CBSE Class 7 Mathematics Perimeter and Area Assignment Set 09

Question 1. The total cost of flooring a room at Rs. 8.50 per square metre is Rs. 510. If the length of the room is 9 metres. Find its width.
Answer: Total cost of flooring = Rs. 510
Rate of flooring = Rs. 8.50 per \(\text{m}^2\)
Area of the room = \(\frac{\text{Total Cost}}{\text{Rate}}\)
Area of the room = \(\frac{510}{8.50} = \frac{51000}{850} = 60 \text{ m}^2\)
Length of the room = 9 m
Area = Length \(\times\) Width
\(60 = 9 \times \text{Width}\)
Width = \(\frac{60}{9} = \frac{20}{3} = 6.67 \text{ m}\) (or \(6\frac{2}{3} \text{ m}\))
In simple words: Divide the total cost by the rate per square metre to find the area, then divide the area by the length to find the width.

Exam Tip: Always check that the units are consistent before performing calculations.

Question 2. A rectangular metal plate is 7 cm long and 5 cm broad. Find the cost of the plate at the rate of Rs. 75 per square cm.
Answer: Length of the metal plate = 7 cm
Breadth of the metal plate = 5 cm
Area of the metal plate = Length \(\times\) Breadth
Area = \(7 \times 5 = 35 \text{ cm}^2\)
Rate = Rs. 75 per \(\text{cm}^2\)
Total cost = Area \(\times\) Rate
Total cost = \(35 \times 75 = \text{Rs. } 2625\)
In simple words: Find the area of the rectangle by multiplying length and breadth, then multiply by the cost per square centimetre.

Exam Tip: Clearly write the formula for the area of a rectangle before substituting values.

Question 3. The length and breadth of a playground are 62 m 60 cm and 25 m 40 cm. Find the cost of turfing it at Rs. 2.50 per square metre. How long will a man take to go three times round the field, if he walks at the rate of 2 metre per second.
Answer: Length of the playground = 62 m 60 cm = 62.60 m
Breadth of the playground = 25 m 40 cm = 25.40 m
Area of the playground = Length \(\times\) Breadth
Area = \(62.60 \times 25.40 = 1590.04 \text{ m}^2\)
Cost of turfing = Area \(\times\) Rate
Cost of turfing = \(1590.04 \times 2.50 = \text{Rs. } 3975.10\)
Perimeter of the playground = \(2 \times (\text{Length} + \text{Breadth})\)
Perimeter = \(2 \times (62.60 + 25.40) = 2 \times 88 = 176 \text{ m}\)
Total distance for 3 rounds = \(3 \times 176 = 528 \text{ m}\)
Speed of the man = 2 m/s
Time taken = \(\frac{\text{Distance}}{\text{Speed}} = \frac{528}{2} = 264 \text{ seconds}\)
Converting time into minutes and seconds: \(264 \div 60 = 4\) minutes and 24 seconds.
In simple words: First find the area and multiply by the rate for turfing cost. Then find the perimeter, multiply by 3 for total distance, and divide by speed to get the time.

Exam Tip: Convert all mixed units (metres and centimetres) into decimals of a single unit before calculating.

Question 4. The length and breadth of a rectangular field are 360 m and 150 m respectively. Find its area in hectares as well as in ares.
Answer: Length = 360 m
Breadth = 150 m
Area = Length \(\times\) Breadth
Area = \(360 \times 150 = 54000 \text{ m}^2\)
We know that \(1 \text{ hectare} = 10000 \text{ m}^2\), and \(1 \text{ are} = 100 \text{ m}^2\).
Area in hectares = \(\frac{54000}{10000} = 5.4 \text{ hectares}\)
Area in ares = \(\frac{54000}{100} = 540 \text{ ares}\)
In simple words: Calculate the area in square metres first, then divide by 10000 for hectares and by 100 for ares.

Exam Tip: Memorize land conversion units: \(1 \text{ are} = 100 \text{ m}^2\) and \(1 \text{ hectare} = 100 \text{ ares} = 10000 \text{ m}^2\).

Question 5. The area of a rectangular field is 5 hectares. If one of the sides of the field is 400 m. Find the other side.
Answer: Area of the field = 5 hectares = \(5 \times 10000 \text{ m}^2 = 50000 \text{ m}^2\)
Given side (Length) = 400 m
Area = Length \(\times\) Other Side (Breadth)
\(50000 = 400 \times \text{Breadth}\)
Breadth = \(\frac{50000}{400} = \frac{500}{4} = 125 \text{ m}\)
In simple words: Convert hectares to square metres, then divide the total area by the given side to find the other side.

Exam Tip: Always convert hectares into square metres before doing division with linear measurements in metres.

Question 6. How many envelopes can be made out of a sheet of paper 324 cm by 172 cm, if each envelop requires a piece of paper of size 18 cm by 12 cm.
Answer: Dimensions of the large sheet of paper = \(324 \text{ cm} \times 172 \text{ cm}\)
Area of the large sheet = \(324 \times 172 = 55728 \text{ cm}^2\)
Dimensions of one envelope piece = \(18 \text{ cm} \times 12 \text{ cm}\)
Area of one envelope piece = \(18 \times 12 = 216 \text{ cm}^2\)
Number of envelopes = \(\frac{\text{Area of large sheet}}{\text{Area of one envelope piece}}\)
Number of envelopes = \(\frac{55728}{216} = 258\)
In simple words: Divide the total area of the large paper sheet by the area required for a single envelope.

Exam Tip: You can also find the number of pieces by dividing the length and breadth separately and multiplying the results only if they divide evenly.

Question 7. Two plots of land have the same perimeter. One is a square of side 64 m and another one is rectangle of length 70 m. Which plot has the greater area and by how much?
Answer: Side of the square plot = 64 m
Perimeter of the square plot = \(4 \times \text{side} = 4 \times 64 = 256 \text{ m}\)
Area of the square plot = \(\text{side}^2 = 64 \times 64 = 4096 \text{ m}^2\)
Perimeter of the rectangular plot = Perimeter of the square plot = 256 m
Length of the rectangle = 70 m
Perimeter = \(2 \times (\text{Length} + \text{Breadth})\)
\(256 = 2 \times (70 + \text{Breadth})\)
\(128 = 70 + \text{Breadth}\)
Breadth = \(128 - 70 = 58 \text{ m}\)
Area of the rectangular plot = \(\text{Length} \times \text{Breadth} = 70 \times 58 = 4060 \text{ m}^2\)
Comparing areas: Square plot area (\(4096 \text{ m}^2\)) > Rectangular plot area (\(4060 \text{ m}^2\))
Difference in area = \(4096 - 4060 = 36 \text{ m}^2\)
In simple words: Find the perimeter of the square to know the perimeter of the rectangle. Then find the breadth and area of the rectangle, and compare both areas.

Exam Tip: For the same perimeter, a square always encloses a larger area than any rectangle.

Question 8. A room is 13 m long and 9 m broad. Find the area of carpeting the room with a carpet 75 cm broad at the rate of Rs. 7.50 per meter.
Answer: Length of the room = 13 m
Breadth of the room = 9 m
Area of the room = Length \(\times\) Breadth = \(13 \times 9 = 117 \text{ m}^2\)
Breadth of the carpet = 75 cm = 0.75 m
Length of the carpet needed = \(\frac{\text{Area of the room}}{\text{Breadth of the carpet}}\)
Length of the carpet = \(\frac{117}{0.75} = 156 \text{ m}\)
Rate of the carpet = Rs. 7.50 per meter
Total cost = Length of carpet \(\times\) Rate
Total cost = \(156 \times 7.50 = \text{Rs. } 1170\)
In simple words: Divide the room's area by the carpet's width to find the total length of carpet required, then multiply by the cost per metre.

Exam Tip: Ensure carpet width and room dimensions are in the same units (metres) before division.

Question 9. What happens to the area of a square if its side is doubled?
Answer: Let the initial side of the square be \(x\).
Initial Area = \(x^2\)
If the side is doubled, new side = \(2x\)
New Area = \((2x)^2 = 4x^2\)
Ratio of New Area to Initial Area = \(\frac{4x^2}{x^2} = 4\)
Therefore, the area of the square becomes 4 times.br /> In simple words: When you double the side of a square, its area increases by a factor of 4 (\(2^2\)).

Exam Tip: Remember the scaling rule: if linear dimensions are multiplied by \(k\), the area is multiplied by \(k^2\).

Question 10. The area of a rectangular field is 24.03 ares and its length is 54 metres. Find its breadth and perimeter.
Answer: Area of the field = 24.03 ares
Since \(1 \text{ are} = 100 \text{ m}^2\),
Area in square metres = \(24.03 \times 100 = 2403 \text{ m}^2\)
Length = 54 m
Area = Length \(\times\) Breadth
\(2403 = 54 \times \text{Breadth}\)
Breadth = \(\frac{2403}{54} = 44.5 \text{ m}\)
Perimeter = \(2 \times (\text{Length} + \text{Breadth})\)
Perimeter = \(2 \times (54 + 44.5) = 2 \times 98.5 = 197 \text{ m}\)
In simple words: Convert ares to square metres, divide by length to find breadth, and then use the perimeter formula.

Exam Tip: Always convert land area units (ares or hectares) into standard square metres before solving for dimensions.