NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Ex 10.1 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Ex 10.1.

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 6 |

Subject |
Maths |

Chapter |
Chapter 10 |

Chapter Name |
Mensuration |

Exercise |
Ex 10.1 |

Number of Questions Solved |
17 |

Category |
NCERT Solutions |

## NCERT Solutions for Class 6 Maths Chapter 10 Mensuration Ex 10.1

Ex 10.1 Class 6 Maths 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

Ex 10.1 Class 6 Maths 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

Ex 10.1 Class 6 Maths 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

Ex 10.1 Class 6 Maths 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.

Ex 10.1 Class 6 Maths 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

Ex 10.1 Class 6 Maths 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

Ex 10.1 Class 6 Maths 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

Ex 10.1 Class 6 Maths 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

Ex 10.1 Class 6 Maths 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

Ex 10.1 Class 6 Maths 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

Ex 10.1 Class 6 Maths 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

Ex 10.1 Class 6 Maths 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.

Ex 10.1 Class 6 Maths 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

Ex 10.1 Class 6 Maths 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

Ex 10.1 Class 6 Maths 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.

Ex 10.1 Class 6 Maths 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.

Ex 10.1 Class 6 Maths 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

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