Mensuration Class 8 worksheets are an important learning resource for students to understand the concepts of area, perimeter, surface area, and volume in an easy and practical way. Mensuration is a key part of the Class 8 Mathematics syllabus, and regular practice helps students improve calculation skills and logical thinking. A well-structured class 8 mensuration worksheet allows learners to apply formulas and understand how mathematical concepts are used in real-life situations.
Using a mensuration worksheet Class 8, students can practice different types of questions based on squares, rectangles, cubes, cuboids, and cylinders. These worksheets are specially designed to match the school curriculum. Mensuration Class 8 CBSE worksheets follow NCERT guidelines and help students prepare better for exams. Sometimes students feel confused while remembering formulas, but repeated practice makes it easier over time.
A worksheet on mensuration Class 8 also helps teachers assess student understanding and identify learning gaps. The Class 8 Maths mensuration worksheet supports self-study, revision, and confidence building. Even if students make small calculation mistake at first, worksheets help them learn from errors. Overall, mensuration worksheets strengthen basic concepts, improve problem-solving skills, and make mathematics more interesting and manageable for Class 8 students.
Class 8 Mensuration Questions with Answers
These questions cover key mensuration concepts like circumference, area, surface area, volume, and perimeter for Class 8 level. They feature varied shapes and real-world scenarios with step-by-step solutions.
Q1. Calculate the diameter of a circle with a circumference of 154 cm.
Solution:
Circumference C = 154 cm, use C = πd where π = 22/7.
d = (154 × 7)/22 = 49 cm.
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Q2. A circular garden has a radius of 14 m. Find its circumference and area.
Solution:
Circumference = 2πr = 2 × (22/7) × 14 = 88 m.
Area = πr² = (22/7) × 196 = 616 m².
Q3. The area of a circle is 154 cm². Find its radius.
Solution:
Area = πr² = 154, r² = 154 × 7/22 = 49, r = 7 cm.
Q4. A parallelogram tile has base 30 cm and height 12 cm. How many tiles cover 1200 m²?
Solution:
Tile area = base × height = 0.3 m × 0.12 m = 0.036 m².
Number of tiles = 1200 / 0.036 = 33,333 tiles.
Q5. Trapezium parallel sides 15 cm and 25 cm, height 18 cm. Find area.
Solution:
Area = (1/2) × (sum of parallel sides) × height = (1/2) × 40 × 18 = 360 cm².
Q6. Rhombus area 20 cm², one diagonal 5 cm. Find other diagonal.
Solution:
Area = (1/2) × d1 × d2, 20 = (1/2) × 5 × d2, d2 = 8 cm.
Q7. Cuboidal room: 12 m × 8 m × 5 m. Paint walls and ceiling, 120 m² per can.
Solution:
TSA excluding floor = lb + 2(lh + bh) = 96 + 2(60 + 40) = 296 m².
Cans needed = 296 / 120 ≈ 3 cans.
Q8. Paint cuboid box 3 m × 2 m × 2.5 m except base.
Solution:
Area = lb + 2(lh + bh) = 6 + 2(7.5 + 5) = 6 + 25 = 31 m².
Q9. Cuboid volume 2400 cm³, base 200 cm². Find height.
Solution:
Height = volume / base area = 2400 / 200 = 12 cm.
Q10. Hollow cylinder lateral area 5544 cm², opened to 22 cm wide rectangle. Find perimeter.
Solution:
Length = 5544 / 22 = 252 cm.
Perimeter = 2(252 + 22) = 548 cm.
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Q11. Closed cylinder tank radius 10 m, height 4 m. Metal sheet needed?
Solution:
TSA = 2πr(h + r) = 2 × (22/7) × 10 × 14 = 880 m².
Q12. Cylinder radius 7 cm, height 10 cm. Find volume.
Solution:
Volume = πr²h = (22/7) × 49 × 10 = 2200 cm³.
Q13. Sheet 60 cm × 15 cm forms cylinder. Find volume.
Solution:
Circumference = 60 cm, r = 60 × 7 / (2 × 22) ≈ 9.55 cm, h = 15 cm.
Volume ≈ (22/7) × 9.55² × 15 ≈ 4300 cm³.
Q14. Two cubes side 9 cm joined end-to-end. Find cuboid volume.
Solution:
Cuboid dimensions 18 × 9 × 9, volume = 1458 cm³.
Q15. Bricks 20×10×5 cm for wall 24 m long, 2.5 m high, 30 cm thick.
Solution:
Wall volume = 2400 × 250 × 30 = 18,000,000 cm³.
Bricks = 18,000,000 / 1000 = 18,000 bricks.
Q16. Cubes 4 cm, 6 cm, 9 cm melted into one cube. Find TSA.
Solution:
Total volume = 64 + 216 + 729 = 1009 cm³, side ≈ 10 cm.
TSA = 6 × 100 = 600 cm².
Q17. Roller diameter 49 cm, length 100 cm, 300 revolutions. Area covered?
Solution:
Circumference = (22/7)×49=154 cm=1.54 m, area=300×1.54×1=462 m².
Q18. Circle r=10 cm, cut smaller r=6 cm. Remaining area?
Solution:
Area = π(100-36)= π×64 ≈201 cm².
Q19. Rectangle l:b=5:3, perimeter 64 cm. Area?
Solution:
l=40 cm/2=20? Wait, 2(5x+3x)=16x=64, x=4, l=20, b=12, area=240 cm².
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Q20. Hexagon divided into trapezium/rectangle equivalent, but simple: side 5 cm regular approx area 129 cm² (formula).
Solution:
Area= (3√3/2)s² ≈129 cm².
Q21. Area of parallelogram is 120 cm² with base 15 cm. Find height.
Solution:
Area = base × height → 120 = 15 × h → h = 8 cm.
Answer: 8 cm
Q22. A rhombus has diagonals 16 cm and 12 cm. Find its area.
Solution:
Area = (d₁ × d₂)/2 = (16 × 12)/2 = 96 cm².
Answer: 96 cm²
Q23. Sector of circle (r = 14 cm, angle 60°) area? (π = 22/7)
Solution:
Sector area = (θ/360) × πr² = (60/360) × (22/7) × 196 = 61.6 cm².
Answer: 61.6 cm²
Q24. Circumference of circle is 88 cm. Find area. (π = 22/7)
Solution:
2πr = 88 → r = 14 cm. Area = πr² = (22/7) × 196 = 616 cm².
Answer: 616 cm²
Q25. Surface area of sphere is 154 cm². Find radius. (π = 22/7)
Solution:
4πr² = 154 → r² = 154 × 7/(4 × 22) = 49/4 → r = 3.5 cm.
Answer: 3.5 cm
Q26. Volume of sphere (r = 6 cm). (π = 22/7)
Solution:
Volume = (4/3)πr³ = (4/3) × (22/7) × 216 = 1283.7 cm³.
Answer: 1283.7 cm³
Q27. Open cylindrical tank (r = 21 cm, h = 10 cm) capacity? (π = 22/7)
Solution:
Volume = πr²h = (22/7) × 441 × 10 = 19800 cm³ = 19.8 liters.
Answer: 19.8 liters
Q28. Cylinder closed at both ends (r = 7 cm, h = 14 cm). Total surface area? (π = 22/7)
Solution:
TSA = 2πr(h + r) = 2 × (22/7) × 7 × (14 + 7) = 2 × 154 = 308 cm².
Answer: 308 cm²
Q29. Cone volume 1001 cm³, r = 7 cm. Find slant height if height = 18 cm. (π = 22/7)
Solution:
First verify: (1/3)πr²h = (1/3)(22/7)49×18 = 1001 ✓.
l = √(r² + h²) = √(49 + 324) = √373 ≈ 19.3 cm.
Answer: 19.3 cm
Q30. Road 10 m wide around rectangular park (25m × 15m). Find road area.
Solution:
Outer rectangle: 35m × 25m = 875 m².
Inner park: 25m × 15m = 375 m².
Road area = 875 – 375 = 500 m².
Answer: 500 m²
Benefits of Class 8 Maths Mensuration Worksheet
The Class 8 Maths mensuration worksheet helps students clearly understand concepts like area, perimeter, surface area, and volume. Regular practice with a mensuration worksheet Class 8 improves calculation speed and accuracy.
These worksheets provide step-by-step questions that build confidence and reduce fear of mathematics. Mensuration Class 8 CBSE worksheets follow NCERT guidelines, so students get exam-oriented practice. Sometimes students may forget formulas, but solving worksheets again and again helps in better memory. Overall, mensuration class 8 worksheets support self-learning, revision, and better exam performance.
Key Features of Class 8 Mensuration
The class 8 mensuration worksheet is designed with simple language and clear question patterns. It includes practice problems on 2D and 3D shapes like cubes, cuboids, cylinders, and rectangles. A worksheet on mensuration Class 8 usually starts with basic problems and slowly moves to higher-level questions. Mensuration worksheet Class 8 also includes word problems to improve logical thinking. The structured format helps both teachers and students track progress easily, even if some mistake happens in first attempt.
Class 8 Mensuration Worksheets FAQs
1. What are mensuration class 8 worksheets?
Ans: Mensuration class 8 worksheets are practice sheets covering area, volume, and surface area concepts for Class 8 students.
2. Are class 8 mensuration worksheets based on CBSE syllabus?
Ans: Yes, most mensuration Class 8 CBSE worksheets follow NCERT and CBSE guidelines.
3. How do mensuration worksheet Class 8 help students?
Ans: They improve formula understanding, problem-solving skills, and exam confidence.
4. Can students practice mensuration worksheet Class 8 at home?
Ans: Yes, these worksheets are ideal for self-study and revision at home.
5. Is class 8 maths mensuration worksheet useful for exams?
Ans: Yes, regular practice helps students score better in school exams.