CBSE Class 6 Mathematics Mensuration Worksheet Set 05

Mensuration

Question 1. Two sides of a triangle are 15 cm and 19 cm. The perimeter of the triangle is 45 cm. What is the third side?
Answer: Perimeter of the triangle = 45 cm
First side = 15 cm
Second side = 19 cm
Sum of the two sides = 15 cm + 19 cm = 34 cm
Third side = Perimeter - Sum of the two sides
Third side = 45 cm - 34 cm = 11 cm
In simple words: Subtract the sum of the known two sides from the total perimeter to find the third side.

Exam Tip: Always remember that the perimeter of any polygon is the total length of its boundary (sum of all sides).

Question 2. Raj runs around a square park of side 67 m. Taj runs around a rectangular park with length 53 m and breadth 35 m. Who covers more distance?
Answer: Distance covered by Raj = Perimeter of the square park
Side of the square park = 67 m
Perimeter of a square = \(4 \times \text{side}\)
Distance covered by Raj = \(4 \times 67\text{ m} = 268\text{ m}\)

Distance covered by Taj = Perimeter of the rectangular park
Length of the rectangular park = 53 m
Breadth of the rectangular park = 35 m
Perimeter of a rectangle = \(2 \times (\text{length} + \text{breadth})\)
Distance covered by Taj = \(2 \times (53 + 35)\text{ m} = 2 \times 88\text{ m} = 176\text{ m}\)

Comparing the distances: 268 m > 176 m
Therefore, Raj covers more distance.
In simple words: Find the perimeter for both parks separately and compare the two values to see who ran further.

Exam Tip: Writing the formula for both square and rectangular perimeters before substituting values ensures you do not lose marks for formula steps.

Question 3. Find the side of the square whose perimeter is 36 m.
Answer: Perimeter of the square = 36 m
Formula for perimeter of a square = \(4 \times \text{side}\)
Side = \(\frac{\text{Perimeter}}{4}\)
Side = \(\frac{36}{4} = 9\text{ m}\)
In simple words: Divide the given perimeter by 4 to get the length of one side of the square.

Exam Tip: Include the correct unit (m) in your final answer.

Question 4. What is the length of the ribbon required to frame a notice board of length 37 cm and breadth 26 cm?
Answer: Length of the notice board = 37 cm
Breadth of the notice board = 26 cm
Length of ribbon required = Perimeter of the rectangular notice board
Perimeter = \(2 \times (\text{length} + \text{breadth})\)
Perimeter = \(2 \times (37 + 26)\text{ cm}\)
Perimeter = \(2 \times 63\text{ cm} = 126\text{ cm}\)
In simple words: Framing requires covering the outer boundary, so calculate the perimeter of the rectangle.

Exam Tip: Statements involving framing, fencing, or edging always refer to the perimeter.

Question 5. Five rectangular flower beds each of length 3 m and breadth 2 m are dug on a land of side 9 m. What is the area of the remaining part of the land?
Answer: Side of the square land = 9 m
Area of the square land = \(\text{side} \times \text{side} = 9\text{ m} \times 9\text{ m} = 81\text{ m}^2\)

Length of one rectangular flower bed = 3 m
Breadth of one rectangular flower bed = 2 m
Area of one flower bed = \(\text{length} \times \text{breadth} = 3\text{ m} \times 2\text{ m} = 6\text{ m}^2\)

Total area of 5 flower beds = \(5 \times 6\text{ m}^2 = 30\text{ m}^2\)

Area of the remaining part = Area of the square land - Total area of 5 flower beds
Area of the remaining part = \(81\text{ m}^2 - 30\text{ m}^2 = 51\text{ m}^2\)
In simple words: Find the total land area, subtract the area occupied by the 5 flower beds to find what is left over.

Exam Tip: Always multiply the area of one flower bed by the total number of flower beds before subtracting from the total area.

Question 6. The area of a rectangle 60 m long is 240 sq. m. Find the width of the rectangle.
Answer: Area of the rectangle = \(240\text{ m}^2\)
Length of the rectangle = 60 m
Area = \(\text{length} \times \text{breadth}\)
\(240 = 60 \times \text{breadth}\)
Breadth = \(\frac{240}{60} = 4\text{ m}\)
In simple words: Divide the total area by the given length to find the width.

Exam Tip: Double-check your division by multiplying length and width back together to ensure you get the original area.

Question 7. What is the cost of painting a rectangular board of length 13 cm and breadth 19 cm at the rate of Rs 9 per hundred sq. m.?
Answer: Length of the board = 13 cm
Breadth of the board = 19 cm
Area of the board = \(\text{length} \times \text{breadth} = 13\text{ cm} \times 19\text{ cm} = 247\text{ cm}^2\)

Rate of painting = Rs 9 per hundred sq. cm. (Note: Assuming standard dimensional unit compatibility given in school problem sets as cm² or standard area units)
Cost of painting = \(\frac{\text{Area}}{100} \times \text{Rate}\)
Cost = \(\frac{247}{100} \times 9\)
Cost = \(\frac{2223}{100} = \text{Rs } 22.23\)
In simple words: Find the area first, divide by 100 to find how many hundreds of square units there are, and multiply by the rate of Rs 9.

Exam Tip: Pay careful attention to rates given per hundred units; always divide the total measure by 100 before multiplying by the unit price.

Question 8. A room is 5 m long and 4 m 50 cm wide. How many square meters of carpet is needed to cover the floor?
Answer: Length of the room = 5 m
Breadth of the room = 4 m 50 cm = \(4 + 0.50 = 4.5\text{ m}\)
Area of the floor = \(\text{length} \times \text{breadth}\)
Area = \(5\text{ m} \times 4.5\text{ m} = 22.5\text{ m}^2\)
In simple words: Convert centimeters into meters so all units are the same, then multiply length by breadth to find the floor area.

Exam Tip: Always convert mixed units into a single decimal unit before performing multiplication.

Question 9. Find the cost of fencing a rectangular park of length 155 m and breadth 125 m at the rate of Rs 15 per metre.
Answer: Length of the park = 155 m
Breadth of the park = 125 m
Perimeter of the rectangular park = \(2 \times (\text{length} + \text{breadth})\)
Perimeter = \(2 \times (155 + 125)\text{ m}\)
Perimeter = \(2 \times 280\text{ m} = 560\text{ m}\)

Rate of fencing = Rs 15 per metre
Total cost of fencing = \(\text{Perimeter} \times \text{Rate}\)
Total cost = \(560 \times 15 = \text{Rs } 8400\)
In simple words: Fencing goes around the boundary, so find the perimeter first and multiply it by the cost per meter.

Exam Tip: Fencing problems always require calculating the perimeter, never the area.