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

CBSE Class 6 Maths Ganita Prakash Chapter 6 Perimeter and Area Practice Paper 16 is a 40-mark test of 20 questions in Sections A to E, solved here step by step. It covers the perimeter of rectangles, squares, triangles and a regular hexagon, how the area of a square or rectangle changes when its sides are halved, doubled or tripled, unit conversion in carpet and fencing problems, the area of a stepped figure split into rectangles, and the area of a path built around a park.

Section A: Multiple Choice Questions (1 Mark Each)

Section B: Short Answer Questions (2 Marks Each)

Question 11. A room is 4 m long and 3 m 50 cm wide. How many square metres of carpet are needed to cover the floor of the room?
Answer:
Breadth = 3 m 50 cm = 3.5 m (since 50 cm = 0.5 m); length = 4 m.
Area of the floor = length × breadth = 4 × 3.5 = 14 m².
So 14 m² of carpet is needed.
In simple words: Change 3 m 50 cm into 3.5 m and multiply by 4 m.

Teacher's Note:
1. Carpet covers the floor, so we need area, not perimeter.
2. 4 × 3.5 = 4 × 3 + 4 × 0.5 = 12 + 2 = 14.

Question 12. Two sides of a triangle are 12 cm and 14 cm. The perimeter of the triangle is 36 cm. What is its third side?
Answer:
Perimeter = sum of the three sides, so 12 + 14 + third side = 36.
26 + third side = 36, so third side = 36 − 26 = 10 cm.
The third side is 10 cm.
In simple words: Take away the two known sides from the perimeter.

Teacher's Note:
1. Check: 12 + 14 + 10 = 36 cm.
2. A missing side = perimeter − sum of the other sides.

Question 13. The lid of a rectangular box, with sides 40 cm by 10 cm, is sealed all around with tape. What is the length of the tape required?
Answer:
The tape goes once around the edge of the lid, so its length is the perimeter of the lid.
Perimeter = 2 × (40 + 10) = 2 × 50 = 100 cm.
Length of tape required = 100 cm (that is, 1 m).
In simple words: Tape all around the lid = perimeter of a 40 cm × 10 cm rectangle.

Teacher's Note:
1. "All around" is a signal word for perimeter.
2. The area 400 cm² is not needed here.

Question 14. A rectangular piece of land measures 0.7 km by 0.5 km. Each side is to be fenced with 4 rows of wires. What is the length of the wire needed?
Answer:
One round of wire = perimeter = 2 × (0.7 + 0.5) = 2 × 1.2 = 2.4 km.
There are 4 rows, so wire needed = 4 × 2.4 = 9.6 km.
The length of wire needed is 9.6 km (= 9600 m).
In simple words: Find the perimeter once and multiply by the 4 rows.

Teacher's Note:
1. Each row of wire goes once around the whole boundary.
2. Keep decimals in line: 4 × 2.4 = 9.6.

Section C: Short Answer Questions (3 Marks Each)

Question 15. Five square flower beds, each of sides 1 m, are dug on a piece of land 5 m long and 4 m wide. What is the area of the remaining part of the land?
Answer:
Area of the land = 5 × 4 = 20 m².
Area of one flower bed = 1 × 1 = 1 m², so area of 5 beds = 5 × 1 = 5 m².
Area of the remaining land = 20 − 5 = 15 m².
In simple words: Whole land 20 m² minus the beds 5 m² leaves 15 m².

Teacher's Note:
1. Find the whole area first, then subtract the parts used.
2. Each 1 m × 1 m bed is exactly one square metre.

Question 16. The length and breadth of the three rectangles are as given below:
(a) 9 m and 6 m
(b) 17 m and 3 m
(c) 4 m and 14 m
Which one has the largest area, and which one has the smallest?
Answer:
(a) Area = 9 × 6 = 54 m².
(b) Area = 17 × 3 = 51 m².
(c) Area = 4 × 14 = 56 m².
Comparing 56 > 54 > 51: rectangle (c) has the largest area (56 m²) and rectangle (b) has the smallest area (51 m²).
In simple words: Multiply length by breadth for each and compare the three answers.

Teacher's Note:
1. The rectangle with the longest side, (b), is not the one with the largest area.
2. Write the units as m² for area.

Question 17. Sweety runs around a square park of side 75 m. Bulbul runs around a rectangular park with a length of 60 m and a breadth of 45 m. Who covers less distance?
Answer:
Distance run by Sweety = perimeter of the square = 4 × 75 = 300 m.
Distance run by Bulbul = perimeter of the rectangle = 2 × (60 + 45) = 2 × 105 = 210 m.
210 m < 300 m, so Bulbul covers less distance, by 300 − 210 = 90 m.
In simple words: One round of each park is its perimeter: 300 m for Sweety and 210 m for Bulbul.

Teacher's Note:
1. Running around a park means covering its perimeter.
2. Always compare the two numbers in the same unit.

Section D: Long Answer Question (5 Marks)

Question 18. What will happen to the area of rectangle if its
(i) Length and breadth are tripled
(ii) Length is doubled and breadth is same
Answer:
Let the original length be l and breadth be b, so the original area = l × b.
(i) New length = 3l and new breadth = 3b.
New area = 3l × 3b = 9 × (l × b).
So the area becomes 9 times the original area.
Example: a 2 cm × 1 cm rectangle (area 2 cm²) becomes 6 cm × 3 cm (area 18 cm²), and 18 = 9 × 2.
(ii) New length = 2l and breadth stays b.
New area = 2l × b = 2 × (l × b).
So the area becomes 2 times (double) the original area.
Example: a 2 cm × 1 cm rectangle (area 2 cm²) becomes 4 cm × 1 cm (area 4 cm²), and 4 = 2 × 2.
In simple words: Tripling both sides multiplies the area by 3 × 3 = 9; doubling only the length multiplies it by 2.

Teacher's Note:
1. The area grows by the product of the two scale factors.
2. Checking with a small example is a good way to be sure.
3. Say "9 times the original", not "9 times more", which can be misread.

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

Question 19. Case Study 1: Kiran wants to find the area of the figure using area of rectangle. She splits the figure into four rectangles and then she coloured the four rectangles with yellow, orange, green and grey colour. (The measures are given in centimetres)
Based on the above, answer the following questions
(a) Find the area of orange colour.
(b) Find the area of yellow colour.
(c) Find the area of grey colour.
(d) Find the total area of the figure.
Answer:
Read the sides of each rectangle from the figure. The left side 4 cm is made of the yellow square (3 cm) and the orange strip (4 − 3 = 1 cm).
(a) Orange rectangle: 2 cm long and 1 cm high, so area = 2 × 1 = 2 cm².
(b) Yellow rectangle: 3 cm by 3 cm, so area = 3 × 3 = 9 cm².
(c) Grey rectangle: 3 cm wide and 2 + 1 = 3 cm high, so area = 3 × 3 = 9 cm².
(d) Green rectangle: the full width of the figure is 3 + 4 = 7 cm (bottom 3 and ledge 4), and also orange 2 + grey 3 + green width, so green width = 7 − 5 = 2 cm. Its height = 3 (grey) + 1 (top step) = 4 cm, so green area = 2 × 4 = 8 cm².
Total area = 2 + 9 + 9 + 8 = 28 cm².
In simple words: Find the area of each coloured rectangle and add them: 2 + 9 + 9 + 8 = 28 cm².

Teacher's Note:
1. Missing sides are found from the given ones by adding or subtracting along a straight edge.
2. Check: the figure fits in a 7 cm × 7 cm square, and 28 cm² is well below 49 cm².

Question 20. Case Study 2: Meet Rohan, an avid morning walker! He loves the rectangular park in his neighborhood, which is 50 m long and 30 m wide. To make his workout more accessible, the city decided to build a 2 m wide walking path constructed along the outside of the park's existing boundary. Rohan is eager to see how much larger the total area is now for his exercise route!
(a) Find the area of the entire park (excluding the path). (1 Mark)
(b) Find the new length and breadth of the park including the path. (1 Mark)
(c) Calculate the area of the path only. (2 Marks)
Answer:
(a) Area of the park = 50 × 30 = 1500 m².
(b) The path adds 2 m on both ends of each side.
New length = 50 + 2 + 2 = 54 m; new breadth = 30 + 2 + 2 = 34 m.
(c) Area of park with path = 54 × 34 = 1836 m².
Area of the path = 1836 − 1500 = 336 m².
In simple words: Path area = big outer rectangle minus the park inside it.

Teacher's Note:
1. A path outside the boundary adds its width twice to each dimension.
2. Check (c) another way: two long strips 54 × 2 = 108 each and two short strips 30 × 2 = 60 each give 216 + 120 = 336 m².