## NCERT Solutions for Ch 10 Mensuration Class 6 Maths

**Exercise 10.1**

**Answer**

(a) Perimeter = Sum of all the sides

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

= 12 cm

(b) Perimeter = Sum of all the sides

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

= 133 cm

(c) Perimeter = Sum of all the sides

= 15 cm + 15 cm + 15 cm + 15 cm

= 60 cm

(d) Perimeter = Sum of all the sides

= 4 cm + 4 cm + 4 cm + 4 cm + 4 cm

= 20 cm

(e) Perimeter = Sum of all the sides

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

= 15 cm

(f) Perimeter = Sum of all the sides

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

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

= 52 cm

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?

**Answer**

Total length of tape required = Perimeter of rectangle

= 2 ( length + breadth)

= 2 ( 40 + 10)

= 2Ã—50

= 100 cm = 1 m

Thus, the total length of tape required is 100 cm or 1 m.

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

**Answer**

Length of table top = 2 m 25 cm = 2.25 m

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

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

= 2 Ã— (2.25 + 1.50)

= 2 Ã— 3.75 = 7.50 m

Thus, perimeter of table top is 7.5 m.

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

**Answer**

Length of wooden strip = Perimeter of photograph

Perimeter of photograph = 2 Ã— (length + breadth)

= 2 (32 + 21)

= 2 Ã— 53 cm = 106 cm

Thus, the length of the wooden strip required is 106 cm.

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?

**Answer**

Since the 4 rows of wires are needed. Therefore the total length of wires is equal to 4 times the perimeter of rectangle.

Perimeter of rectangular piece of land = 2 Ã— (length + breadth)

= 2 Ã— (0.7 + 0.5)

= 2 Ã— 1.2

= 2.4 km

= 2.4 Ã— 1000 m

= 2400 m

Thus, the length of wire = 4 Ã— 2400 = 9600 m = 9.6 km

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.

**Answer**

(a) Perimeter of Î”ABCÎ”ABC = AB + BC + CA

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

(b) Perimeter of equilateral Î”ABCÎ”ABC = 3Ã—side

= 3Ã—9 cm = 27 cm

(c) Perimeter of Î”ABCÎ”ABC = AB + BC + CA

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

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

**Answer**

Perimeter of triangle = Sum of all three sides

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

Thus, perimeter of triangle is 39 cm.

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

**Answer**

Perimeter of Hexagon = 6 Ã— length of one side

= 6 x 8 m = 48 m

Thus, the perimeter of hexagon is 48 m.

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

**Answer**

Perimeter of square = 4 Ã— side

â‡’ 20 = 4 Ã— side

â‡’ side = 20/4 = 5 cm

Thus, the side of square is 5 cm.

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

**Answer**

Perimeter of regular pentagon = 100 cm

â‡’ 5 Ã— side = 100 cm

â‡’ side = 100/5 = 20 cm

Thus, the side of regular pentagon is 20 cm.

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?

**Answer**

Length of string = Perimeter of each figure

(a) Perimeter of square = 30 cm

â‡’ 4 Ã— side = 30 cm

â‡’ side = 30/4 = 7.5 cm

Thus, the length of each side of square is 7.5 cm.

(b) Perimeter of equilateral triangle = 30 cm

â‡’ 3 Ã— side = 30 cm

â‡’ side = 30/3 = 10 cm

Thus, the length of each side of equilateral triangle is 10 cm.

(c) Perimeter of hexagon = 30 cm

â‡’ 6 Ã— side = 30 cm

â‡’ side = 30/6 = 5 cm

Thus, the length of each side of hexagon is 5 cm.

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

**Answer**

Let the length of third side be xx cm.

Length of other two side are 12 cm and 14 cm.

Now, Perimeter of triangle = 36 cm

â‡’ 12+14+x = 36

â‡’ 26+x = 6

â‡’ x = 36âˆ’26

â‡’ x=10

Thus, the length of third side is 10 cm.

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

**Answer**

Side of square = 250 m

Perimeter of square = 4 Ã— side

= 4 Ã— 250 = 1000 m

Since, cost of fencing per meter = Rs. 20

Therefore, cost of fencing of 1000 meters = 20Ã—1000 = Rs. 20,000

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

**Answer**

Length of rectangular park = 175 m

Breadth of rectangular park = 125 m

Perimeter of park = 2 x (length + breadth)

= 2 Ã— (175 + 125)

= 2Ã—300 = 600 m

Since, cost of fencing park per meter = Rs. 12

Therefore, cost of fencing park of 600 m = 12Ã—600 = Rs. 7,200

15. 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?

**Answer**

Distance covered by Sweety = Perimeter of square park

Perimeter of square = 4Ã—side

= 4Ã—75 = 300 m

Thus, distance covered by Sweety is 300 m.

Now, distance covered by Bulbul = Perimeter of rectangular park

Perimeter of rectangular park = 2Ã—(length + breadth)

= 2Ã—(60 + 45)

= 2Ã—105 = 210 m

Thus, Bulbul covers the distance of 210 m.

So, Bulbul covers less distance.

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

**Answer**

(a) Perimeter of square = 4Ã—side

= 4Ã—25 = 100 cm

(b) Perimeter of rectangle = 2Ã—(length + breadth)

= 2Ã—(40 + 10)

= 2Ã—50 = 100 cm

(c) Perimeter of rectangle = 2Ã—(length + breadth)

= 2Ã—(30 + 20)

= 2Ã—50 = 100 cm

(d) Perimeter of triangle = Sum of all sides

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

Thus, all the figures have same perimeter.

17. Avneet buys 9 square paving slabs, each with a side 1212 m. He lays them in the form of a square

(a) What is the perimeter of his arrangement?

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

perimeter of her arrangement?

(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.)

**Answer**

(a) 6 m

(b) 10 m

(c) Second arrangement has greater perimeter.

(d) Yes, if all the squares are arranged in row, the perimeter be 10 cm.

**Exercise 10.2**

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

**Answer**

(a) Number of filled square = 9

Area covered by squares = 9 Ã— 1 = 9 sq. units

(b) Number of filled squares = 5

Area covered by filled squares = 5 Ã— 1 = 5 sq. units

(c) Number of full filled squares = 2

Number of half filled squares = 4

Area covered by full filled squares = 2 Ã— 1 = 2 sq. units

And, Area covered by half filled squares = 1/2 Ã—4 = 2 sq. units

Total area = 2 + 2 = 4 sq. units

(d) Number of filled squares = 8

Area covered by filled squares = 8 Ã— 1 = 8 sq. units

(e) Number of filled squares = 10

Area covered by filled squares = 10 Ã— 1 = 10 sq. units

(f) Number of full filled squares = 2

Number of half filled squares = 4

Area covered by full filled squares = 2 Ã— 1 = 2 sq. units

And Area covered by half filled squares = 1/2 Ã—4 = 2 sq. units

Total area = 2 + 2 = 4 sq. units

(g) Number of full filled squares = 4

Number of half filled squares = 4

Area covered by full filled squares = 4 Ã— 1 = 4 sq. units

And, Area covered by half filled squares = 1/2 Ã—4 = 2 sq. units

Total area = 4 + 2 = 6 sq. units

(h) Number of filled squares = 5

Area covered by filled squares = 5 Ã— 1 = 5 sq. units

(i) Number of filled squares = 9

Area covered by filled squares = 9 Ã— 1 = 9 sq. units

(j) Number of full filled squares = 2

Number of half filled squares = 4

Area covered by full filled squares = 2 Ã— 1 = 2 sq. units

And Area covered by half filled squares = 1/2 Ã—4 = 2 sq. units

Total area = 2 + 2 = 4 sq. units

(k) Number of full filled squares = 4

Number of half filled squares = 2

Area covered by full filled squares = 4 Ã— 1 = 4 sq. units

And Area covered by half filled squares = 1/2 Ã—2 = 1 sq. units

Total area = 4 + 1 = 5 sq. units

(l) Number of full filled squares = 3

Number of half filled squares = 10

Area covered by full filled squares = 3 Ã— 1 = 3 sq. units

And Area covered by half filled squares = 1/2 Ã—10 = 5 sq. units

Total area = 3 + 5 = 8 sq. units

(m) Number of full filled squares = 7

Number of half filled squares = 14

Area covered by full filled squares = 7 Ã— 1 = 7 sq. units

And Area covered by half filled squares = 1/2 Ã—14 = 7 sq. units

Total area = 7 + 7 = 14 sq. units

(n) Number of full filled squares = 10

Number of half filled squares = 16

Area covered by full filled squares = 10 Ã— 1 = 10 sq. units

And Area covered by half filled squares = 1/2 Ã—16 = 8 sq. units

Total area = 10 + 8 = 18 sq. units

**Exercise 10.3**

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

**Answer**

(a) Area of rectangle = length Ã— breadth

= 3 cm Ã— 4 cm = 12 cm

^{2}
(b) Area of rectangle = length Ã— breadth

= 12 m Ã— 21 m = 252 m

^{2}
(c) Area of rectangle = length Ã— breadth

= 2 km Ã— 3 km = 6 km

^{2}
(d) Area of rectangle = length Ã— breadth

= 2 m Ã— 70 cm = 2 m Ã— 0.7 m = 1.4 m

^{2}
2. Find the areas of the squares whose sides are:

(a) 10 cm (b) 14 cm (c) 5 m

**Answer**

(a) Area of square = side Ã— side

= 10 cm Ã— 10 cm = 100 cm

^{2}
(b) Area of square = side x side

= 14 cm Ã— 14 cm = 196 cm

^{2}
(c) Area of square = side x side

= 5 m Ã— 5 m = 25 m

^{2}
3. 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?

**Answer**

(a) Area of rectangle = length Ã— breadth

= 9 m Ã— 6 m = 54 m

^{2}
(b) Area of rectangle = length Ã— breadth

= 3 m Ã— 17 m = 51 m

^{2}
(c) Area of rectangle = length x breadth

= 4 m Ã— 14 m = 56 m

^{2}
Thus, the rectangle (c) has largest area, i.e. 56 m

^{2}and rectangle (b) has smallest area, i.e., 51 m^{2}.4. The area of a rectangular garden 50 m long is 300 m

^{2}, find the width of the garden.

**Answer**

Length of rectangle = 50 m and Area of rectangle = 300 m

^{2}
Since, Area of rectangle = length x breadth

Therefore, Breadth = =300/50 = 6 m

Thus, the breadth of the garden is 6 m.

5. What is the cost of tiling a rectangular plot of land 500 m long and 200 m wide at the rate of Rs. 8 per hundred sq. m?

**Answer**

Length of land = 500 m and Breadth of land = 200 m

Area of land = length x breadth = 500 m Ã— 200 m = 1,00,000 m

^{2}
âˆµ Cost of tiling 100 sq. m of land = Rs. 8

âˆ´ Cost of tilling 1,00,000 sq. m of land = 8/100 Ã—100000 = Rs. 8000

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

**Answer**

Length of table = 2 m and breadth of table = 1 m 50 cm = 1.50 m

Area of table = length Ã— breadth

= 2 m Ã— 1.50 m = 3 m

^{2}7. A room is 4 m long and 3 m 50 cm wide. How many square meters of carpet is needed to cover the floor of the room?

**Answer**

Length of room = 4 m and breadth of room = 3 m 50 cm = 3.50 m

Area of carpet = length Ã— breadth

= 4 Ã— 3.50 = 14m

^{2}
Therefore, 14m

^{2}of carpet required to cover the floor.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.

**Answer**

Length of floor = 5 m and breadth of floor = 4 m

Area of floor = length Ã— breadth

= 5 m Ã— 4 m = 20 m

^{2}
Now, Side of square carpet = 3 m

Area of square carpet = side x side = 3 Ã— 3 = 9 m

^{2}
Area of floor that is not carpeted = 20 m

^{2}â€“ 9 m^{2}= 11 m^{2}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?

**Answer**

Side of square bed = 1 m

Area of square bed = side Ã— side = 1 m Ã— 1 m = 1 m

^{2}
âˆ´ Area of 5 square beds = 1 Ã— 5 = 5 m

^{2}
Now, Length of land = 5 m and breadth of land = 4 m

âˆ´ Area of land = length Ã— breadth = 5 m Ã— 4 m = 20 m

^{2}
Area of remaining part = Area of land â€“ Area of 5 flower beds

= 20 m

^{2}â€“ 5 m^{2}= 15 m^{2}10. By splitting the following figures into rectangles, find their areas. (The measures are given in centimeters)

**Answer**

(a) The given figure can be broken into rectangles as

Area of 1st rectangle = 12 Ã— 2 = 24 cm

^{2}
Area of 2nd rectangle = 8 Ã— 2 = 16 cm

^{2}
Total area of the figure = 24 + 16 = 40 cm

^{2}
(b) The given figure can be broken into rectangles as follow

Area of 1st rectangle = 21 Ã— 7 = 147 cm

^{2}
Area of 1st square = 7 Ã— 7 = 49 cm

^{2}
Area of 2nd square = 7 Ã— 7 = 49 cm

^{2}
Total area of the figure = 147 + 49 + 49 = 245 cm

^{2}
(c) The given figure can be broken into rectangles as follow

Area of 1st rectangle = 5 Ã— 1 = 5 cm

^{2}
Area of 2nd rectangle = 4 Ã— 1 = 4 cm

^{2}
Total area of the figure = 5 + 4 = 9 cm

^{2}(a)100 cm and 144 cm

(b)70 cm and 36 cm

**Answer**