NCERT Solutions Class 6 Maths Chapter 10 Mensuration

• Exercise 10.1: In this exercise, students will learn the perimeter of Simple Shapes. Understand and calculate the perimeter of basic geometric shapes. Calculation of perimeter for rectangles, squares, and triangles. Application of perimeter in real-life scenarios.

• Exercise 10.2: In this exercise, students will learn how to calculate the area of rectangles and squares using specific formulas. Introduction to the area. Solving problems related to finding the area.

• Exercise 10.3: In this exercise, students will learn Mixed Problems on Perimeter and Area. The objective is to solve problems involving both perimeter and area calculations. Combining the concepts of perimeter and area. Solving real-life application problems. Enhancing problem-solving skills with mixed questions.

Access NCERT Solutions for Class 6 Maths Chapter 10 – Mensuration

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

Ans:  According to this figure:

As we know Perimeter= Sum of all the sides

=$1+2+4+5$ 

=$3+8$ 

=$11$ 

So, the perimeter= $11$ cm

2. Find the perimeter of each of the following figures:

Ans: According to this figure:

 As we know Perimeter= Sum of all the sides

=$23+35+35+40$ 

=$58+75$ 

=$133$ 

So, the perimeter= $133$ cm

3. Find the perimeter of each of the following figures:

Ans:  According to this figure:

As we know Perimeter= Sum of all the sides

=$15+15+15+15$ 

=$30+30$ 

=$60$ 

So, the perimeter= $60$ cm

4. Find the perimeter of each of the following figures:

Ans: According to this figure:

As we know Perimeter= Sum of all the sides

=$4+4+4+4+4$ 

=$8+8+4$ 

=$20$ 

So, the perimeter= $20$ cm

5. Find the perimeter of each of the following figures:

Ans: According to this figure:

As we know Perimeter= Sum of all the sides

=\[\text{1 + 4 + 0}\text{.5 + 2}\text{.5 + 2}\text{.5 + 0}\text{.5 + 4 }\] 

=$5+3+3+4$ 

=$8+7$ 

So, the perimeter= $15$ cm

6. Find the perimeter of each of the following figures:

Ans: According to this figure:

As we know Perimeter= Sum of all the sides

=\[\text{4 + 1 + 3 + 2 + 3 + 4 + 1 + 3 + 2 + 3 + 4 + 1 + 3 + 2 + 3 + 4 + 1 + 3 + 2 + 3 }\] 

=$5+5+7+4+5+5+5+7+4+5$ 

=$10+11+10+12+9$ 

=$21+22+9$ 

=$52$ 

So, the perimeter= $52$ cm

7. 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?

Ans:

Here,

Required tape length = Perimeter of rectangle

Given, Length=$40$cm

Breadth=$10$cm

=$2(length+breadth)$ 

= $2(40+10)$

=$2\times 50$

=$100$

Required tape length = $100$ cm.

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

Ans: Here

$25$cm = $0.25$cm

$50$cm = $0.5$cm

So, Length of table top= $2+0.25$= $2.25$m

Breadth of table top = $1+0.5$=$1.5$m

Perimeter of table top = $2(length+breadth)$

=$2(2.25+1.5)$

=$2\times 3.75$

=$7.50$m

So, Perimeter = $7.5$m

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

Ans: Here

Given, Length=$32$cm

Breadth=$21$cm

Required length of wooden strip  = Perimeter of photograph

=$2(length+breadth)$

=$2(32+21)$

=$2\times 53$cm

=$106$cm

Required Length of wooden strip = $106$cm

10. 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: As we know,

Perimeter of field = $2(length+breadth)$

Given, Length=$0.7$km

Breadth=$0.5$km

=$2(length+breadth)$

=\[2(0.7+0.5)\]

=$2\times 1.2$km

=$2.4$km

A $4$row fence is required on each side =$4\times 2.4$= $9.6$km

So, Total length of the required field =  $9.6$km

11. Find the perimeter of each of the following shapes: A triangle of sides $3$ cm,  $4$cm and $5$ cm. 

Ans: We know,

Perimeter of Triangle=\[3+4+5\]

=\[12\]cm

So, Perimeter of Triangle=\[12\]cm

12. Find the perimeter of each of the following shapes:  An equilateral triangle of side $9$ cm

Ans: We know,

Perimeter of Equilateral Triangle=\[3\times side\]

=\[3\times 9\]cm

= \[27\]cm

So, Perimeter of Equilateral  Triangle=\[27\]cm

13. Find the perimeter of each of the following shapes: An isosceles triangle with equal sides $8$ cm each and third side $6$ cm.  

Ans: We know,

Perimeter of isosceles Triangle=\[8+8+6\]

=\[22\]cm

So, Perimeter of isosceles Triangle=\[22\]cm

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

Ans: We know,

Perimeter of Triangle=\[10+14+15\]

=\[39\]cm

So, Perimeter of Triangle=\[39\]cm

15. Find the perimeter of a regular hexagon with each side measuring $8$ cm. 

Ans: We know,

Perimeter of Hexagon=\[6\times 8\]

=\[48\]m

So, Perimeter of Hexagon=\[48\]m

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

Ans: We know,

Perimeter of square=\[4\times side\]

\[20=4\times side\]

Side=\[\frac{20}{4}\]

=$5$ m

So, Perimeter of Square=\[5\]m

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

Ans: We know,

Perimeter of regular pentagon=\[100\]cm

\[100=5\times side\]

Side=\[\frac{100}{5}\]

=$20$ cm

So, Perimeter of Square=\[20\]cm

18. A piece of string is $30$ cm long. What will be the length of each side if the string is used to form: a square 

Ans: We know,

Perimeter of square=\[4\times side\]

\[30=4\times side\]

Side=\[\frac{30}{4}\]

=$7.5$ cm

So, Perimeter of Square=\[7.5\]cm

19. A piece of string is $30$ cm long. What will be the length of each side if the string is used to form: an equilateral triangle  

Ans: We know,

Perimeter of Equilateral Triangle=\[3\times side\]

\[30=3\times side\]

Side=\[\frac{30}{3}\] cm

=$10$ cm

So, Perimeter of Equilateral Triangle=\[10\]cm

20. A piece of string is $30$ cm long. What will be the length of each side if the string is used to form: a regular hexagon

Ans: We know,

Perimeter of Hexagon=\[30\]cm

\[6\times side=30\]cm

Side=\[\frac{30}{6}\] cm

=$5$ cm

So, Perimeter of Hexagon=\[5\]cm

21. Two sides of a triangle are $12$ cm and $14$ cm. The perimeter of the triangle is $36$ cm. What is the third side? 

Ans: Assume third side to be $x$ cm

We know,

Perimeter of Triangle=\[36\]

\[12+14+x=36\]cm

\[26+x=36\]cm

\[x=36-26\]

\[x=10\]cm

So, third side =\[10\]cm

22. Find the cost of fencing a square park of side $250$ m at the rate of ` $20$ per meter.

Ans: Given, Side of square =\[250\]m

We know,

Perimeter of square=\[4\times side\]

\[=4\times 250\]

=\[1000\]m

=$7.5$ cm

So, Perimeter of Square=\[7.5\]cm

Now, Cost of fencing= Rs.$20$per m

Than, for $1000$m cost of fencing = $20\times 1000$

=Rs.$20000$

23. Find the cost of fencing a rectangular park of length $175$ m and breadth $125$ m at the rate of ` $12$ per meter. 

Ans: Given, Length=$175$m

Breadth=$125$m

We know, Perimeter of park=$2(length+breadth)$

=$2(175+125)$

=$2\times 300$

=$600$m

Cost of fencing = $12\times 600$=Rs.$7200$

Hence, Cost of fencing = Rs.$7200$

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

Ans: We know,

Perimeter of square=\[4\times side\]

=\[4\times 75=300\]m

Distance covered by Sweety =\[300\]m

Perimeter of rectangular park=$2(length+breadth)$

=$2(60+45)$

=$2\times 105$

=$210$m

Distance covered by Bulbul =\[210\]m

So, distance covered by Bulbul is less.

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

Ans:

(a) We know,

Perimeter of square=\[4\times side\]

=\[4\times 25\]

=\[100\]cm

(b) We know, 

Perimeter of rectangle=$2(length+breadth)$

=$2(40+10)$

=$2\times 50$

=$100$cm

(c) We know, 

Perimeter of rectangle=$2(length+breadth)$

=$2(30+20)$

=$2\times 50$

=$100$cm

(d) We know,

Perimeter of Triangle=\[30+30+40\]

=\[100\]cm

The perimeter of each fig is the same. 

26. Avneet buys $9$ square paving slabs, each with a side $\frac{1}{2}$m He lays them in the form of a square  

(a) What is the perimeter of his arrangement?

Ans: Side of Square= $3\times side$

=$3\times \frac{1}{2}$

=$\frac{3}{2}$m

Perimeter of Square = $4\times \frac{3}{2}$

=$2\times 3$

=$6$m

(b) Shari does not like his arrangement. She gets him to lay them out like a cross. What is the perimeter of her arrangement? 

Ans: Perimeter = \[\text{ 0}\text{.5 + 1 + 1 + 0}\text{.5 + 1 + 1+ 0}\text{.5 + 1 + 1 + 0}\text{.5 + 1 + }1\]

=\[10\]m

(c) Which has greater perimeter? 

Ans: Cross arrangement 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.) 

Ans: Yes, the perimeter will be 10 cm if all of the squares are lined up in a straight line.

Exercise 10.2

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

Ans:  Here,

The number of squares is $9$

So, Area = $9$ sq. units

2. Find the areas of the following figures by counting squares: 

Ans:  Here,

The number of squares is $5$

   So, Area = $5$ sq. units

3. Find the areas of the following figures by counting squares:  

Ans:  Here,

The number of complete squares is $2$

The number of half squares is $4$

   So,  Total Area = $2+4(0.5)=4$ sq. units 

4. Find the areas of the following figures by counting squares:  

Ans: Here,

The number of squares is $8$

   So, Area = $8$ sq. units 

5. Find the areas of the following figures by counting squares:  

Ans: Here,

 The number of squares is $10$

   So, Area = $10$ sq. units 

6. Find the areas of the following figures by counting squares:  

Ans: Here,

 The number of complete squares is $2$

The number of half squares is $4$

   So,  Total Area = $2+4(0.5)=4$ sq. units 

7. Find the areas of the following figures by counting squares: 

Ans: Here,

The number of complete squares is $4$

The number of half squares is $4$

   So,  Total Area = $4+4(0.5)=6$ sq. units 

8. Find the areas of the following figures by counting squares:  

Ans: Here,

 The number of squares is $5$

   So, Area = $5$ sq. units 

9. Find the areas of the following figures by counting squares: 

Ans: Here,

The number of squares is $9$

   So, Area = $9$ sq. units 

10. Find the areas of the following figures by counting squares: 

Ans: Here,

The number of complete squares is $2$

The number of half squares is $4$

So,  Total Area = $2+4(0.5)=4$ sq. units 

11. Find the areas of the following figures by counting squares:  

Ans: Here,

The number of complete squares is $4$

The number of half squares is $2$

   So,  Total Area = $4+2(0.5)=5$ sq. units 

12. Find the areas of the following figures by counting squares:  

Ans: Here,

The number of complete squares is $3$

The number of half squares is $10$

   So,  Total Area = $3+10(0.5)=8$ sq. units 

13. Find the areas of the following figures by counting squares:  

Ans: Here,

The number of complete squares is $7$

The number of half squares is $14$

So, Total Area = $7+14(0.5)=14$ sq. units 

14. Find the areas of the following figures by counting squares:  

Ans:  Here,

The number of complete squares is $10$

The number of half squares is $16$

 So,  Total Area = $10+16(0.5)=18$ sq. units 

Exercise 10.3

1. Find the areas of the rectangles whose sides are:$3$cm and $4$cm

Ans: We know, the area of triangle = Length $\times $ Breadth

Here,

Length=$3$cm

Breadth=$4$cm

Area=$3\times 4$= $12$cm2

2. Find the areas of the rectangles whose sides are:$12$m and $21$m

Ans: We know, the area of triangle = Length $\times $ Breadth

Here,

Length=$12$m

Breadth=$21$m

Area=$12\times 21$= $252$m2

3. Find the areas of the rectangles whose sides are:$2$km and $3$km

Ans: We know, the area of triangle = Length $\times $ Breadth

Here,

Length=$2$km

Breadth=$3$km

Area=$2\times 3$= $6$km2

4. Find the areas of the rectangles whose sides are:$2$m and $70$cm

Ans: We know, the area of triangle = Length $\times $ Breadth

Here,

Length=$2$m

Breadth=$70$cm = $0.7$m

Area=$2\times 0.7$= $1.4$m2

5. Find the areas of the squares whose sides are: $10$cm

Ans: We know,

 Area of square = Side2

=${{10}^{2}}$

=$100$ cm2 

6. Find the areas of the squares whose sides are: $14$cm

Ans: We know,

 Area of square = Side2

=\[{{14}^{2}}\]

=$196$ cm2

7. Find the areas of the squares whose sides are: $5$cm

Ans: We know,

 Area of square = Side2

=${{5}^{2}}$

=$25$ m2

8. The length and the 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?  

Ans:

(a) We know, the area of rectangle = Length $\times $ Breadth

Here,

Length=$9$m

Breadth=$6$m

Area=$9\times 6$= $54$m2

(b) We know, the area of rectangle = Length $\times $ Breadth

Here,

Length=$3$m

Breadth=$17$m 

Area=$3\times 17$= $51$m2

(c) We know, the area of rectangle = Length $\times $ Breadth

Here,

Length=$4$m

Breadth=$14$m 

Area=$4\times 14$= $56$m2

Rectangle (c) has the largest size, $56$ m2, whereas rectangle (b) has the smallest area, $51$ m2. 

9. The area of a rectangle garden $50$ m long is $300$ m2, find the width of the garden. 

Ans: We know, the area of rectangle = Length $\times $ Breadth

Here,

Area of rectangle =$300$ m2

Length= $50$m

$300=50\times width$

Width=$6$m

10. What is the cost of tilling a rectangular plot of land $500$ m long and $200$ m wide at the rate of ` $8$ per hundred sq. m? 

Ans: Area of the land can be calculated as

=$500\times 200$

=$1,00,000$m2

Cost of tiling $1,00,000$m2 of land will be

=$\frac{8\times 100000}{100}$ = Rs.$8000$

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

Ans: Given,

L=$2$m

B=$1$m$50$cm= $1.5$m

Area = L$\times $B

=$2\times 1.50=3$m2

12. A room us $4$ m long and $3$ m $50$ cm wide. How many square meters of carpet is needed to cover the floor of the room?

Ans:: Given,

L=$4$m

B=$3$m$50$cm= $3.5$m

Area = L$\times $B

=$4\times 3.50=14$m2

13. 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:  Here,

Area of floor = L$\times $B=$5$ m $\times $ $4$ m = $20$ m2

Area of square carpet = $3$ $\times $ $3$ = $9$ m2

Area of floor that is not carpeted =  $20-9$=$11$ m2

14. 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: Here,

Area of flower square bed =$1\times 1=1$ m2

 Area of 5 square beds = $1\times 5=5$ m2

Area of land = $5\times 4=20$m2

Remaining Area = Area of land – Area of 5 flower beds = $20-5=15$ m2

Therefore, remaining part of the land= $15$m2

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

Ans: 

Here,

Area of 1st region =$3\times 3=9$ cm2

 Area of 2nd region = $1\times 2=2$cm2

 Area of 3rd region = $3\times 3=9$ cm2  

 Area of 4th region = $2\times 4=8$ cm2

 Hence, Total area = $9+2+9+8$ = $28$ cm2   

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

Ans: 

Area of 1st region =$3\times 1=3$ cm2

 Area of 2nd region = $3\times 1=3$cm2

 Area of 3rd region = $3\times 1=3$ cm2  

 Hence, Total area = $3+3+3$ = $9$ cm2   

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

Ans:

 Here,

Area of 1st rectangle= $12\times 2$

Area of 2nd rectangle= $8\times 2$

Total area of the figure = $12\times 2+8\times 2$ = $40$ cm2

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

Ans: Here,

From figure it is clear that there are five square with each side  $7$.

Area of one square=$7\times 7=49$

So, area of five square= $49\times 5=245$cm2

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

Ans:

 Area of 1st region = $5\times 1=5$cm2

Area of 2nd region =$2\times 1=2$ cm2

 Area of 3rd region = $2\times 1=2$ cm2  

 Hence, Total area = $2+5+2$ = $9$ cm2   

20. 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:$100$ cm and $144$ cm

Ans: Here,

Area of Rectangle= $100\times 144=14400$cm2 

 Area of one tile = $5\times 12=60$cm2 

We know, Number of tiles = Area of rectangle / Area of one tile  

=$\frac{14400}{60}$ 

= $240$cm2   

$240$ tiles are required.

21. 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:$70$ cm and $36$ cm

Ans: Here,

Area of Rectangle= $70\times 36=2520$cm2 

 Area of one tile = $5\times 12=60$cm2 

We know, Number of tiles = Area of rectangle / Area of one tile  

=$\frac{2520}{60}$

= $42$cm2   

$42$ tiles are required. 

Class 6 Maths Chapter 10: Exercises Breakdown

Conclusion

Class 6 Maths Chapter Mensuration provides students with essential tools to measure and understand the perimeter and area of various geometric shapes like rectangles and squares. Through this exercise, students learn to apply fundamental formulas, enhancing their ability to solve practical problems involving space and boundaries. By mastering these concepts, students gain the ability to solve practical problems and apply their knowledge to real-world scenarios, such as calculating the amount of material needed for different projects. The exercises and problems in this chapter reinforce practical application and critical thinking, laying a strong foundation for future geometrical studies.

Other Study Material for CBSE Class 6 Physics Chapter 10

Chapter-Specific NCERT Solutions for Class 6 Maths

Given below are the chapter-wise NCERT Solutions for Class 6 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.