CBSE Class 6 Mathematics Perimeter and Area Practice Paper Assignment Set 02 with Answers

Practice Paper 17 for CBSE Class 6 Maths Ganita Prakash Chapter 6 Perimeter and Area has 20 questions worth 40 marks, arranged in Sections A to E, each followed by a worked answer. Questions focus on the perimeter of joined squares and of a park with corner flower beds, the cost of fencing at a given rate, finding a missing side from the area or the perimeter, rectangles made from a fixed length of wire, tiling a wall, an exhibition-board framing problem and the area of two crossing roads in a park.

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Question 11. A wire of length 20 m is to be folded in the form of a rectangle. How many rectangles can be formed by folding the wire if the sides are positive integers in metres?
Answer:
The whole wire becomes the boundary, so perimeter = 20 m and length + breadth = 20 ÷ 2 = 10 m.
Pairs of whole numbers adding to 10: 1 and 9, 2 and 8, 3 and 7, 4 and 6, 5 and 5.
(9 and 1 is the same rectangle as 1 and 9, turned round, so it is not counted again.)
So 5 rectangles can be formed (the 5 m × 5 m one is a square, which is a special rectangle).
In simple words: Length and breadth must add up to 10 m, and there are five such whole-number pairs.

Teacher's Note:
1. Halve the perimeter first to get length + breadth.
2. These rectangles have the same perimeter but different areas: 9, 16, 21, 24 and 25 m².

Question 12. The dimensions of a photograph are 30 cm × 20 cm. What length of wooden frame is needed to frame the picture?
Answer:
The frame goes around the edge of the photograph, so its length equals the perimeter.
Perimeter = 2 × (30 + 20) = 2 × 50 = 100 cm.
Length of wooden frame needed = 100 cm (= 1 m).
In simple words: Frame length = distance around the photo = 100 cm.

Teacher's Note:
1. Framing, fencing and edging are perimeter problems.
2. The area 600 cm² is not needed.

Question 13. The area of a rectangle is 49 cm² and its breadth is 2.8 cm. Find the length of the rectangle.
Answer:
Area = length × breadth, so length = area ÷ breadth = 49 ÷ 2.8.
49 ÷ 2.8 = 490 ÷ 28 = 17.5.
The length of the rectangle is 17.5 cm.
In simple words: Divide the area by the breadth to get the length.

Teacher's Note:
1. Multiply both numbers by 10 to divide without a decimal in the divisor.
2. Check: 17.5 × 2.8 = 49.

Question 14. A rectangle has the area equal to that of a square of side 80 cm. If the breadth of the rectangle is 20 cm, find its length.
Answer:
Area of the square = 80 × 80 = 6400 cm², so the area of the rectangle is also 6400 cm².
Length = area ÷ breadth = 6400 ÷ 20 = 320 cm.
The length of the rectangle is 320 cm (= 3.2 m).
In simple words: Same area as the square, 6400 cm², shared out over a breadth of 20 cm.

Teacher's Note:
1. Equal areas do not mean equal shapes: 320 × 20 and 80 × 80 are very different.
2. Check: 320 × 20 = 6400.

Section C: Short Answer Questions (3 Marks Each)

Question 15. Shikha runs around a square of side 75 m. Priya runs around a rectangle with length 60 m and breadth 45 m. Who covers the smaller distance?
Answer:
Shikha's distance = perimeter of the square = 4 × 75 = 300 m.
Priya's distance = perimeter of the rectangle = 2 × (60 + 45) = 2 × 105 = 210 m.
210 m < 300 m, so Priya covers the smaller distance, 90 m less than Shikha.
In simple words: One round is the perimeter: 300 m for Shikha, 210 m for Priya.

Teacher's Note:
1. Running around a shape once = its perimeter.
2. Finish with a clear comparison sentence.

Question 16. To fix fence wires in a garden, 70 m long and 50 m wide, Arvind bought metal pipes for posts. He fixed a post every 5 metres apart. Each post was 2 m long. What is the total length of the pipes he bought for the posts?
Answer:
Perimeter of the garden = 2 × (70 + 50) = 2 × 120 = 240 m.
Posts are 5 m apart all the way around a closed boundary, so the number of posts = 240 ÷ 5 = 48.
Total length of pipes = 48 × 2 = 96 m.
In simple words: Count the posts around the boundary (48) and multiply by 2 m each.

Teacher's Note:
1. On a closed loop, the number of posts equals the number of gaps, because the last gap ends at the first post.
2. On an open straight line you would need one extra post.

Question 17. A marble tile measures 15 cm × 20 cm. How many tiles will be required to cover a wall of size 4 m × 6 m?
Answer:
Change the wall to centimetres: 4 m × 6 m = 400 cm × 600 cm.
Area of the wall = 400 × 600 = 240000 cm²; area of one tile = 15 × 20 = 300 cm².
Number of tiles = 240000 ÷ 300 = 800.
So 800 tiles are required.
(Check by rows: 400 ÷ 20 = 20 tiles one way and 600 ÷ 15 = 40 tiles the other way, and 20 × 40 = 800.)
In simple words: Wall area divided by one tile's area gives 800 tiles.

Teacher's Note:
1. Both areas must be in the same unit before dividing.
2. Here the tiles fit exactly with no cutting, so dividing areas gives the right count.

Section D: Long Answer Question (5 Marks)

Question 18. The below figure has been split into rectangles. Find their areas. (The measures are given in centimetres)

Answer:
Draw two dashed lines down from the top to the inner corners. This gives three rectangles:

Rectangle I (left part): 1 cm wide and 3 cm high, area = 1 × 3 = 3 cm².
Rectangle II (right part): 1 cm wide and 3 cm high, area = 1 × 3 = 3 cm².
Rectangle III (top middle): 3 cm long; its height = 3 − 2 = 1 cm, area = 3 × 1 = 3 cm².
Total area = 3 + 3 + 3 = 9 cm².
Check: the whole 5 cm × 3 cm rectangle has area 15 cm² and the gap below is 3 cm × 2 cm = 6 cm², and 15 − 6 = 9 cm².
In simple words: Split the shape into two 1 × 3 legs and a 3 × 1 top piece: 3 + 3 + 3 = 9 cm².

Teacher's Note:
1. Missing side: total height 3 cm − gap height 2 cm = 1 cm.
2. Checking by subtracting the gap from the full rectangle is a quick way to confirm.
3. The right leg is also 1 cm wide since 1 + 3 + 1 = 5 cm.

Section E: Case Study Based Questions (4 Marks Each)

Question 19. Case Study 1: The students of Global Reach International School are buzzing with excitement for their annual exhibition! They've chosen the grand exhibition hall to showcase their projects. For this, they have 24 green display boards, each measuring 1 m 50 cm long and 1 m wide. To give a professional finish, a 100 m long aluminium strip has been provided to frame all these boards. The organizing committee needs to figure out if there's enough strip for every single board!
Based on the above, answer the following questions
(a) What is the perimeter of one display board? (1)
(b) Find the number of boards that will be framed using the available aluminium strip. (1)
(c) Find the length of the aluminium strip required for the remaining boards. (1)
(d) What is the area of one display board? (1)
Answer:
Length of a board = 1 m 50 cm = 1.5 m; breadth = 1 m.
(a) Perimeter = 2 × (1.5 + 1) = 2 × 2.5 = 5 m.
(b) Each board needs 5 m of strip, so 100 ÷ 5 = 20 boards can be framed.
(c) Remaining boards = 24 − 20 = 4, so extra strip needed = 4 × 5 = 20 m (all 24 boards need 24 × 5 = 120 m, and 120 − 100 = 20 m).
(d) Area = 1.5 × 1 = 1.5 m².
In simple words: Each board needs 5 m of frame, so 100 m frames 20 boards and 4 more boards need 20 m.

Teacher's Note:
1. Framing uses the perimeter, but the board's surface uses the area.
2. So the strip is not enough: it is short by 20 m.

Question 20. Case Study 2: A local council member, Mr. Sharma, proposed building new roads to improve access within the city's 100 m × 100 m square central park. They planned two perpendicular, 2 m wide paths right through the middle, one running north-south, and the other east-west. Before starting construction, the project manager needs to precisely calculate the area for the roads and the central intersection to ensure the budget is accurate.
(a) What is the total area of the entire square park? (1 Mark)
(b) What is the area of the road running parallel to the length? (1 Mark)
(c) What is the area of the central square region where the two roads overlap? (1 Mark)
(d) Find the total area covered by the two roads. (1 Mark)
Answer:
(a) Area of the park = 100 × 100 = 10000 m².
(b) This road is 100 m long and 2 m wide, so its area = 100 × 2 = 200 m².
(c) The overlap is a square of side 2 m, area = 2 × 2 = 4 m².
(d) The other road also has area 100 × 2 = 200 m², but the 4 m² square in the middle is counted in both.
Total area of the roads = 200 + 200 − 4 = 396 m².
In simple words: Add the two roads and remove the middle square once, because it was counted twice.

Teacher's Note:
1. Whenever two strips cross, subtract the overlap once.
2. Remaining grass area = 10000 − 396 = 9604 m².