Contents

NCERT Solutions for Class 6 Maths Chapter 10 Mensuration are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 10 Mensuration.

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 6 |

Subject |
Maths |

Chapter |
Chapter 10 |

Chapter Name |
Mensuration |

Exercise |
Ex 10.1, Ex 10.2, Ex 10.3 |

Number of Questions Solved |
30 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 6 Maths Chapter 10 Mensuration

### Chapter 10 Mensuration Exercise 10.1

Question 1.

Find the perimeter of each of the following figures:

Solution:

**(a)** Perimeter = the sum of the lengths of sides

= 5 cm +1 cm +2 cm + 4 cm = 12cm

**(b)** Perimeter = the sum of the lengths of sides

= 40 cm +35 cm + 23 cm +35 cm = 133 cm

**(c)** Perimeter = 4x the length of one side

= 4 x 15 cm = 60 cm

**(d)** Perimeter = 5 x the length of one side

4 = 5 x 4 cm = 20 cm

**(e)** Perimeter = the sum of the lengths of sides

= 4 cm + 0.5 cm + 2.5 cm + 2.5 cm + 0.5 cm + 4 cm + 1 cm = 15 cm

**(f)** Perimeter = the sum of the lengths of sides

= 4 cm + 3 cm + 2 cm + 3 cm + 1 cm + 4 cm + 3 cm + 2 cm + 3 cm + 1 cm + 4 cm + 3 cm + 2 cm +3 cm +1 cm + 4 cm + 3 cm + 2 cm + 3 cm + 1 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?

Solution:

Length of the tape required

= Perimeter of the lid of a rectangular box = 2 x ( length + breadth)

= 2 x (40 cm +10 cm)

= 2 x 50 cm = 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 table-top?

Solution:

Perimeter of the table-top

= 2 x (length + breadth)

= 2 x (2 m 25 cm +1 m 50 cm)

= 2 x (3 m 75 cm)

= 2 x 3.75 m = 7.50 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?

Solution:

The length of the wooden strip required to frame a photograph is the perimeter of the photograph.

Perimeter of the photograph = 2 x (length + breadth)

= 2 x (32 cm +21 cm)

= 2 x 53 cm = 106 cm

∴ 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:

We have to cover 4 times the perimeter of the land measuring 0.7 km by 0.5 km.

∴ Total length of wire required is 4 times its perimeter.

Perimeter of the land = 2 x (length + breadth)

= 2x (0.7 km+0.5 km)

= 2 x 1.2 km = 2.4 km

∴ Total length of wire required = 4 x 2.4 km = 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 each and third side 6 cm.

Solution:

**(a)** Perimeter = the sum of the sides

= 3 cm + 4 cm + 5cm = 12 cm

**(b)** Perimeter = 3 x the length of one side

= 3 x 9 cm = 27 cm

**(c)** Perimeter = the sum of the lengths of the sides

= 8 cm + 8 cm + 6 cm = 22 cm

Question 7.

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

Solution:

Perimeter of the triangle = the sum of the lengths of its sides

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

Question 8.

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

Solution:

A regular hexagon has 6 sides, so its perimeter

= 6 x length of its one side = 6 x 8 m

= 48 m

Question 9.

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

Solution:

Perimeter = 20 m

A square has 4 equal sides, so we can divide the perimeter by 4 to get the length of one side.

One side of the square = 20 m + 4 = 5m

Question 10.

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

Solution:

Perimeter = 100 cm

A regular pentagon has 5 equal sides, so we can divide the perimeter by

5 to get the length of one side

∴ length of one side = 100 cm + 5 = 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 forms

**(a)** a square?

**(b)** an equilateral triangle?

**(c)** a regular hexagon?

Solution:

**(a)** Perimeter = Length of the string = 30 cm

A square has 4 equal sides, so we can divide the perimeter by 4 to get the length of one side.

One side of the square = 30 cm + 4 = 7.5 cm

**(b)** Perimeter = Length of the string = 30 cm

An equilateral triangle has 3 equal sides, so we can divide the perimeter by 3 to get the length of one side.

∴ One side of an equilateral triangle = 30 cm ÷ 3 = 10 cm

**(c)** Perimeter = Length of the string = 30 cm

A regular hexagon has 6 equal sides, so we can divide the perimeter by

6 to get the length of one side.

One side of a regular hexagon = 30 cm + 6 = 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?

Solution:

Let ABC be the given triangle such that AB =12 cm, BC = 14 cm and its perimeter = 36 cm.

i.e., AB + BC + CA = 36 cm

or 12 cm + 14 cm + CA = 36 cm

or 26 cm + CA = 36 cm

or CA =36 cm – 26 cm = 10 cm

∴ The third side of triangle is 10 cm.

Question 13.

Find the cost offencing a square park ofside 250 mat the rate of ? 20 per metre.

Solution:

Side of the square park = 250 m

∴ Perimeter of the square park

= 4 x side

= 4 x 250 m = 1000 m

∴ Cost of fencing = ₹ (1000 x 20)

= ₹ 20000

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 the rectangular park = 175 m Breadth of the rectangular park 125 m .-. Perimeter of the park = 2 x (length + breadth)

= 2 x (175m + 125m)

= 2 x 300 m = 600 m

∴ Cost of fencing = ₹ (600 x 12) = ₹ 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?

Solution:

The distance each girl covers in one round is the same as the perimeter of the respective field. Therefore, the distance that Sweety covers in one round

= 4 x side

= 4 x 75 m = 300 m

Also, the distance that Bulbul covers in one round

= 2 x (length + breadth)

= 2 x (60 m + 45 m)

= 2 x 105 m = 210 m

This shows that Bulbul covers less distance than Sweety.

Question 16.

What is the perimeter of each of the following figures? What do you infer from the answer?

Solution:

**(a)** Perimeter = 4 x side

= 4 x 25 cm

= 100 cm

**(b)** Perimeter = 2 x (length + breadth)

= 2 x (40 cm +10 cm)

= 2 x 50 cm = 100 cm

**(c)** Perimeter = 2 x (length + breadth)

2 x (30 cm +20 cm)

= 2 x 50 cm = 100 cm

**(d)** Perimeter = 30 cm +30 cm +40 cm = 100 cm

Thus, we observe that the perimeter of each figure is 100 cm i.e., they have equal perimeters.

Question 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. (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.)

Solution:

**(a)** **In case of Avneet’s arrangement:**

**(b)** **In case of Shari’s arrangement:**

**(c)** Clearly, perimeter in case of Shari is greater.

**(d)** Yes, there is a way shown in the figure in which we get a greater perimeter

Perimeter = 2 x (9 + 1) units = 2 x 10 units = 20 units

### Chapter 10 Mensuration Exercise 10.2

Question 1.

Find the areas of the following figures by counting square:

Solution:

Placing these figures on the centimetre square, we have

**(a)** Full squares = 9

∴ Area covered by the figure = 9 x 1 sq. cm = 9 sq. cm

**(b)** Full squares = 5

∴ Area covered by the figure = 5 x 1 sq. cm = 5 sq. cm

**(c)** Full squares = 2

Half squares = 4

∴ Area covered by the figure sq. cm

= (2 + 2) sq cm = 4 sq. cm

**(d)** Full squares = 10

∴ Area covered by the figure = 10 x 1 sq. cm = 10 sq. cm

**(e)** Full squares = 10

∴ Area covered by the figure = 10 x 1 sq. cm = 10 sq. cm

**(f)** Full squares = 2

Half squares = 4

∴ Area covered by the figure = sq. cm

= (2 + 2)sq. cm = 4 sq. cm

**(g)** Full squares = 4

Half squares = 4

∴ Area covered by the figure = sq. cm

= (4 + 2)sq. cm = 6 sq. cm

**(h)** Full squares = 5

∴ Area covered by the figure = 5 x 1 sq. cm

= 5 sq. cm

**(i)** Full squares = 9

∴ Area covered by the figure = 9 x 1 sq. cm

= 9 sq. cm

**(j)** Full squares = 2

Half squares = 4

∴ Area covered by the figure = sq. cm

= (2 + 2)sq. cm = 4 sq. cm

**(k)** Full squares = 4

Half squares = 2

∴ Area covered by the figure = sq. cm

= (4 + 1)sq. cm = 5 sq. cm

**(l)** Full squares = 4

More than half squares = 3

Half square = 2

∴ Area covered by the figure = sq. cm

= (4 + 3 + 1)sq. cm

= 8 sq. cm

**(m)** Full squares = 7

More than half squares = 7

Half square = 0

∴ Area covered by the figure = sq. cm

= (7 + 7 + 0)sq. cm

= 14 sq. cm

**(n)** Full squares = 10

More than half squares = 8

Half square = 0

∴ Area covered by the figure = sq. cm

= (10 + 8 + 0)sq. cm

= 18 sq. cm

### Chapter 10 Mensuration 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

Solution:

**(a)** Length of the rectangle = 4 cm

Breadth of the rectangle = 3 cm

Area = Length x Breadth

= 4 cm x 3 cm

= 12 sq. cm

**(b)** Length of the rectangle = 21 m

Breadth of the rectangle = 12 m

Area = Length x Breadth

= 21 m xl2 m

= 252 sq. m.

**(c)** Length of the rectangle = 3 km

Breadth of the rectangle =2 km

Area = Length x Breadth

= 3 km x 2km

= 6 sq. km

**(d)** Length of the rectangle =2m = 2 x l00= 200 cm

Breadth of the rectangle = 70 cm

Area = Length x Breadth

= 200 cm x 70 cm

= 14000 sq. cm

Question 2.

Find the areas of the squares whose sides are:

**(a)** 10 cm

**(b)** 14 cm

**(c)** 5 m

Solution:

**(a)** Side of the square = 10 cm

Area = (side)^{2}

= (10)^{2} sq. cm

= 100 sq. cm

**(b)** Side of the square = 14 cm

Area = (side)^{2}

= (14)^{2} sq. cm

= 196 sq. cm

**(c)** Side of the square = 5 m

Area = (side)^{2}

= (5)^{2} sq. m = 25sq. m

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?

Solution:

In rectangle (a), Area = (9 x 6) sq. m

= 54 sq. m

In rectangle (b), Area =(3 x 17) sq. m = 51 sq. m

In rectangle (c), Area = (4 x 14) sq. m = 56 sq. m

Clearly, 56 > 54 > 51 i.e., 56 is the largest number and 51 is the smallest number.

∴ The rectangle (c) has the largest area and the rectangle (a) has the smallest area.

Question 4.

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

Solution:

We use the formula A = l x b, where l is the length and b is the breadth of the rectangle in metre and its area in m2.

Here, A = 300 m^{2}, and l = 50 m

Therefore, 300 =50 x b [ ∵ A = l x b]

or

Hence the breadth (width) of the rectangle 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:

Area of rectangular piece of land = 500 x 200 sq. m

= 100000 sq. m

Cost of tiling the land at the rate of ₹ 8 per hundred sq. m

= ₹ = ₹ 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’s top = 2 m

Breadth of table’s top = 1 m 50 cm = 1.50 m

∴ Area = Length x Breadth

= (2 x 1.50) sq m = 3 sq m

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 the room = 4 m

Breadth of the room = 3 m 50 cm = 3.50 m

Carpet needed to cover the floor of the room

= Area of the floor of the room

= Length x Breadth

= (4 x 3.50) sq. m = 14 sq. m

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:

Area of the floor of the room = (5 x 4) sq. m

= 20 sq. m

Area of the carpet = (3 x 3) sq. m = 9 sq. m

Area of the floor not carpeted = (20 – 9) sq. m

= 11 sq. m

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:

Area of the piece of land =(5 x 4) sq. m

= 20 sq. m

Area of square flower bed =(1 x 1) sq. m

= 1 sq. m

Area of 5 such flower beds =(5 x 1) sq. m

= 5 sq. m

∴ Area of the remaining part of land

= (20 – 5) sq. m = 15 sq. m

Question 10.

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

Solution:

**(a)** Let the figure may be divided into rectangles marked as A, B, C etc. Area of rectangle (A) = (3 x 3) sq. cm = 9 sq. cm

Area of rectangle (B) = (1 x 2) sq. cm = 2 sq. cm

Area of rectangle (C) = (3 x 3) sq. cm = 9 sq. cm

Area of rectangle (D) = (4 x 2) sq. cm = 8 sq. cm

∴ The total area of the figure = (9 + 2 + 9 + 8) sq. cm = 28 sq. cm

**(b)** Let the figure may be divided into rectangle marked as shown.

Area of rectangle (A) = (2 x 1) sq cm = 2 sq cm

Area of rectangle (B) = (5 x 1) sq cm = 5 sq cm

Area of rectangle (C) = (2 x 1) sq cm = 2 sq cm

∴ Total area of the figure = (2 + 5 + 2) sq cm = 9 sq cm

Question 11.

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

Solution:

**(a)** Splitting the given shape into rectangles as marked A and B.

Area of the rectangle (A) = (10 x 2) sq cm

= 20 sq cm

Area of the rectangle (B) = (10 x 2) sq cm

= 20 sq cm

∴ Total area of given shape = (20 +20) sq cm

= 40 sq cm

**(b)** Splitting the given shape into five squares each of side 7 cm.

Area of the given shape

= 5 x Area of one square of side 7 cm = 5 x (7 x 7) sq. cm

= (5 x 49)sq cm 7

= 245 sq cm

**(c)** Splitting the given shape into two rectangles named A and B. Area of rectangle (A) = (5 x 1) sq cm = 5 sq cm

Area of rectangle (B) = (4 x 1) sq cm = 4 sq cm

∴ Total area of given shape = (5 + 4) sq cm = 9 sq cm

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

Solution:

**(a)** Area of the rectangular region

= (100 x 144) sq cm

= 14400 sq cm

Area of one tile = (5 x 12) sq cm

= 60 sq cm

**(b)** Area of the rectangular region

= (70 x 36) sq cm

= 2520 sq cm

We hope the NCERT Solutions for Class 6 Maths Chapter 10 Mensuration help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 10 Mensuration, drop a comment below and we will get back to you at the earliest.

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