Surface Area and Volume in Real Life
Surface area and volume formulas are not just textbook exercises — they solve practical problems involving material estimation, liquid capacity, packaging design, and construction costs.
This topic applies the formulas from the chapter "Surface Areas and Volumes" (Class 10 NCERT) to real-world contexts: painting walls, filling tanks, wrapping gifts, digging wells, and manufacturing containers.
Understanding which formula to use (SA vs volume), which type of SA (CSA vs TSA), and how to handle unit conversions is the key skill in these problems.
What is Surface Area and Volume in Real Life?
When to use Surface Area:
- Painting, polishing, wrapping, or covering an object → use surface area.
- Only curved surface (e.g., label on a can) → use CSA.
- All surfaces (e.g., gift wrapping a box) → use TSA.
When to use Volume:
- Filling, capacity, material needed to make a solid → use volume.
- Pouring liquid from one container to another → volumes are equal.
- Melting and recasting → volumes are conserved.
Surface Area and Volume in Real Life Formula
Key formulas for real-life problems:
Material cost = Surface Area × Rate per unit area
Capacity = Volume (in cm³) ÷ 1000 = litres
Rise in level = Volume of object ÷ Base area of container
Common conversions:
- 1 m³ = 1,000 litres = 10⁶ cm³
- 1 litre = 1000 cm³
- 1 m = 100 cm
Solved Examples
Example 1: Painting a Cylindrical Pillar
Problem: A cylindrical pillar has diameter 1.4 m and height 4 m. Find the cost of painting its curved surface at Rs. 15 per m². (Use π = 22/7)
Solution:
- r = 0.7 m, h = 4 m
- CSA = 2πrh = 2 × (22/7) × 0.7 × 4 = 17.6 m²
- Cost = 17.6 × 15 = Rs. 264
Answer: The painting cost is Rs. 264.
Example 2: Water Tank Capacity
Problem: A cylindrical water tank has radius 1.4 m and height 2 m. How many litres of water can it hold? (Use π = 22/7)
Solution:
- Volume = πr²h = (22/7) × 1.96 × 2 = (22/7) × 3.92 = 12.32 m³
- 1 m³ = 1000 litres
- Capacity = 12.32 × 1000 = 12,320 litres
Answer: The tank can hold 12,320 litres.
Example 3: Making Conical Tent Canvas
Problem: A conical tent has base radius 7 m and slant height 25 m. Find the area of canvas needed. If canvas costs Rs. 70 per m², find the total cost. (Use π = 22/7)
Solution:
- Canvas needed = CSA of cone = πrl = (22/7) × 7 × 25 = 550 m²
- Cost = 550 × 70 = Rs. 38,500
Answer: Canvas needed = 550 m². Cost = Rs. 38,500.
Example 4: Digging a Well and Spreading Earth
Problem: A well of diameter 7 m and depth 10 m is dug. The earth taken out is spread evenly on a rectangular plot 28 m × 22 m. Find the rise in the level of the plot. (Use π = 22/7)
Solution:
- Volume of earth = πr²h = (22/7) × 3.5² × 10 = (22/7) × 12.25 × 10 = 385 m³
- Area of plot = 28 × 22 = 616 m²
- Rise in level = Volume / Area = 385/616 = 0.625 m = 62.5 cm
Answer: The level of the plot rises by 62.5 cm.
Example 5: Gift Wrapping a Cuboidal Box
Problem: A gift box has dimensions 30 cm × 20 cm × 15 cm. Find the area of wrapping paper needed (assume no overlap). If paper costs Rs. 5 per 100 cm², find the cost.
Solution:
- TSA = 2(lb + bh + hl) = 2(600 + 300 + 450) = 2 × 1350 = 2700 cm²
- Cost = (2700/100) × 5 = 27 × 5 = Rs. 135
Answer: Paper needed = 2700 cm². Cost = Rs. 135.
Example 6: Ice Cream Cone with Hemisphere
Problem: An ice cream cone has radius 3.5 cm and height 12 cm, with a hemispherical scoop on top. Find the total volume of ice cream. (Use π = 22/7)
Solution:
- Volume of cone = (1/3)πr²h = (1/3)(22/7)(12.25)(12) = (1/3)(462) = 154 cm³
- Volume of hemisphere = (2/3)πr³ = (2/3)(22/7)(42.875) = (2/3)(134.75) = 89.83 cm³
- Total = 154 + 89.83 = 243.83 cm³
Answer: Total volume of ice cream = 243.83 cm³.
Example 7: Metal Sheet for a Bucket
Problem: A bucket is in the shape of a frustum with top radius 20 cm, bottom radius 8 cm, and slant height 24 cm. Find the area of metal sheet needed to make it (open at the top). (Use π = 3.14)
Solution:
- LSA of frustum = π(r₁ + r₂)l = 3.14(20 + 8)(24) = 3.14 × 672 = 2110.08 cm²
- Bottom circle area = πr₂² = 3.14 × 64 = 200.96 cm²
- Total metal = 2110.08 + 200.96 = 2311.04 cm²
Answer: Metal sheet needed = 2311.04 cm².
Example 8: Number of Balls from a Cylinder
Problem: A cylindrical rod of radius 2 cm and length 12 cm is melted into spherical balls of radius 1 cm. How many balls can be made?
Solution:
- Volume of cylinder = πr²h = π(4)(12) = 48π cm³
- Volume of one ball = (4/3)πr³ = (4/3)π(1) = (4/3)π cm³
- Number of balls = 48π ÷ (4/3)π = 48 × 3/4 = 36
Answer: 36 balls can be made.
Real-World Applications
Real-life applications of SA and volume:
- Construction: Estimating cement, bricks, paint for buildings.
- Manufacturing: Calculating metal sheet for cans, boxes, containers.
- Agriculture: Estimating water storage in tanks, reservoirs.
- Packaging: Designing containers with minimum material for maximum volume.
- Civil Engineering: Calculating earth removed during excavation.
- Cooking: Estimating how much a pot or container can hold.
Key Points to Remember
- Painting/wrapping → use surface area. Filling/capacity → use volume.
- Open containers (bucket, tank without lid) → TSA minus top circle.
- Rise in level = Volume added / Base area of container.
- Melting and recasting → volume of original = volume of new shape.
- Number of objects = Total volume ÷ Volume of one object.
- Cost of material = Area × Rate per unit area.
- Always ensure consistent units before calculation.
- 1 m³ = 1000 litres = 10⁶ cm³.
- For combined shapes, add volumes; for surface area, add only exposed surfaces.
- These are among the highest-scoring questions in CBSE board exams.
Practice Problems
- A room is 10 m long, 8 m wide, and 3.5 m high. Find the cost of painting four walls at Rs. 25 per m² (excluding a door 2 m × 1.5 m and two windows each 1.5 m × 1 m).
- A hemispherical bowl of internal radius 10.5 cm is full of soup. How many bowls of radius 3.5 cm can be filled from it?
- An oil drum is a cylinder of radius 35 cm and height 1 m. How many litres of oil can it store?
- A solid copper sphere of radius 3 cm is melted and drawn into a wire of radius 0.2 cm. Find the length of the wire.
- A conical tent must cover a circular ground area of radius 10 m and be 8 m high. Find the canvas needed.
- A well is dug 28 m deep with radius 3 m. The earth is used to raise a platform 22 m × 14 m. Find the height of the platform.
Frequently Asked Questions
Q1. How do I decide between CSA and TSA in a real-life problem?
If only the curved/lateral surface needs material (e.g., a label on a can, paint on a pillar), use CSA. If all surfaces need covering (e.g., wrapping a box, total tin for a closed can), use TSA. For open containers, use TSA minus the open face.
Q2. How do I convert m³ to litres?
1 m³ = 1000 litres. Multiply the volume in m³ by 1000 to get litres. Or 1 litre = 1000 cm³.
Q3. What happens to volume when a solid is melted?
Volume is conserved. The total volume of the new objects equals the volume of the original solid. This principle is used to find dimensions or count how many smaller objects can be made.
Q4. How do you find how much the water level rises?
Divide the volume of the object placed in the container by the cross-sectional area of the container: Rise = Volume / Base area.
Q5. What if a container is open at the top?
Subtract the area of the open face from the TSA. For example, an open cylindrical tank: surface area = CSA + one circular base = 2πrh + πr².
Q6. How do you handle combined shapes in real life?
Identify each component shape. For volume, add all volumes. For surface area, add only the exposed surfaces — subtract the areas where shapes are joined.
Related Topics
- Combination of Solids
- Conversion of Solids
- Frustum of a Cone
- Surface Area and Volume Formulas Summary
- Surface Area of Cone
- Volume of Cone
- Surface Area of Sphere
- Volume of Sphere
- Slant Height of Cone
- Surface Area of Hemisphere
- Volume of Hemisphere
- Surface Area of Combined Solids
- Volume Word Problems
- Volume of Frustum of Cone
