CBSE Class 6 Mathematics Mensuration Assignment Set 05

Question. A rectangular field is 60m long and 40m broad. Find the cost of fencing at the rate of Rs 10 per metre.
Answer:
Given:
Length of the rectangular field (\( l \)) = \( 60\text{ m} \)
Breadth of the rectangular field (\( b \)) = \( 40\text{ m} \)

First, find the perimeter of the rectangular field:
\( \text{Perimeter} = 2(l + b) \)
\( \text{Perimeter} = 2(60\text{ m} + 40\text{ m}) = 2 \times 100\text{ m} = 200\text{ m} \)

Now, find the cost of fencing:
Rate of fencing = \( \text{Rs } 10\text{ per metre} \)
\( \text{Total Cost} = \text{Perimeter} \times \text{Rate} \)
\( \text{Total Cost} = 200\text{ m} \times \text{Rs } 10/\text{m} = \text{Rs } 2000 \)

Question. How many envelops can be made from a sheet of paper 125cm by 75cm, the size of one envelop is 25cm by 5cm.
Answer:
Given:
Dimensions of the paper sheet = \( 125\text{ cm} \times 75\text{ cm} \)
\( \text{Area of the paper sheet} = 125\text{ cm} \times 75\text{ cm} = 9375\text{ cm}^2 \)

Dimensions of one envelop = \( 25\text{ cm} \times 5\text{ cm} \)
\( \text{Area of one envelop} = 25\text{ cm} \times 5\text{ cm} = 125\text{ cm}^2 \)

Number of envelops that can be made:
\( \text{Number of envelops} = \frac{\text{Area of the paper sheet}}{\text{Area of one envelop}} \)
\( \text{Number of envelops} = \frac{9375\text{ cm}^2}{125\text{ cm}^2} = 75 \)

Question. How many times the area of a square changes if each of its side is halved?
Answer:
Let the original side of the square be \( s \).
\( \text{Original Area} = s^2 \)

If each of its side is halved, the new side becomes \( \frac{s}{2} \).
\( \text{New Area} = \left(\frac{s}{2}\right)^2 = \frac{s^2}{4} = \frac{1}{4} \times \text{Original Area} \)

Thus, the area becomes \( \frac{1}{4}\text{th} \) times the original area.

Question. Find the cost of fencing a rectangular field 34m long and 18m wide at Rs 2.25 per metre. What is the cost of cultivating the field at Rs 4.50 per square metre?
Answer:
Given:
Length of the rectangular field (\( l \)) = \( 34\text{ m} \)
Width of the rectangular field (\( w \)) = \( 18\text{ m} \)

1. Cost of Fencing:
Fencing is done along the perimeter of the field.
\( \text{Perimeter} = 2(l + w) = 2(34\text{ m} + 18\text{ m}) = 2 \times 52\text{ m} = 104\text{ m} \)
Rate of fencing = \( \text{Rs } 2.25\text{ per metre} \)
\( \text{Cost of fencing} = 104\text{ m} \times \text{Rs } 2.25/\text{m} = \text{Rs } 234 \)

2. Cost of Cultivating:
Cultivating is done over the entire area of the field.
\( \text{Area} = l \times w = 34\text{ m} \times 18\text{ m} = 612\text{ m}^2 \)
Rate of cultivating = \( \text{Rs } 4.50\text{ per square metre} \)
\( \text{Cost of cultivating} = 612\text{ m}^2 \times \text{Rs } 4.50/\text{m}^2 = \text{Rs } 2754 \)

Question. A room 9.68 m long and 6.2 m wide. Its floor is to be covered with glazed tiles of 22 cm by 10 cm each. If rate of tiles is Rs 25/tile. Find the total cost of tiles.
Answer:
Given:
Length of the room = \( 9.68\text{ m} = 968\text{ cm} \)
Width of the room = \( 6.2\text{ m} = 620\text{ cm} \)
\( \text{Area of the floor} = 968\text{ cm} \times 620\text{ cm} = 600,160\text{ cm}^2 \)

Dimensions of each tile = \( 22\text{ cm} \times 10\text{ cm} \)
\( \text{Area of each tile} = 22\text{ cm} \times 10\text{ cm} = 220\text{ cm}^2 \)

Number of tiles required:
\( \text{Number of tiles} = \frac{\text{Area of the floor}}{\text{Area of one tile}} = \frac{600,160}{220} = 2728\text{ tiles} \)

Cost of tiles:
Rate of tiles = \( \text{Rs } 25\text{ per tile} \)
\( \text{Total Cost} = 2728 \times \text{Rs } 25 = \text{Rs } 68,200 \)

Question. Find in square metres, the area of a square of side 16.5 dm.
Answer:
Given:
Side of the square = \( 16.5\text{ dm} \)
Since \( 1\text{ dm} = 0.1\text{ m} \),
Side of the square in metres = \( 16.5 \times 0.1\text{ m} = 1.65\text{ m} \)

\( \text{Area of the square} = \text{side} \times \text{side} \)
\( \text{Area} = 1.65\text{ m} \times 1.65\text{ m} = 2.7225\text{ m}^2 \)

Question. The length and breadth of a rectangular field are 360m and 150m. Find its area in hectares also in acres.
Answer:
Given:
Length of the field (\( l \)) = \( 360\text{ m} \)
Breadth of the field (\( b \)) = \( 150\text{ m} \)
\( \text{Area of the rectangular field} = 360\text{ m} \times 150\text{ m} = 54,000\text{ m}^2 \)

1. Area in Hectares:
Since \( 1\text{ hectare} = 10,000\text{ m}^2 \),
\( \text{Area in hectares} = \frac{54,000}{10,000} = 5.4\text{ hectares} \)

2. Area in Acres (intended as "ares" in standard conversion):
Note: While standard questions of this type request "ares" (\( 1\text{ are} = 100\text{ m}^2 \)), the text specifies "acres". Following the conversion corresponding to the original answer sheet's value (\( 540 \)):
\( \text{Area in ares} = \frac{54,000}{100} = 540\text{ ares (written as 540 acres in the answer key)} \)

Question. Each side of a square measures 65m 5dm. Find its area. By how many square metres does it fall short of a hectare?
Answer:
Given:
Side of the square = \( 65\text{ m } 5\text{ dm} \)
Since \( 1\text{ dm} = 0.1\text{ m} \),
Side of the square in metres = \( 65 + 0.5 = 65.5\text{ m} \)

1. Area of the square:
\( \text{Area} = 65.5\text{ m} \times 65.5\text{ m} = 4290.25\text{ m}^2 \)

2. Difference from one hectare:
Since \( 1\text{ hectare} = 10,000\text{ m}^2 \),
\( \text{Difference} = 10,000\text{ m}^2 - 4290.25\text{ m}^2 = 5709.75\text{ m}^2 \)

Question. A Carpet measures 30m 75cm by 80cm. Find its cost at the rate of Rs 15 per square metre.
Answer:
Given:
Length of the carpet = \( 30\text{ m } 75\text{ cm} = 30.75\text{ m} \)
Breadth of the carpet = \( 80\text{ cm} = 0.8\text{ m} \)

\( \text{Area of the carpet} = 30.75\text{ m} \times 0.8\text{ m} = 24.6\text{ m}^2 \)

Rate of carpet = \( \text{Rs } 15\text{ per square metre} \)
\( \text{Total Cost} = 24.6\text{ m}^2 \times \text{Rs } 15/\text{m}^2 = \text{Rs } 369 \)

Question. The area of a rectangular field is 0.5 hectares. If one of the sides of the field is 40m. Find the other side.
Answer:
Given:
Area of the rectangular field = \( 0.5\text{ hectares} \)
Since \( 1\text{ hectare} = 10,000\text{ m}^2 \),
\( \text{Area in square metres} = 0.5 \times 10,000 = 5000\text{ m}^2 \)

One of the sides = \( 40\text{ m} \)
Let the other side be \( x \).

\( \text{Area} = \text{side}_1 \times \text{side}_2 \)
\( 5000 = 40 \times x \)
\( x = \frac{5000}{40} = 125\text{ m} \)