Class 6 Maths Chapter 6 Perimeter and Area Practice Paper 1

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 perimeter of a square with side 8 cm is:
(a) 16 cm
(b) 32 cm
(c) 64 cm
(d) 24 cm

2. The formula for the perimeter of a rectangle is:
(a) 4 × side
(b) length × breadth
(c) 2 × (length + breadth)
(d) length + breadth

3. A rectangular park is 40 m long and 30 m wide. What is its perimeter?
(a) 70 m
(b) 140 m
(c) 1200 m
(d) 100 m

4. The area of a square with side 10 cm is:
(a) 40 sq cm
(b) 100 sq cm
(c) 20 sq cm
(d) 50 sq cm

5. A rectangle has length 12 m and breadth 8 m. Its area is:
(a) 40 sq m
(b) 96 sq m
(c) 20 sq m
(d) 48 sq m

6. The perimeter of an equilateral triangle with side 6 cm is:
(a) 12 cm
(b) 18 cm
(c) 24 cm
(d) 36 cm

7. If the perimeter of a square is 24 cm, its side is:
(a) 4 cm
(b) 6 cm
(c) 8 cm
(d) 12 cm

8. The area of a rectangle is 48 sq m and its length is 8 m. Its breadth is:
(a) 6 m
(b) 8 m
(c) 10 m
(d) 12 m

9. Which of the following has the largest perimeter?
(a) A square of side 5 cm
(b) A rectangle of length 7 cm and breadth 3 cm
(c) An equilateral triangle of side 7 cm
(d) A rectangle of length 6 cm and breadth 4 cm

10. The formula for the area of a triangle is:
(a) base × height
(b) ½ × base × height
(c) 2 × base × height
(d) ½ × perimeter

SECTION B - Short Answer Questions (2 marks each)

11. A wire of length 32 cm is bent to form a square. Find the side of the square and also find its area.

12. The length and breadth of a rectangular field are 50 m and 30 m respectively. Find the cost of fencing the field at the rate of ₹15 per metre.

13. A rectangular garden is 25 m long and 15 m wide. Find the length of the wire required to fence it with 2 rounds.

14. Find the perimeter of a triangle with sides 7 cm, 8 cm and 9 cm. If the triangle is made using a wire and then the wire is straightened to form a square, what will be the side of the square?

SECTION C - Short Answer Questions (3 marks each)

15. A rectangular plot of land measures 80 m by 60 m. A path of width 5 m runs all around inside the plot. Find the area of the path.

16. The perimeter of a rectangular field is 240 m. If the length is 70 m, find its breadth and area.

17. A square and a rectangle have the same perimeter of 40 cm. If the rectangle has length 12 cm, which figure has the greater area and by how much?

SECTION D - Long Answer Question (5 marks)

18. A rectangular hall is 18 m long and 12 m wide. A square carpet of side 8 m is laid on the floor. Four square flower beds, each of side 2 m, are placed at the four corners of the hall.
(a) Find the area of the carpet.
(b) Find the total area of the four flower beds.
(c) Find the area of the floor that is not covered by the carpet and flower beds.
(d) Find the perimeter of the hall. (e) Find the cost of polishing the uncovered floor at ₹25 per sq m.

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

19. Running Track Raj and Simran are running on two different rectangular tracks. Raj's track is 70 m long and 40 m wide. Simran's track is 60 m long and 50 m wide.
(a) Find the perimeter of Raj's track. (1 mark)
(b) Find the perimeter of Simran's track. (1 mark)
(c) If Raj completes 5 rounds, what total distance does he cover? (1 mark)
(d) Whose track has a larger area and by how much? (1 mark)

20. Garden Design Mrs. Sharma has a rectangular garden that is 30 m long and 20 m wide. She wants to create a walking path of width 2 m all around inside the garden and plant grass on the remaining area.
(a) What will be the perimeter of the garden? (1 mark)
(b) What will be the dimensions of the inner rectangle (after the path)? (1 mark)
(c) What is the area of the walking path? (1 mark)
(d) Find the cost of planting grass at ₹20 per sq m. (1 mark) DETAILED ANSWER KEY - PAPER 01

SECTION A - Answers to MCQs

1.
(b) 32 cm Perimeter of square = 4 × side = 4 × 8 = 32 cm

2.
(c) 2 × (length + breadth) The perimeter of a rectangle is twice the sum of its length and breadth.

3.
(b) 140 m Perimeter = 2 × (40 + 30) = 2 × 70 = 140 m

4.
(b) 100 sq cm Area of square = side × side = 10 × 10 = 100 sq cm

5.
(b) 96 sq m Area of rectangle = length × breadth = 12 × 8 = 96 sq m

6.
(b) 18 cm Perimeter of equilateral triangle = 3 × side = 3 × 6 = 18 cm

7.
(b) 6 cm Side = Perimeter ÷ 4 = 24 ÷ 4 = 6 cm

8.
(a) 6 m Breadth = Area ÷ length = 48 ÷ 8 = 6 m

9.
(c) An equilateral triangle of side 7 cm
(a) Square: 4 × 5 = 20 cm
(b) Rectangle: 2 × (7 + 3) = 20 cm
(c) Triangle: 3 × 7 = 21 cm (Largest)
(d) Rectangle: 2 × (6 + 4) = 20 cm

10.
(b) ½ × base × height Area of triangle is half the product of base and height.

SECTION B - Answers to Short Answer Questions

11. Given: Length of wire = 32 cm Step 1: The wire forms a square, so perimeter of square = 32 cm Step 2: Side of square = 32 ÷ 4 = 8 cm Step 3: Area of square = 8 × 8 = 64 sq cm Answer: Side = 8 cm, Area = 64 sq cm 12. Given: Length = 50 m, Breadth = 30 m, Rate = ₹15 per m Step 1: Perimeter = 2 × (50 + 30) = 2 × 80 = 160 m Step 2: Cost = 160 × 15 = ₹2400 Answer: Cost of fencing = ₹2400 13. Given: Length = 25 m, Breadth = 15 m, Rounds = 2 Step 1: Perimeter = 2 × (25 + 15) = 2 × 40 = 80 m Step 2: Wire needed = 80 × 2 = 160 m Answer: Length of wire = 160 m 14.

Given: Sides of triangle = 7 cm, 8 cm, 9 cm Step 1: Perimeter = 7 + 8 + 9 = 24 cm Step 2: Same wire forms a square, so perimeter = 24 cm Step 3: Side of square = 24 ÷ 4 = 6 cm Answer: Perimeter = 24 cm, Side of square = 6 cm

SECTION C - Answers to Short Answer Questions

15. Given: Plot dimensions = 80 m × 60 m, Path width = 5 m Step 1: Area of plot = 80 × 60 = 4800 sq m Step 2: Inner dimensions (after path) = (80 - 2×5) × (60 - 2×5) = 70 × 50 Step 3: Area of inner rectangle = 70 × 50 = 3500 sq m Step 4: Area of path = 4800 - 3500 = 1300 sq m Answer: Area of the path = 1300 sq m 16. Given: Perimeter = 240 m, Length = 70 m Step 1: 2 × (length + breadth) = 240 Step 2: 70 + breadth = 120 Step 3: Breadth = 120 - 70 = 50 m Step 4: Area = 70 × 50 = 3500 sq m Answer: Breadth = 50 m, Area = 3500 sq m 17.

Given: Both have perimeter = 40 cm, Rectangle length = 12 cm For Square: Side = 40 ÷ 4 = 10 cm Area = 10 × 10 = 100 sq cm For Rectangle: 2 × (12 + breadth) = 40 12 + breadth = 20 Breadth = 8 cm Area = 12 × 8 = 96 sq cm Answer: Square has greater area by 100 - 96 = 4 sq cm

SECTION D - Answer to Long Answer Question

18. Given: Hall = 18 m × 12 m, Carpet = 8 m × 8 m, Flower beds = 2 m × 2 m (4 beds)
(a) Area of carpet: Area = 8 × 8 = 64 sq m
(b) Total area of flower beds: Area of one bed = 2 × 2 = 4 sq m Total area = 4 × 4 = 16 sq m
(c) Uncovered floor area: Total area of hall = 18 × 12 = 216 sq m Covered area = 64 + 16 = 80 sq m Uncovered area = 216 - 80 = 136 sq m
(d) Perimeter of hall: Perimeter = 2 × (18 + 12) = 2 × 30 = 60 m (e) Cost of polishing: Cost = 136 × 25 = ₹3400

Answers: (a) 64 sq m, (b) 16 sq m, (c) 136 sq m, (d) 60 m, (e) ₹3400

SECTION E - Answers to Case Study Based Questions

19.
(a) Perimeter of Raj's track: Perimeter = 2 × (70 + 40) = 2 × 110 = 220 m
(b) Perimeter of Simran's track: Perimeter = 2 × (60 + 50) = 2 × 110 = 220 m
(c) Distance covered by Raj in 5 rounds: Distance = 220 × 5 = 1100 m
(d) Comparing areas: Raj's track area = 70 × 40 = 2800 sq m Simran's track area = 60 × 50 = 3000 sq m Simran's track is larger by 3000 - 2800 = 200 sq m

Answers: (a) 220 m, (b) 220 m, (c) 1100 m, (d) Simran's by 200 sq m

20.
(a) Perimeter of garden: Perimeter = 2 × (30 + 20) = 2 × 50 = 100 m
(b) Dimensions of inner rectangle: Length = 30 - 2 × 2 = 26 m Breadth = 20 - 2 × 2 = 16 m Dimensions: 26 m × 16 m
(c) Area of walking path: Area of garden = 30 × 20 = 600 sq m Area of inner rectangle = 26 × 16 = 416 sq m Area of path = 600 - 416 = 184 sq m
(d) Cost of planting grass: Cost = 416 × 20 = ₹8320

Answers: (a) 100 m, (b) 26 m × 16 m, (c) 184 sq m, (d) ₹8320