Mensuration
Class 6th Mathematics CBSE Solution
- Find the perimeter of each of the following figures:
- The lid of a rectangular box of sides 40 cm by 10 cm is sealed all round with…
- A table-top measures 2 m 25 cm by 1 m 50 cm. What is the perimeter of the top…
- What is the length of the wooden strip required to frame a photograph of length…
- A rectangle piece of land measures 0.7 km by 0.5 km. Each side is to be fenced…
- Find the perimeter of each of the following shapes: (a) A triangle of sides 3…
- Find the perimeter of a triangle with sides measuring 10 cm, 14 cm and 15 cm.…
- Find the perimeter of a regular hexagon with each side measuring 8 m.…
- Find the side of the square whose perimeter is 20 m.
- The perimeter of a regular pentagon is 100 cm. How long is its each side?…
- A piece of string is 30 cm long. What will be the length of each side, if the…
- Two sides of a triangle are 12 cm and 14 cm. The perimeter of the triangle is…
- Find the cost of fencing a square park of side 250 m at the rate of Rs. 20 per…
- Find the cost of fencing a rectangular park of length 175 m and breadth 125 m…
- Sweety runs around a square park of side 75 m. Bulbul runs around a…
- What is the perimeter of each of the following figures? What do you infer from…
- Avneet buys 9 square paving slabs, each with a side of 1/2 m. He lays them in…
- Find the areas of the rectangles whose sides are: (a) 3 cm and 4 cm (b) 12 m…
- Find the areas of the squares whose sides are : (a) 10 cm (b) 14 cm (c) 5 m…
- The length and breadth of three rectangles are as given below : Which one has…
- The area of a rectangular garden 50 m long is 300 sq. m. Find the width of the…
- What is the cost of tiling a rectangular plot of land 500 m long and 200 m wide…
- A table top measures 2 m by 1 m 50 cm. What is its area in square metres?…
- A room is 4 m long and 3 m 50 cm wide. How many square metres of carpet is…
- A floor is 5 m long and 4 m wide. A square carpet of side 3 m is laid on the…
- Five square flower beds each of side 1 m are dug on a piece of land 5 m long…
- By splitting the following figures into rectangles, find their areas (The…
- Split the following shapes into rectangles and find their areas. (The measures…
- How many tiles whose length and breadth are 12 cm and 5 cm respectively will…
Exercise 10.1
Question 1.Find the perimeter of each of the following figures:
Answer:
(a)
We know that,
Perimeter of a polygon = Sum of the lengths of all sides of that polygon
In this question, we have:
Perimeter = (4 + 2 + 1 + 5) cm = 12 cm
Hence, perimeter is 12 cm
(b)
We know that,
Perimeter of a polygon = Sum of the lengths of all sides of that polygon
In this question, we have:
Perimeter = (23 + 35 + 40 + 35) cm
= 133 cm
(c)
We know that,
Perimeter of a polygon = Sum of the lengths of all sides of that polygon
In this question, we have:
Perimeter = (15 + 15 + 15 + 15) cm
= 60 cm
(d)
We know that,
Perimeter of a polygon = Sum of the lengths of all sides of that polygon
In this question, we have:
Perimeter = (4 + 4 + 4 + 4 + 4) cm
= 20 cm
(e)
We know that,
Perimeter of a polygon = Sum of the lengths of all sides of that polygon
In this question, we have:
Perimeter = (1 + 4 + 0.5 + 2.5 + 2.5 + 0.5 + 4) cm
= 15 cm
(f)
We know that,
The perimeter of a polygon = Sum of the lengths of all sides of that polygon
In this question, we have:
Perimeter = (1 + 3 + 2 + 3 + 4 + 1 + 3 + 2 + 3 + 4 + 1 + 3 + 2 + 3 + 4 + 1 + 3 + 2 + 3 + 4) 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?
Answer:
It is given in the question that,
Length of the rectangular box = 40 cm
Breadth of the rectangular box = 10 cm
Therefore,
Length of tape required = Perimeter of the rectangular box
= 2 (l + b)
= 2 (40 + 10)
= 100 cm
Question 3.
A table-top measures 2 m 25 cm by 1 m 50 cm. What is the perimeter of the top of the table-top?
Answer:
It is given in the question that,
Length of table top, l = 2 m 25 cm
= 2 + 0.25 (Because 25 cm = 0.25m)
= 2.25m
Also,
Breadth of the table top, b = 1 m 50 cm
= 1 + 0.50 (Because 50 cm = 0.50m)
= 1.50 m
We know that,
Perimeter of table-top = 2 (l + b)
= 2 (2.25 + 1.50)
= 2 × 3.75
= 7.5 m
Hence, 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?
Answer:
It is given in the question that,
Length of photograph, l = 32 cm
Breadth of photograph, b= 21 cm
Therefore,
Length of wooden strip required = Perimeter of Photograph
= 2 × (l + b)
= 2 × (32 + 21)
= 2 × 53
= 106 cm
Question 5.
A rectangle 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:
It is given in the question that,
Length of land, l = 0.7 km
Breadth of land, b = 0.5 km
We know that,
Perimeter = 2 × (l + b)
= 2 × (0.7 + 0.5)
= 2 × 1.2
= 2.4 km
Now, one round of fencing the piece of land will require 2.4 km of wire.
Thus, four rounds of fencing will require = 4 × 2.4 = 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 and third side 6 cm.
Answer:
(a) We know that,
Perimeter of a triangle = Sum of all sides
= (3 + 4 + 5) cm
= 12 cm
(b) We know that in equilateral triangle all sides are equal.
Perimeter of an equilateral triangle = 3 × Side of triangle
= (3 × 9) cm
= 27 cm
(c) We need to find out perimeter of an isosceles triangle:
We know in isosceles triangle two sides are equal.
Perimeter = (2 × 8) + 6
= 16+ 6= 22 cm
Question 7.
Find the perimeter of a triangle with sides measuring 10 cm, 14 cm and 15 cm.
Answer:
We know that,
Perimeter of triangle = Sum of the lengths of all sides of the triangle
Perimeter = 10 + 14 + 15
= 39 cm
Question 8.
Find the perimeter of a regular hexagon with each side measuring 8 m.
Answer:
We know that,
The perimeter of regular hexagon = 6 × Side of the regular hexagon
Side of hexagon = 8 m (Given)
Therefore, Perimeter = 6 × 8
= 48 m
Question 9.
Find the side of the square whose perimeter is 20 m.
Answer:
We know that,
Perimeter of square = 4 × Side
20 = 4 × Side
Therefore, side = 
= 5 m
Thus, the side of the square is 5m.
Question 10.
The perimeter of a regular pentagon is 100 cm. How long is its each side?
Answer:
It is given that,
Perimeter of a regular pentagon = 100 cm
We know that,
Perimeter of a regular pentagon = 5 × Length of side
100 = 5 × Side
Side = 
= 20 cm
Thus, the side of the 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?
Answer:
(a) Given that,
Perimeter of square = 30 cm
We know that,
Perimeter of square = 4 × Side
30 = 4 × Side
Side = 
= 7.5 cm
Therefore, side of square is 7.5 cm
(b) Given that,
Perimeter of equilateral triangle = 30 cm
We know that,
Perimeter of equilateral triangle = 3 × Side
30 = 3 × Side
Side = 
= 10 cm
Therefore, side of equilateral triangle is 10 cm
(c) Given that,
Perimeter of a regular hexagon = 30 cm
We know that,
Perimeter of regular hexagon = 6 × side
30 = 6 × Side
Side = 
= 5 cm
Therefore, side of regular 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 its third side?
Answer:
Given that,
Two sides of triangle are 12 cm and 14 cm respectively
Also,
Perimeter of triangle = 36 cm
We know that,
Perimeter of triangle = Sum of all sides of a triangle
36 = 12 + 14 + Side
36 = 26 + Side
Side = 36 – 26
Side = 10 cm
Therefore, the third side of the triangle be 10 cm.
Question 13.
Find the cost of fencing a square park of side 250 m at the rate of Rs. 20 per metre.
Answer:
We have,
Length of fence covered = Perimeter of the square park
We know that,
Perimeter of square = 4 × Side of square park
= 4 × 250 m
= 1000 m
It is given that,
Cost of fencing of a square park = Rs. 20
Therefore,
Cost of fencing 1000 m of square park = 1000 × 20
= Rs. 20,000
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:
It is given in the question that,
Length of rectangular park, l = 175 m
Breadth of rectangular park, b = 125 m
Therefore,
Length of wire required for fencing the park = Perimeter of the park
= 2 × (l + b)
= 2 × (175 + 125)
= 2 × (300)
= 600 m
It is given in the question that,
Cost of fencing 1 m of the park = Rs. 12
Therefore, the cost of fencing 600 m of the rectangular park = 600 × Rs.12 = Rs 7200
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:
Given that,
Side of square park = 75 m
And,
Length of rectangular park, l = 60 m
Breadth of rectangular park, b = 45 m
Therefore,
Distance covered by Sweety = Perimeter of square park
We know that,
Perimeter of square park = 4 × Side
= 4 × 75 =300 m
Also,
Distance covered by Bulbul = Perimeter of rectangular park
We know that,
Perimeter of rectangular park = 2 × (l + b)
= 2 × (60 + 45) = 2 × 105
= 210 m
Therefore, Bulbul covers less distance as compared to Sweety.
Question 16.
What is the perimeter of each of the following figures? What do you infer from the answers?
Answer:
(a) We know that,
Perimeter of square = 4 × Side of square
= 4 × 25 cm
= 100 cm
(b) We know that,
Perimeter of rectangle = 2 × (l + b)
Here, l is 40 cm and b is 10 cm.
= 2 × (40 + 10)
= 2 × (50)
= 100 cm
(c) We know that,
Perimeter of rectangle = 2 × (l + b)
Here, l is 30 cm and b is 20 cm.
= 2 × (30 + 20)
= 2 × (50)
= 100 cm
(d) We know that,
Perimeter of triangle = Sum of all sides
= 30 + 30 + 40
= 100 cm
The inference
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 [Fig. (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 [Fig. (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) From the given figure, the side of one slab is 
m
Therefore, the side of square = (3 × 
) m = 
m
Now,
We know that the perimeter of a square = 4 × Side
= 4 × = 6 m
(b) We know that,
Perimeter = Sum of all sides
Therefore Perimeter = 0.5 + 1 + 1 + 0.5 + 1 + 1 + 0.5 + 1 + 1 + 0.5 + 1 + 1
= 10 m
(c) The arrangement which is in the shape of cross has a greater perimeter (i.e. 10 m)
(d) If we put all 9 slabs in a line then, the perimeter will be 10 m, as shown below.
Thus, from the given arrangements, arrangements with perimeters greater than 10 m cannot be determined.
Exercise 10.2
Question 1.Find the areas of the following figures by counting squares:
Answer:
(a) By observing the figure, we see that it contains 9 fully filled squares.
If the area of one such square is taken to be as 1 square unit.
Then,
The area of the figure = 9 square units.
(b)
By observing the figure, we see that it contains 5 fully filled squares.
If the area of one such square is taken to be as 1 square unit.
Then,
The area of the figure = 5 square units.
(c)
By observing the figure, we see that it contains 2 fully filled squares and 4 half- filled squares.
If the area of one such square is taken to be as 1 square unit.
Then,
The area of the figure = 2 + (4× 0.5) = 2+ 2 = 4 square units.
(d)
By observing the figure, we see that it contains 8 fully filled squares.
If the area of one such square is taken to be as 1 square unit.
Then,
The area of the figure = 8 square units.
(e)
By observing the figure, we see that it contains 10 fully filled squares.
If the area of one such square is taken to be as 1 square unit.
Then,
The area of the figure = 10 square units.
(f)
By observing the figure, we see that it contains 2 fully filled squares and 4 half- filled squares.
If the area of one such square is taken to be as 1 square unit.
Then,
The area of the figure = 2 + (4× 0.5) = 2+ 2 = 4 square units.
(g)
By observing the figure, we see that it contains 4 fully filled squares and 4 half- filled squares.
If the area of one such square is taken to be as 1 square unit.
Then,
The area of the figure = 4 + (4× 0.5) = 4 + 2 = 6 square units.
(h)
By observing the figure, we see that it contains 5 fully filled squares.
If the area of one such square is taken to be as 1 square unit.
Then,
The area of the figure = 5 square units.
(i)
By observing the figure, we see that it contains 9 fully filled squares.
If the area of one such square is taken to be as 1 square unit.
Then,
The area of the figure = 9 square units.
(j)
By observing the figure, we see that it contains 2 fully filled squares and 4 half- filled squares.
If the area of one such square is taken to be as 1 square unit.
Then,
The area of the figure = 2 + (4× 0.5) = 2+ 2 = 4 square units.
(k)
By observing the figure, we see that it contains 4 fully filled squares and 2 half- filled squares.
If the area of one such square is taken to be as 1 square unit.
Then,
The area of the figure = 4 + (2× 0.5) = 4 + 1 = 5 square units.
(l)
In the figure given, we find that there are variations in the filling of the squares.
Thus, we form a table to organise the data by observing the above figure.
Therefore,
Total Area = 2 + 6 = 8 square units
(m)
In the figure given, we find that there are variations in the filling of the squares.
Thus, we form a table to organise the data by observing the above figure.
Therefore,
Total Area = 5 + 9 = 14 square units
(n)
By observing above figure, we get
Therefore,
Total Area = 8 + 10 = 18 square 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
Answer:
(a) We know that,
Area of rectangle = Length × Breadth
Length of rectangle, l = 3 cm
Breadth of rectangle, b = 4 cm
Therefore,
Area of rectangle = Length × Breadth = 3 × 4 = 12 cm2
(b) We know that,
Area of rectangle = Length × Breadth
It is given in the question that,
Length of rectangle, l = 12 m
Breadth of rectangle, b = 21 m
Therefore,
Area of rectangle = Length × Breadth = 12 × 21 = 252 m2
(c) We know that,
Area of rectangle = Length × Breadth
It is given in the question that,
Length of rectangle, l = 2 km
Breadth of rectangle, b = 3 km
Therefore,
Area of rectangle = Length × Breadth = 2 × 3 = 6 km2
(d) We know that,
Area of rectangle = Length × Breadth
It is given in the question that,
Length of rectangle, l = 2 m
Breadth of rectangle, b = 70 cm = 0.70 m
Therefore,
Area of rectangle = Length × Breadth
= 2 × 0.70
= 1.40 m2
Question 2.
Find the areas of the squares whose sides are :
(a) 10 cm
(b) 14 cm
(c) 5 m
Answer:
(a) We know that,
Area of square = (Side)2
It is given in the question that,
Side of square = 10 cm
Therefore, Area of square = (Side)2
= (10)2 = 100 cm2
(b) We know that,
Area of square = (Side)2
It is given in the question that,
Side of square = 14 cm
Therefore, Area of square = (Side)2
= (14)2 = 196 cm2
(c) We know that,
Area of square = (Side)2
It is given in the question that,
Side of square = 5 m
Therefore,
Area of square = (Side)2 = (5)2 = 25 m2
Question 3.
The length and breadth of three rectangles are as given below :
Which one has the largest area and which one has the smallest?
(a) 9 m and 6 m
(b) 17 m and 3 m
(c) 4 m and 14 m
Answer:
(a) We know that,
Area of rectangle = Length × Breadth
It is given in the question that,
Length of rectangle = 9 m
Breadth of rectangle = 6 m
Therefore,
Area of rectangle = Length × Breadth
= 9 × 6
= 54 m2
(b) We know that,
Area of rectangle = Length × Breadth
It is given in the question that,
Length of rectangle = 17 m
Breadth of rectangle = 3 m
Therefore,
Area of rectangle = Length × Breadth
= 17 × 3
= 51 m2
(c) We know that,
Area of rectangle = Length × Breadth
It is given in the question that,
Length of rectangle = 4 m
Breadth of rectangle = 14 m
Therefore,
Area of rectangle = Length × Breadth
= 4 × 14
= 56 m2
Therefore, from above three results it is clear rectangle (c) has the largest area i.e. 56m2 while rectangle (b) has the smallest area i.e.51 m2
Question 4.
The area of a rectangular garden 50 m long is 300 sq. m. Find the width of the garden.
Answer:
We know that,
Area of rectangle = Length × Breadth
It is given in the question that,
Length of rectangle = 50 m
Area of rectangle = 300 m2
We have to find out the width of the rectangle
As,
Area of rectangle = Length × Breadth
300 = 50 × Breadth
Breadth = 
Breadth = 6 m
Therefore, width of the garden is 6 m.
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:
We know that,
Area of rectangle = Length × Breadth
It is given in the question that,
Length of rectangle = 500 m
Breadth of rectangle = 200 m
Therefore, Area of rectangular plot = Length × Breadth
= 500 × 200
= 100000 m2
It is also given that,
Cost of tiling per 100 m2 = Rs 8
Therefore, Cost of tiling per 1m2 = Rs 
Cost of tiling per 100000 m2 = × 100000
= Rs 8000
Question 6.
A table top measures 2 m by 1 m 50 cm. What is its area in square metres?
Answer:
We know that,
Area of rectangle = Length × Breadth
It is given in the question that,
Length of rectangle = 2 m
Breadth of rectangle = 1 m 50 cm
= (1 + 
) m
= 1.5 m
Therefore,
Area = Length × Breadth
= 2 × 1.5
= 3 m2
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:
We know that,
Area of rectangle = Length × Breadth
It is given in the question that,
Length of rectangle = 4 m
Breadth of rectangle = 3 m 50 cm
As 1 cm = 1/100 m
50 cm = 50/100 m
= 0.5 m
3 m 50 cm = 3.5 m
Therefore,
Area of floor = Length × Breadth
= 4 × 3.5
= 14 m2
Question 8.
A floor is 5 m long and 4 m wide. A square carpet of side 3 m is laid on the floor. Find the area of the floor that is not carpeted.
Answer:
The figure is given below:
We know that,
Area of rectangle = Length × Breadth
It is given in the question that,
Length of rectangle = 5 m
Breadth of rectangle = 4 m
Therefore,
Area of floor = Length × Breadth
= 5 × 4
= 20 m2
Also,
Area covered by the square carpet = (Side)2
= (3 m)2
= 9 m2
Therefore,
Area of the floor that is not carpeted = Area of Rectangle - Area of square
= 20 m2 – 9m2
= 11 m2
Question 9.
Five square flower beds each of side 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:
We know that,
Area of rectangle = Length × Breadth
It is given in the question that,
Length of rectangle = 5 m
Breadth of rectangle = 4 m
Therefore,
Area of land = Length × Breadth
= 5 × 4
= 20 m2
Also,
Area occupied by 5 flower beds = 5 × (Side)2
= 5 × (1)2
= 5 m2
Therefore,
Area of remaining part = 20 – 5 = 15 m2
Question 10.
By splitting the following figures into rectangles, find their areas (The measures are given in centimetres).
Answer:
(a) We know that,
Area of rectangle = Length × Breadth
Therefore,
Therefore, we complete the figure as shown below to find the area:
Area of 1st rectangle(square) = 3 × 3 = 9 cm2
Area of 2nd rectangle = 2 × 1 = 2 cm2
Area of 3rd rectangle = 1 × 1 = 1 cm2
Area of 4th rectangle = 1 × 2 = 2 cm2
Area of 5th rectangle = 3 × 2 = 6 cm2Area of 6th rectangle = 4 × 2 = 8 cm2
Hence,
Total Area = 9 + 2 + 1 + 2 + 6 + 8 =
= 28 cm2
(b) We know that,
Area of rectangle = Length × Breadth
Therefore,
Area of 1st rectangle = 2 × 1 = 2 cm2
Area of 2nd rectangle = 5 × 1 = 5 cm2
Area of 3rd rectangle = 2 × 1 = 2 cm2
Hence,
Total Area = 2 + 5 + 2
= 9 cm2
Question 11.
Split the following shapes into rectangles and find their areas. (The measures are given in centimetres).
Answer:
The area of the above given figure can be calculated as:
(a)
We know that,
Area of rectangle = Length × Breadth
Therefore,
Atrea of 1st rectangle = 12 × 2 = 24 cm2
Also,
Area of 2nd rectangle = 8 × 2 = 16 cm2
Thereore,
Total area= 24 + 16 cm2
= 40 cm2
(b)
The area of the above given figure can be calculated as:
We know that,
Area of rectangle = Length × Breadth
Also,
Area of square = (Side)2
Therefore,
Area of one square = 7 × 7
= 49 cm2
As there are 5 squares in total of equal size.
Area of 5 squares = 5× 49 cm2
= 245 cm2
(c)
The area of the above given figure can be calculated as:
We know that,
Area of rectangle = Length × Breadth
Therefore,
Area of 1st recatngle = 5 × 1 = 5 cm2
Also,
Area of 2nd rectangle = 4 × 1 = 4 cm2
Therefore,
Total area = 5 + 4 = 9 cm2
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) We know that,
Area of rectangle = Length × Breadth
It is given in the question that,
Length = 100 cm
Breadth = 144 cm
Therefore,
Area = Length × Breadth
= 100 × 144
= 14400 cm2
Therefore,
Area of one tile = 12 × 5
= 60 cm2
Hence,
Number of tiles required = 
= 240
Therefore, total 240 tiles are required
(b) We know that,
Area of rectangle = Length × Breadth
It is given in the question that,
Length = 100 cm
Breadth = 144 cm
Therefore,
Area = Length × Breadth
= 70 × 36
= 2520 cm2
Area of one tile = 60 cm2
Hence,
Number of tiles required = 
= 42
Therefore, total 42 tiles are required.