Get the most accurate GSEB Solutions for Class 6 Mathematics Chapter 10 Mensuration here. Updated for the 2026-27 academic session, these solutions are based on the latest GSEB textbooks for Class 6 Mathematics. Our expert-created answers for Class 6 Mathematics are available for free download in PDF format.
Detailed Chapter 10 Mensuration GSEB Solutions for Class 6 Mathematics
For Class 6 students, solving GSEB textbook questions is the most effective way to build a strong conceptual foundation. Our Class 6 Mathematics solutions follow a detailed, step-by-step approach to ensure you understand the logic behind every answer. Practicing these Chapter 10 Mensuration solutions will improve your exam performance.
Class 6 Mathematics Chapter 10 Mensuration GSEB Solutions PDF
Question 1. Find the areas of the rectangles whose sides are:
(a) 3 cm and 4 cm
(b) 12 m and 21 m
(c) 2 km and 3 km
(d) 2 m and 70 cm
Answer:
(a) The length of the rectangle is 4 cm. Its breadth measures 3 cm. Therefore, the area equals length multiplied by breadth, which calculates to 4 cm times 3 cm, giving 12 sq. cm.
(b) The length of the rectangle is 21 m. Its breadth measures 12 m. So, the area means length multiplied by breadth, resulting in 21 m by 12 m, which makes 252 sq. m.
(c) The length of the rectangle is 3 km. Its breadth is 2 km. The area equals length multiplied by breadth, which is 3 km times 2 km, resulting in 6 sq. km.
(d) The length of the rectangle is 2 m, which means 200 cm. Its breadth is 70 cm. The area calculates as length multiplied by breadth, so 200 cm times 70 cm equals 14000 sq. cm.
Exam Tip: Always make sure the units for length and breadth are consistent before calculating the area. Convert different units (like m and cm) to a common unit first.
Question 2. Find the areas of the squares whose sides are:
(a) 10 cm
(b) 14 cm
(c) 5 m
Answer:
(a) Each side of the square measures 10 cm. Therefore, the square's area is found by side multiplied by side, which means 10 cm times 10 cm, giving 100 sq. cm.
(b) The square's side length is 14 cm. So, the area of the square is side times side, equalling 14 cm multiplied by 14 cm, which gives 196 sq. cm.
(c) The side of the square is 5 m. The area of the square means side multiplied by side, so 5 m times 5 m results in 25 sq. m.
Exam Tip: Remember that for a square, both length and breadth are equal to the side length, simplifying the area formula to \( \text{side} \times \text{side} \).
Question 3. The length and breadth of the three rectangles are as given below:
(a) 9 m and 6m
(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) The area of this rectangle is found by length multiplied by breadth, which is 9 m times 6 m, equalling 54 sq. m.
(b) The rectangle's area equals length multiplied by breadth, calculated as 17 m times 3 m, resulting in 51 sq. m.
(c) The area of the rectangle is length multiplied by breadth, which is 4 m times 14 m, giving 56 sq. m.
Because 56 is greater than 54 and also greater than 51, rectangle (c) possesses the largest area, and rectangle (b) has the smallest area.
Exam Tip: To compare areas, always calculate each area first, and then arrange them in ascending or descending order to identify the largest and smallest. Write down all intermediate calculations.
Question 4. The area of a rectangular garden 50 m long is 300 sq. m. Find the width of the garden.
Answer: The rectangular garden's area is 300 sq. m. The garden's length is 50 m. Therefore, the breadth of the garden equals Area divided by Length, which is \( \frac{300 \text{ sq. m}}{50 \text{ m}} \), resulting in 6 m.
Exam Tip: If you know the area and one side of a rectangle, you can always find the other side by dividing the area by the known side length.
Question 5. What is the cost of tiling a rectangular plot of land 500 m long and 200 m wide at the rate of Rs 8 per hundred m sq?
Answer: The rectangular plot's length is 500 m. Its breadth measures 200 m. So, the plot's area is length multiplied by breadth, equalling 500 m times 200 m, which gives 100000 sq. m. The cost to tile for every 100 sq. m area is Rs 8. The total cost to tile the plot is calculated as \( \left( \frac{100000}{100} \right) \times 8 \), which results in Rs 1000 times 8, making it Rs 8000.
Exam Tip: When given a cost per unit area (like per hundred sq. m), first find the total area, then determine how many "units" of area you have, and finally multiply by the cost per unit.
Question 6. A table-top measures 2 m by 1 m 50 cm. What is its area in square metres?
Answer: A measurement of 2 m by 1 m 50 cm implies: The length is 2 m, which means 200 cm. The breadth is 1 m 50 cm, or 150 cm. Therefore, the area is length multiplied by breadth, calculated as 200 cm times 150 cm, giving 30000 sq. cm. So, the table top's area is 30000 sq. cm, which converts to \( \frac{30000}{100 \times 100} \) or 3 sq. m. To change square centimetres into square metres, we need to divide by \( 100 \times 100 \).
Exam Tip: Always convert all measurements to the required final unit (e.g., meters) *before* performing calculations to avoid errors in unit conversion for the area.
Question 7. A room is 4 m long and 3 m 50 cm wide. How many square metres of carpet is needed to cover the floor of the room?
Answer: The room's length is 4 m, which is 400 cm. Its breadth measures 3 m 50 cm, or 350 cm. So, the floor's area equals length multiplied by breadth, calculating to 400 cm times 350 cm, giving 140000 sq. cm. This converts to \( \frac{140000}{100 \times 100} \) or 14 sq. m. Hence, 14 sq. m of carpet is needed to cover the room's floor.
Exam Tip: Ensure all dimensions are in the same unit (e.g., cm) before multiplying to find the area, and then convert the final area to the desired unit (e.g., sq. m).
Question 8. A floor is 5 m long and 4 m wide. A square carpet of sides 3 m is laid on the floor. Find the area of the floor that is not carpeted.
Answer: The floor's area is length multiplied by breadth, which is 5 m times 4 m, giving 20 sq. m. The square carpet's area is side multiplied by side, equalling 3 m times 3 m, resulting in 9 sq. m. Therefore, the uncovered floor area is 20 sq. m minus 9 sq. m, which is 11 sq. m. So, 11 sq. m of the floor is not covered by the carpet.
Exam Tip: When finding an "uncovered" or "remaining" area, always calculate the total area first, then the area of the covered part, and subtract the latter from the former.
Question 9. 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: The piece of land has a length of 5 m. Its breadth measures 4 m. So, the land's area is found by length multiplied by breadth, equalling 5 m times 4 m, which gives 20 sq. m. A single 'square flower bed' has an area of side multiplied by side, meaning 1 m times 1 m, making it 1 sq. m. Therefore, the area of 5 square flower beds is 5 multiplied by 1 sq. m, resulting in 5 sq. m. The remaining part's area is 20 sq. m minus 5 sq. m, which means 15 sq. m.
Exam Tip: For problems involving multiple identical objects (like flower beds), calculate the area of one object first, then multiply by the number of objects to find their total area.
Question 10. By splitting the following figures into rectangles, find their areas. (The measures are given in centimetres.)
(a)
(b)
Answer:
(a) When we split the given figure into various squares and rectangles as shown in the adjoining figure:Now, the area of rectangle I is found by length multiplied by breadth, which is 4 cm times 2 cm, equalling 8 sq. cm. The area of square II is side multiplied by side, meaning 3 cm times 3 cm, giving 9 sq. cm. The area of rectangle III is length multiplied by breadth, which is 2 cm times 1 cm, making it 2 sq. cm. The area of square IV is side multiplied by side, meaning 3 cm times 3 cm, resulting in 9 sq. cm. Therefore, the total area of the figure is 8 sq. cm plus 9 sq. cm plus 2 sq. cm plus 9 sq. cm, which sums up to 28 sq. cm.
(b) When we split the given figure into different rectangles, we get:The area of rectangle I is length multiplied by breadth, equalling 3 cm times 1 cm, which is 3 sq. cm. The area of rectangle II is length multiplied by breadth, 3 cm times 1 cm, giving 3 sq. cm. The area of rectangle III is length multiplied by breadth, 3 cm times 1 cm, also resulting in 3 sq. cm. Therefore, the total area of the figure is the sum of Area of rectangle I plus Area of rectangle II plus Area of rectangle III, which is 3 sq. cm plus 3 sq. cm plus 3 sq. cm, summing up to 9 sq. cm.
Exam Tip: When splitting complex figures, ensure that all parts are simple rectangles or squares and that there are no overlaps or uncounted areas. Clearly label each sub-figure in your working.
Question 11. Split the following shapes into rectangles and find their areas. (The measures are given in centimetres.)
(a)
(b)
(c)
Answer:
(a) Splitting the given figure into rectangles we have:The area of rectangle I is length multiplied by breadth, equalling 10 cm times 2 cm, which gives 20 sq. cm. The area of rectangle II is also length multiplied by breadth, 10 cm times 2 cm, resulting in 20 sq. cm. Therefore, the total area of the figure is 20 sq. cm plus 20 sq. cm, adding up to 40 sq. cm.
(b)The area of square I is 7 cm multiplied by 7 cm, giving 49 sq. cm. The area of rectangle II is (7 + 7 + 7) cm multiplied by 7 cm, which means 21 cm times 7 cm, equalling 147 sq. cm. The area of square III is 7 cm multiplied by 7 cm, also giving 49 sq. cm. Therefore, the total area of the given figure is the sum of Area of I plus Area of II plus Area of III, which is 49 sq. cm plus 147 sq. cm plus 49 sq. cm, totalling 245 sq. cm.
(c)The area of rectangle I is 5 cm multiplied by 1 cm, giving 5 sq. cm. The area of rectangle II is 4 cm multiplied by 1 cm, equalling 4 sq. cm. Therefore, the total area of the given figure is the sum of Area of rectangle I plus Area of rectangle II, which is 5 sq. cm plus 4 sq. cm, resulting in 9 sq. cm.
Exam Tip: For complex shapes, break them into the smallest number of non-overlapping rectangles possible. Label each part and its dimensions to simplify calculation and avoid errors.
Question 12. How many tiles whose length and breadth are 12 cm and 5 cm respectively will be needed o fit in a rectangular region whose length and breadth are respectively:
(a) 100 cm and 144 cm
(b) 70 cm and 36 cm
Answer: The area of each rectangular tile is 12 cm multiplied by 5 cm, equalling 60 sq. cm.
(a) The rectangular region's area is 100 cm multiplied by 144 cm, making it 14400 sq. cm. Therefore, the number of tiles needed to cover this 14400 sq. cm area is found by dividing the region's area by the tile's area, which is \( \frac{14400}{60} \), resulting in 240 tiles.
(b) The rectangular region's area is length multiplied by breadth, which is 70 cm times 36 cm, equalling 2520 sq. cm. The tile's area is 60 sq. cm. Therefore, the number of tiles required is the region's area divided by 60, which is \( \frac{2520}{60} \), resulting in 42 tiles.
Exam Tip: To find the number of smaller units needed to cover a larger area, calculate both areas separately and then divide the larger area by the smaller area. Ensure both areas are in the same units.
Free study material for Mathematics
GSEB Solutions Class 6 Mathematics Chapter 10 Mensuration
Students can now access the GSEB Solutions for Chapter 10 Mensuration prepared by teachers on our website. These solutions cover all questions in exercise in your Class 6 Mathematics textbook. Each answer is updated based on the current academic session as per the latest GSEB syllabus.
Detailed Explanations for Chapter 10 Mensuration
Our expert teachers have provided step-by-step explanations for all the difficult questions in the Class 6 Mathematics chapter. Along with the final answers, we have also explained the concept behind it to help you build stronger understanding of each topic. This will be really helpful for Class 6 students who want to understand both theoretical and practical questions. By studying these GSEB Questions and Answers your basic concepts will improve a lot.
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The complete and updated GSEB Class 6 Maths Solutions Chapter 10 Mensuration Exercise 10.3 is available for free on StudiesToday.com. These solutions for Class 6 Mathematics are as per latest GSEB curriculum.
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