CBSE Class 6 Mathematics Mensuration Worksheet Set 03

Question 1. Find the perimeter of following shapes given below; if the distance between two consecutive dots is \(2\text{ cm}\).
Answer: For each shape, the perimeter is found by counting the total number of grid units along the outer boundary and multiplying by the distance between two consecutive dots (\(2\text{ cm}\)).
(i) Number of boundary segments = \(20\)
Perimeter = \(20 \times 2\text{ cm} = 40\text{ cm}\)
(ii) Number of boundary segments = \(16\)
Perimeter = \(16 \times 2\text{ cm} = 32\text{ cm}\)
(iii) Number of boundary segments = \(16\)
Perimeter = \(16 \times 2\text{ cm} = 32\text{ cm}\)
(iv) Number of boundary segments = \(16\)
Perimeter = \(16 \times 2\text{ cm} = 32\text{ cm}\)
(v) Number of boundary segments = \(24\)
Perimeter = \(24 \times 2\text{ cm} = 48\text{ cm}\)
(vi) Number of boundary segments = \(24\)
Perimeter = \(24 \times 2\text{ cm} = 48\text{ cm}\)
In simple words: Count the outer grid units of each figure and multiply by 2 cm to get the total perimeter.

Exam Tip: Be careful not to miss any inner corners or indentations when counting boundary segments on dot grids.

Question 2. Find the perimeter of the following figures:
Answer:
(i) Rectangle with length \(7\text{ cm}\) and breadth \(2\text{ cm}\).
Perimeter = \(2 \times (\text{length} + \text{breadth}) = 2 \times (7 + 2) = 2 \times 9 = 18\text{ cm}\)
(ii) Scalene triangle with sides \(4\text{ cm}\), \(4\text{ cm}\), and \(3\text{ cm}\).
Perimeter = \(4 + 4 + 3 = 11\text{ cm}\)
(iii) Regular hexagon with each side \(3\text{ cm}\).
Perimeter = \(6 \times \text{side} = 6 \times 3 = 18\text{ cm}\)
(iv) Triangle with sides \(5\text{ cm}\), \(4\text{ cm}\), and \(2\text{ cm}\).
Perimeter = \(5 + 4 + 2 = 11\text{ cm}\)
(v) Square with each side \(5\text{ cm}\).
Perimeter = \(4 \times \text{side} = 4 \times 5 = 20\text{ cm}\)
(vi) Rectangle with length \(6\text{ cm}\) and breadth \(2\text{ cm}\).
Perimeter = \(2 \times (6 + 2) = 2 \times 8 = 16\text{ cm}\)
(vii) L-shaped polygon with side dimensions summing up appropriately.
Perimeter = \(7 + 4 + 5 + 2 + 2 + 2 = 22\text{ cm}\)
(viii) Right-angled triangle with sides \(5\text{ cm}\), \(6\text{ cm}\), and \(4\text{ cm}\).
Perimeter = \(5 + 6 + 4 = 15\text{ cm}\)
In simple words: Add the lengths of all outer sides of each shape to find its perimeter.

Exam Tip: For regular shapes like squares and hexagons, multiply the length of one side by the total number of sides.

Question 3. Find the missing lengths:
Answer:
(i) Perimeter = \(28\text{ cm}\). Given sides = \(4\text{ cm}\), \(9\text{ cm}\), \(8\text{ cm}\).
Missing side = \(28 - (4 + 9 + 8) = 28 - 21 = 7\text{ cm}\)
(ii) Perimeter = \(35\text{ cm}\). Given sides = \(2\text{ cm}\), \(8\text{ cm}\), \(5\text{ cm}\).
Missing side = \(35 - (2 + 8 + 5) = 35 - 15 = 20\text{ cm}\)
(iii) Perimeter = \(28\text{ cm}\). Given sides = \(10\text{ cm}\), \(12\text{ cm}\).
Missing side = \(28 - (10 + 12) = 28 - 22 = 6\text{ cm}\)
In simple words: Subtract the sum of the known sides from the total perimeter to find the missing side.

Exam Tip: Always verify your result by adding all sides together to check if they equal the given perimeter.

Question 4. Find the perimeter of following figure by using formula:
Answer:
Length (\(l\)) = \(123\text{ dm}\)
Breadth (\(b\)) = \(115\text{ dm}\)
Perimeter of rectangle = \(2 \times (l + b)\)
Perimeter = \(2 \times (123 + 115)\)
Perimeter = \(2 \times 238 = 476\text{ dm}\)
In simple words: Add the length and breadth, then multiply the sum by 2.

Exam Tip: Do not forget to include the unit (\(\text{dm}\)) in your final answer.

Question 5. Fill in the blanks in the following table by calculating the perimeter in each case:
Answer:
Formula for perimeter of a square = \(4 \times \text{side}\).
For side \(3\text{ cm}\): \(4 \times 3 = 12\)
For side \(5\text{ cm}\): \(4 \px 5 = 20\) [Wait, \(4 \times 5 = 20\)]
Completing the table rows:

In simple words: Multiply each given side length by 4 to get the perimeter.

Exam Tip: A square has four equal sides, so perimeter is always four times the side length.

Question 6. Find the perimeter of these rectangles:
Answer:
Formula: \(p = 2 \times (l + b)\).
1. \(l = 2\text{ cm}\), \(b = 3\text{ cm} \implies p = 2 \times (2 + 3) = 10\text{ cm}\)
2. \(l = 2\text{ cm}\), \(b = 4\text{ cm} \implies p = 2 \times (2 + 4) = 12\text{ cm}\)
3. \(l = 3\text{ cm}\), \(b = 4\text{ cm} \implies p = 2 \times (3 + 4) = 14\text{ cm}\)
4. \(l = 5\text{ cm}\), \(b = 4\text{ cm} \implies p = 2 \times (5 + 4) = 18\text{ cm}\)
5. \(l = 5\text{ cm}\), \(b = 2\text{ cm} \implies p = 2 \times (5 + 2) = 14\text{ cm}\)
6. \(l = 3\text{ cm}\), \(b = 6\text{ cm} \implies p = 2 \times (3 + 6) = 18\text{ cm}\)
7. \(l = 4\text{ cm}\), \(b = 6\text{ cm} \implies p = 2 \times (4 + 6) = 20\text{ cm}\)
8. \(l = 5\text{ cm}\), \(b = 6\text{ cm} \implies p = 2 \times (5 + 6) = 22\text{ cm}\)
9. \(l = 8\text{ cm}\), \(b = 6\text{ cm} \implies p = 2 \times (8 + 6) = 28\text{ cm}\)
10. \(l = 7\text{ cm}\), \(b = 9\text{ cm} \implies p = 2 \times (7 + 9) = 32\text{ cm}\)
In simple words: Sum length and breadth, then multiply by 2 for each column.

Exam Tip: Double-check addition inside the brackets before multiplying by 2.

Question 7. The perimeter is given. Find the side of these squares:
Answer:
Formula: \(\text{Side of square} = \text{Perimeter} \div 4\).
- Perimeter \(28 \implies 28 \div 4 = 7\text{ cm}\)
- Perimeter \(40 \implies 40 \div 4 = 10\text{ cm}\)
- Perimeter \(64 \implies 64 \div 4 = 16\text{ cm}\)
- Perimeter \(96 \implies 96 \div 4 = 24\text{ cm}\)
- Perimeter \(120 \implies 120 \div 4 = 30\text{ cm}\)
- Perimeter \(172 \implies 172 \div 4 = 43\text{ cm}\)
- Perimeter \(232 \implies 232 \div 4 = 58\text{ cm}\)
- Perimeter \(300 \implies 300 \div 4 = 75\text{ cm}\)
- Perimeter \(384 \implies 384 \div 4 = 96\text{ cm}\)
- Perimeter \(556 \implies 556 \div 4 = 139\text{ cm}\)
In simple words: Divide the given perimeter by 4 to get the length of one side.

Exam Tip: Division by 4 can be done by dividing the number by 2 twice.

Question 8. The breadth of a garden is \(13.2\text{ m}\) and its length is \(15.8\text{ m}\). What is the perimeter of the garden?
Answer:
Length (\(l\)) = \(15.8\text{ m}\)
Breadth (\(b\)) = \(13.2\text{ m}\)
Perimeter = \(2 \times (l + b)\)
Perimeter = \(2 \times (15.8 + 13.2)\)
Perimeter = \(2 \times 29.0 = 58.0\text{ m}\)
In simple words: Add length and breadth, then multiply by 2.

Exam Tip: Align decimal points carefully when adding decimals.

Question 9. Mr. Chintu has a rectangular backyard that is \(45\text{ m}\) wide and \(46\text{ m}\) long. How much fencing will he need to enclose the yard? If fencing costs Rs \(52\) per metre, how much will he need to pay?
Answer:
Length (\(l\)) = \(46\text{ m}\)
Breadth (\(b\)) = \(45\text{ m}\)
Fencing needed = Perimeter of backyard = \(2 \times (l + b)\)
Perimeter = \(2 \times (46 + 45) = 2 \times 91 = 182\text{ m}\)
Cost per metre = Rs \(52\)
Total cost = \(182 \times 52 = \text{Rs } 9,464\)
In simple words: Find the perimeter to get the total length of fencing, then multiply by the cost per metre.

Exam Tip: Make sure to state units clearly for both length (\(\text{m}\)) and cost (\(\text{Rs}\)).

Question 10. Mr. Ranjan goes on a morning walk everyday in a rectangular park which is \(200\text{ m}\) long and \(175\text{ m}\) wide. One day Mr. Ranjan walked \(5\) times around the park. Find the total distance ran by him.
Answer:
Length (\(l\)) = \(200\text{ m}\)
Breadth (\(b\)) = \(175\text{ m}\)
Perimeter of park = \(2 \times (200 + 175) = 2 \times 375 = 750\text{ m}\)
Total distance for \(5\) rounds = \(5 \times 750 = 3,750\text{ m}\) or \(3.75\text{ km}\)
In simple words: Multiply the perimeter of the park by 5 to find the total distance covered in 5 rounds.

Exam Tip: Always calculate the perimeter for one complete round before multiplying by the number of rounds.

Question 11. Shivani wants to frame \(10\) pictures, each with length \(4\text{ m}\) and breadth \(3\text{ m}\). What length of wood she need to make for the frame?
Answer:
Length of one picture (\(l\)) = \(4\text{ m}\)
Breadth of one picture (\(b\)) = \(3\text{ m}\)
Perimeter of one picture frame = \(2 \times (4 + 3) = 2 \times 7 = 14\text{ m}\)
Total length of wood for \(10\) pictures = \(10 \times 14 = 140\text{ m}\)
In simple words: Find the perimeter of one picture and multiply by 10.

Exam Tip: Check whether dimensions are in meters or centimeters to avoid unit mismatch.

Question 12. The perimeter of a regular pentagon is \(100\text{ cm}\). How long is each side?
Answer:
Number of sides in a regular pentagon = \(5\)
Perimeter = \(100\text{ cm}\)
Length of each side = \(\text{Perimeter} \div 5 = 100 \div 5 = 20\text{ cm}\)
In simple words: Divide the total perimeter by 5 since a pentagon has 5 equal sides.

Answer:
Side of square field = \(100\text{ m}\)
Perimeter of square = \(4 \times 100 = 400\text{ m}\)
Length of wire for \(3\) layers = \(3 \times 400 = 1,200\text{ m}\)
In simple words: Multiply the perimeter of the square field by 3 to account for 3 layers of wire.

Exam Tip: When multiple rounds of wire or fencing are required, multiply the single perimeter by the number of rounds.

Question 14. The length and breadth of a rectangular field are \(150\text{ m}\) and \(125\text{ m}\), respectively. Find the perimeter of the field.
Answer:
Length (\(l\)) = \(150\text{ m}\)
Breadth (\(b\)) = \(125\text{ m}\)
Perimeter = \(2 \times (l + b)\)
Perimeter = \(2 \times (150 + 125) = 2 \times 275 = 550\text{ m}\)
In simple words: Add length and breadth, then multiply by 2.

Exam Tip: Always write the formula first before substituting values.

Question 15. Find the perimeter of each of the following shapes:
(i) A triangle of sides \(4\text{ cm}\), \(5\text{ cm}\), and \(6\text{ cm}\)
(ii) An equilateral triangle of side \(6\text{ cm}\)
(iii) An isosceles triangle with equal sides \(6\text{ cm}\) each and third side \(7\text{ cm}\).
Answer:
(i) Perimeter = \(4 + 5 + 6 = 15\text{ cm}\)
(ii) Perimeter = \(3 \times 6 = 18\text{ cm}\)
(iii) Perimeter = \(6 + 6 + 7 = 19\text{ cm}\)
In simple words: Sum all the sides of each triangle.

Exam Tip: An equilateral triangle has 3 equal sides, and an isosceles triangle has 2 equal sides.

Question 16. In the morning walk, Abhishek took \(3\) rounds of a rectangular park of length \(320\text{ m}\) and breadth \(210\text{ m}\). Find the total distance covered by Abhishek in the morning walk.
Answer:
Length (\(l\)) = \(320\text{ m}\)
Breadth (\(b\)) = \(210\text{ m}\)
Perimeter of park = \(2 \times (320 + 210) = 2 \times 530 = 1,060\text{ m}\)
Total distance for \(3\) rounds = \(3 \times 1,060 = 3,180\text{ m}\)
In simple words: Multiply the perimeter of the park by 3.

Exam Tip: Ensure correct addition inside the brackets before multiplying.

Question 17. Prachi wants to frame a photograph of length \(40\text{ cm}\) and breadth \(25\text{ cm}\). Find the perimeter of the frame required for this purpose.
Answer:
Length (\(l\)) = \(40\text{ cm}\)
Breadth (\(b\)) = \(25\text{ cm}\)
Perimeter = \(2 \times (40 + 25) = 2 \times 65 = 130\text{ cm}\)
In simple words: Add length and breadth and double the sum.

Exam Tip: The perimeter of the photograph equals the length of the frame required.

Question 18. Veena wants to fix a border on a quilt which is \(2\text{ m } 50\text{ cm}\) long and \(1\text{ m } 30\text{ cm}\) broad. Find the length of the border.
Answer:
Length (\(l\)) = \(2\text{ m } 50\text{ cm} = 2.50\text{ m}\)
Breadth (\(b\)) = \(1\text{ m } 30\text{ cm} = 1.30\text{ m}\)
Perimeter = \(2 \times (l + b)\)
Perimeter = \(2 \times (2.50 + 1.30) = 2 \times 3.80 = 7.60\text{ m}\) or \(7\text{ m } 60\text{ cm}\)
In simple words: Convert centimeters to meters or vice versa, then calculate the perimeter.

Exam Tip: Convert mixed units into a single unit (like all meters or all centimeters) to make calculation easier.

Question 19. The edges of a carpet \(5\text{ m } 20\text{ cm}\) long and \(3\text{ m } 20\text{ cm}\) broad are to be surrounded by a lace. Find the length of the lace required.
Answer:
Length (\(l\)) = \(5\text{ m } 20\text{ cm} = 5.20\text{ m}\)
Breadth (\(b\)) = \(3\text{ m } 20\text{ cm} = 3.20\text{ m}\)
Length of lace required = Perimeter = \(2 \times (5.20 + 3.20)\)
Perimeter = \(2 \times 8.40 = 16.80\text{ m}\) or \(16\text{ m } 80\text{ cm}\)
In simple words: Find the perimeter of the carpet to get the length of the lace.

Exam Tip: Lace goes around the edges, which is the definition of perimeter.

Question 20. Find the perimeter of the rectangle whose breadth is \(53.5\text{ cm}\) and its length is double of the breadth.
Answer:
Breadth (\(b\)) = \(53.5\text{ cm}\)
Length (\(l\)) = \(2 \times 53.5 = 107.0\text{ cm}\)
Perimeter = \(2 \times (l + b) = 2 \times (107.0 + 53.5) = 2 \times 160.5 = 321.0\text{ cm}\)
In simple words: First find the length by doubling the breadth, then calculate the perimeter.

Answer:
Total length of string = Perimeter = \(36\text{ cm}\)
(i) For a square (\(4\) sides): Side = \(36 \div 4 = 9\text{ cm}\)
(ii) For an equilateral triangle (\(3\) sides): Side = \(36 \div 3 = 12\text{ cm}\)
(iii) For a regular hexagon (\(6\) sides): Side = \(36 \div 6 = 6\text{ cm}\)
In simple words: Divide the total length of the string by the number of sides of the shape.

Exam Tip: The total length of the string always represents the perimeter of the formed shape.

Question 22. Find the perimeter of regular hexagon with each side measuring \(12\text{ cm}\).
Answer:
Number of sides in a regular hexagon = \(6\)
Length of each side = \(12\text{ cm}\)
Perimeter = \(6 \times \text{side} = 6 \times 12 = 72\text{ cm}\)
In simple words: Multiply the side length by 6 since a regular hexagon has 6 equal sides.

Exam Tip: A regular hexagon's perimeter is always \(6\) times its side length.

Please refer to attached file for CBSE Class 6 Mensuration Worksheet (2)