Class 6 Maths Chapter 6 Perimeter and Area Practice Paper 4

Class: VI Subject: Mathematics Session: 2025-26 Chapter: 06 - Perimeter and Area Time: 1½ Hours Max. Marks: 40

General Instructions:

1. All questions are compulsory.

2. This question paper contains 20 questions divided into five sections A, B, C, D and E.

3. Section A contains 10 MCQs of 1 mark each.

4. Section B contains 4 questions of 2 marks each.

5. Section C contains 3 questions of 3 marks each.

6. Section D contains 1 question of 5 marks.

7. Section E contains 2 Case Study Based questions of 4 marks each.

SECTION A - Multiple Choice Questions (1 mark each)

1. The area of a square plot is 2025 sq m. Its perimeter is:
(a) 45 m
(b) 90 m
(c) 180 m
(d) 135 m

2. A string of length 144 cm can form:
(a) A square of side 36 cm
(b) A rectangle of dimensions 40 cm × 32 cm
(c) An equilateral triangle of side 48 cm
(d) All of the above

3. The perimeter of a rectangle is 100 m. If the length is 30 m more than the breadth, the length is:
(a) 20 m
(b) 35 m
(c) 40 m
(d) 50 m

4. Which of the following cannot be the perimeter of a triangle with sides as whole numbers?
(a) 15 cm
(b) 20 cm
(c) 5 cm
(d) 12 cm

5. A room 5 m long and 4 m wide is surrounded by a verandah. If verandah's width is 2 m, the area of verandah is:
(a) 48 sq m
(b) 36 sq m
(c) 52 sq m
(d) 60 sq m

6. The cost of fencing a square field at ₹18 per metre is ₹1872. The length of each side is:
(a) 26 m
(b) 52 m
(c) 104 m
(d) 208 m

7. If the perimeter and area of a square are numerically equal, the side of square is:
(a) 1 unit
(b) 2 units
(c) 4 units
(d) 8 units

8. A wire can form a rectangle of length 12 cm and breadth 8 cm. If the same wire is bent to form a circle, its circumference will be:
(a) 20 cm
(b) 40 cm
(c) 32 cm
(d) 48 cm

9. A rectangular field has area equal to 150 sq m and perimeter equal to 50 m. Its length is:
(a) 10 m
(b) 12 m
(c) 15 m
(d) 20 m

10. Among all rectangles with same perimeter, which has the maximum area?
(a) The one with length = 2 × breadth
(b) The one with length = 3 × breadth
(c) The one with length = breadth (square)
(d) All have equal area

SECTION B - Short Answer Questions (2 marks each)

11. A rectangular garden is 90 m long and 75 m wide. Two paths run through the center of the garden - one parallel to length (2 m wide) and one parallel to breadth (3 m wide). Find the area of the paths.

12. A square and a rectangle have equal areas. If the side of the square is 16 cm and the length of rectangle is 32 cm, find the breadth of rectangle and the perimeter of both figures.

13. Find the perimeter of a regular pentagon if each side measures 8 cm. If a wire of this length is bent to form a rectangle with length 12 cm, find the area of rectangle.

14. The perimeter of a rectangular plot is 300 m. If the length is increased by 10 m and breadth is decreased by 10 m, the perimeter remains unchanged. Find the original dimensions.

SECTION C - Short Answer Questions (3 marks each)

15. A rectangular field is 48 m long and 30 m wide. Three square flower beds each of side 6 m are prepared inside it. Find the area of remaining field. If it costs ₹50 per sq m to plant grass, find the total cost.

16. Two rectangles have the same perimeter 56 cm. Rectangle A measures 18 cm by 10 cm. If rectangle B is a square, compare their areas.

17. A 36 m long and 24 m wide lawn has a path 3 m wide all around it on the outside. Find the cost of constructing the path at ₹60 per sq m.

SECTION D - Long Answer Question (5 marks)

18. A rectangular hall 20 m long and 16 m wide is to be paved with square tiles.
(a) What is the largest size of such a tile? (Hint: Tile size must divide both dimensions)
(b) How many tiles will be required if each tile is 4 m × 4 m?
(c) Find the cost of tiles at ₹500 per tile.
(d) A carpet is placed in the center leaving 2 m margin from all sides. Find the carpet's dimensions. (e) Find the area not covered by the carpet.

SECTION E - Case Study Based Questions (4 marks each)

19. Photo Frame Rahul wants to make a photo frame. The photo is 20 cm long and 15 cm wide. He wants a uniform border of 2.5 cm width all around the photo.
(a) Find the dimensions of the frame with border. (1 mark)
(b) Find the area of the photo. (1 mark)
(c) Find the total area of the frame with border. (1 mark)
(d) Find the area of the border alone. (1 mark)

20. Comparative Plots Mr. Sharma has to choose between two plots for building a house. Plot X is square with side 50 m. Plot Y is rectangular with length 60 m and breadth 40 m. The cost per sq m is same for both.
(a) Which plot has greater area and by how much? (1 mark)
(b) Mr. Sharma decides to fence the larger plot with 4 rounds of barbed wire. How much wire is needed? (1 mark)
(c) Find the cost of wire at ₹12 per metre. (1 mark)
(d) If he builds a house occupying 30% of the plot area, how much area remains for a garden? (1 mark) DETAILED ANSWER KEY - PAPER 04

SECTION A - Answers to MCQs

1.
(c) 180 m Side = √2025 = 45 m Perimeter = 4 × 45 = 180 m

2.
(d) All of the above Square: 4 × 36 = 144 cm ✓ Rectangle: 2 × (40 + 32) = 144 cm ✓ Triangle: 3 × 48 = 144 cm ✓

3.
(c) 40 m Let breadth = x, length = x + 30 2(x + x + 30) = 100 4x + 60 = 100, x = 10 Length = 40 m

4.
(c) 5 cm Triangle inequality: sum of two sides must be greater than third side. For perimeter 5 cm, no valid whole number sides exist. (e.g., 1+1<3, 1+2=3)

5.
(c) 52 sq m Room area = 5 × 4 = 20 sq m Total area with verandah = 9 × 8 = 72 sq m Verandah area = 72 - 20 = 52 sq m

6.
(a) 26 m Perimeter = 1872 ÷ 18 = 104 m Side = 104 ÷ 4 = 26 m

7.
(c) 4 units 4a = a², where a is side a = 4 units

8.
(b) 40 cm Perimeter of rectangle = 2(12 + 8) = 40 cm Same wire forms circle, so circumference = 40 cm

9.
(c) 15 m 2(l + b) = 50, so l + b = 25 l × b = 150 Solving: l = 15 m, b = 10 m

10.
(c) The one with length = breadth (square) For a fixed perimeter, a square encloses maximum area. This is a fundamental geometric property.

SECTION B - Answers to Short Answer Questions

11. Given: Garden = 90 m × 75 m, Path widths = 2 m and 3 m Step 1: Path parallel to length = 90 × 3 = 270 sq m Step 2: Path parallel to breadth = 75 × 2 = 150 sq m Step 3: Intersection = 2 × 3 = 6 sq m Step 4: Total = 270 + 150 - 6 = 414 sq m Answer: Area of paths = 414 sq m 12. Given: Square side = 16 cm, Rectangle length = 32 cm, Equal areas Step 1: Area = 16 × 16 = 256 sq cm Step 2: Breadth = 256 ÷ 32 = 8 cm Step 3: Perimeter of square = 64 cm Step 4: Perimeter of rectangle = 2(32 + 8) = 80 cm Answer: Breadth = 8 cm, Perimeters: 64 cm and 80 cm 13.

Given: Pentagon side = 8 cm, Rectangle length = 12 cm Step 1: Perimeter of pentagon = 5 × 8 = 40 cm Step 2: Wire length = 40 cm Step 3: 2(12 + b) = 40, b = 8 cm Step 4: Area = 12 × 8 = 96 sq cm Answer: Perimeter = 40 cm, Area = 96 sq cm 14. Given: Perimeter = 300 m, Remains unchanged after changes Step 1: 2(l + b) = 300, so l + b = 150 Step 2: After changes: 2(l+10 + b-10) = 300 (satisfied for any l, b) Note: Any dimensions with l + b = 150 work Example: 80 m × 70 m, 90 m × 60 m, etc. Answer: Any dimensions where l + b = 150 m

SECTION C - Answers to Short Answer Questions

15. Given: Field = 48 m × 30 m, 3 flower beds of 6 m × 6 m, Rate = ₹50/sq m Step 1: Area of field = 48 × 30 = 1440 sq m Step 2: Area of one bed = 6 × 6 = 36 sq m Step 3: Total bed area = 3 × 36 = 108 sq m Step 4: Remaining area = 1440 - 108 = 1332 sq m Step 5: Cost = 1332 × 50 = ₹66,600 Answer: Remaining area = 1332 sq m, Cost = ₹66,600 16. Given: Both perimeter = 56 cm, A = 18 cm × 10 cm, B = square Verification: 2(18 + 10) = 56 ✓ Step 1: Side of B = 56 ÷ 4 = 14 cm Step 2: Area of A = 18 × 10 = 180 sq cm Step 3: Area of B = 14 × 14 = 196 sq cm Step 4: Difference = 196 - 180 = 16 sq cm Answer: Square B has 16 sq cm more area 17.

Given: Lawn = 36 m × 24 m, Path = 3 m wide (outside), Rate = ₹60/sq m Step 1: Outer dimensions = (36+6) × (24+6) = 42 × 30 Step 2: Total area = 42 × 30 = 1260 sq m Step 3: Lawn area = 36 × 24 = 864 sq m Step 4: Path area = 1260 - 864 = 396 sq m Step 5: Cost = 396 × 60 = ₹23,760 Answer: Cost = ₹23,760

SECTION D - Answer to Long Answer Question

18. Given: Hall = 20 m × 16 m
(a) Largest tile size: HCF of 20 and 16 = 4 m Largest tile = 4 m × 4 m
(b) Number of tiles: Area of hall = 320 sq m Area of one tile = 16 sq m Tiles needed = 320 ÷ 16 = 20 tiles
(c) Cost of tiles: Cost = 20 × 500 = ₹10,000
(d) Carpet dimensions: Length = 20 - 2 × 2 = 16 m Breadth = 16 - 2 × 2 = 12 m Dimensions: 16 m × 12 m (e) Uncovered area: Carpet area = 16 × 12 = 192 sq m Uncovered = 320 - 192 = 128 sq m

Answers: (a) 4 m × 4 m, (b) 20 tiles, (c) ₹10,000, (d) 16 m × 12 m, (e) 128 sq m

SECTION E - Answers to Case Study Based Questions

19.
(a) Dimensions with border: Length = 20 + 2 × 2.5 = 25 cm Width = 15 + 2 × 2.5 = 20 cm Dimensions: 25 cm × 20 cm
(b) Area of photo: Area = 20 × 15 = 300 sq cm
(c) Total area with border: Area = 25 × 20 = 500 sq cm
(d) Area of border alone: Border area = 500 - 300 = 200 sq cm

Answers: (a) 25 cm × 20 cm, (b) 300 sq cm, (c) 500 sq cm, (d) 200 sq cm

20.
(a) Comparing areas: Plot X: 50 × 50 = 2500 sq m Plot Y: 60 × 40 = 2400 sq m Plot X is greater by 100 sq m
(b) Wire needed: Perimeter of X = 4 × 50 = 200 m Wire for 4 rounds = 200 × 4 = 800 m
(c) Cost of wire: Cost = 800 × 12 = ₹9600
(d) Garden area: House area = 30% of 2500 = 750 sq m Garden = 2500 - 750 = 1750 sq m

Answers: (a) Plot X by 100 sq m, (b) 800 m, (c) ₹9600, (d) 1750 sq m