# Solution of CLASS 6TH MATHEMATICS CHAPTER 10 Mensuration (2023-24)

- Posted by doubtout
- Categories Class 6th Math
- Date October 4, 2023
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#### Exercise 10.1

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

**Solutions:**

(a) Perimeter = Sum of all the sides

= 1 + 2 + 4 + 5

= 12 cm

(b) Perimeter = Sum of all the sides

= 23 + 35 + 35 + 40

= 133 cm

(c) Perimeter = Sum of all the sides

= 15 + 15 + 15 + 15

= 60 cm

(d) Perimeter = Sum of all the sides

= 4 + 4 + 4 + 4 + 4

=20 cm

(e) Perimeter = Sum of all the sides

= 1 + 4 + 0.5 + 2.5 + 2.5 + 0.5 + 4

= 15 cm

(f) Perimeter = Sum of all the sides

= 4 + 1 + 3 + 2 + 3 + 4 + 1 + 3 + 2 + 3 + 4 + 1 + 3 + 2 + 3 + 4 + 1 + 3 + 2 + 3

= 52 cm

**2. The lid of a rectangular box, with sides 40 cm by 10 cm, is sealed all around with tape. What is the length of the tape required?**

**Solutions:**

Length of required tape = Perimeter of rectangle

= 2 (Length + Breadth)

= 2 (40 + 10)

= 2 (50)

= 100 cm

∴ The required length of tape is 100 cm.

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

**Solutions:**

Length of tabletop = 2 m 25 cm = 2.25 m

Breadth of tabletop = 1 m 50 cm = 1.50 m

Perimeter of tabletop = 2 (Length + Breadth)

= 2 (2.25 + 1.50)

= 2 (3.75)

= 2 × 3.75

= 7.5 m

∴ The perimeter of the table top is 7.5 m.

**4. What is the length of the wooden strip required to frame a photograph of length and breadth, 32 cm and 21 cm, respectively?**

**Solutions:**

The required length of the wooden strip = Perimeter of the photograph

= 2 (Length + Breadth)

= 2 (32 + 21)

= 2 (53)

= 2 × 53

= 106 cm

∴ The required length of the wooden strip is 106 cm.

**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?**

**Solutions:**

Perimeter of the field = 2 (Length + Breadth)

= 2 (0.7 + 0.5)

= 2 (1.2)

= 2 × 1.2

= 2.4 km

Each side is to be fenced with 4 rows = 4 × 2.4

= 9.6 km

∴ The total length of the required wire is 9.6 km.

**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 of 8 cm each and a third side of 6 cm.**

**Solutions:**

(a) Perimeter of triangle = 3 + 4 + 5

= 12 cm

(b) Perimeter of an equilateral triangle = 3 × side

= 3 × 9

= 27 cm

(c) Perimeter of isosceles triangle = 8 + 8 + 6

= 22 cm

**7. Find the perimeter of a triangle with sides measuring 10 cm, 14 cm and 15 cm.**

**Solutions:**

Perimeter of triangle = 10 + 14 + 15

= 39 cm

∴ The perimeter of the triangle is 39 cm.

**8. Find the perimeter of a regular hexagon with each side measuring 8 m.**

**Solutions:**

Perimeter of hexagon = 6 × 8

= 48 m

∴ The perimeter of the regular hexagon is 48 m.

**9. Find the side of the square whose perimeter is 20 m.**

**Solutions:**

Perimeter of square = 4 × side

20 = 4 × side

Side = 20 / 4

Side = 5 m

∴ The side of the square is 5 m.

**10. The perimeter of a regular pentagon is 100 cm. How long is it on each side?**

**Solutions:**

The perimeter of the regular pentagon = 100 cm

5 × side = 100 cm

Side = 100 / 5

Side = 20 cm

∴ The side of the pentagon is 20 cm.

**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?**

**Solutions:**

(a) Perimeter of square = 30 cm

4 × side = 30

Side = 30 / 4

Side = 7.5 cm

(b) Perimeter of equilateral triangle = 30 cm

3 × side = 30

Side = 30 / 3

Side = 10 cm

(c) Perimeter of regular hexagon = 30 cm

6 × side = 30

Side = 30 / 6

Side = 5 cm

**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?**

**Solutions:**

Let x cm be the third side

Perimeter of triangle = 36 cm

12 + 14 + x = 36

26 + x = 36

x = 36 – 26

x = 10 cm

∴ The third side is 10 cm.

**13. Find the cost of fencing a square park of side 250 m at the rate of ₹ 20 per metre.**

**Solutions:**

Side of square = 250 m

Perimeter of square = 4 × side

= 4 × 250

= 1000 m

Cost of fencing = ₹ 20 per m

Cost of fencing for 1000 m = ₹ 20 × 1000

= ₹ 20,000

∴ The cost of fencing the square park is ₹ 20,000.

**14. Find the cost of fencing a rectangular park of length 175 cm and breadth 125 m at the rate of ₹ 12 per metre.**

**Solutions:**

Length = 175 cm

Breadth = 125 m

Perimeter of rectangular park = 2 (Length + Breadth)

= 2 (175 + 125)

= 2 (300)

= 2 × 300

= 600 m

Cost of fencing = 12 × 600

= 7200

∴ The cost of fencing is ₹ 7,200.

**15. Sweety runs around a square park of side 75 m. Bulbul runs around a rectangular park with a length of 60 m and a breadth of 45 m. Who covers less distance?**

**Solutions:**

Perimeter of square = 4 × side

= 4 × 75

= 300 m

∴ The distance covered by Sweety is 300 m

Perimeter of the rectangular park = 2 (Length + Breadth)

= 2 (60 + 45)

= 2 (105)

= 2 **× **105

= 210 m

∴ The distance covered by Bulbul is 210 m

Hence, Bulbul covers less distance than Sweety.

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

**Solutions:**

(a) Perimeter of square = 4 × side

= 4 × 25

= 100 cm

(b) Perimeter of rectangle = 2 (40 + 10)

= 2 × 50

= 100 cm

(c) Perimeter of rectangle = 2 (Length + Breadth)

= 2 (30 + 20)

= 2 (50)

= 2 × 50

= 100 cm

(d) Perimeter of triangle = 30 + 30 + 40

= 100 cm

∴ All the figures have the same perimeter.

**17. Avneet buys 9 square paving slabs, each with a side of 1 / 2 m. He lays them in the form of a square.**

**(a) What is the perimeter of his arrangement [fig 10.7(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 10.7 (ii)]?**

**(c) Which has a 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.)**

**Solutions:**

(a) Side of square = 3 × side

= 3 × 1 / 2

= 3 / 2 m

Perimeter of Square = 4 × 3 / 2

= 2 × 3

= 6 m

(b) Perimeter = 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) Perimeters greater than 10 m cannot be determined.

#### Exercise 10.2

**1. Find the areas of the following figures by counting squares:**

(a) The figure contains only 9 fully filled squares. Hence, the area of this figure will be 9 square units.

(b) The figure contains only 5 fully filled squares. Hence, the area of this figure will be 5 square units.

(c) The figure contains 2 fully filled squares and 4 half filled squares. Hence, the area of this figure will be 4 square units.

(d) The figure contains only 8 fully filled squares. Hence, the area of this figure will be 8 square units.

(e) The figure contains only 10 fully filled squares. Hence, the area of this figure will be 10 square units.

(f) The figure contains only 2 fully filled squares and 4 half filled squares. Hence, the area of this figure will be 4 square units.

(g) The figure contains 4 fully filled squares and 4 half filled squares. Hence, the area of this figure will be 6 square units.

(h) The figure contains 5 fully filled squares. Hence, the area of this figure will be 5 square units.

(i) The figure contains 9 fully filled squares. Hence, the area of this figure will be 9 square units.

(j) The figure contains 2 fully filled squares and 4 half filled squares. Hence, the area of this figure will be 4 square units.

(k) The figure contains 4 fully filled squares and 2 half filled squares. Hence, the area of this figure will be 5 square units.

(l) From the given figure, we observe

Covered Area | Number | Area Estimate (square units) |

Fully filled squares | 2 | 2 |

Half filled squares | – | – |

More than half filled squares | 6 | 6 |

Less than half filled squares | 6 | 0 |

Therefore total area = 2 + 6

= 8 square units.

(m) From the given figure, we observe

Covered Area | Number | Area Estimate (square units) |

Fully filled squares | 5 | 5 |

Half filled squares | – | – |

More than half filled squares | 9 | 9 |

Less than half filled squares | 12 | 0 |

Therefore total area = 5 + 9

= 14 square units

(n) From the given figure, we observe

Covered Area | Number | Area estimate (square units) |

Fully filled squares | 8 | 8 |

Half filled squares | – | – |

More than half filled squares | 10 | 10 |

Less than half filled squares | 9 | 0 |

Therefore total area = 8 + 10 = 18 square units

#### Exercise 10.3

**1. Find the area 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**

**Solutions:**

We know that

Area of rectangle = Length × Breadth

(a) l = 3 cm and b = 4 cm

Area = l × b = 3 × 4

= 12 cm^{2}

(b) l = 12 m and b = 21 m

Area = l × b = 12 × 21

= 252 m^{2}

(c) l = 2 km and b = 3 km

Area = l × b = 2 × 3

= 6 km^{2}

(d) l = 2 m and b = 70 cm = 0.70 m

Area = l × b = 2 × 0.70

= 1.40 m^{2}

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

**(a) 10 cm**

**(b) 14 cm**

**(c) 5 m**

**Solutions:**

**(a) **Area of square = side^{2}

= 10^{2}

= 100 cm^{2}

(b) Area of square = side^{2}

= 14^{2}

= 196 cm^{2}

(c) Area of square = side^{2}

= 5^{2}

=25 cm^{2}

**3. The length and breadth of the 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?**

**Solutions:**

**(a) **Area of rectangle = l × b

= 9 × 6

= 54 m^{2}

(b) Area of rectangle = l × b

= 17 × 3

= 51 m^{2}

(c) Area of rectangle = l × b

= 4 × 14

= 56 m^{2}

The area of rectangle 56 m^{2}, i.e. (c), is the largest area and the area of rectangle 51 m^{2}, i.e. (b), is the smallest area

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

**Solutions:**

Area of rectangle = length × width

300 = 50 × width

width = 300 / 50

width = 6 m

∴ The width of the garden is 6 m.

**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?**

**Solutions:**

Area of land = length × breadth

= 500 × 200

= 1,00,000 m^{2}

∴ Cost of tiling 1,00,000 sq m of land = (8 × 1,00,000) / 100

= ₹ 8000

∴ The cost of tiling a rectangular plot of land is ₹ 8000.

**6. A tabletop measures 2 m by 1 m 50 cm. What is its area in square metres?**

**Solutions:**

Given

l = 2m

b = 1m 50 cm = 1.50 m

Area = l × b = 2 × 1.50

= 3 m^{2}

∴ The area of the tabletop is 3 m^{2}.

**7. A room is 4 m long and 3 m 50 cm wide. How many square metres of carpet are needed to cover the floor of the room?**

**Solutions:**

Given

l = 4m

b = 3 m 50 cm = 3.50 m

Area = l × b = 4 × 3.50

= 14 m^{2}

∴ The carpet required to cover the floor is 14 m^{2}.

**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.**

**Solutions:**

Area of floor = l × b = 5 × 4

= 20 m^{2}

Area of square carpet = 3 × 3

= 9 m^{2}

Area of floor that is not carpeted = 20 – 9

= 11 m^{2}

∴ The area of the floor that is not carpeted is 11 m^{2}.

**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?**

**Solutions:**

Area of flower square bed = 1 × 1

= 1 m^{2}

Area of 5 square bed = 1 × 5

= 5 m^{2}

Area of land = 5 × 4

= 20 m^{2}

Remaining part of the land = Area of land – Area of 5 square bed

= 20 – 5

= 15 m^{2}

∴ The remaining part of the land is 15 m^{2}.

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

**Solutions:**

**(a)**

Area of yellow region = 3 × 3

= 9 cm^{2}

Area of orange region = 1× 2

= 2 cm^{2}

Area of grey region = 3 × 3

= 9 cm^{2}

Area of brown region = 2 × 4

= 8 cm^{2}

Total area = 9 + 2 + 9 + 8

= 28 cm^{2}

∴ The total area is 28 cm^{2}.

(b)

Area of brown region = 3 × 1

= 3 cm^{2}

Area of orange region = 3 × 1

= 3 cm^{2}

Area of grey region = 3 × 1

= 3 cm^{2}

Total area = 3 + 3 + 3

= 9 cm^{2}

∴ The total area is 9 cm^{2}.

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

**Solutions:**

**(a)**

Total area of the figure = 12 × 2 + 8 × 2

= 40 cm^{2}

(b)

There are 5 squares, and each side is 7 cm.

Area of 5 squares = 5 × 7^{2}

= 245 cm^{2}

(c)

Area of grey rectangle = 2 × 1

= 2 cm^{2}

Area of brown rectangle = 2 × 1

= 2 cm^{2}

Area of orange rectangle = 5 × 1

= 5 cm^{2}

Total area = 2 + 2 + 5

= 9 cm^{2}

**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?**

**Solutions:**

(a) Area of rectangle = 100 × 144

= 14400 cm

Area of one tile = 5 × 12

= 60 cm^{2}

Number of tiles = (Area of rectangle) / (Area of one tile)

= 14400 / 60

= 240

Hence, 240 tiles are needed

(b) Area of rectangle = 70 × 36

= 2520 cm^{2}

Area of one tile = 5 × 12

= 60 cm^{2}

Number of tiles = (Area of rectangle) / (Area of one tile)

= 2520 / 60

= 42

Hence, 42 tiles are neede