CBSE Class 6 Mathematics Perimeter and Area Practice Paper Worksheet Set 03 with Answers

Here is CBSE Class 6 Maths Ganita Prakash Chapter 6 Perimeter and Area Practice Paper 18: 20 questions, 40 marks, Sections A to E, with a full step-by-step solution under every question. This paper tests the perimeter of a six-sided field and of rectilinear figures, finding the side of a square from its area, the effect of doubling the sides on area, costs of grass planting and levelling, the ratio of length to breadth from the perimeter, splitting L-shaped and stepped figures into rectangles, reshaping a square wire into a rectangle, and counting floor tiles.

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Question 11. Find the perimeter of the following figure:

Answer:
Name the corners A, B, C, D, E, F, G, H going round the figure.

Given: AB = 10 m, BC = 10 m, CD = 20 m, DE = 30 m and AH = 20 m.
All corners are right angles, and in the figure F is directly below C, so EF = CD = 20 m.
Going up from F to G and from H to A must balance the drops BC + DE = 10 + 30 = 40 m: FG + HA = 40, so FG = 40 − 20 = 20 m.
Going left from E to F and G to H must balance AB + CD = 10 + 20 = 30 m: GH = 30 − 20 = 10 m.
Perimeter = 10 + 10 + 20 + 30 + 20 + 20 + 10 + 20 = 140 m.
In simple words: Find the three unmarked sides, then add all eight sides to get 140 m.

Teacher's Note:
1. In a figure with only right-angle corners, total rightward lengths = total leftward lengths, and total down = total up.
2. Shortcut: such a staircase shape has the same perimeter as its bounding rectangle, 2 × (30 + 40) = 140 m.

Question 12. One side of a square plot is 250 m, find the cost of levelling it at the rate of Rs 2 per square metre.
Answer:
Levelling is done over the whole surface, so find the area.
Area = 250 × 250 = 62500 m².
Cost = 62500 × 2 = Rs 1,25,000.
In simple words: Area 62500 m² at Rs 2 per m² costs Rs 1,25,000.

Teacher's Note:
1. Rate per square metre means multiply by the area.
2. 250 × 250 = 25 × 25 × 100 = 625 × 100.

Question 13. If the cost of fencing a rectangular field at Rs. 7.50 per metre is Rs. 600, and the length of the field is 24 m, then find the breadth of the field.
Answer:
Perimeter = total cost ÷ rate = 600 ÷ 7.50 = 80 m.
Length + breadth = 80 ÷ 2 = 40 m.
Breadth = 40 − 24 = 16 m.
In simple words: Cost ÷ rate gives the fence length 80 m; half of it minus the length gives the breadth.

Teacher's Note:
1. Check: 2 × (24 + 16) × 7.50 = 600.
2. 600 ÷ 7.50 = 6000 ÷ 75 = 80.

Question 14. Split the following shapes into rectangles and find the area of each. (The measures are given in centimetres)

Answer:
Cut the L-shape along the line where the upright part meets the base, as shown.

Upright rectangle (blue): 2 cm wide and 12 cm high, area = 2 × 12 = 24 cm².
Base rectangle (yellow): 8 cm long (10 − 2) and 2 cm high, area = 8 × 2 = 16 cm².
Total area = 24 + 16 = 40 cm².
In simple words: Split into a 2 × 12 strip and an 8 × 2 strip: 24 + 16 = 40 cm².

Teacher's Note:
1. The 10 cm inner side plus the 2 cm base thickness makes 12 cm.
2. Another split, 2 × 10 + 10 × 2 = 40 cm², gives the same total.

Section C: Short Answer Questions (3 Marks Each)

Question 15. A square sheet of side 5 cm is cut out from a rectangular piece of an aluminium sheet of length 9 cm and breadth 6 cm. What is the area of the aluminium sheet left over?
Answer:
Area of the rectangular sheet = 9 × 6 = 54 cm².
Area of the square cut out = 5 × 5 = 25 cm².
Area left over = 54 − 25 = 29 cm².
In simple words: Whole sheet 54 cm² minus the 25 cm² square leaves 29 cm².

Teacher's Note:
1. Area left = area of whole − area removed.
2. The 5 cm square fits since 5 < 6 and 5 < 9.

Question 16. If the ratio between the length and the perimeter of a rectangular plot is 1 : 3, then find the ratio between the length and breadth of the plot.
Answer:
Length : perimeter = 1 : 3 means perimeter = 3 × length.
Perimeter = 2 × length + 2 × breadth, so 3 × length = 2 × length + 2 × breadth.
Taking away 2 × length from both sides: length = 2 × breadth.
So length : breadth = 2 : 1.
Check: length 2 m and breadth 1 m give perimeter 6 m, and 2 : 6 = 1 : 3.
In simple words: The perimeter is three lengths, which leaves one length to equal two breadths.

Teacher's Note:
1. Write the ratio as a sentence (perimeter is 3 times the length) before using it.
2. Always test the final ratio with simple numbers.

Question 17. A rectangular piece of land measures 0.7 km by 0.5 km. Each side is to be fenced with four rows of wires. What length of the wire is needed?
Answer:
One row of wire = perimeter = 2 × (0.7 + 0.5) = 2 × 1.2 = 2.4 km.
Four rows = 4 × 2.4 = 9.6 km.
Length of wire needed = 9.6 km (9600 m).
In simple words: One round is 2.4 km; four rounds need 9.6 km of wire.

Teacher's Note:
1. Each row goes all the way round once.
2. Converting to metres (700 m by 500 m) gives the same answer, 9600 m.

Section D: Long Answer Question (5 Marks)

Question 18. By splitting the following figures into rectangles, find their areas (The measures are given in centimetres).

Answer:
Split the figure into four rectangles as shown, and find each missing side from the given ones.

Yellow: 3 cm by 3 cm, area = 3 × 3 = 9 cm².
Orange: the left side is 4 cm and the yellow takes 3 cm, so the orange is 4 − 3 = 1 cm high and 2 cm long, area = 2 × 1 = 2 cm².
Grey: 3 cm long and 2 + 1 = 3 cm high, area = 3 × 3 = 9 cm².
Brown: the whole width is 3 + 4 = 7 cm, so its width = 7 − (2 + 3) = 2 cm; its height = 3 + 1 = 4 cm, area = 2 × 4 = 8 cm².
Total area = 9 + 2 + 9 + 8 = 28 cm².
In simple words: Cut the shape into four rectangles of 9, 2, 9 and 8 cm² and add them to get 28 cm².

Teacher's Note:
1. Missing sides come from adding or subtracting along one straight edge.
2. Any other correct splitting also gives 28 cm².
3. Write cm² for each area.

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

Question 19. Case Study 1: Anu wants to fence the garden in front of her house (see the figure in the answer), on three sides with lengths 20 m, 12 m and 12 m. For fencing she went to shop to purchase wire. She brought a wire which is in the shape of a square of side 10 m. She changed the shape of the wire into a rectangle of length 12 m.
Based on the above, answer the following questions
(a) Find the cost of fencing at the rate of Rs 150 per metre.
(b) What is the perimeter of the wire?
(c) What is the breadth of the rectangle she formed?
(d) Find the difference between the area of the square and the area of the rectangle.
Answer:

(a) Fence needed = 20 + 12 + 12 = 44 m (the side along the house is not fenced). Cost = 44 × 150 = Rs 6,600.
(b) The wire is a square of side 10 m, so its length (perimeter) = 4 × 10 = 40 m.
(c) The same 40 m wire makes the rectangle: 2 × (12 + breadth) = 40, so 12 + breadth = 20 and breadth = 8 m.
(d) Area of the square = 10 × 10 = 100 m²; area of the rectangle = 12 × 8 = 96 m². Difference = 100 − 96 = 4 m².
In simple words: Same wire, same perimeter (40 m), but the square encloses 4 m² more than the 12 m × 8 m rectangle.

Teacher's Note:
1. Reshaping a wire keeps the perimeter but changes the area.
2. For a fixed perimeter, the square gives the largest area among rectangles.

Question 20. Case Study 2: Maya is renovating her family's grand entrance hall. The floor needs new, elegant marble tiles, each measuring 10 cm × 12 cm. The entire floor area to be covered is a large rectangle measuring 3 m × 4 m. Maya needs to know the exact number of tiles required and the total budget for the tiles, which cost Rs 2 per tile, to ensure the project stays on track.
(a) What is the area of a single marble tile?
(b) What is the area of the floor?
(c) Find the number of tiles required to cover the floor.
(d) Find the cost of tiles at the rate of Rs. 2 per tile.
Answer:
(a) Area of one tile = 10 × 12 = 120 cm².
(b) Floor = 3 m × 4 m = 300 cm × 400 cm, so area = 300 × 400 = 1,20,000 cm² (= 12 m²).
(c) Number of tiles = 120000 ÷ 120 = 1000 tiles.
(Check by rows: with the 12 cm side along the 3 m side, 300 ÷ 12 = 25 tiles and 400 ÷ 10 = 40 tiles, and 25 × 40 = 1000.)
(d) Cost = 1000 × Rs 2 = Rs 2,000.
In simple words: Floor area ÷ tile area gives 1000 tiles, costing Rs 2,000.

Teacher's Note:
1. Change metres to centimetres before dividing areas.
2. Laying the tiles the right way round makes them fit exactly, so no tile is cut.