## NCERT Solutions For Class 6 Maths Chapter 10 Mensuration

Here at MentorAtHome, We provide NCERT Solutions For Class 6 Maths Chapter 10 Mensuration. NCERT solutions help students to improve their performance, and attain better results. In NCERT Solutions Class 6 Maths Chapter 10 there is 3 exercise.

NCERT Solutions For Class 6 Maths Chapter 10 Mensuration, will discuss perameter, area and volume. NCERT Solutions For Class 6 Maths Chapter is important for basic understanding.

### NCERT Solutions For Class 6 Maths Chapter 10 Mensuration Exercises 10.1

Class 6th Maths Chapter 10 Solutions ex 10.1

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

Ans:-**(a) **Perimeter = Sum of all the sides

= 4 cm + 2 cm + 1 cm + 5 cm = 12 cm**(b) **Perimeter = Sum of all the sides

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

= 133 cm or 1.33 m**(c) **Perimeter = Sum of all the sides

= 15 cm + 15 cm + 15 cm + 15 cm = 15 cm x 4 = 60 cm**(d) **Perimeter = Sum of all the sides

= 4 cm + 4 cm + 4 cm + 4 cm + 4 cm – 4 cm x 5 = 20 cm**(e) **Perimeter = Sum of all the sides

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

= 15 cm**(f) **Perimeter = Sum of all the sides

1 cm + 3 cm + 2 cm + 3 cm + 4 cm + 1 cm + 3 cm + 2 cm + 3 cm

= 52 cm

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

Ans:-

Total Length of required tape = Perimeter of rectangle lid

= 2 (Length + Breadth)

= 2 (40 + 10) = 2 (50)

= 2 x 50 = 100 cm

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

Ans:-

Length of table-top = 2 m 25 cm

Breadth of table-top = 1 m 50 cm

Perimeter of the table top = 2 [length + breadth]

= 2 (2.25 + 1.50) = 2 (3.75)

= 2 × 3.75 = 7.5 m

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

Ans:-

Length of the strip = 32 cm

Breadth of the strip = 21 cm

Perimeter = 2 [length + breadth]

= 2 (32 + 21) = 2 (53)

= 2 × 53 = 106 cm

**Q.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?**

Ans:-

Length of the rectangular piece of land = 0.7 km

Breadth of the rectangular piece of land = 0.5 km

Perimeter = 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

**Q.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.**

Ans:-

**(a)**the perimeter of the given triangle

= The sum of all sides of the triangle

= 3 + 4 + 5 = 12 cm

**(b)**the perimeter of the equilateral triangle

= 3 × side

= 3 x 9 = 27 cm

**(c)**Perimeter of the isosceles triangle

= Sum of all the sides of the triangle

= (8 + 8 + 6) = 22 cm

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

Ans:-

The perimeter of a triangle

= Sum of all the sides of the triangle

= 10 + 14 + 15 = 39 cm

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

Ans:-

The perimeter of a regular hexagon

= 6 x side

= 6 x 8 = 48 m.

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

Ans:-

The perimeter of a square = 4 x side

20 = 4 x side

side = 20 ÷ 4 = 5 m

The side of the square is 5 m

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

Ans:-

The perimeter of the regular pentagon = 100 cm

Number of sides in regular pentagon = 5

Length of each side = Perimeter ÷ Number of sides

= 100 ÷ 5 = 20 cm.

**Q.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?**

Ans:-**(a) **Length of string = 30 cm

Number of equal sides in a square = 4

Length of each side

= 30 ÷ 4 = 7.50 cm.**(b)** Length of string = 30 cm

Number of equal sides in equilateral triangle = 3

Length of each side

= 30 ÷ 3 = 10 cm**(c) **Length of string = 30 cm

Number of equal sides in regular hexagon = 6

Length of each side

= 30 cm ÷ 6 = 5 cm

**Q.12. The two sides of a triangle are 12 cm and 14 cm. The perimeter of the triangle is 36 cm. What is its third side?**

Ans:-

Let the third side of the triangle = x

Given two sides of triangle 12cm and 14cm

The perimeter of triangle = 36 cm

The perimeter of triangle = Sum of all the sides of the triangle

12 + 14 + x = 36

26 + x = 36

x = 36 – 26

x = 10 cm

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

Ans:-

Side of square = 250 m

The 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

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

Ans:-

Length of the rectangular park = 175 m

Breadth of the rectangular park = 125 m

Perimeter of the park = 2 [length + breadth]

= 2 (175 + 125) = 2 (300)

= 2 × 300 = 600 m

Cost of fencing = ₹ 12 per m

Cost of fencing for 600 m

12 x 600 = ₹ 7200

**Q.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?**

Ans:-

Side of the square park = 75 m

its perimeter = 4 x 75 = 300 m

Perimeter of the rectangular park = 2 [length + breadth]

= 2 [60 + 45 ]

= 2 x 105 = 210 m.

Since 210 m < 300 m.

So, Bulbul covers less distance.

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

Ans:-**(a)** Perimeter of the square = = 4 × side

= 4 x 25 cm = 100 cm**(c)** Perimeter of the rectangle

= 2 [40 cm + 10 cm] = 2 x 50 cm = 100 cm**(b)** Perimeter of the rectangle

= 2 [30 cm + 20 cm] = 2 x 50 cm = 100 cm **(d)** Perimeter of the triangle = Sum of all sides

= 30 cm + 30 cm + 40 cm = 100 cm

All the figures have 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.)**

Ans:-**(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 the cross has a greater perimeter**(d)** Perimeters greater than 10 m cannot be determined.

### NCERT Solutions For Class 6 Maths Chapter 10 Mensuration Exercises 10.2

NCERT Class 6 Maths Chapter 10 Solutions ex 10.2

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

Ans:-**(a)** Number of full squares = 9

Area of 1 square = 1

∴ Area of 9 squares = 9 x 1 = 9 sq. units.**(b)** Number of full squares = 5

Area of 1 square = 1

∴ Area of the figure = 5 x 1 = 5 sq. units**(c)** Number of full squares = 2

Number of half squares = 4

∴ Area of the covered figure = 2 x 1 + ( 4 x 1 / 2)

= 2 + 2 = 4 sq. units**(d)** Number of full squares = 8

Area of 1 square = 1

∴ Area of the covered figure = 8 x 1 = 8 sq. units.**(e)** Number of full squares = 10

Area of 1 square = 1

Area covered by the figure = 10 x 1 = 10 sq. units.**(f)** Number of full squares = 2

Number of half squares = 4

∴ Area of the covered figure = 2 x 1 + (4 x 1 / 2)

= 2 + 2 = 4 sq. units.**(g) **Number of full squares = 4

Number of half squares = 4

∴ Area of the covered figure = 4 x 1 + (4 x 1 / 2)

= 4 + 2 = 6 sq. units.**(h)** Number of full squares = 5

Area of 1 square = 1

∴ Area of the covered figure = 5 x 1 = 5 sq. units.**(i)** Number of full squares = 9

Area of 1 square = 1

∴ Area of the covered figure = 9 x 1 = 9 sq. units.**(j)** Number of full squares = 2

Number of half squares = 4

∴ Area of the covered figure = 2 x 1 + (4 x 1 / 2)

= 2 + 2 = 4 sq. units.**(k)** Number of full squares = 4

Number of half squares = 2

∴ Area of the covered figure = 4 x 1 + (2 x 1 / 2)

= 4 + 1 = 5 sq. units**(l)** Number of full squares = 4

Number of squares more than half = 3

Number of half squares = 2

∴ Area of the covered figure = 4 x 1 + 3 x 1 + (2 x 1 / 2)

= (4 + 3 + 1) = 8 sq. units.**(m)** Number of full squares = 6

Number of more than half squares = 8

Area of the covered figure = (6 x 1 + 8 x 1)

= 6 + 8 = 14 sq. units.**(n)** Number of full squares = 9

Number of more than half squares = 9

∴ Area of the covered figure= (9 x 1 + 9 x 1)

= 9 + 9 = 18 sq. units.

### NCERT Solutions For Class 6 Maths Chapter 10 Mensuration Exercises 10.3

Ch 10 Class 6 Maths Solutions ex 10.3

**Q.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**

Ans:-

We know that Area of rectangle = Length × Breadth**(a)** Length of the rectangle = 3 cm

Breadth of the rectangle = 4 cm

= 3 cm x 4 cm = 12 cm^{2} **(b)** Length of the rectangle = 12 m

breadth of the rectangle = 21 m

= 12 m x 21 m = 252 m^{2} **(c)** Length of the rectangle = 2 km

breadth of the rectangle 3 km

= 2 km x 3 km = 6 km^{2}

(d) Length of the rectangle = 2 m

breadth of the rectangle = 70 cm or 0.70 m

= 2 m x 0.70 m = 1.40 m^{2}

**Q.2. Find the areas of the squares whose sides are:****(a) 10 cm****(b) 14 cm****(c) 5 m**

Ans:-

We know that Area of the square = Side x Side**(a) **10 cm

= 10 cm x 10 cm = 100 cm^{2} **(b) **14 cm

= 14 cm x 14 cm = 196 cm^{2} **(c) **5 m

= 5 m x 5 m = 25 m^{2}

**Q.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?**

Ans:-

We know that Area of rectangle = Length × Breadth**(a)** Length of the rectangle = 9 m

breadth of the rectangle = 6 m

= 9 m x 6 m = 54 m^{2} **(b)** Length of the rectangle = 17 m

breadth of the rectangle = 3m

= 17 m x 3 m 51 m^{2} **(c)** Length of the rectangle = 4 m

breadth of the rectangle = 14 m

= 4 m x 14 m = 56 m^{2}

Rectangle (c) has the largest area, i.e., 56 m^{2} and (b) has the smallest area, i.e., 51m^{2}

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

Ans:-

Length of the rectangular garden = 50 m

Area of the rectangular garden = 300 m^{2}

Area of rectangle = length × width

300 = 50 × width

width = 300 / 50

width = 6 m

**Q.5. What are 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.?**

Ans:-

Length of the rectangular plot = 500 m

and the breadth = 200 m

Area of the plot = length x breadth

= 500 m x 200 m = 100000 sq m

Now rate of tiling the plot = ₹8 per 100 sq m

= (8 × 1,00,000) / 100 = ₹ 8000

Hence the required cost = ₹8000

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

Ans:-

Length of the table-top = 2 m

breadth of the table-top = 1 m 50 cm or 1.50 m

Area of the table-top = length x breadth

= 2 m x 1.50 m = 3 m^{2}

**Q.7. A room is 4 m long and 3 m 50 cm wide. Howe,** many square meters** of carpet is needed to cover the floor of the room?**

Ans:-

Length of the room = 4 m

breadth of the room = 3 m 50 cm = 3.5 m

Area of the room = length x breadth

= 4 m x 3.5 m = 14 m^{2}

**Q.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.**

Ans:-

Length of the floor = 5 m

breadth of the floor = 4 m

Area of the floor = length x breadth

= 5m x 4m = 20 m^{2}

Side of the carpet = 3 m

Area of the square carpet = side x side

= 3 m x 3 m = 9 m^{2}

Area of the floor which is not carpeted

= 20 m^{2} – 9 m^{2} = 11 m^{2}

**Q.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?**

Ans:-

Side of the square flower bed = 1 m.

Area of the flower bed = side x side

Area of 1 square flower bed = 1 m x 1 m = 1 m^{2}

Area of 5 square flower beds = 1 m x 5 m = 5 m^{2}

Now length of the land = 5 m

breadth of the land = 4 m

Area of the land = length x breadth

=5m x 4m = 20 m^{2}

Area of the remaining part of the land

= 20 m^{2} – 5 m^{2} = 15 m^{2}

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

Ans:-

(A)

Area of the rectangle I = length x breadth

= 4 cm x 3 cm = 12 cm^{2}

Area of the rectangle II = length x breadth

= 3 cm x 2 cm = 6 cm^{2}

Area of the rectangle III = length x breadth

= 4 cm x 1 cm = 4 cm^{2}

Area of the rectangle IV = length x breadth

= 3 cm x 2 cm = 6 cm^{2}

Total area of the given figure

= 12 cm^{2} + 6 cm^{2} + 4 cm^{2} + 6 cm^{2} = 28 cm^{2} **(b)**

Area of the rectangle I length x breadth

= 12 cm x 2 cm = 24 cm^{2}

Area of the rectangle II length x breadth

= 8 cm x 2 cm = 16 cm^{2}

Area of rectangle III length x breadth

= 3 cm x 1 cm = 3 cm^{2}

∴ Total area of the given figure

= 3 cm^{2} + 3 cm^{2} + 3 cm^{2} = 9 cm^{2}

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

Ans:-**(a)** Splitting the given figure into the rectangles I and II, we get

Area of the rectangle I

= 12 cm x 2 cm = 24 cm^{2}

Area of the rectangle II

= 8 cm x 2 cm = 16 cm^{2}

∴ Total area of the figure

= 24 cm^{2} + 16 cm^{2} = 40 cm^{2} **(b)** Splitting the given figure into the rectangles I, II and III, we get

Area of the rectangle I

= 7 cm x 7 cm = 49 cm^{2}

Area of the rectangle II

= 21 cm x 7 cm = 147 cm^{2}

Area of the rectangle III

= 7 cm x 7 cm = 49 cm^{2}

∴ Total area of the figure

= 49 cm^{2} + 147 cm^{2} + 49 cm^{2} = 245 cm^{2} **(c)** Splitting the given figure into the rectangles I, and II, we get

Area of the rectangle I

= 5 cm x 1 cm = 5 cm^{2}

Area of the rectangle II

= 4 cm x 1 cm = 4 cm^{2}

∴ Total area of the figure

= 5 cm^{2} + 4 cm^{2} = 9 cm^{2}

**Q.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**

Ans:-

Length of one tile = 12 cm

Breadth of one tile = 5 cm

Area of 1 tile = length x breadth

= 12 cm x 5 cm = 60 cm^{2}

**(a)** Length of the region = 144 cm

Breadth of the region = 100 cm

Area of the region = length x breadth

= 144 cm x 100 cm = 14400 cm^{2}

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

= 14400 cm^{2} ÷ 60 cm^{2} = 240 tiles

**(b)** Length of the region = 70 cm

Breadth of the region = 36 cm

Area of the rectangular region = length x breadth

= 70 cm x 36 cm = 2520 cm^{2}

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

= 2520 cm^{2} ÷ 60 cm^{2} = 42 tiles.

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NCERT Solutions For Class 6 Maths Chapter 10 mensuration are based on NCERT Books. You can download ncert book for class 6 maths chapter 10. Ch 10 class 6 maths solutions are designed by our subject expert team.

## What We Learn?

- Perimeter is the distance covered along the boundary forming a closed figure

when you go round the figure once. - (a) Perimeter of a rectangle = 2 × (length + breadth)

(b) Perimeter of a square = 4 × length of its side

(c) Perimeter of an equilateral triangle = 3 × length of a side - Figures in which all sides and angles are equal are called regular closed figures.
- The amount of surface enclosed by a closed figure is called its area.
- To calculate the area of a figure using a squared paper, the following conventions

are adopted :

(a) Ignore portions of the area that are less than half a square.

(b) If more than half a square is in a region. Count it as one square.

(c) If exactly half the square is counted, take its area as 1/2 sq units. - (a) Area of a rectangle = length × breadth

(b) Area of a square = side × side