Case Study Questions for Class 9 Surface Area and Volume

Case Study Questions for Class 9 Surface Area and Volume

Case Study Questions for Class 9 Surface Area and Volume

Case Study Questions for Class 9 Surface Area and Volume

Preparing well for board exams requires practice on case study questions for class 9 surface area and volume. These questions test conceptual clarity and real-life application skills. Students often search for class 9 maths surface area and volume case study to improve their confidence and problem-solving approach. Moreover, cbse class 9 surface area and volume case study questions are frequently included in sample papers, making them essential for exam preparation.

Class 9 Mensuration Case Study Questions

In mensuration, students face math case study questions on surface area and volume for class 9. These involve cubes, cuboids, cylinders, and cones. Additionally, class 9 mensuration case study questions strengthen the ability to apply formulas. Therefore, solving these regularly ensures better time management. Furthermore, combining NCERT examples with online resources makes practice more effective and result-oriented.

Online Practice for Better Results

Free tests and solved examples are widely available. With proper revision and guided practice, students can enhance accuracy. Hence, consistent attempts at cbse class 9 surface area and volume case study questions will boost exam performance significantly.

Case Study 3: Surface Areas and Volumes

Case Study 3: Surface Areas and Volumes

A small engineering firm manufactures a decorative hollow cube lamp for a design fair. The lamp is made from a solid wooden cube of side 1.40 m (i.e. 140 cm). A vertical cylindrical hole of radius 0.35 m (i.e. 35 cm) is drilled centrally through the cube from the top face to the bottom face so that the axis of the cylinder passes through the centre of the cube. The wood removed by drilling will be used to make small dowel pieces, and the interior curved surface exposed by the drilling will be finished with a satin polish. The outer six faces of the cube are to be painted except the two circular openings on the top and bottom (i.e. those circular areas are left open). The client asks the firm to provide: (i) the volume of wood remaining in the cube after drilling (in cubic centimetres and in litres), (ii) the area of the interior curved cylindrical surface exposed by drilling (in square centimetres), (iii) the total external painted area (in square metres) after excluding the two circular openings, and (iv) the total material removed (in cubic centimetres). Use π = 22/7 wherever numerical evaluation is required. Carefully convert units where necessary. This problem requires combining formulae for cubes and cylinders, unit conversion, and attention to which faces/areas are included or excluded when computing painted area.

1. The volume of the original solid cube (before drilling) in cubic centimetres is:

Answer: (a) 2744000 cm³

Solution: Side a = 140 cm. Volume of cube = a³ = 140³. Calculate digit by digit: 140² = 19600, then 19600 × 140 = 19600 × (100 + 40) = 1960000 + 784000 = 2744000 cm³.

2. The volume of wood removed by drilling the cylindrical hole (in cubic centimetres) is:

Answer: (a) 539000 cm³

Solution: Cylinder radius r = 35 cm, height h = 140 cm. Using π = 22/7, Vcyl = πr²h = (22/7) × 35² × 140. Compute stepwise: 35² = 1225. Divide 1225 by 7 gives 175. Then 175 × 140 = 175 × (100 + 40) = 17500 + 7000 = 24500. Multiply by 22: 24500 × 22 = 24500 × (20 + 2) = 490000 + 49000 = 539000 cm³.

3. The remaining volume of the wooden cube after drilling (in litres) is:

Answer: (a) 2205 L

Solution: Remaining volume = Vcube – Vcyl = 2744000 – 539000 = 2205000 cm³. Convert to litres by dividing by 1000: 2205000 ÷ 1000 = 2205 L.

4. The area of the interior curved cylindrical surface exposed by drilling (in square centimetres) is:

Answer: (a) 30800 cm²

Solution: Curved surface area of cylinder = 2πrh. With r = 35 cm and h = 140 cm, Acyl,curved = 2 × (22/7) × 35 × 140. Compute: 35 ÷ 7 = 5. Thus 2 × 22 × 5 × 140 = 44 × 5 × 140 = 220 × 140 = 220 × (100 + 40) = 22000 + 8800 = 30800 cm².

5. The total external painted area (in square metres) when the two circular openings (top and bottom) are left open and the interior curved surface is NOT painted, is:

Answer: (d) 10.990 m²

Solution: Surface area of cube = 6a² = 6 × 140² = 6 × 19600 = 117600 cm². The two circular openings (top and bottom) together have area 2πr². Compute 2πr² = 2 × (22/7) × 35². Since 35² = 1225 and 1225 ÷ 7 = 175, we get 2 × 22 × 175 = 44 × 175 = 7700 cm². The painted external area (excluding these two circular holes) is Apaint = 117600 – 7700 = 109900 cm². Convert to square metres: 109900 ÷ 10000 = 10.99 m². Rounded to three decimal places: 10.990 m².

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