 Perimeter is the distance covered along the boundary forming a closed figure when you go around the figure once.
 The amount of surface enclosed by a closed figure is called its area.
 Figures in which all sides and angles are equal are called regular closed figures.
1. Perimeter of a rectangle = 2 × (length + breadth)
2. Perimeter of square = 4 × length of a side
3. Perimeter of an equilateral triangle = 3 × length of a side
Next part of the chapter emphasises on area.
The first method explained is Calculation of area by counting squares.
 To calculate the area of a figure using a squared paper, the following conventions are adopted:
(b) Ignore portions of the area that are less than half a square.
(c) If more than half a square is in a region. Count it as one square.
(d) If exactly half the square is counted, take its area as 1/2 sq units.
Another method is direct formula based method.
 Area of a rectangle = length × breadth
 Area of a square = side × side
Assessment of these concepts can be done by solving exercises 10.1, 10.2 and 10.3. Exercise 10.2 is a short exercise containing only one question.
Solved examples are covered in the chapter to make the students understand each topic in a crystal clear way.
Summarization of all important points is done at the end of the chapter Mensuration.
Page No 212:
Question 1:
Find the perimeter of each of the following figures:
(a) 
(b) 
(c) 
(d) 
(e) 
(f) 
Answer:
Perimeter of a polygon is equal to the sum of the lengths of all sides of that polygon.
(a) Perimeter = (4 + 2 +1 + 5) cm = 12 cm
(b) Perimeter = (23 + 35 + 40 + 35) cm = 133 cm
(c) Perimeter = (15 + 15 + 15 + 15) cm = 60 cm
(d) Perimeter = (4 + 4 + 4 + 4 + 4) cm = 20 cm
(e) Perimeter = (1 + 4 + 0.5 + 2.5 + 2.5 + 0.5 + 4) cm = 15 cm
(f) Perimeter = (1 + 3 + 2 + 3 + 4 + 1 + 3 + 2 + 3 + 4 + 1 + 3 + 2 + 3 + 4 +
1 + 3 + 2 + 3 + 4) = 52 cm
Page No 212:
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?
Answer:
Length (l) of rectangular box = 40 cm
Breadth (b) of rectangular box = 10 cm
Length of tape required = Perimeter of rectangular box
= 2 (l + b) = 2(40 + 10) = 100 cm
Page No 212:
Question 3:
A tabletop measures 2 m 25 cm by 1 m 50 cm. What is the perimeter of the tabletop?
Answer:
Length (l) of tabletop = 2 m 25 cm = 2 + 0.25 = 2.25 m
Breadth (b) of tabletop = 1 m 50 cm = 1 + 0.50 = 1 .50 m
Perimeter of tabletop = 2 (l + b)
= 2 × (2.25 + 1.50)
= 2 × 3.75 = 7.5 m
Page No 212:
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?
Answer:
Length (l) of photograph = 32 cm
Breadth (b) of photograph = 21 cm
Length of wooden strip required = Perimeter of Photograph
= 2 × (l + b)
= 2 × (32 + 21) = 2 × 53 = 106 cm
Video Solution for mensuration (Page: 212 , Q.No.: 4)
NCERT Solution for Class 6 maths  mensuration 212 , Question 4
Page No 212:
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?
Answer:
Length (l) of land = 0.7 km
Breadth (b) of land = 0.5 km
Perimeter = 2 × (l + b)
= 2 × (0.7 + 0.5) = 2 × 1.2 = 2.4 km
Length of wire required = 4 × 2.4 = 9.6 km
Video Solution for mensuration (Page: 212 , Q.No.: 5)
NCERT Solution for Class 6 maths  mensuration 212 , Question 5
Page No 213:
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.
Answer:
(a) Perimeter = (3 + 4 + 5) cm = 12 cm
(b) Perimeter of an equilateral triangle = 3 × Side of triangle
= (3 × 9) cm = 27 cm
(c) Perimeter = (2 × 8) + 6 = 22 cm
Page No 213:
Question 7:
Find the perimeter of a triangle with sides measuring 10 cm, 14 cm and 15 cm.
Answer:
Perimeter of triangle = Sum of the lengths of all sides of the triangle
Perimeter = 10 + 14 + 15 = 39 cm
Page No 213:
Question 8:
Find the perimeter of a regular hexagon with each side measuring 8 m.
Answer:
Perimeter of regular hexagon = 6 × Side of regular hexagon
Perimeter of regular hexagon = 6 × 8 = 48 m
Page No 213:
Question 9:
Find the side of the square whose perimeter is 20 m.
Answer:
Perimeter of square = 4 × Side
20 = 4 × Side
Side =
Page No 213:
Question 10:
The perimeter of a regular pentagon is 100 cm. How long is its each side?
Answer:
Perimeter of regular pentagon = 5 × Length of side
100 = 5 × Side
Side = = 20 cm
Page No 213:
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?
Answer:
(a) Perimeter = 4 × Side
30 = 4 × Side
Side =
(b) Perimeter = 3 × Side
30 = 3 × Side
Side =
(c) Perimeter = 6 × Side
30 = 6 × Side
Side =
Video Solution for mensuration (Page: 213 , Q.No.: 11)
NCERT Solution for Class 6 maths  mensuration 213 , Question 11
Page No 213:
Question 12:
Two sides of a triangle are 12 cm and 14 cm. The perimeter of the triangle is 36 cm. What is its third side?
Answer:
Perimeter of triangle = Sum of all sides of the triangle
36 = 12 + 14 + Side
36 = 26 + Side
Side = 36 − 26 = 10 cm
Hence, the third side of the triangle is 10 cm.
Page No 213:
Question 13:
Find the cost of fencing a square park of side 250 m at the rate of Rs 20 per metre.
Answer:
Length of fence required = Perimeter of the square park
= 4 × Side
= 4 × 250 = 1000 m
Cost for fencing 1 m of square park = Rs 20
Cost for fencing 1000 m of square park = 1000 × 20
= Rs 20000
Page No 213:
Question 14:
Find the cost of fencing a rectangular park of length 175 m and breadth 125 m at the rate of Rs 12 per metre.
Answer:
Length (l) of rectangular park = 175 m
Breadth (b) of rectangular park = 125 m
Length of wire required for fencing the park = Perimeter of the park
= 2 × (l + b)
= 2 × (175 + 125)
= 2 × 300
= 600 m
Cost for fencing 1 m of the park = Rs 12
Cost for fencing 600 m of the square park = 600 × 12
= Rs 7200
Page No 213:
Question 15:
Sweety runs around a square park of side 75 m. Bulbul runs around a rectangular park with length 60 m and breadth 45 m. Who covers less
distance?
Answer:
Distance covered by Sweety = 4 × Side of square park
= 4 × 75 = 300 m
Distance covered by Bulbul = 2 × (60 + 45)
= 2 × 105 = 210 m
Therefore, Bulbul covers less distance.
Video Solution for mensuration (Page: 213 , Q.No.: 15)
NCERT Solution for Class 6 maths  mensuration 213 , Question 15
Page No 213:
Question 16:
What is the perimeter of each of the following figures? What do you infer from the answers?
(a) 
(b) 
(c) 
(d) 
Answer:
(a) Perimeter of square = 4 × 25 = 100 cm
(b) Perimeter of rectangle = 2 × (10 + 40) = 100 cm
(c) Perimeter of rectangle = 2 × (20 + 30) = 100 cm
(d) Perimeter of triangle = 30 + 30 + 40 = 100 cm
It can be inferred that all the figures have the same perimeter.
Page No 213:
Question 17:
Avneet buys 9 square paving slabs, each with a side of m. He lays them in the form of a square.
(a) What is the perimeter of his arrangement [figure (i)]?
(b) Shari does not like his arrangement. She gets him to lay them out like a cross. What is the perimeter of her arrangement [figure (ii)]?
(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.)
Answer:
(a) Side of square =
Perimeter of square =
(b) Perimeter of cross = 0.5 + 1 + 1 + 0.5 + 1 + 1 + 0.5 + 1 + 1
+ 0.5 + 1 + 1 = 10 m
(c) The arrangement in the form of a cross has a greater perimeter.
(d) Arrangements with perimeters greater than 10 m cannot be determined.
Page No 216:
Question 1:
Find the areas of the following figures by counting square:
(a) 
(b) 
(c) 
(d) 
(e) 
(f) 
(g) 
(h) 
(i) 
(j) 
(k) 
(l) 


(m) 
(n) 
Answer:
(a) The figure contains 9 fully filled squares only. Therefore, the area of
this figure will be 9 square units.
(b) The figure contains 5 fully filled squares only. Therefore, the area of this figure will be 5 square units.
(c) The figure contains 2 fully filled squares and 4 halffilled squares. Therefore, the area of this figure will be 4 square units.
(d) The figure contains 8 fully filled squares only. Therefore, the area of this figure will be 8 square units.
(e) The figure contains 10 fully filled squares only. Therefore, the area of this figure will be 10 square units.
(f) The figure contains 2 fully filled squares and 4 halffilled squares. Therefore, the area of this figure will be 4 square units.
(g) The figure contains 4 fully filled squares and 4 halffilled squares. Therefore, the area of this figure will be 6 square units.
(h) The figure contains 5 fully filled squares only. Therefore, the area of this figure will be 5 square units.
(i) The figure contains 9 fully filled squares only. Therefore, the area of this figure will be 9 square units.
(j) The figure contains 2 fully filled squares and 4 halffilled squares. Therefore, the area of this figure will be 4 square units.
(k) The figure contains 4 fully filled squares and 2 halffilled squares. Therefore, the area of this figure will be 5 square units.
(l) From the given figure, it can be observed that,

Covered Area
Number
Area estimate (sq units)
Fully filled squares
2
2
Half filled squares
−
−
More than half  filled squares
6
6
Less than half  filled squares
6
0
Total area = 2 + 6 = 8 square units
(m) From the given figure, it can be observed that,

Covered Area
Number
Area estimate (sq units)
Fully filled squares
5
5
Halffilled squares
−
−
More than halffilled squares
9
9
Less than halffilled squares
12
0
Total area = 5 + 9 = 14 square units
(n) From the given figure, it can be observed that,

Covered Area
Number
Area estimate (sq units)
Fully filled squares
8
8
Halffilled squares
−
−
More than halffilled squares
10
10
Less than halffilled squares
9
0
Total area = 8 + 10 = 18 square units
Page No 219:
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
Answer:
It is known that,
Area of rectangle = Length × Breadth
(a) l = 3 cm
b = 4 cm
Area = l × b = 3 × 4 = 12 cm^{2}
(b) l = 12 m
b = 21 m
Area = l × b = 12 × 21 = 252 m^{2}
(c) l = 2 km
b = 3 km
Area = l × b = 2 × 3 = 6 km^{2}
(d) l = 2 m
b = 70 cm = 0.70 m
Area = l × b = 2 × 0.70 = 1.40 m^{2}
Page No 219:
Question 2:
Find the areas of the squares whose sides are:
(a) 10 cm (b) 14 cm (c) 5 m
Answer:
It is known that,
Area of square = (Side)^{2}
(a) Side = 10 cm
Area = (10)^{2} =100 cm^{2}
(b) Side = 14 cm
Area = (14)^{2} = 196 cm^{2}
(c) Side = 5 m
Area = (5)^{2} = 25 m^{2}
Page No 219:
Question 3:
The length and 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?
Answer:
It is known that,
Area of rectangle = Length × Breadth
(a) l = 9 m
b = 6 m
Area = l × b = 9 × 6 = 54 m^{2}
(b) l = 17 m
b = 3 m
Area = l × b = 17 × 3 = 51 m^{2}
(c) l = 4 m
b = 14 m
Area = l × b = 4 × 14 = 56 m^{2}
It can be seen that rectangle (c) has the largest area and rectangle (b) has the smallest area.
Page No 219:
Question 4:
The area of a rectangular garden 50 m long is 300 sq m. Find the width of the garden.
Answer:
Let the breadth of the rectangular garden be b.
l = 50 m
Area = l × b = 300 square m
50 × b = 300
b =
Page No 219:
Question 5:
What is the cost of tiling a rectangular plot of land 500 m long and 200 m wide at the rate of Rs 8 per hundred sq m?
Answer:
Area of rectangular plot = 500 × 200 = 100000 m^{2}
Cost of tiling per 100 m^{2} = Rs 8
Cost of tiling per 100000 m^{2} = = Rs 8000
Page No 219:
Question 6:
A tabletop measures 2 m by 1 m 50 cm. What is its area in square metres?
Answer:
Length (l) = 2 m
Breadth (b) = 1 m 50 cm =
Area = l × b = 2 × 1.5 = 3 m^{2}
Video Solution for mensuration (Page: 219 , Q.No.: 6)
NCERT Solution for Class 6 maths  mensuration 219 , Question 6
Page No 219:
Question 7:
A room is 4 m long and 3 m 50 cm wide. How many square metres of carpet is needed to cover the floor of the room?
Answer:
Length (l) = 4 m
Breadth (b) = 3 m 50 cm = 3.5 m
Area = l × b = 4 × 3.5 = 14 m^{2}
Page No 219:
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.
Answer:
Length (l) = 5 m
Breadth (b) = 4 m
Area of floor = l × b = 5 × 4 = 20 m^{2}
Area covered by the carpet = (Side)^{2 }= (3)^{2} = 9 m^{2}
Area not covered by the carpet = 20 − 9 = 11 m^{2}
Video Solution for mensuration (Page: 219 , Q.No.: 8)
NCERT Solution for Class 6 maths  mensuration 219 , Question 8
Page No 219:
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?
Answer:
Area of the land = 5 × 4 = 20 m^{2}
Area occupied by 5 flower beds = 5 × (Side)^{2 }= 5 × (1)^{2} = 5 m^{2}
∴ Area of the remaining part = 20 − 5 = 15 m^{2}
Video Solution for mensuration (Page: 219 , Q.No.: 9)
NCERT Solution for Class 6 maths  mensuration 219 , Question 9
Page No 219:
Question 10:
By splitting the following figures into rectangles, find their areas (The measures are given in centimetres).
(a) (b)
Answer:
(a) The given figure can be broken into rectangles as follows.
Area of 1^{st} rectangle = 4 × 2 = 8 cm^{2}
Area of 2^{nd} rectangle = 6 × 1 = 6 cm^{2}
Area of 3^{rd} rectangle = 3 × 2 = 6 cm^{2}
Area of 4^{th} rectangle = 4 × 2 = 8 cm^{2}
Total area of the complete figure = 8 + 6 + 6 + 8 = 28 cm^{2}
(b) The given figure can be broken into rectangles as follows.
Area of 1^{st} rectangle = 3 × 1 = 3 cm^{2}
Area of 2^{nd} rectangle = 3 × 1 = 3 cm^{2}
Area of 3^{rd} rectangle = 3 × 1 = 3 cm^{2}
Total area of the complete figure = 3 + 3 + 3 = 9 cm^{2}
Page No 220:
Question 11:
Split the following shapes into rectangles and find their areas. (The measures are given in centimetres)
Answer:
(a) The given figure can be broken into rectangles as follows.
Area of 1^{st} rectangle = 12 × 2 = 24 cm^{2}
Area of 2^{nd} rectangle = 8 × 2 = 16 cm^{2}
Total area of the complete figure = 24 + 16 = 40 cm^{2}
(b) The given figure can be broken into rectangles as follows.
Area of 1^{st} rectangle = 21 × 7 = 147 cm^{2}
Area of 2^{nd} square = 7 × 7 = 49 cm^{2}
Area of 3^{rd} square = 7 × 7 = 49 cm^{2}
Total area of the complete figure = 147 + 49 + 49 = 245 cm^{2}
(c) The given figure can be broken into rectangles as follows.
Area of 1^{st} rectangle = 5 × 1 = 5 cm^{2}
Area of 2^{nd} rectangle = 4 × 1 = 4 cm^{2}
Total area of the complete figure = 5 + 4 = 9 cm^{2}
Page No 220:
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
Answer:
(a) Total area of the region = 100 × 144 = 14400 cm^{2}
Area of one tile = 12 × 5 = 60 cm^{2}
Number of tiles required =
Therefore, 240 tiles are required.
(b) Total area of the region = 70 × 36 = 2520 cm^{2}
Area of one tile = 60 cm^{2}
Number of tiles required =
Therefore, 42 tiles are required.
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