Solution :
From fig, length of inner rectangle length l = 70 m
Breadth of inner rectangle b = 40 m
Thus, area of inner rectangle
= l × b
= 70 × 40
= 2800 m
Now, length of outer rectangle L = 80 m
Breadth of outer rectangle B = 50 m
Thus, area of outer rectangle
= 80 × 50
= 4000 m
Therefore, the area of the pathway = area of outer rectangle – area of inner rectangle
= 4000 – 2200
= 1200 m
Therefore, the area of the pathway is 1200 m.
(6) A rectangular garden has dimensions 11 m m × 8 . A path of 2 m wide has to be constructed along its sides. Find the area of the path.
Solution :
Given, rectangular garden has dimensions 11 m × 8 m.
Area of garden
= 11 × 8
= 88 m
Since, a path of 2 m wide has to be constructed along its sides.
Thus, dimension of construction side
= (11 – 2 × 2 ) × (8 – 2 × 2)
= 7 × 4
Therefore, area
= (7 × 4)
= 28 m
So, area of path
= 88 – 28
= 60 m
Therefore, area of the path is 60 m.
(7) A picture is painted on a ceiling of a marriage hall whose length and breadth are 18 m and 7 m respectively. There is a border of 10 cm along each of its sides. Find the area of the border.
Solution :
Given, L = 18 m, B = 7 m
Area of hall = L × B
= 18 × 7
= 126 m
Since, there is a border of 10 cm along each of its sides.
Thus, l = 18 – 2 × 0.1
= 18 – 0.2 = 17.8 m
b = 7 – 2 × 0.1
= 7 – 0.2
= 6.8 m
Now, area including border
= l × b = 17.8 × 6.8
= 121.04
Therefore, the area of border will be,
= 126 – 121.04
= 4.96 sq.m
(8) A canal of width 1 m is constructed all along inside the field which is 24 m long and 15 m wide. Find (i) the area of the canal (ii) the cost of constructing the canal at the rate of ₹12 per sq.m.
Solution :
(i) Find the area of the canal
Here, L = 24 m, B = 15 m
Area field = 24 × 15 = 360 sq.m
Now, l = 24 – 2 × 1 = 22 m ,
b = 15 – 2 × 1
= 15 – 2
= 13 m
Area of canal including field
= l × b
= 22 × 13
= 286 sq.m
Therefore, area of canal
= 360 – 286
= 74 sq.m
(ii) Find the cost of constructing the canal at the rate of ₹12 per sq.m.
Since cost of construction for 1 sq.m = ₹12
Therefore, for 74 sq.m
= ₹(12 × 74)
= ₹888
Therefore, the cost of constructing the canal is ₹888.
Objective Type Question Solutions :
(9) The formula to find the area of the circular path is
(i) π (R² – r²) sq. units
(ii) πr² sq. units
(iii) 2πr² sq. units
(iv) πr²+2r sq. units
Solution :
Correct Option → (i)
The formula to find the area of the circular path is π (R² – r²) sq. units.
(10) The formula used to find the area of the rectangular path is
(i) π (R² -r²) sq. units
(ii) (L × B) – (l × b) sq. units
(iii) LB sq. units
(iv) lb sq. units
Solution :
Correct Option → (ii)
The formula used to find the area of the rectangular path is (L × B) – (l × b) sq. units.
(11) The formula to find the width of the circular path is
(i) ( L− l ) units
(ii) ( B −b ) units
(iii) ( R − r ) units
(iv) ( r − R ) units
Solution :
Correct Option → (iii)
The formula to find the width of the circular path is (R − r) units.
Measurements Exercise 2.4 Solutions :
(1) A wheel of a car covers a distance of 3520 cm in 20 rotations. Find the radius of the wheel?
Solution :
Given, a wheel of a car covers a distance of 3520 cm in 20 rotations.
Thus, for 20 rotation = 3520 cm
So, for 1 rotation = 3520/20 = 176 cm
That is circumference of wheel = 176 cm
We know, circumference of circle = 2πr
176 = 2 × 22/7 × r
r = 176 × 7/44
r = 4 × 7
r = 28 cm
Therefore, the radius of wheel will be 28 cm.
(2) The cost of fencing a circular race course at the rate of ₹ 8 per metre is ₹ 2112. Find the diameter of the race course.
Solution :
Since, for 1m = ₹8
Thus, for xm = ₹2112
Therefore,
x = 2112/8
= 264 m
That is, circumference of race course is 264 m
We know, circumference of circle = 2πr
264 = 2 × 22/7 × r
r = 264 × 7/44
r = 6 × 7
r = 42 m
Now, d = 2r = 2 × 42
= 84 m
Hence, diameter of race course is 84 m.
(4) The cost of decorating the circumference of a circular lawn of a house at the rate of ₹ 55 per metre is ₹ 16940. What is the radius of the lawn?
Solution :
Let cost of x m is ₹ 16940.
Thus, cost of 1m = ₹ 55
So, cost of xm = ₹ 16940
Therefore, x = 16940/55 = 308 m
That is circumference of circular lawn is 308 m
Now, circumference of circular lawn = 2πr
308 = 2 × 22/7 × r
r = 308 × 7/2 × 22
r = 14 × 7/2
r = 7 × 7
r = 49
Therefore, the radius of circular lawn will be 49 m.
Challenge Problem Solutions :
(6) A circular path has to be constructed around a circular lawn. If the outer and inner circumferences of the path are 88 cm and 44 cm respectively, find the width and area of the path.
Solution :
Outer circumference = 2πR
88 = 2 × 22/7 × R
R = 88 × 7/44
R = 2 × 7
R = 14 cm
Also, inner circumference = 2πr
44 = 2 × 22/7 × r
r = 44 × 7/44
r = 7 cm
Width of path = 14 – 7 = 7 cm
Now, area of path = π (R² – r²)
= 22/7 (14 × 14 – 7 × 7)
= 22/7 × (196 – 49)
= 22/7 × (147)
= 22 × 21
= 462 sq.cm
Hence, width and area of path is 7 cm and 462 sq.cm respectively.
(7) A cow is tethered with a rope of length 35 m at the centre of the rectangular field of length 76 m and breadth 60 m. Find the area of the land that the cow cannot graze?
Solution :
Given, length of rectangle = 76 m and breadth of rectangle = 60 m
Area of rectangle = l × b
= 76 × 60
= 4560 sq.m
Since, rope of length = r = 35 m
Now, area of circle = π r²
= 22/7 × 35 × 35
= 22 × 5 × 35
= 3850 sq.m
Thus, area of cow cannot graze
= 4560 – 3850
= 710 sq.m
Hence, the area of the land that the cow cannot graze is 710 sq.m.
(10) A goat is tethered with a rope of length 45 m at the centre of the circular grass land whose radius is 52 m. Find the area of the grass land that the goat cannot graze.
Solution :
Given, R = 52 m and r = 45 m
Thus, area goat cannot graze = π (R² – r²)
= 22/7 × (52 × 52 – 45 × 45)
= 22/7 × (2704 – 2025)
= 22/7 × 679
= 22 × 97
= 2134 sq.m
Hence, the area of the grass land that the goat cannot graze is 2134 sq.m.
(11) A strip of 4 cm wide is cut and removed from all the sides of the rectangular cardboard with dimensions 30 cm × 20 cm . Find the area of the removed portion and area of the remaining cardboard.
Solution :
Area of cardboard,
= 30 cm × 20 cm
= 600 sq.cm
Since, a strip of 4 cm wide cut remove
Thus, dimension of removed portion
= 30 – 8 × 20 – 8
= 22 cm × 12 cm
Therefore, area of removed portion
= 22 cm × 12 cm
= 264 sq.cm
Now, area of the remaining cardboard = area of cardboard – area of removed portion
= 600 – 264
= 334 sq.cm
Therefore, the area of removed portion and area of remaining cardboard will be 264 sq.cm and 334 sq.cm resp.
(12) A rectangular field is of dimension 20 m × 15 m . Two paths run parallel to the sides of the rectangle through the centre of the field. The width of the longer path is 2 m and that of the shorter path is 1 m. Find (i) the area of the paths (ii) the area of the remaining portion of the field (iii) the cost of constructing the roads at the rate of ₹ 10 per sq.m.
Solution :
As per the given question,
Each side of the square field = 20 m
∴ Perimeter of the square field,
= (20 × 4)
= 80 m
∴ Length of the rectangular field,
= 80/2 × 3/4
= 30 m
∴ Breadth of the rectangular field,
= 80/2 × 1/4
= 10 m.
Next Chapter Solution :
👉 Algebra
