CBSE Solutions For Class 6 Maths Chapter 10 Mensuration

Mensuration Exercise – 10.1

Question 1. Find the perimeter of each of the following figures:

Solution:

1. Perimeter = Sum of all the sides = 4 cm + 2 cm +1 cm + 5 cm = 12 cm
2. Perimeter = Sum of all the sides = 23 cm + 35 cm + 40 cm + 35 cm = 133 cm
3. Perimeter = Sum of all the sides = 15 cm + 15 cm + 15 cm + 15 cm = 60 cm
4. Perimeter = Sum of all the sides =4 cm+4 cm+4 cm+4 cm+ 4 cm=20 cm
5. 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
6. Perimeter = Sum of all the sides = 4cm +lcm + 3cm + 2cm + 3cm + 4cm +lcm + 3cm + 2cm + 3cm + 4cm + lcm + 3cm + 2cm + 3cm + 4cm +lcm + 3cm + 2cm + 3cm = 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) cm = 2 x 50 cm = 100 cm =1 m

Thus, the total length of tape required is 100 cm or 1 m.

Question 3. A table-top measures 2 m 25 cm by 1 m 50 cm. What is the perimeter of the tabletop?
Solution:

Length of table-top = 2m 25 cm = 2.25 m

Breadth of table-top =1 m 50 cm = 1.50 m

Perimeter of table-top = 2 x (length + breadth)

= 2 x (2.25 + 1.50) m

= 2 x 3.75 m = 7.50 m

Thus, the perimeter of the 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 of 32 cm and 21 cm respectively?
Solution:

Length of the wooden strip

= Perimeter of photograph

= 2 x (length + breadth)

= 2 (32 + 21) cm = 2 x 53 cm = 106 cm

Thus, the length of the wooden strip required is 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 4 rows of wires are needed.

Therefore, the total length of the wire is equal to 4 times the perimeter of land.

Perimeter of land = 2 x (length + breadth)

= 2 x (0.7 + 0.5) km = 2 x 1.2 km = 2.4 km

= 2.4 x 1000 m = 2400 m

Thus, the length of wire = 4 x 2400 m = 9600 m = 9.6 km

Question 6. Find the perimeter of each of the following shapes :

1. A triangle of sides 3 cm, 4 cm and 5 cm.
2. An equilateral triangle of side 9 cm.
3. An isosceles triangle with equal sides 8 cm each and third side 6 cm.

Solution:

1. Perimeter of ΔABC

=AB+BC+CA

= 3cm+5cm+4cm

= 12 cm

2. Perimeter of equilateral

ΔABC

= 3 x side

= 3 x 9 cm

= 27 cm

3. Perimeter of ΔABC

= AB + BC + CA

= 8cm + 6cm + 8cm

= 22 cm

Question 7. Find the perimeter of a triangle with sides measuring 1 0 cm, 1 4 cm and 1 5 cm.
Solution:

Perimeter of triangle = Sum of all three sides

= 10 cm + 14 cm + 15 cm = 39 cm

Thus, the perimeter of the triangle is 39 cm.

Question 8. Find the perimeter of a regular hexagon with each side measuring 8 m.
Solution:

The perimeter of a regular hexagon

= 6 x length of one side

= 6 x 8m = 48m

Thus, the perimeter of a regular hexagon is 48 m.

Question 9. Find the side of the square whose perimeter is 20 m,
Solution:

The perimeter of the square = 4 x side

20 m = 4 x side ⇒ side = \(=\frac{20}{4} \mathrm{~m}=5 \mathrm{~m}\)

Thus, the side of the square is 5 m.

Question 10. The perimeter of a regular pentagon is 1 00 cm. How long is it on each side?
Solution:

The perimeter of a regular pentagon = 5 x side

100 cm = 5 x side ⇒ side = \(\frac{100}{5} \mathrm{~cm}=20 \mathrm{~cm}\)

Thus, the side of the 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:

1. A square?
2. An equilateral triangle?
3. A regular hexagon?

Solution:

Length of string= Perimeter of each shape

The perimeter of the square = 4 x side

30 cm = 4 x side ⇒ side = \(\frac{30}{4} \mathrm{~cm}=7.5 \mathrm{~cm}\)

Thus, the length of each side of the square will be 7.5 cm

The perimeter of equilateral triangle = 3 x side

30 cm = 3 x side ⇒ side =\(\frac{30}{3} \mathrm{~cm}=10 \mathrm{~cm}\)

Thus, each side of the equilateral triangle will be 10 cm long.

The perimeter of a regular hexagon = 6 x side

30 cm = 6 x side ⇒ side = \(\frac{30}{6} \mathrm{~cm}=5 \mathrm{~cm}\)

Thus, each side of a regular hexagon will be 5 cm long.

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?
Solution:

Let the length of the third title be x cm. The length of the other two sides are 12 cm and 14 cm

Now, the perimeter of the triangle = 36 cm

12 + 14 + x = 36

26 + x = 36

X = 36- 26 ⇒ x = 10

Thus, the length of the third side is 10 cm

Question 13. Find the cost of fencing a square park of side 250 m at the rat of? 20 per metre.
Solution:

Side of square park = 250 m

The perimeter of the square park = 4 x side

= 4 x 250 m = 1000 m

Sin ce, cost of fencing for metre = ₹ 20

Therefore, the cost of fencing for 1000 metres

= ₹ 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 metre
Solution:

Length of rectangular park = 175 m

The breadth of the rectangular park = 125 m

Perimeter of park = 2 x (length + breadth)

= 2 x (175 + 125) m

= 2 x 300 m = 600 m

Since, the cost of fencing a park for metre = ? 12

Therefore, the cost of fencing the park for 600 m

= ₹ 12 x 600 = ₹ 7,200

Question 15. Sweety runs around a square park of side 75 m. Bulbul runs around a rectangular park with a leaf of 60 m a and breadth of h 45 m. Who covers less distance?
Solution:

Distance covered by Sweety

= Perimeter of square park = 4 x side

= 4 x 75 m = 300 m

Thus, the distance covered by Sweety is 300 m.

Nthe ow, distance covered by Bulbul

= Perimeter of a rectangular park

= 2 x (length + breadth)

= 2 x (60 + 45) m

= 2 x 105 m = 210 m

Thus, Bulbul covers a distance of 210 m.

So, Bulbul covers less distance.

Question 16. What is the perimeter of each of the following figures? What do you infer from the answers?

Solution:

Perimeter of square = 4 x side = 4 x 25 cm = 100 cm

Perimeter of rectangle = 2 x (length + breadth)

= 2 x (40 + 10) cm = 2 x 50 cm = 100 cm

Perimeter of rectangle = 2 x (length + breadth)

= 2 x (30 + 20) cm = 2 x 50 cm = 100 cm

Perimeter of triangle = Sum of all sides

= 30 cm + 30 cm + 40 cm

= 100 cm

Thus, all the figures have the same perimeter

Question 17. Avneet buys 9 square paving slabs, each with a side of \(\frac{1}{2}\) m. He lays them in the form of a square.

1. What is the perimeter of his arrangement?
2. Shari does not like his arrangement. She gets him to lay them out like a cross. What is the perimeter of her arrangement? ent.
3. Which a has greater perimeter?
4. 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:

1. Side of one small square = \(\frac{1}{2}\) m

Side of given square ⇒ \(\frac{3}{2}\) perimetereter of square = 4 x side

⇒\(4 \times \frac{3}{2} \mathrm{~m}=6 \mathrm{~m}\)

2. The perimeter of the given figure

⇒ Sum of all sides = \(20 \times \frac{1}{2} \mathrm{~m}=10 \mathrm{~m}\)

3. The arrangement crosses a greater perimeter.

4. It is not possible to determine the arrangement with a perimeter greater than 10 m.

Mensuration Exercise – 10.2

Question 1. Find the areas of the following figures by countsquaresuare:

Solution:

1. Number of filled squares = 9

∴ Area covered by filled squares

= (9 x 1) sq units = 9 sq units

2. Number of filled squares = 5

∴ Area covered by filled squares

= (5 x 1) sq units = 5 sq units

3. Number of fully-filled squares = 2

Number of half-filled squares = 4

∴ Area covered by fully-filled squares

= (2 x 1) sq units = 2 sq units

Area covered by half-filled squares

⇒ \(\left(4 \times \frac{1}{2}\right) \text { sq units }=2 \text { sq units }\)

Total area = (2 + 2) sq units = 4 sq units

4. Number of filled squares = 8

∴ Area covered filled squares

= (8 x 1) sq units = 8 sq units

5. Number of filled squares = 10

∴ Area covered by filled squares

= (10 x 1) sq units = 10 sq units

6. Number of fully-filled squares = 2

Number of half-filled squares = 4

∴ Area covered by fully-filled squares

= (2 x 1) sq units = 2 sq units

Area covered by half-filled squares

⇒ \(\left(4 \times \frac{1}{2}\right) \text { sq units }=2 \text { sq units }\)

Total area = (2 + 2) sq units = 4 sq units

7. Number of fully-filled squares = 4

Number of half-filled squares = 4

∴ Area covered by fully-filled squares

= (4 x 1) sq units = 4 sq units

Area covered by half-filled squares

⇒ \(\left(4 \times \frac{1}{2}\right) \text { sq units }=2 \text { sq units }\)

∴ Total area = (4 + 2) sq units = 6 sq units

8. Number of filled squares = 5

∴ Area covered by filled squares

= (5 x 1) sq units = 5 sq units

9. Number of filled squares = 9

∴ Area covered by filled squares

= (9 x 1) sq units = 9 sq units

10. Number of fully-filled squares = 2

Number of half-filled squares = 4

∴ Area covered by fully-filled squares

= (2 x 1) sq units = 2 sq units

Area covered by half-filled squares

⇒ \(\left(4 \times \frac{1}{2}\right) \text { sq units }=2 \text { sq units }\)

∴ Total area = (2 + 2) sq units = 4 sq units

11. Number of fully-filled squares = 4

Number of half-filled squares = 2

∴ Area covered by fully-filled squares

= (4 x 1) sq units = 4 sq units

Area covered by half-filled squares

⇒ \(\left(2 \times \frac{1}{2}\right) \text { sq units }=1 \text { sq units }\)

Total area = (4+1) sq units = 5 sq units

12. Number of fully-filled squares = 3,

Number of half-filled squares* 2,

Number of more than half-filled

squares = 4

and number of less than half-filled squares = 4.

Now, estimated area covered by fully-filled squares = 3 sq units, half-filled squares = \(\left(2 \times \frac{1}{2}\right) \text { sq units }\)

=1 sq unit

more than half-filled squares = 4 sq units and less than half-filled squares = 0 sq unit

∴ Total area- (3 +1 + 4 + 0) sq units = 8 sq units

13. Number of fully-filled squares = 7,

Number of more than half-filled

squares = 7

The number of less than half-filled

squares = 5

Estimated area covered by

fully-filled squares = 7 sq units,

more than half-filled squares = 7 sq units

and less than half-filled squares = 0 sq unit

∴ Total area = (7 + 7 + 0) sq units = 14 sq units

14. Number of fully-filled squares = 10,

Number of more than half-filled squares = 8

and number of less than half-filled squares = 5

Estimated area covered by

fully-filled squares = 10 sq units,

more than half-filled squares = 8 sq units

less than half-filled squares = 0 squint

∴ Total area= (10 + 8 + 0) sq units = 18 sq units

Mensuration Exercise – 10.3

Question 1. Find the areas of the rectangles whose sides are:

1. 3 cm and 4 cm
2. 12 m and 21 m
3. 2 km and 3 km
4. 2 m and 70 cm

Solution:

1. Area of rectangle = length x breadth = 3 cm x 4 cm = 12 cm²
2. Area of rectangle- length x breadth = 12 m x 21 m = 252 m²
3. Area of rectangle = length x breadth = 2 km x 3km = 6 km²
4. Area of rectangle = length x breadth = 2 m x 70 cm = 2 m x 0.7 m = 1.4 m²

Question 2. Find the areas of the squares whose sides are:

1. 10 cm
2. 14 cm
3. 5 m

Solution:

1. Area of square = side x side = 10 cm x 10 cm = 100 cm²
2. Area of square = side x side = 14 cm x 14 cm = 196 cm²
3. Area of square = side x side =5mx5m = 25m²

Question 3. The length and breadth  of the three rectangles are as given below:

1. 9 m and 6 m
2. 17 m and 3 m
3. 4 m and 14 m

Which one has the largest area and which one has the smallest?
Solution:

1. Area of rectangle = length x breadth =9mx6m= 54m²
2. Area of rectangle = length x breadth = 17m x 3 m = 51 m²
3. Area of rectangle = length x breadth = 4 m x 14 m = 56 m²

Thus, rectangle (c) has the largest area, therefore 56 m2 and rectangle (b) has the smallest area, therefore, 51 m²

Question 4. The area of a rectangular garden 50 m long is 300 sq m. Find the width of the garden.
Solution:

Length of rectangle = 50 m

Area of rectangle = 300 m²

Since, area other rectangle = length x breadth

Therefore, breadth  \(=\frac{\text { area of rectangle }}{\text { length }}\)

⇒ \(=\frac{300}{50} \mathrm{~m}=6 \mathrm{~m}\)

Thus, the breadth 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 ₹ 8 per hundred sq m?
Solution:

Length of land = 500 m

Breadth of land = 200 m

Area of land = length x breadth = 500 m x 200 m = 1,00,000 sqm

Cost of tiling 100 sq m of land = ₹ 8

Cost of tiling 1,00,000 sq m of land

⇒ \(₹ \frac{8 \times 100000}{100}=₹ 8000\)

Question 6. A table-top measures 2 m by 1 m 50 cm. What is its area in square metres?
Solution:

Length of table-top = 2 m

Breadth of table-top =1 m 50 cm = 1.50m

Area of table-top = length x breadth = 2 m x 1.50 m = 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?
Solution:

Length of room = 4 m

And breadth of room = 3 m 50 cm = 3.50 m

Area of carpet = length x breadth = 4 m x 3.50 m = 14 m²

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

The d breadth of the floor = 4 m

Area of floor = length x breadth = 5mx4m = 20m²

Now, side the f square carpet = 3 m

Area of square carpet = side x side = 3mx3m = 9m²

Area of floor that is not carpeted = 20 m²- 9 m² = 11 m²

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 flower bed =1 m

Area of square flower bed = side x side =lm x lm = lm²

Area of 5 square flower beds = (1×5) m²

= 5 m²

Now, length of land = 5 m

The d breadth of land = 4 m

Area of land = length x breadth – 5 m x 4 m

=20 m²

Area of remaining part

= Area of land- Area of 5 flower beds

= 20 m²- 5 m² = 15 m²

Question 10. By splitting the following figures into rectangles, find their areas (The measures are given in centimetres).

Solution:

1. we have,

Area of square HKLM = 3×3 cm²= 9 cm²

Area of rectangle IJGH-1×2 cm² = 2 cm²

Area of square FEDG = 3×3 cm² = 9 cm²

Area of rectangle ABCD = 2×4 cm² = 8 cm²

Total area of the figure = (9 + 2 + 9 + 8) cm² = 28 cm²

2. we have,

Area of rectangle ABCD =3×1 cm² = 3 cm²

Area of rectangle BJEF = 3×1 cm² = 3 cm²

Area of rectangle FGHI = 3×1 cm² = 3 cm²

Total area of the figure = (3 + 3 + 3) cm² = 9 cm²

Question 11. Split the following shapes into rectangles and find their areas. (The measures are given in centimetres).

Solution:

1.  We have,

Area of rectangle ABCD = 2 x 10 cm² = 20 cm²

Area of rectangle DEFG = 10 x 2 cm² = 20 cm²

Total area of the figure = (20 + 20) cm²

= 40 cm²

2.  We have,

There are 5 squares each of side 7 cm.

Area of one square = 7×7 cm² = 49 cm²

Area of 5 squares = 5 x 49 cm² = 245 cm²

3.  We have,

Area of rectangle ABCD = 5 x 1 = 5 cm²

Area of rectangle EFGH = 4 x 1 = 4 cm²

Total area of the figure = (5+4) cm²

= 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:

1. 100 cm and 144 cm
2. 70 cm and 36 cm.

Solution:

1.  The area of the rectangular region

=length x breadth= 100 cm x 144 cm= 14400 cm²

Area of one tile = 12 cm x 5 cm = 60 cm²

Number of tiles = \(=\frac{\begin{array}{r}
\text { Area of rectangular region }
\end{array}}{\text { Area of one tile }}\)

\(=\frac{14400}{60}=240\)

Thus, 240 tiles are required.

2.  The area of the rectangular region

= length x breadth = 70 cm x 36 cm = 2520 cm²

Area of one tile = 12 cm x 5 cm- 60 cm²

Number of tiles \(=\frac{\text { Area of rectangular region }}{\text { Area of one tile }}=\frac{2520}{60}=42\)

Thus, 42 tiles are required.