# NCERT Solutions Class 6 Maths Chapter 10 Mensuration | class 6th maths

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# NCERT Solutiontions Class 6 Maths Chapter 10 Mensuration Class 6th maths

**NCERT Solutiontions Class 6 Maths Chapter 10 Mensuration class 6th maths**

Ex 10.1

Ex 10.1

**Exercise**** 10.1**

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

**Solutions :**

**Perimeter :** Perimeter is the sum of all side of a figure.

(a) Perimeter of figure (a) = 4 + 2 + 1 + 5 = 12 cm

(b) Perimeter of figure (b) = 23 + 35 + 40 + 35 = 133 cm

(c) Perimeter of figure (c) = 15 + 15 + 15 + 15 = 60 cm

(d) Perimeter of figure (d) = 4 + 4 + 4 + 4 + 4 = 20 cm

(e) Perimeter of figure (e) = 1 + 4 + 0.5 + 2.5 + 2.5 + 0.5 + 4 = 15 cm

(f) Perimeter of figure (f) = 1 + 3 + 2 + 3 + 4 + 1 + 3 + 2 + 3 + 4 + 1 + 3 + 2 + 3 + 4 + 1 + 3 + 2 + 3 + 4 = 52 cm

**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 :**

Rectangular box length = 40 cm

Rectangular box breadth = 10 cm

Perimeter of rectangular box = 2 × (Length + Breadth)

= 2 × (40 + 10)

= 2 × 50

= 100 cm

So, the length of the tape required is 100 cm or 1 meter.

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

**Solution :**

A table-top measures are = 2 m 25 cm by 1 m 50 cm

Perimeter of table = Perimeter of rectangle (because table is a rectangle shape)

Perimeter of table = 2 × (Length + Breadth)

= 2 × (2.25 + 1.50)

= 2 × 3.75

= 7.5 meter

So, the perimeter of the table-top is 7.5 meters.

**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 photograph = 32 cm

Breadth of photograph = 21 cm

the wooden strip perimeter = 2 × (Length + Breadth)

= 2 × (32 + 21)

= 2 × 53

= 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?**

**Solution :**

Length of rectangular piece of land = 0.7 km

Breadth of rectangular piece of land = 0.5 km

Perimeter of piece of land = Perimeter of rectangle

Perimeter of piece of land = 2 × (Length + Breadth)

= 2 × (0.7 + 0.5)

= 2 × 1.2

= 2.4 km

∵ one row length = 2.4 km

∴ four rows length = 4 × 2.4 = 9.6 km

So, the length of the wire needed 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 8 cm each and third side 6 cm.

**Solution :**

**(a) A triangle of sides 3 cm, 4 cm and 5 cm.**

∵ All sides of tringle is different so this is a scalene triangle.

Perimeter of a scalene triangle = Sum of all three sides

= 3 + 4 + 5 = 12 cm

**(b) An equilateral triangle of side 9 cm.**

Perimeter of an equilateral triangle = 3 × side

= 3 × 9

= 36 cm

**(c) An isosceles triangle with equal sides 8 cm each and third side 6 cm.**

Perimeter of an isosceles triangle = (2 × equal side) + third side

= (2 × 8) + 6

= 16 + 6

= 22 cm

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

**Solution :**

∵ All sides of tringle is different so this is a scalene triangle.

Perimeter of a scalene triangle = Sum of all three sides

= 10 + 14 + 15

= 39 cm

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

**Solution :**

Perimeter of a regular hexagon = Sum of all sides

= 6 × 8

= 48 meter

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

**Solution :**

Perimeter of square = 20 meter

4 × side = 20

Side = \(\displaystyle \frac{{20}}{4}\)

Side = 5 m

So, the side of the square is 5 meter.

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

**Solution :**

The perimeter of a regular pentagon is = 100 cm

5 × side = 100

Side = \(\displaystyle \frac{{100}}{5}\)

Side = 20 cm

So, its each side is 20 cm long.

**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 :**

**(a) a square?**

If a square is made from a thread, then the perimeter of that square is = 30 cm

4 × side = 30

Side = \(\displaystyle \frac{{30}}{4}\)

Side = 7.5 cm

Therefore, if a square is made with string, then the side of that square will be 7.5 cm.

**(b) an equilateral triangle?**

If an equilateral triangle is made from the string, then the perimeter of that equilateral triangle is = 30 cm

3 × side = 30

Side = \(\displaystyle \frac{{30}}{3}\)

Side = 10 cm

Therefore, if an equilateral triangle is made from the string, then the side of that equilateral triangle will be 10 cm.

**(c) a regular hexagon?**

If a regular hexagon is made with string, then the perimeter of that regular hexagon = 30 cm

6 × side = 30

Side = \(\displaystyle \frac{{30}}{6}\)

Side = 5 cm

Therefore, if a regular hexagon is made with string, then the side of that regular hexagon will be 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?**

**Solution :**

Let the third side of the triangle be = *x* cm

Perimeter of triangle = 36 cm

sum of three sides of a triangle = 36

So, 12 + 14 + *x* + = 36

26 + *x* = 36

*x* = 36 − 26

*x* = 10 cm

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

**Solution :**

Side of a square park is = 250 m

Perimeter of a square park is = 4 × Side

= 4 × 250

= 1000 m

∵ Fencing cost on one meter area = ₹20

∴ Fencing cost on 1000 meters area = 20 × 1000

= ₹20000

Hence, the cost of fencing a square garden of side 250 m will be ₹ 20000.

**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 a rectangular park = 175 m

Breadth of a rectangular park = 125 m

Perimeter of a rectangular park = 2 × (Length + Breadth)

Perimeter of a rectangular park = 2 × (175 + 125)

= 2 × 300

= 600 m

∵ one meter fencing cost = ₹20

∴ cost of fencing 600 meters = 20 × 600

= ₹12000

Hence, the cost of fencing all around will be ₹ 12000.

**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?**

**Solution :**

Side of a square park is = 75 m

Perimeter of square park = 4 × side

= 4 × 75

= 300 m

Length of rectangular park is = 60 m

Breadth of rectangular park is = 45 m

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

= 2 × (60 + 45)

= 2 × 105

= 210 m

So, Sweety cover less distance.

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

**Solution :**

Perimeter of figure (a) = Perimeter of square

Perimeter of square = 4 × Side

= 4 × 25

= 100 cm

Perimeter of figure (b) = Perimeter of rectangle

Length of rectangle = 30 cm

Breadth of rectangle = 20 cm

Perimeter of rectangle = 2 × (Length + Breadth)

= 2 × (30 + 20)

= 2 × 50

= 100 cm

Perimeter of figure (c) = Perimeter of rectangle

Length of rectangle = 40 cm

Breadth of rectangle = 10 cm

Perimeter of rectangle = 2 × (Length + Breadth)

= 2 × (40 + 10)

= 2 × 50

= 100 cm

Perimeter of figure (d) = Perimeter of isosceles triangle

Length of equal side of an isosceles triangle = 30 cm

Length of third side of isosceles triangle = 40 cm

Perimeter of isosceles triangle = (2 × equal side) + third side

Perimeter of equilateral triangle = (2 × equal side) + third side

= (2 × 30) + 40

= 60 + 40

= 100 cm

Conclusion: Finding the perimeter of all the above figures, it can be concluded that all the figures have the same perimeter.

**17. Avneet buys 9 square paving slabs, each with a side of \(\displaystyle \frac{{1}}{2}\) 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 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) What is the perimeter of his arrangement [Fig 10.7(i)]?**

Side of the new square = 1.5 m

Perimeter of new square = 4 × side

= 4 × 1.5

= 6.0 m

So the perimeter of the new square will be 6 metres.

**(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)]?**

Perimeter of the figure made by Shari = Sum of all sides

= 0.5 + 1 + 1 + 0.5 + 1 + 1 + 0.5 + 1 + 1 + 0.5 + 1 + 1

= 10 meters

Hence, the perimeter of the figure made by Sherry is 10 m.

**(c) Which has greater perimeter?**

Fig. 10.7 (b) has a greater perimeter.

**NCERT Solutions for Class 6 Maths Chapter 10 Mensuration class 6th maths**

Ex 10.2

Ex 10.2

1. Find the areas of the following figures by counting square:

Solution :

(a) 9 squre unit

(b) 5 squre unit

(c) 4 squre unit

(d) 8 squre unit

(e) 10 squre unit

(f) 4 squre unit

(g) 6 squre unit

(h) 5 squre unit

(i) 9 squre unit

(j) 4 squre unit

(k) 5 squre unit

(l) 8 squre unit

(m) 14 squre unit

(n) 18 squre unit

**NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Class 6th maths**

Ex 10.3

Ex 10.3

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) 3 cm and 4 cm

Area of Rectangle = Length × Breadth

Area of Rectangle = 3 × 4

= 12 Sq cm

(b) 12 m and 21 m

Area of Rectangle = Length × Breadth

Area of Rectangle = 12 × 21

= 252 Sq meter

(c) 2 km and 3 km

Area of Rectangle = Length × Breadth

Area of Rectangle = 2 × 3

= 6 sq km

(d) 2 m and 70 cm

Area of Rectangle = Length × Breadth

Area of Rectangle = 2 × 0.70 (∵ 70 cm = 0.70 m)

= 1.40 sq meter

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

(a) 10 cm | (b) 14 cm | (c) 5 m |

Solution :

**(a) 10 cm**

Area of Square = (side)^{2}

= (10)^{2}

= 100 sq cm

**(b) 14 cm**

Area of Square = (side)^{2}

= (14)^{2}

= 196 sq cm

**(c) 5 मी**

Area of Square = (side)^{2}

= (5)^{2}

= 25 sq meter

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?

Solutions :

(a) 9 m and 6 m

Area of Rectangle = Length × Breadth

Area of Rectangle = 9 × 6

= 54 sq meter

(b) 17 m and 3 m

Area of Rectangle = Length × Breadth

Area of Rectangle = 3 × 14

= 42 sq meter

(c) 4 m and 14 m

Area of Rectangle = Length × Breadth

Area of Rectangle = 4 × 14

= 56 sq meter

(c) has the largest area and (b) has the smallest.

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

Solutions :

Rectangular garden length = 50 m

Area of Rectangular garden = 300 sq m

Length × Breadth = 300

50 × Breadth = 300

Breadth = \(\displaystyle \frac{{300}}{50}\)

Breadth = 6 m

The width of the garden is 6 meter.

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 :

Length of Rectangular plot = 500 m

Breadth of Rectangular plot = 200 m

Area of Rectangular plot = Length × Breadth

= 500 × 200

= 100000 sq m

∵ Rate pf per hundred sq m is = ₹ 8

∴ Rate pf per 1 sq m is = \(\displaystyle \frac{{8}}{100}\)

∴ Rate pf per 100000 sq m is = \(\displaystyle \frac{{8}}{100}\) × 10000

= ₹ 8000

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

Solutions :

A table-top measures 2 m by 1 m 50 cm

Area of a table-top = Length × Breadth

= 2 × 1.5

= 3.0 sq m

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?

Solutions :

Length of Room = 4 m

Breadth of Room = 3 m 50 cm

Area of Room = Length × Breadth

= 4 × 3.5

= 14.0 sq meter

So, 14 square metres of carpet is needed to cover the floor of the room?

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 :

Carpet length = 5 m

Carpet breadth = 4 m

Area of carpet = Length × Breadth

= 5 × 4

= 20 sq m

Square carpet side = 3 m

Area of Square carpet = (side)^{2}

= (3)^{2}

= 9 sq meter

The area of the floor that is not carpeted = Area of carpet − Area of Square carpet

= 20 − 9

= 11 sq meter

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 :

Length of rectangular piece of land = 5 m

Breadth of rectangular piece of land = 4 m

Area of rectangular piece of land = Length × Breadth

= 5 × 4

= 20 sq meter

Square flower bed side = 1 m

Area of Square flower bed = (side)^{2}

= (1)^{2}

= 1 sq m

Five square flower bed area = 1 × 5 = 5 sq m

The area of the remaining part of the land = Area of rectangular piece of land − Five square flower bed area

= 20 − 5

= 15 sq m

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

Solutions :

**Area of figures (a)**

First part area = 1 × 1 = 1

Second part area = 4 × 2 = 8

Third part area = 4 × 1 = 4

Fourth part area = 4 × 2 = 8

Fifth part area = 3 × 1 = 3

Total Area of figures (a) = 1 + 8 + 4 + 8 + 3 = 24 sq cm

Area of figures (b)

First part area = 5 × 1 = 5

Second part area = 2 × 1 = 2

Third part area = 2 × 1 = 2

Total Area of figures (b) = 5 + 2 + 2 = 9 sq cm

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

Solutions :

Area of figures (a)

First part area = 12 × 2 = 24 sq cm

Second part area = 8 × 2 = 16 sq cm

Total Area of figures (a) = 24 + 16 = 40 sq cm

Area of figures (b)

First part area = 7 × 7 = 49 sq cm

Second part area = 21 × 7 = 147 sq cm

Third part area = 7 × 7 = 49 sq cm

Total Area of figures (b) = 49 + 147 + 49 = 245 sq cm

Area of figures (c)

First part area = 5 × 1 = 5 sq cm

Second part area = 4 × 1 = 4 sq cm

Total Area of figures (c) = 5 + 4 = 9 sq cm

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 :

Area of tiles = 5 × 12 = 60 sq cm

(a) 100 cm and 144 cm

Area of rectangular region = Length × Breadth

= 144 × 100

= 14400 sq cm

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

(b) 70 cm and 36 cm

Area of rectangular region = Length × Breadth

= 70 × 36

= 2520 sq cm

= \(\displaystyle \frac{{2520}}{60}\)
= 42 tiles