Class 9 Maths Surface Area and Volume Case Study

Class 9 Maths Surface Area and Volume Case Study

Case Study Questions for Class 9 Surface Area and Volume

Class 9 Maths Surface Area and Volume Case Study

Preparing well for board exams requires practice on Class 9 Maths Surface Area and Volume Case Study. 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 2: Surface Areas and Volumes

Case Study 2: Surface Areas and Volumes

A local craftsman makes a decorative trophy for a school event. The trophy is made by fixing a solid formed by joining a hemisphere to a right circular cone (they are joined along their circular bases so they share the same radius). This composite solid is placed on a wooden cuboidal pedestal whose top has a cylindrical recess to accommodate the hemisphere so that the hemisphere sits flush with the pedestal top. The dimensions are as follows: the cuboidal pedestal measures 84 cm × 84 cm × 60 cm (length × breadth × height). The hemisphere and cone each have radius 21 cm. The cone has height 63 cm. The recess in the top of the pedestal is a right circular cylinder of radius 21 cm and depth 21 cm (so that the hemisphere rim fits exactly). The craftsman needs the following: (i) volume of wood removed to make the recess, (ii) volume of the entire trophy (wooden pedestal with the composite solid mounted in the recess), (iii) the total external surface area to be painted (assume the bottom face of the pedestal is not painted), and (iv) convert the final volume into litres. Use π = 22/7 wherever required. Apply unit conversion carefully where needed.

1. The volume of wood removed to make the cylindrical recess (in cubic centimetres) is:

Answer: (a) 29106 cm³

Solution: Recess is a cylinder of radius r = 21 cm and depth h = 21 cm. Thus Vrecess = πr²h = (22/7) × 21² × 21 = (22/7) × 441 × 21 = 22 × 63 × 21 = 29106 cm³.

2. The volume of the cone (in cubic centimetres) is:

Answer: (b) 29106 cm³

Solution: For the cone, r = 21 cm and h = 63 cm. So Vcone = (1/3)πr²h = (1/3) × (22/7) × 21² × 63. Simplify stepwise: 21² = 441, 441 × 63 = 27783, and dividing 441 by 7 gives 63, hence Vcone = (1/3) × 22 × 63 × 63 = 22 × 63 × 21 = 29106 cm³.

3. The total volume of the assembled trophy (pedestal minus recess plus hemisphere plus cone) in litres is:

Answer: (b) 442.764 L

Solution: Compute each part in cubic centimetres.
Vpedestal = 84 × 84 × 60 = 423360 cm³,
Vhemisphere = (2/3)πr³ = (2/3) × (22/7) × 21³ = (2/3) × (22/7) × 9261 = 19404 cm³,
Vcone = 29106 cm³ (from Q2),
Vrecess = 29106 cm³ (from Q1).
Total volume (pedestal minus recess plus hemisphere plus cone):
Vtotal = 423360 – 29106 + 19404 + 29106 = 423360 + 19404 = 442764 cm³.
Convert to litres: 442764 cm³ = 442.764 L. Thus option (b) is correct.

4. The total external surface area to be painted (excluding the bottom face of the pedestal) in square metres (rounded to three decimal places) is:

Answer: (a) 3.299 m²

Solution: Compute external painted area in cm² first.
1. Lateral surface area of the cuboidal pedestal:
Alat, pedestal = 2(lh + bh) = 4 × 84 × 60 = 20160 cm².
2. Top surface area of pedestal excluding the circular hole:
Atop = 84 × 84 – π × 21² = 7056 – (22/7) × 441 = 7056 – 1386 = 5670 cm².
3. Curved surface area of hemisphere:
Ahem,curved = 2πr² = 2 × (22/7) × 441 = 2772 cm².
4. Curved (lateral) surface area of cone: slant height l = √(r² + h²) = √(21² + 63²) = 21√10, hence
Acone,lat = πrl = (22/7) × 21 × 21√10 = 1386√10 cm².
Numerical value 1386√10 ≈ 4382.0 cm².
Now total painted area in cm²:
Atotal = 20160 + 5670 + 2772 + 1386√10 ≈ 32984.9168 cm².
Convert to m²:
Atotal = 32984.9168 / 10000 ≈ 3.298492 m² ≈ 3.299 m².
Therefore option (a) is correct.

5. If painting costs Rs. 60 per square metre, the approximate cost to paint the trophy’s outside (rounded to nearest rupee) is:

Answer: (a) Rs. 198

Solution: Using Atotal ≈ 3.298492 m²,
Cost = 60 × 3.298492 ≈ 197.9095 Rs ≈ 198 Rs.
Hence option (a) is the correct rounded value.

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