NCERT Solutions Class 6th Maths Chapter – 10 Mensuration All Examples

Example 1. Shabana wants to put a lace border all around a rectangular table cover, 3 m long and 2 m wide. Find the length of the lace required by Shabana.

Solution –

Length of the rectangular table cover = 3 m
Breadth of the rectangular table cover = 2 m

Shabana wants to put a lace border all around the table cover. Therefore, the length of the lace required will be equal to the perimeter of the rectangular table cover.

Now, perimeter of the rectangular table cover
= 2 × (length + breadth) = 2 × (3 m + 2 m) = 2 × 5 m = 10 m
So, length of the lace required is 10 m.

Example 2. An athlete takes 10 rounds of a rectangular park, 50 m long and 25 m wide. Find the total distance covered by him.

Solution –

Length of the rectangular park = 50 m
Breadth of the rectangular park = 25 m

Total distance covered by the athlete in one round will be the perimeter of the park.
Now, perimeter of the rectangular park
= 2 × (length + breadth)= 2 × (50 m + 25 m)
= 2 × 75 m = 150 m

So, the distance covered by the athlete in one round is 150 m.
Therefore, distance covered in 10 rounds = 10 × 150 m = 1500m
The total distance covered by the athlete is 1500 m.

Solution –

Length = 150 cm
Breadth = 1m = 100 cm
Perimeter of the rectangle
= 2 × (length + breadth)
= 2 × (150 cm + 100 cm)
= 2 × (250 cm) = 500 cm = 5 m.

Example 4. A farmer has a rectangular field of length and breadth 240 m and 180 m respectively. He wants to fence it with 3 rounds of rope as shown in figure 10.4. What is the total length of rope he must use?

Solution –

The farmer has to cover three times the perimeter of that field. Therefore, total length of rope required is thrice its perimeter.

Perimeter of the field = 2 × (length + breadth)
= 2 × ( 240 m + 180 m)
= 2 × 420 m = 840 m
Total length of rope required = 3 × 840 m = 2520 m.

Example 5. Find the cost of fencing a rectangular park of length 250 m and breadth 175 m at the rate of ` 12 per metre.

Solution –

Length of the rectangular park = 250 m
Breadth of the rectangular park = 175 m

To calculate the cost of fencing we require perimeter.
Perimeter of the rectangle = 2 × (length + breadth)
= 2 × (250 m + 175 m)
= 2 × (425 m) = 850 m
Cost of fencing 1m of park = ` 12

Therefore, the total cost of fencing the park
= ` 12 × 850 = ` 10200

Example 6. Find the distance travelled by Shaina if she takes three rounds of a square park of side 70 m.

Solution –

Perimeter of the square park
= 4 × length of a side = 4 × 70 m = 280m
Distance covered in one round = 280 m
Therefore, distance travelled in three rounds=3×280m = 840m.

Solution –

Distance covered by Pinky in one round
= Perimeter of the square
= 4 × length of a side
= 4 × 75 m = 300 m

Distance covered by Bob in one round
= Perimeter of the rectangle
= 2 × (length + breadth)
= 2 × (160 m + 105 m)
= 2 × 265 m = 530 m

Difference in the distance covered = 530 m – 300 m = 230 m.
Therefore, Bob covers more distance by 230 m.

Example 8. Find the perimeter of a regular pentagon with each side measuring 3 cm.

Solution –

This regular closed figure has 5 sides, each with a length of 3 cm. Thus, we get Perimeter of the regular pentagon
= 5 × 3 cm = 15 cm

Solution –

Perimeter = 18 cm
A regular hexagon has 6 sides, so we can divide the perimeter by 6 to get the length of one side.

One side of the hexagon = 18 cm ÷ 6 = 3 cm
Therefore, length of each side of the regular
hexagon is 3 cm.

Example 10. Find the area of the shape shown in the figure 10.10.

Solution –

This figure is made up of line-segments. Moreover, it is covered by full squares and half squares only. This makes our job simple.

(i) Fully-filled squares = 3
(ii)Half-filled squares = 3

Area covered by full squares
= 3 × 1 sq units = 3 sq units
Total area = 4 1/2 sq units.

Example 11. By counting squares, estimate the area of the figure 10.9 b.

Soultion –

Make an outline of the figure on a graph sheet.

Total area = 11 + 3 × 1/2 + 7 = 19 1/2 sq units.

Example 12. By counting squares, estimate the area of the figure 10.9 a.

Soultion –

Make an outline of the figure on a graph sheet. This is how the squares cover the figure (Fig 10.12).

Total area = 1 + 7 = 8 sq units.

Example 13. Find the area of a rectangle whose length and breadth are 12 cm and 4 cm respectively.

Solution –

Length of the rectangle = 12 cm
Breadth of the rectangle = 4 cm
Area of the rectangle = length × breadth
= 12 cm × 4 cm = 48 sq cm.

Example 14. Find the area of a square plot of side 8 m.

Solution –

Side of the square = 8 m
Area of the square = side × side
= 8 m × 8 m = 64 sq m.

Example 15. The area of a rectangular piece of cardboard is 36 sq cm and its length is 9 cm. What is the width of the cardboard?

Solution –

Area of the rectangle = 36 sq cm
Length = 9 cm
Width = ?

Area of a rectangle = length × width
So, width = Area/Length = 36/9 = 4 cm

Thus, the width of the rectangular cardboard is 4 cm.

Example 16. Bob wants to cover the floor of a room 3 m wide and 4 m long by squared tiles. If each square tile is of side 0.5 m, then find the number of tiles required to cover the floor of the room.

Solution –

Total area of tiles must be equal to the area of the floor of the room.

Length of the room = 4 m
Breadth of the room = 3 m
Area of the floor = length × breadth
= 4 m × 3 m = 12 sq m
Area of one square tile = side × side
= 0.5 m × 0.5 m
= 0.25 sq m

Number of tiles required = Area of the floor/Area of one tile
= 12/0.25 = 1200/25 = 48 tiles.

Example 17. Find the area in square metre of a piece of cloth 1m 25 cm wide and 2 m long.

Solution –

Length of the cloth = 2 m
Breadth of the cloth = 1 m 25 cm = 1 m + 0. 25 m = 1.25 m
(since 25 cm = 0.25m)

Area of the cloth = length of the cloth × breadth of the cloth
= 2 m × 1.25 m = 2.50 sq m.