NCERT Solutions for Chapter 10 Mensuration Class 6 Maths
Mensuration Questions and Answers
Chapter Name  Mensuration NCERT Solutions 
Class  CBSE Class 6 
Textbook Name  Maths 
Related Readings 

Exercise 10.1
Question 1: Find the perimeter of each of the following figures:
Solution
(a) Perimeter = Sum of all the sides
= 4 cm + 2 cm + 1 cm + 5 cm = 12 cm
(b) Perimeter = Sum of all the sides
= 23 cm + 35 cm + 40 cm + 35 cm = 133 cm
(c) Perimeter = Sum of all the sides
= 15 cm + 15 cm + 15 cm + 15 cm = 60 cm
(d) Perimeter = Sum of all the sides
= 4 cm + 4 cm + 4 cm + 4 cm + 4 cm = 20 cm
(e) Perimeter = Sum of all the sides
= 1 cm + 4 cm + 0.5 cm + 2.5 cm + 2.5 cm + 0.5 cm + 4 cm = 15 cm
(f) Perimeter = Sum of all the sides
= 4 cm + 1 cm + 3 cm + 2 cm + 3 cm + 4 cm + 1 cm + 3 cm + 2 cm + 3 cm + 4 cm + 1 cm + 3 cm + 2 cm + 3 cm + 4 cm + 1 cm + 3 cm + 2 cm + 3 cm = 52 cm
Question 2: The lid of a rectangular box of sides 40 cm by 10 cm is sealed all round with tape. What is the length of the tape required?
Solution
Total length of tape required = Perimeter of rectangle
= 2 (length + breadth)
= 2 (40 + 10)
= 2 × 50
= 100 cm
= 1 m
Thus, the total length of tape required is 100 cm or 1 m.
Question 3: A tabletop measures 2 m 25 cm by 1 m 50 cm. What is the perimeter of the tabletop?
Solution
Length of table top = 2 m 25 cm = 2.25 m
Breadth of table top = 1 m 50 cm = 1.50 m
Perimeter of table top = 2 × (length + breadth)
= 2 × (2.25 + 1.50)
= 2 × 3.75
= 7.50 m
Thus, the perimeter of table top is 7.5 m.
Question 4: What is the length of the wooden strip required to frame a photograph of length and breadth 32 cm and 21 cm respectively?
Solution
Length of wooden strip = Perimeter of photograph
Perimeter of photograph = 2 × (length + breadth)
= 2 (32 + 21)
= 2 × 53 cm
= 106 cm
Thus, the length of the wooden strip required is equal to 106 cm.
Question 5: A rectangular piece of land measures 0.7 km by 0.5 km. Each side is to be fenced with 4 rows of wires. What is the length of the wire needed?
Solution
Since the 4 rows of wires are needed.
Therefore the total length of wires is equal to 4 times the perimeter of rectangle.
Perimeter of field = 2 × (length + breadth)
= 2 × (0.7 + 0.5)
= 2 × 1.2
= 2.4 km
= 2.4 × 1000 m
= 2400 m
Thus, the length of wire = 4 × 2400 = 9600 m = 9.6 km
Question 6: Find the perimeter of each of the following shapes:
(a) A triangle of sides 3 cm, 4 cm and 5 cm.
(b) An equilateral triangle of side 9 cm.
(c) An isosceles triangle with equal sides 8 cm each and third side 6 cm
Solution
(a) Perimeter of ABC
= AB + BC + CA
= 3 cm + 5 cm + 4 cm
= 12 cm
(b) Perimeter of equilateral ABC = 3 × side
= 3 × 9 cm
= 27 cm
(c) Perimeter of ABC
= AB + BC + CA
= 8 cm + 6 cm + 8 cm
= 22 cm
Question 7: Find the perimeter of a triangle with sides measuring 10 cm, 14 cm and 15 cm.
Solution
Perimeter of triangle = Sum of all three sides
= 10 cm + 14 cm + 15 cm
= 39 cm
Thus, the perimeter of triangle is 39 cm.
Question 8: Find the perimeter of a regular hexagon with each side measuring 8 cm.
Solution
Perimeter of Hexagon = 6 × length of one side
= 6 × 8 m
= 48 m
Thus, the perimeter of hexagon is 48 m.
Question 9: Find the side of the square whose perimeter is 20 m.
Solution
Perimeter of square = 4 × side
= 20 = 4 × side
= Side = 20/4 = 5 cm
Thus, the side of square is 5 cm.
Question 10: The perimeter of a regular pentagon is 100 cm. How long is its each side?
Solution
Perimeter of regular pentagon = 100 cm
= 5 × side = 100 cm
Side = 100/5 = 20 cm
Thus, the side of regular pentagon is 20 cm.
Question 11: A piece of string is 30 cm long. What will be the length of each side if the string is used to form:
(a) a square
(b) an equilateral triangle
(c) a regular hexagon?
Solution
Length of string = Perimeter of each figure
(a) Perimeter of square = 30 cm
= 4 × side = 30 cm
= Side = 30/4 = 7.5 cm
Thus, the length of each side of square is 7.5 cm.
(b) Perimeter of equilateral triangle = 30 cm
= 3 × side = 30 cm
= Side = 30/3 = 10 cm
Thus, the length of each side of equilateral triangle is 10 cm.
(c) Perimeter of hexagon = 30 cm
= 6 × side = 30 cm
= Side = 30/6 = 5 cm
Thus, the side of each side of hexagon is 5 cm.
Question 12: Two sides of a triangle are 12 cm and 14 cm. The perimeter of the triangle is 36 cm.
What is the third side?
Solution
Let the length of third side be x cm.
Length of other two side are 12 cm and 14 cm.
Now, Perimeter of triangle = 36 cm
= 12+14+x = 36
= 26+x = 36
= x = 3626
= x = 10 cm
Thus, the length of third side is 10 cm.
Question 13: Find the cost of fencing a square park of side 250 m at the rate of ₹20 per meter.
Solution
Side of square = 250 m
Perimeter of square = 4 × side
= 4 × 250
= 1000 m
Since, cost of fencing of per meter = ₹ 20
Therefore, the cost of fencing of 1000 meters = 20 × 1000 = ₹20,000
Question 14: Find the cost of fencing a rectangular park of length 175 m and breadth 125 m at the rate of ₹12 per meter.
Solution
Length of rectangular park = 175 m
Breadth of rectangular park = 125 m
Perimeter of park = 2 × (length + breadth)
= 2 × (175 + 125)
= 2 × 300 = 600 m
Since, the cost of fencing park per meter = ₹ 12
Therefore, the cost of fencing park of 600 m = 12 × 600 = ₹ 7,200
Question 15: Sweety runs around a square park of side 75 m. Bulbul runs around a rectangular park with length of 60 m and breadth 45 m. Who covers less distance?
Solution
Distance covered by Sweety = Perimeter of square park
Perimeter of square = 4 × side
= 4 × 75 = 300 m
Thus, distance covered by Sweety is 300 m.
Now, distance covered by Bulbul = Perimeter of rectangular park
Perimeter of rectangular park = 2 × (length + breadth)
= 2 × (60 + 45)
= 2 × 105 = 210 m
Thus, Bulbul covers the distance of 210 m and Bulbul covers less distance.
Question 16: What is the perimeter of each of the following figures? What do you infer from the answer?
Solution
(a) Perimeter of square = 4 × side
= 4 × 25 = 100 cm
(b) Perimeter of rectangle = 2 × (length + breadth)
= 2 × (40 + 10)
= 2 × 50
= 100 cm
(c) Perimeter of rectangle = 2 × (length + breadth)
= 2 × (30 + 20)
= 2 × 50
= 100 cm
(d) Perimeter of triangle = Sum of all sides
= 30 cm + 30 cm + 40 cm
= 100 cm
Thus, all the figures have same perimeter.
Question 17: Avneet buys 9 square paving slabs, each with a side 1/2 m. He lays them in the form of a square
(a) What is the perimeter of his arrangement?
(b) Shari does not like his arrangement. She gets him to lay them out like a cross.
What is the perimeter of her arrangement?
(c) Which has greater perimeter?
(d) Avneet wonders, if there is a way of getting an even greater perimeter. Can you find a way of doing this? (The paving slabs must meet along complete edges, i.e., they cannot be broken.)
Solution
(a) 6 m
(b) 10 m
(c) Second arrangement has greater perimeter.
(d) Yes, if all the squares are arranged in row, the perimeter be 10 cm.
Exercise 10.2
Question 1: Find the areas of the following figures by counting squares:
Solution
(a) Number of filled square = 9
∴ Area covered by squares = 9 × 1 = 9 sq. units
(b) Number of filled squares = 5
∴ Area covered by filled squares = 5 × 1 = 5 sq. units
(c) Number of full filled squares = 2
Number of halffilled squares = 4
∴ Area covered by full filled squares = 2 × 1 = 2 sq. units
∴ Total area = 2 + 2 = 4 sq. units
(d) Number of filled squares = 8
∴ Area covered by filled squares = 8 × 1 = 8 sq. units
(e) Number of filled squares = 10
∴ Area covered by filled squares = 10 × 1 = 10 sq. units
(f) Number of full filled squares = 2
Number of halffilled squares = 4
∴ Area covered by full filled squares = 2 × 1 = 2 sq. units
∴ Total area = 2 + 2 = 4 sq. units
(g) Number of full filled squares = 4
Number of halffilled squares = 4
∴ Area covered by full filled squares = 4 × 1 = 4 sq. units
∴ Total area = 4 + 2 = 6 sq. units
(h) Number of filled squares = 5
∴ Area covered by filled squares = 5 × 1 = 5 sq. units
(i) Number of filled squares = 9
∴ Area covered by filled squares = 9 × 1 = 9 sq. units
(j) Number of full filled squares = 2
Number of halffilled squares = 4
∴ Area covered by full filled squares = 2 × 1 = 2 sq. units
∴ Total area = 2 + 2 = 4 sq. units
(k) Number of full filled squares = 4
Number of halffilled squares = 2
∴ Area covered by full filled squares = 4 × 1 = 4 sq. units
∴ Total area = 4 + 1 = 5 sq. units
(l) Number of full filled squares = 3
Number of halffilled squares = 10
∴ Area covered by full filled squares = 3 × 1 = 3 sq. units
∴ Total area = 3 + 5 = 8 sq. units
(m) Number of full filled squares = 7
Number of halffilled squares = 14
∴ Area covered by full filled squares = 7 × 1 = 7 sq. units
∴ Total area = 7 + 7 = 14 sq. units
(n) Number of full filled squares = 10
Number of halffilled squares = 16
∴ Area covered by full filled squares = 10 × 1 = 10 sq. units
∴ Total area = 10 + 8 = 18 sq. units
Exercise 10.3
Question 1: Find the areas of the rectangles whose sides are:
(a) 3 cm and 4 cm
(b) 12 m and 21 m
(c) 2 km and 3 km
(d) 2 m and 70 cm
Solution
(a) Area of rectangle = length × breadth
= 3 cm × 4 cm = 12 cm^{2}
(b) Area of rectangle = length × breadth
= 12 m × 21 m = 252 m^{2}
(c) Area of rectangle = length × breadth
= 2 km × 3 km = 6 km^{2}
(d) Area of rectangle = length × breadth
= 2 m × 70 cm = 2 m × 0.7 m = 1.4 m^{2}
^{}
Question 2: Find the areas of the squares whose sides are:
(a) 10 cm
(b) 14 cm
(c) 5 cm
Solution
(a) Area of square = side × side = 10 cm × 10 cm = 100 cm^{2}
(b) Area of square = side × side = 14 cm × 14 cm = 196 cm^{2}
(c) Area of square = side × side = 5 m × 5 m = 25 m^{2}
^{}
Question 3: The length and the breadth of three rectangles are as given below:
(a) 9 m and 6 m
(b) 17 m and 3 m
(c) 4 m and 14 m
Which one has the largest area and which one has the smallest?
Solution
(a) Area of rectangle = length × breadth = 9 m × 6 m = 54 m^{2}
(b) Area of rectangle = length × breadth= 3 m × 17 m = 51 m^{2}
(c) Area of rectangle = length × breadth= 4 m × 14 m = 56 m^{2}
Thus, the rectangle (c) has largest area, and rectangle (b) has smallest area.
Question 4: The area of a rectangle garden 50 m long is 300 m2, find the width of the garden.
Solution
Length of rectangle = 50 m and Area of rectangle = 300 m^{2}
Since, Area of rectangle = length × breadth
Thus, the breadth of the garden is 6 m.
Question 5: What is the cost of tilling a rectangular plot of land 500 m long and 200 m wide at the rate of ₹8 per hundred sq. m?
Solution
Length of land = 500 m and Breadth of land = 200 m
Area of land = length × breadth = 500 m × 200 m = 1,00,000 m^{2}
∵ Cost of tilling 100 sq. m of land = ₹ 8
Question 6: A tabletop measures 2 m by 1 m 50 cm. What is its area in square meters?
Solution
Length of table = 2 m
Breadth of table = 1 m 50 cm = 1.50 m
Area of table = length × breadth
= 2 m × 1.50 m = 3 m^{2}
^{}
Question 7: A room us 4 m long and 3 m 50 cm wide. How many square meters of carpet is needed to cover the floor of the room?
Solution
Length of room = 4 m
Breadth of room = 3 m 50 cm = 3.50 m
Area of carpet = length × breadth
= 4 × 3.50 = 14m^{2}
^{}
Question 8: A floor is 5 m long and 4 m wide. A square carpet of sides 3 m is laid on the floor. Find the area of the floor that is not carpeted.
Solution
Length of floor = 5 m and breadth of floor = 4 m
Area of floor = length × breadth
= 5 m × 4 m = 20 m^{2}
Now, Side of square carpet = 3 m
Area of square carpet = side × side = 3 × 3 = 9 m^{2}
Area of floor that is not carpeted = 20 m^{2} – 9 m^{2 }= 11 m^{2}
^{}
Question 9: Five square flower beds each of sides 1 m are dug on a piece of land 5 m long and 4 m wide. What is the area of the remaining part of the land?
Solution
Side of square bed = 1 m
∴ Area of square bed = side × side = 1 m × 1 m = 1 m^{2}
Area of 5 square beds = 1 × 5 = 5 m^{2}
Now, Length of land = 5 m
Breadth of land = 4 m
∴ Area of land = length × breadth
= 5 m × 4 m = 20 m^{2}
Area of remaining part = Area of land – Area of 5 flower beds
= 20 m^{2} – 5 m^{2} = 15 m^{2}
^{}
Question 10: By splitting the following figures into rectangles, find their areas. (The measures are given in centimetres)
Solution
(a) Area of HKLM = 3 × 3 = 9 cm^{2}
Area of IJGH = 1 × 2 = 2 cm^{2}
Area of FEDG = 3 × 3 = 9 cm^{2}
Area of ABCD = 2 × 4 = 8 cm^{2}
Total area of the figure = 9 + 2 + 9 + 8 = 28 cm^{2}
(b) Area of ABCD = 3 × 1 = 3 cm^{2}
Area of BDEF = 3 × 1 = 3 cm^{2}
Area of FGHI = 3 × 1 = 3 cm^{2}
Total area of the figure = 3 + 3 + 3 = 9 cm^{2 }
Question 11: Split the following shapes into rectangles and find their areas. (The measures are given in centimetres)
Solution
(a) Area of rectangle ABCD = 2 × 10 = 20 cm2
Area of rectangle DEFG = 10 × 2 = 20 cm^{2}
Total area of the figure = 20 + 20 = 40 cm^{2}
(b) There are 5 squares each of side 7 cm.
Area of one square = 7 × 7 = 49 cm^{2}
Area of 5 squares = 49 × 5 = 245 cm^{2}
(c) Area of rectangle ABCD = 5 × 1 = 5 cm^{2}
Area of rectangle EFGH = 4 × 1 = 4 cm2
Total area of the figure = 5 + 4 cm^{2}
Question 12: How many tiles whose length and breadth are 12 cm and 5 cm respectively will be needed
to fit in a rectangular region whose length and breadth are respectively?
(a) 100 cm and 144 cm
(b) 70 cm and 36 cm
Solution
(a) Area of region = 100 cm × 144 cm = 14400 cm^{2}
Area of one tile = 5 cm × 12 cm = 60 cm^{2}
Thus, 240 tiles are required.
(b) Area of region = 70 cm × 36 cm = 2520 cm^{2}
Area of one tile = 5 cm × 12 cm = 60 cm^{2}
Thus, 42 tiles are required.