Selina Solutions Concise Mathematics Class 6 Chapter 32 Perimeter and Area of Plane Figures contains vital formula and shortcut methods, described in simple language. The main objective of providing chapter wise solutions is to boost the exam preparation of students. Students, who want to overcome exam fear, are advised to practice these solutions on a daily basis. For enhancing problem solving skills, Selina Solutions Concise Mathematics Class 6 Chapter 32 Perimeter and Area of Plane Figures, PDF links are provided below, with a free download option.
Chapter 32 covers important concepts such as finding the perimeter and area of figures like rectangle, square and triangle. The important formulas and the method of solving problems are given in a descriptive manner, for a better conceptual knowledge among students.
Selina Solutions Concise Mathematics Class 6 Chapter 32 Perimeter and Area of Plane Figures Download PDF
Exercises of Selina Solutions Concise Mathematics Class 6 Chapter 32 Perimeter and Area of Plane Figures
Access Selina Solutions Concise Mathematics Class 6 Chapter 32 Perimeter and Area of Plane Figures
Exercise 32(A)
1. What do you understand by a plane closed figure?
Solution:
Any geometrical plane figure which is bounded by straight or curved lines in a plane is called a plane closed figure
Following figures is a plane closed figure
2. The interior of a figure is called region of the figure. Is this statement true?
Solution:
The interior of the figure along with its boundary is called region of the figure. Hence, the given statement is true.
3. Find the perimeter of each of the following closed figures:
Solution:
The perimeter of the closed figures can be calculated as follows
Perimeter = Sum of all the sides
Perimeter = AB + AC + CD + DG + GH + FH + EF + BE
= 15 + 5 + 10 + 25 + 15 + 5 + 10 + 25
= 110 cm
Therefore, the perimeter of the closed figure = 110 cm
(ii) The perimeter of the closed figures can be calculated as follows
Perimeter = Sum of all the sides
Perimeter = AB + AC + CD + DG + BF + EF + EH + GH
= 20 + 4 + 8 + 20 + 4 + 8 + 20 + 4
= 88 cm
Therefore, the perimeter of the closed figure = 88 cm
4. Find the perimeter of a rectangle whose:
(i) length = 40 cm and breadth = 35 cm
(ii) length = 10 m and breadth = 8 m
(iii) length = 8 m and breadth = 80 cm
(iv) length = 3.6 m and breadth = 2.4 m
Solution:
(i) Given
Length of the rectangle = 40 cm
Breadth of the rectangle = 35 cm
Hence, the perimeter of the rectangle is calculated as follows
Perimeter = 2 (Length + Breadth)
Perimeter = 2 (40 cm + 35 cm)
= 2 × 75 cm
= 150 cm
= 150 / 100 m
We get,
= 1.5 m
Hence, the perimeter of the rectangle = 1.5 m
(ii) Given
Length of the rectangle = 10 m
Breadth of the rectangle = 8 m
Hence, the perimeter of the rectangle is calculated as follows:
Perimeter = 2 (Length + Breadth)
Perimeter = 2 (10 m + 8 m)
= 2 × 18 m
We get,
= 36 m
Hence, the perimeter of the rectangle = 36 m
(iii) Given
Length of the rectangle = 8 m
Breadth of the rectangle = 80 cm
We know that,
100 cm = 1 metre
Hence, the breadth can be converted into metre from centimetre as below
Breadth = 80 cm
= 80 / 100 m
We get,
= 0.8 m
Hence, the perimeter of the rectangle is calculated as follows:
Perimeter = 2 (Length + Breadth)
Perimeter = 2 (8 m + 0.8 m)
= 2 × 8.8 m
We get,
= 17.6 m
Hence, the perimeter of the rectangle = 17.6 m
(iv) Given
Length of the rectangle = 3.6 m
Breadth of the rectangle = 2.4 m
Hence, the perimeter of the rectangle is calculated as follows:
Perimeter = 2 (Length + Breadth)
Perimeter = 2 (3.6 m + 2.4 m)
= 2 × 6 m
We get,
= 12 m
Hence, the perimeter of the rectangle = 12 m
5. If P denotes perimeter of a rectangle, l denotes its length and b denotes its breadth, find:
(i) l, if P = 38 cm and b = 7 cm
(ii) b, if P = 3.2 m and l = 100 cm
(iii) P, if l = 2 m and b = 75 cm
Solution:
(i) Given
P = 38 cm and b = 7 cm
Length of the rectangle (l) can be calculated as below
Length (l) = P / 2 – b
= 38 / 2 – 7 cm
We get,
= 19 cm – 7 cm
= 12 cm
Hence, the length of the rectangle = 12 cm
(ii) Given
P = 3.2 m and l = 100 cm
The breadth of the rectangle (b) can be calculated as follows
Here, the length is in centimetre. Hence, it can be converted into metre from centimetre as below
Length = 100 cm
= 100 / 100 m
Breadth (b) = P / 2 – 1
= 3.2 / 2 – 1 m
= 1.6 m – 1 m
We get,
= 0.6 m
Hence, the breadth (b) of the rectangle = 0.6 m
(iii) Given
L = 2 m and b = 75 cm
The perimeter of the rectangle can be calculated as below:
We know that,
100 cm = 1 m
Here, the breadth is in centimetre. Hence, it can be converted into metre from centimetre as below
Breadth = 75 cm
= 75 / 100 m
We get,
= 0.75 m
Perimeter = 2 (Length + Breadth)
Perimeter = 2 (2 m + 0.75 m)
= 2 × 2.75 m
We get,
= 5.5 m
Hence, the perimeter of the rectangle = 5.5 m
6. Find the perimeter of a square whose each side is 1.6 m.
Solution:
Given
Each side of a square = 1.6 m
Hence, the perimeter of a square can be calculated as follow:
Perimeter = 4 × side
= 4 × 1.6 m
We get,
= 6.4 m
Hence, the perimeter of a square = 6.4 m
7. Find the side of the square whose perimeter is 5 m.
Solution:
Given
The perimeter of the square = 5 m
Hence, the side of the square can be calculated as below
side = Perimeter / 4
= (5 / 4) m
We get,
= 1.25 m
Hence, the side of the square = 1.25 m
8. A square field has each side 70 m whereas a rectangular field has length = 50 m and breadth = 40 m. Which of the two fields has greater perimeter and by how much?
Solution:
Given
The side of square field = 70 m
Length of a rectangular field = 50 m
Breadth of a rectangular field = 40 m
So,
The Perimeter of the square field = 4 × side
= 4 × 70 m
We get,
= 280 m
The Perimeter of the rectangular field = 2 (length + breadth)
= 2 (50 m + 40 m)
= 2 × 90 m
We get,
= 180 m
Therefore, the perimeter of the square field is greater than the perimeter of rectangular field by 280 m – 180 m = 100 m
9. A rectangular field has length = 160 m and breadth = 120 m. Find:
(i) the perimeter of the field
(ii) the length of fence required to enclose the field
(iii) the cost of fencing the field at the rate of 80 per metre
Solution:
(i) Given
Length of the rectangular field = 160 m
Breadth of the rectangular field = 120 m
Hence, the perimeter of the rectangular field can be calculated as below:
Perimeter = 2 (Length + Breadth)
Perimeter = = 2 (160 m + 120 m)
= 2 × 280 m
We get,
= 560 m
Therefore, the perimeter of the field = 560 m
(ii) The length of fence required to enclose the field is equal to the perimeter of the rectangular field. Therefore, the length of the fence is 560 m
(iii) Given
The cost of fencing the field per metre = Rs 80
Hence, total cost of fencing the field can be calculated as follows:
Total Cost = Length of fence × Rate of fence
= 560 m × Rs 80 per metre
We get,
= Rs 44, 800
Therefore, total cost of fencing the field = Rs 44,800
10. Each side of a square plot of land is 55 m. Find the cost of fencing the plot at the rate of Rs 32 per metre.
Solution:
Given
Each side of a square field = 55 m
Hence, the perimeter of square field can be calculated as follows:
Perimeter = 4 × side
= 4 × 55 m
We get,
= 220 m
We know that, the length of fence required to enclose the field is the perimeter of the square field.
Therefore, the length of fence = 220 m
Given
Cost of fence per metre = Rs 32
Hence, total cost of fencing the field can be calculated as follows:
Total cost of fencing the field = Length of fence × Rate of fence
= 220 m × 32
We get,
= Rs 7040
Therefore, total cost of fencing the field = Rs 7040
11. Each side of a square field is 70 cm. How much distance will a boy walk in order to make?
(i) one complete round of this field?
(ii) 8 complete rounds of this field?
Solution:
(i) Given
Each side of a square fields = 70 m
Distance covered to complete one round of the field by the boy is equal to the perimeter of the field
Hence, the perimeter of square field can be calculated as below
Perimeter = 4 × side
= 4 × 70 m
We get,
= 280 m
Therefore, the distance covered by the boy to complete one round of the field is 280 m
(ii) The distance covered to complete 8 rounds of the field can be calculated as follows:
Distance = 8 × Perimeter
= 8 × 280
We get,
= 2240 m
Therefore, the distance covered by the boy to complete 8 rounds of the field is 2240 m
12. A school playground is rectangular in shape with length = 120 m and breadth = 90 m. Some school boys run along the boundary of the play-ground and make 15 complete rounds in 45 minutes. How much distance they run during this period?
Solution:
Given
Length of rectangular playground = 120 m
Breadth of rectangular playground = 90 m
Hence, the perimeter of rectangular playground can be calculated as follows:
Perimeter = 2 (Length + Breadth)
Perimeter = 2 (120 m + 90 m)
= 2 × 210 m
We get,
= 420 m
Hence, distance covered by the boy in one round = 420 m
Therefore, total distance covered by the boy to complete 15 rounds can be calculated as follows:
Distance = 15 × Perimeter
= 15 × 420
We get,
= 6300 m
Therefore, distance covered by the boys to completer 15 rounds is 6300 m
13. Mohit makes 8 full rounds of a rectangular field with length = 120 m and breadth = 75 m. John makes 10 full rounds of a square field with each side 100 in. Find who covers larger distance and by how much?
Solution:
Given
Length of the rectangular field = 120 m
Breadth of the rectangular field = 75 m
Side of square field = 100 m
Distance covered by Mohit in one round is equal to the perimeter of the rectangular field
Hence, the perimeter of the rectangular field can be calculated as follows:
Perimeter = 2 (Length + Breadth)
= 2 × 195 m
We get,
= 390 m
Now, the total distance covered by the Mohit to complete the 8 rounds can be calculated as follows:
Distance = 8 × Perimeter
= 8 × 390
We get,
= 3120 m
Distance covered by John in one round is equal to the perimeter of the square field
Hence, the perimeter of square field can be calculated as follows:
Perimeter = 4 × side
= 4 × 100 m
We get,
= 400 m
Therefore the total distance covered by John to complete 10 rounds can be calculated as follows:
Distance = 10 × Perimeter
= 10 × 400 m
We get,
= 4000 m
Difference between the distance covered by Mohit and John is,
= 4000 m – 3120 m
We get,
= 880 m
Therefore, John covers larger distance by 880 m than the distance covered by Mohit.
14. The length of a rectangle is twice of its breadth. If its perimeter is 60 cm, find its length
Solution:
Given
The perimeter of the rectangle = 60 cm
Length of rectangle is twice of its breadth
Let us consider the breadth of the field = x
Hence, length of the field = 2x
Perimeter = 2 (Length + Breadth)
Perimeter = 2 (2x + x)
60 = 2 × 3x
60 = 6x
x = 60 / 6
We get,
x = 10 cm
So, the breadth of the rectangle = 10 cm
Then the length of the rectangle = 2x
= 2 × 10
= 20 cm
Therefore, the length of the rectangle is 20 cm
15. Find the perimeter of:
(i) an equilateral triangle of side 9.8 cm.
(ii) an isosceles triangle with each equal side = 13 cm and the third side = 10 cm.
(iii) a regular pentagon of side 8.2 cm.
(iv) a regular hexagon of side 6.5 cm.
Solution:
(i) Equilateral triangle side = 9.8 cm
Hence, the perimeter of the equilateral triangle can be calculated as follows:
Perimeter = 3 × side
= 3 × 9.8 cm
We get,
= 29.4 cm
Therefore, the perimeter of the equilateral triangle = 29.4 cm
(ii) Given
Each side of isosceles triangle = 13 cm
Third side of isosceles triangle = 10 cm
Hence, the perimeter of isosceles triangle can be calculated as follows:
Perimeter = Sum of all sides
= 13 + 13 + 10
We get,
= 36 cm
Therefore, the perimeter of isosceles triangle = 36 cm
(iii) Given
The side of regular pentagon = 8.2 cm
Hence, the perimeter of the regular pentagon can be calculated as follows:
Perimeter = 5 × side
= 5 × 8.2 cm
We get,
= 41 cm
Therefore, the perimeter of the regular pentagon = 41 cm
(iv) Given
The side of the regular hexagon = 6.5 cm
Hence, the perimeter of the regular hexagon can be calculated as follows:
Perimeter = 6 × side
= 6 × 6.5 cm
We get,
= 39 cm
Therefore, the perimeter of the regular hexagon = 39 cm
Exercise 32(B)
1. Find the area of a rectangle whose:
(i) Length = 15 cm breadth = 6.4 cm
(ii) Length = 8.5 m breadth = 5 m
(iii) Length = 3.6 m breadth = 90 cm
(iv) Length = 24 cm breadth = 180 mm
Solution:
Given
Length of the rectangle = 15 cm
Breadth of the rectangle = 6.4 cm
Hence, the area of the rectangle is calculated as follow
Area = Length × Breadth
Area = 15 cm × 6.4 cm
We get,
= 96 cm2
Hence, the area of the rectangle = 96 cm2
(ii) Given
Length of the rectangle = 8.5 m
Breadth of the rectangle = 5 m
Hence, the area of the rectangle is calculated as follows:
Area = Length × Breadth
Area = 8.5 m × 5 m
We get,
= 42.5 m2
Hence, the area of the rectangle = 42.5 m2
(iii) Given
Length of the rectangle = 3.6 m
Breadth of the rectangle = 90 cm
We know that,
100 cm = 1 m
Hence, the breadth can be converted into metre from centimetre as follows
Breadth = 90 cm
= 90 / 100 m
We get,
= 0.9 m
Hence, the area of the rectangle is calculated as follows:
Area = Length × Breadth
Area = 3.6 m × 0.9 m
We get,
= 3.24 m2
Hence, the area of the rectangle = 3.24 m2
(iv) Given
Length of the rectangle = 24 cm
Breadth of the rectangle = 180 mm
We know that,
10 mm = 1 cm
Hence, the breadth can be converted into centimetre from millimetre as below
Breadth = 180 mm
= 180 / 10 cm
We get,
= 18 cm
Hence, the area of the rectangle is calculated as follows:
Area = Length × Breadth
Area = 24 cm × 18 cm
We get,
= 432 cm2
Therefore, the area of the rectangle = 432 cm2
2. Find the area of a square, whose each side is:
(i) 7.2 cm
(ii) 4.5 m
(iii) 4.1 cm
Solution:
(i) Given
Each side of square = 7.2 cm
Hence, the area of square can be calculated as below
Area = (side)2
Area = (7.2 cm)2
= 7.2 cm × 7.2 cm
We get,
= 51. 84 cm2
Hence, the area of the square = 51.84 cm2
(ii) Given
Each side of a square = 4.2 m
Hence, the area of square can be calculated as below
Area = (side)2
Area = (4.5 m)2
= 4.5 m × 4.5 m
We get,
= 20.25 m2
Hence, the area of the square = 20.25 m2
(iii) Given
Each side of a square = 4.1 cm
Hence, the area of square can be calculated as below:
Area = (side)2
Area = (4.1 cm)2
= 4.1 cm × 4.1 cm
We get,
= 16.81 cm2
Hence, the area of the square = 16.81 cm2
3. If A denotes area of a rectangle, l represents its length and b represents its breadth, find:
(i) l, if A = 48 cm2 and b = 6 cm
(ii) b, if A = 88 m2 and l = 8 m
Solution:
(i) A = 48 cm2 and b = 6 cm
The length of the rectangle can be calculated as below:
Area = Length × Breadth
A = l × b
l = A / b
Substituting the value of A and b, we get,
l = 48 / 6
l = 8 cm
Therefore, the length of the rectangle = 8 cm
(ii) Given
A = 88 m2 and l = 8 m
The breadth of the rectangle is calculated as below:
Area = Length × Breadth
A = l × b
b = A / l
Substituting the values of A and l, we get,
b = 88 m2 / 8 m
b = 11 m
Therefore, the breadth of the rectangle = 11 m
4. Each side of a square is 3.6 cm; find its
(i) perimeter
(ii) area
Solution:
(i) Given
Each side of a square = 3.6 cm
Therefore, the perimeter of square can be calculated as follows:
Perimeter = 4 × side
= 4 × 3.6 cm
We get,
= 14.4 cm
Hence, the perimeter of the square = 14.4 cm
(ii) Given
Each side of a square = 3.6 cm
Hence, the area of square can be calculated as below:
Area = (side)2
Area = (3.6 cm)2
= 3.6 cm × 3.6 cm
We get,
= 12.96 cm2
Hence, the area of the square = 12.96 cm2
5. The perimeter of a square is 60 m, find:
(i) its each side its area
(ii) its new area obtained on increasing
(iii) each of its sides by 2 m
Solution:
(i) Given
Perimeter of square = 60 m
Hence, the side of square can be calculated as follows:
side = Perimeter / 4
= 60 m / 4
We get,
= 15 m
Therefore, the side of the square = 15 m
(ii) Given
Each side of a square = 15 m
Hence, the area of square can be calculated as below:
Area = (side)2
Area = (15 m)2
= 15 m × 15 m
We get,
= 225 m2
Therefore, the area of the square = 225 m2
(iii) By increasing each side of the square by 2m,
New side = 17 m
Hence, the new area of square can be calculated as below:
Area = (side)2
Area = (17 m)2
= 17 m × 17 m
We get,
= 289 m2
Therefore, the new area of the square = 289 m2
6. Each side of a square is 7 m. If its each side be increased by 3 m, what will be the increase in its area?
Solution:
Given
The side of square = 7 m
Hence, the area of square can be calculated as follows:
Area = (side)2
Area = (7 m)2
= 7 m × 7 m
We get,
= 49 m2
Given each side is increased by 3 m,
So, the new length of side will be = 3 m + 7 m
= 10 m
Hence, the area of square can be calculated as follows:
Area = (side)2
Area = (10 m)2
= 10 m × 10 m
We get,
= 100 m2
Increase in area can be calculated as below:
Increase in area = 100 m2 – 49 m2
= 51 m2
Therefore, the increase in area of square = 51 m2
7.The perimeter of a square field is numerically equal to is area. Find each side of the square.
Solution:
Given
Area of the square is equal to the perimeter of the square
Let us consider each side of square is a
Perimeter of square = Area of square
Hence,
4 × a = a2
a2 / a = 4
We get,
a = 4
Therefore, each side of square = 4
8. A rectangular piece of paper has area = 24 cm2 and length = 5 cm. Find its perimeter.
Solution:
Given
Area of rectangular piece of paper = 24 cm2
Length of the rectangle = 5 cm
We know that,
Area of rectangle = length × breadth
24 cm2 = 5 cm × breadth
breadth = 24 cm2 / 5 cm
breadth = 4.8 cm
The perimeter of a rectangular piece of paper can be calculated as follows:
Perimeter = 2 (Length + Breadth)
Perimeter = 2 (5 cm + 4.8 cm)
= 2 × 9.8 cm
We get,
= 19.6 cm
Therefore, the perimeter of the paper = 19.6 cm
9. Find the perimeter of a rectangle whose area = 2600 m2 and breadth = 50 m.
Solution:
Given
The area of rectangle = 2600 m2
Breadth of the rectangle = 50 m
The length of the rectangle can be calculated as below:
Length = Area / breadth
= 2600 m2 / 50 m
We get,
= 52 m
Hence, the perimeter of rectangle can be calculated as below:
Perimeter = 2 (Length + Breadth)
Perimeter = 2 (52 m + 50 m)
= 2 × 102 m
We get,
= 204 m
Therefore, the perimeter of the rectangle = 204 m
10. What will happen to the area of a rectangle, if its length and breadth both are trippled?
Solution:
Let us consider the length of the rectangle is l and breadth of the rectangle is b
Area = length × breadth
Since,
Length and breadth of the rectangle are trippled
So, new length and breadth will be = 3l × 3b
Hence, the new area of the rectangle is,
Area = 3l × 3b
We get,
Area = 9 × l × b
Therefore, the new area will be 9 times than the original area of the rectangle
11. Length of a rectangle is 30 m and its breadth is 20 m. Find the increase in its area if its length is increased by 10 m and its breadth is doubled.
Solution:
Given
The length of a rectangle = 30 m
The breadth of rectangle = 20 m
The area of the rectangle can be calculated as follows:
Area = length × breadth
= 30 m × 20 m
= 600 m2
Also, given that the length is increased by 10 m and its breadth is doubled
Hence, the new length and breadth is as follows:
New length = 30 m + 10 m
= 40 m
New breadth = 20 m × 2
= 40 m
Therefore, the new area of the rectangle is,
Area = 40 m × 40 m
= 1600 m2
Increase in area = 1600 m2 – 600 m2
Increase in area = 1000 m2
Therefore, the increase in area = 1000 m2
12. The side of a square field is 16 m. What will be increase in its area, if:
(i) each of its sides is increased by 4 m
(ii) each of its sides is doubled
Solution:
Given
Side of the square field = 16 m
Hence, area of square field can be calculated as follows:
Area = (side)2
= (16 m)2
= 16 m × 16 m
We get,
= 256 m2
(i) Since the side is increased by 4 m, then the length of new side will be
= 16 + 4
= 20 m
Hence, the new area of square can be calculated as follows:
Area = (side)2
Area = (20 m)2
= 20 m × 20 m
We get,
= 400 m2
Increase in the area of the square field will be = 400 m2 – 256 m2
= 144 m2
Hence, the increase in the area of the square field = 144 m2
(ii) Given
Each side of its length is doubled
Hence, the new side will be = 16 m × 2
= 32 m
The area of square can be calculated as follows:
Area = (side)2
Area = (32)2
= 32 m × 32 m
We get,
= 1024 m2
Thus, the increase in area of the square field will be = 1024 m2 – 256 m2
= 768 m2
Therefore, the increase in the area of the square field = 768 m2
13. Each rectangular tile is 40 cm long and 30 cm wide. How many tiles will be required to cover the floor of a room with length = 4.8 m and breadth = 2.4 m
Solution:
Given
Length of the rectangular tile = 40 cm
Breadth of the rectangular tile = 30 cm
Length of the floor = 4.8 m
Breadth of the floor = 2.4 m
We know that,
100 cm = 1 m
Hence,
Length = 40 cm
= 40 / 100 m
We get,
= 0.4 m
Breadth = 30 cm
= 30 / 100 cm
We get,
= 0.3 m
The area of rectangular tile is calculated as,
Area = Length × Breadth
Area = 0.4 m × 0.3 m
We get,
= 0.12 m2
The area of the floor is calculated as,
Area = Length × Breadth
Area = 4.8 m × 2.4 m
We get,
= 11.52 m2
Total number of tiles required = (Area of the floor) / (Area of each tile)
= 11.52 m2 / 0.12 m2
We get,
= 96
Therefore, total number of tiles required to cover the floor = 96
14. Each side of a square tile is 60 cm. How many tiles will be required to cover the floor of a hall with length = 50 m and breadth = 36 m.
Solution:
Given
The length of the each side of square = 60 cm
Length of the floor = 50 m
Breadth of the floor = 36 m
We know that,
100 cm = 1 m
side = 60 cm
= 60 / 100 m
We get,
= 0.6 m
Hence, the area of the square tile is calculated as follows:
Area = (side)2
= (0.6 m)2
= 0.6 m × 0.6 m
We get,
= 0.36 m2
The area of the floor is calculated as follows:
Area = Length × Breadth
Area = 50 m × 36 m
We get,
= 1800 m2
Hence, number of tiles required = (Area of the floor) / (Area of each tile)
= 1800 m2 / 0.36 m2
We get,
= 5000
Therefore, the number of tiles required to cover the floor = 5000
15. The perimeter of a square plot = 360 m. Find:
(i) its area
(ii) cost of fencing its boundary at the rate of Rs 40 per metre
(iii) cost of levelling the plot at Rs 60 per square metre
Solution:
Given
The perimeter of square plot = 360 m
Hence,
Side of square plot = Perimeter / 4
= 360 m / 4
We get,
= 90 m
Therefore, the side of square plot = 90 m
(i) Area of square plot can be calculated as follows:
Area = (side)2
= (90 m)2
= 90 m × 90 m
We get,
= 8100 m2
Therefore, the area of the square plot = 8100 m2
(ii) Given
Cost of fencing = Rs 40 per meter
Cost of fencing can be calculated as follows:
Cost = Perimeter × Rate
= 360 m × 40
We get,
= Rs 14400
Therefore, cost of fencing = Rs 14400
(iii) Given
The cost of leveling = Rs 60 per square meter
Cost of leveling can be calculated as follows:
Cost = Area × Rate
= 8100 m2 × 60
We get,
= Rs 486000
Therefore, the cost of leveling the plot = Rs 486000
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