CBSE Class 7 Mathematics Perimeter and Area Assignment Set 10

VII - Mathematics Assignment No - 01 - Perimeter and Area

Q1. Find the area and perimeter of a triangle whose base is 6 cm, each equal side is 5 cm, and height 4 cm.
Answer: Base (\(b\)) = 6 cm
Height (\(h\)) = 4 cm
Equal sides (\(a\)) = 5 cm, 5 cm
Area of the triangle = \(\frac{1}{2} \times \text{base} \times \text{height}\)
Area = \(\frac{1}{2} \times 6 \times 4 = 12 \text{ cm}^2\)
Perimeter of the triangle = Sum of all sides = \(6 + 5 + 5 = 16 \text{ cm}\)
In simple words: The area is found using base and height, while the perimeter is the total length around the triangle.

Exam Tip: Be careful not to confuse height with the side lengths when calculating the area of a triangle.

Q2. Find the area and perimeter of an equilateral triangle whose each side is 5 cm.
Answer: Side (\(a\)) = 5 cm
Area of an equilateral triangle = \(\frac{\sqrt{3}}{4} a^2\)
Area = \(\frac{\sqrt{3}}{4} \times (5)^2 = \frac{25\sqrt{3}}{4} \text{ cm}^2\)
Perimeter = \(3 \times a\)
Perimeter = \(3 \times 5 = 15 \text{ cm}\)
In simple words: An equilateral triangle has all three sides equal, making perimeter calculation very straightforward.

Exam Tip: Remember the standard formula for the area of an equilateral triangle: \(\frac{\sqrt{3}}{4} a^2\).

Q3. Find the area of a rectangle whose length and breadth are 45 cm and 16 cm respectively. Also, find the perimeter of the rectangle.
Answer: Length (\(l\)) = 45 cm
Breadth (\(b\)) = 16 cm
Area of the rectangle = \(l \times b\)
Area = \(45 \times 16 = 720 \text{ cm}^2\)
Perimeter of the rectangle = \(2(l + b)\)
Perimeter = \(2(45 + 16) = 2 \times 61 = 122 \text{ cm}\)
In simple words: Multiply length and breadth for area, and add length and breadth then double it for perimeter.

Q4. Find the area and perimeter of a rectangle whose length and breadth are 15.4 cm and 6.5 cm respectively.
Answer: Length (\(l\)) = 15.4 cm
Breadth (\(b\)) = 6.5 cm
Area of the rectangle = \(l \times b\)
Area = \(15.4 \times 6.5 = 100.10 \text{ cm}^2\)
Perimeter of the rectangle = \(2(l + b)\)
Perimeter = \(2(15.4 + 6.5) = 2 \times 21.9 = 43.8 \text{ cm}\)
In simple words: Perform decimal multiplication carefully to find the area and addition followed by multiplication for the perimeter.

Exam Tip: Align decimal points properly when adding lengths for the perimeter.

Q5. Find the area of a rectangle which is 30 cm long and 25 cm broad.
Answer: Length (\(l\)) = 30 cm
Breadth (\(b\)) = 25 cm
Area of the rectangle = \(l \times b\)
Area = \(30 \times 25 = 750 \text{ cm}^2\)
In simple words: Multiply the length and breadth directly to get the total surface area.

Exam Tip: Check that both dimensions are in the same unit before multiplying.

Q6. Find the area of a square whose each side is 2.5 cm. Also find the perimeter.
Answer: Side (\(a\)) = 2.5 cm
Area of the square = \(a^2\)
Area = \(2.5 \times 2.5 = 6.25 \text{ cm}^2\)
Perimeter of the square = \(4 \times a\)
Perimeter = \(4 \times 2.5 = 10 \text{ cm}\)
In simple words: Square the side length for area and multiply by 4 for perimeter.

Exam Tip: Squaring a decimal number like 2.5 results in two decimal places (\(6.25\)).

Q7. The area of a rectangle is \(540 \text{ cm}^2\). If its length is 27 cm. Find its width also its perimeter.
Answer: Area = \(540 \text{ cm}^2\)
Length (\(l\)) = 27 cm
Width (\(b\)) = \(\frac{\text{Area}}{l}\)
Width = \(\frac{540}{27} = 20 \text{ cm}\)
Perimeter = \(2(l + b) = 2(27 + 20) = 2 \times 47 = 94 \text{ cm}\)
In simple words: Divide the total area by the given length to find the width, then use both to calculate the perimeter.

Exam Tip: Use division to reverse the area formula when one dimension is missing.

Q8. The area of a rectangle is \(650 \text{ cm}^2\). And one of its side is 13 cm. Find the other side. Also find its perimeter.
Answer: Area = \(650 \text{ cm}^2\)
Given side (\(l\)) = 13 cm
Other side (\(b\)) = \(\frac{\text{Area}}{\text{given side}}\)
Other side = \(\frac{650}{13} = 50 \text{ cm}\)
Perimeter = \(2(l + b) = 2(13 + 50) = 2 \times 63 = 126 \text{ cm}\)
In simple words: Finding the missing side of a rectangle requires dividing the area by the known side.

Exam Tip: Clearly label length and breadth after finding the unknown side.

Q9. Find the cost of fencing a rectangular field 34 m long and 18 m wide at Rs. 2.50 per metre. What is the cost of cultivating the field at Rs. 4.25 per square metre?
Answer: Length (\(l\)) = 34 m
Breadth (\(b\)) = 18 m
Perimeter of the field = \(2(l + b) = 2(34 + 18) = 2 \times 52 = 104 \text{ m}\)
Cost of fencing = Perimeter \(\times\) Rate per metre
Cost of fencing = \(104 \times 2.50 = \text{Rs. } 260.00\)
Area of the field = \(l \times b = 34 \times 18 = 612 \text{ m}^2\)
Cost of cultivating = Area \(\times\) Rate per square metre
Cost of cultivating = \(612 \times 4.25 = \text{Rs. } 2601.00\)
In simple words: Fencing uses the perimeter, while cultivation uses the area. Multiply each by its respective rate.

Exam Tip: Always associate fencing with perimeter and cultivation/painting/flooring with area.

Q10. A room is 9.68 m long and 6.2 m wide. Its floor is to be covered with rectangular tiles of size 22 cm by 10 cm. Find the total cost of the tiles at the rate of Rs. 2.75 per tile.
Answer: Length of room = \(9.68 \text{ m} = 9.68 \times 100 = 968 \text{ cm}\)
Breadth of room = \(6.2 \text{ m} = 6.2 \times 100 = 620 \text{ cm}\)
Area of the room floor = \(968 \times 620 = 600160 \text{ cm}^2\)
Length of one tile = 22 cm
Breadth of one tile = 10 cm
Area of one tile = \(22 \times 10 = 220 \text{ cm}^2\)
Number of tiles required = \(\frac{\text{Area of the room}}{\text{Area of one tile}}\)
Number of tiles = \(\frac{600160}{220} = 2728\)
Total cost of tiles = Number of tiles \(\times\) Rate per tile
Total cost = \(2728 \times 2.75 = \text{Rs. } 7502.00\)
In simple words: Convert all measurements to the same unit (cm), find both areas, divide to get the total number of tiles, and multiply by the cost per tile.

Exam Tip: Convert meters to centimeters first so that all unit dimensions match before calculating areas.