NCERT Solutions for Class 6 Maths Chapter 10 - Mensuration

Exercise 10.1

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. 

NCERT Solutions Class 6 Maths Chapter 10 Mensuration - Download PDF for Free

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NCERT Solution Class 6 Maths of Chapter 10 All Exercises

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Preparation Tips For Class 6 Maths Mensuration Chapter

To help students get a better understanding of the Mensuration concepts, we provide the NCERT Solutions for Class 6 Mensuration Chapter PDF free of cost. Inculcate the following tips to score higher-

• Understand the fundamentals of Mensuration by going through the definitions.

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• Solve the NCERT Class 6 Mensuration questions on a regular basis. Take the help of the solutions PDF to learn various approaches.

• Include the questions of all sub-topics related to chapter so as to not miss out on any topic.

• Take the aid of revision notes to refresh the concepts and retain the formulas.