ICSE Class 10 Maths Mensuration Step-by-Step Solved
ICSE Class 10 Maths: Selina Chapter 20 Mensuration
ICSE Class 10 Maths mensuration in Selina Concise Chapter 20 covers the surface area and volume of cylinders, cones and spheres. To solve these questions, first identify the solid, then decide whether the question needs area or volume, convert all units, substitute in the correct formula and write the final answer with the correct unit.
This page supports Concise Mathematics Selina Solutions Class 10 ICSE Chapter 20 Cylinder, Cone and Sphere (Surface Area and Volume) with formulas, worked solutions, examiner notes and common mistakes. Exact exercise numbering may vary by edition, so the worked examples focus on the standard Selina-style question types used in this chapter.
Concept snapshot: Surface area is the covering needed on a solid; volume is the space occupied or material used. Words such as paint, canvas, sheet metal and plaster usually point to area. Words such as capacity, water, dug out, melted, recast and displaced usually point to volume.
Formula reference for cylinder, cone and sphere
| Solid or situation | Formula | Use |
|---|---|---|
| Cylinder curved surface area | 2\pi rh | Side surface only |
| Cylinder total surface area | 2\pi r(h+r) | Closed cylinder |
| Cylinder volume | \pi r^2h | Capacity or material |
| Hollow cylinder volume | \pi h(R^2-r^2) | Pipe or tube material |
| Open pipe total surface area | 2\pi RH+2\pi rH+2\pi(R^2-r^2) | Outer curved surface, inner curved surface and two annular ends |
| Cone curved surface area | \pi rl | Curved canvas or wrapper |
| Cone total surface area | \pi r(l+r) | Cone with circular base included |
| Cone volume | \frac{1}{3}\pi r^2h | Capacity or material of a cone |
| Sphere surface area | 4\pi r^2 | Outer surface of a ball |
| Sphere volume | \frac{4}{3}\pi r^3 | Material of a sphere |
| Hemisphere total surface area | 3\pi r^2 | Curved part plus circular base |
| Hemisphere volume | \frac{2}{3}\pi r^3 | Half of sphere volume |
Method for Chapter 20 questions
Most Maths problems in this chapter combine two skills: choosing the correct formula and keeping units consistent. Use this order in your solution.
- Write the given values and convert all lengths to one unit.
- Draw a small labelled figure if the solid is hollow or combined.
- Use area formulas for covering, painting, plastering and sheet metal.
- Use volume formulas for water, capacity, earth removed, displacement, melting and recasting.
- Check the final unit: surface area is in \text{cm}^2 or \text{m}^2, while volume is in \text{cm}^3 or \text{m}^3.
Worked Selina-style solutions
The following worked examples are original, syllabus-aligned models for common Selina Chapter 20 question types.
Example 1: Capacity of a cylindrical pipe
Question: The inner radius of a pipe is 2.1\text{ cm}. How much water can 12\text{ m} of this pipe hold?
Step 1: The water inside the pipe forms a cylinder with radius r=2.1\text{ cm}.
Step 2: Convert the length: 12\text{ m}=1200\text{ cm}. Hence h=1200\text{ cm}.
Step 3: Use V=\pi r^2h.
V=\frac{22}{7}\times(2.1)^2\times1200
V=\frac{22}{7}\times4.41\times1200=16632\text{ cm}^3
Step 4: Convert to litres if required: 1000\text{ cm}^3=1\text{ litre}.
Final answer: The pipe can hold 16632\text{ cm}^3, or 16.632\text{ litres}, of water.
Example 2: Well excavation and plastering
Question: How many cubic metres of earth must be dug out to make a well 28\text{ m} deep and 2.8\text{ m} in diameter? Also find the cost of plastering its inner curved surface at ₹ 4.50 per square metre.
Step 1: The well is cylindrical. Diameter =2.8\text{ m}, so r=1.4\text{ m}. Depth h=28\text{ m}.
Step 2: Earth dug out equals the volume of the cylindrical well.
V=\pi r^2h=\frac{22}{7}\times1.4\times1.4\times28=172.48\text{ m}^3
Step 3: Plastering is done on the inner curved surface.
\text{Curved surface area}=2\pi rh=2\times\frac{22}{7}\times1.4\times28=246.4\text{ m}^2
Step 4: Multiply the area by the rate.
\text{Cost}=246.4\times4.50=\text{₹ }1108.80
Final answer: Earth dug out =172.48\text{ m}^3; plastering cost =\text{₹ }1108.80.
Example 3: Recasting a cylinder into a hollow cylinder
Question: What length of a solid cylinder of diameter 2\text{ cm} must be taken to recast it into a hollow cylinder of external diameter 20\text{ cm}, thickness 0.25\text{ cm} and length 15\text{ cm}?
Step 1: For the hollow cylinder, outer radius R=10\text{ cm}. Since thickness is 0.25\text{ cm}, inner radius r=10-0.25=9.75\text{ cm}. Height h=15\text{ cm}.
Step 2: Volume of metal in the hollow cylinder is V=\pi h(R^2-r^2).
V=\pi\times15(10^2-9.75^2)
V=15\pi(100-95.0625)=74.0625\pi\text{ cm}^3
Step 3: For the solid cylinder, radius =1\text{ cm}. Let its length be L\text{ cm}. Its volume is \pi(1)^2L=\pi L.
Step 4: Recasting keeps volume the same.
\pi L=74.0625\pi
L=74.0625\text{ cm}
Final answer: The required length is 74.0625\text{ cm}, or about 74.06\text{ cm}.
Examiner’s mindset for mensuration answers
In a CISCE-style ICSE Class 10 Maths solution, the working should show the formula, substitution, simplification and final unit. Exact mark splits vary by paper, so avoid writing only the final answer. A correct method with clear units is safer than a number without explanation.
Underline the action word in the question. Plaster, paint and canvas usually require area. Capacity, dug out, displaced, melted and recast usually require volume.
Common mistakes students make
- Using diameter as radius: If diameter is 20\text{ cm}, radius is 10\text{ cm}.
- Mixing units: Do not combine 12\text{ m} with 2.1\text{ cm}; convert first.
- Adding a missing base: An open cylinder or open tent does not include the missing circular base.
- Confusing hollow radius: If the outer radius is R and thickness is t, the inner radius is R-t.
- Rounding too early: Keep working values accurate and round only at the end.
Quick answer index
| Type | Final answer |
|---|---|
| Pipe capacity | 16632\text{ cm}^3=16.632\text{ litres} |
| Well excavation and plastering | 172.48\text{ m}^3, cost =\text{₹ }1108.80 |
| Recasting | 74.0625\text{ cm}, about 74.06\text{ cm} |
Related ICSE Class 10 Maths study links
For more chapter-wise practice, use the ICSE solutions section and the Class 10 ICSE solutions page. To revise connected topics, practise Selina Arithmetic Progression solutions and Selina Circles solutions.
Sources referenced
This page follows the standard ICSE Class 10 treatment of mensuration and Selina Concise Mathematics Chapter 20. For the official syllabus and board framework, refer to CISCE. NCERT school-level mensuration concepts are useful as a supporting concept reference.
Frequently Asked Questions
Which formulas are most important in ICSE Class 10 Maths mensuration?
The key formulas are cylinder volume \pi r^2h, cone volume \frac{1}{3}\pi r^2h, sphere volume \frac{4}{3}\pi r^3, cylinder curved surface area 2\pi rh, cone curved surface area \pi rl, and sphere surface area 4\pi r^2.
How do I know whether to use curved surface area or total surface area?
Use curved surface area when only the side surface is covered, painted or plastered. Use total surface area when the circular base or bases are included. If the solid is open, subtract the missing face.
Why do recasting questions use equal volumes?
In recasting, the material changes shape but not amount. So the original volume equals the total volume of the new solid or solids, unless wastage is mentioned.
What unit should I write for surface area and volume?
Surface area uses square units such as \text{cm}^2 or \text{m}^2. Volume uses cubic units such as \text{cm}^3 or \text{m}^3. Convert all lengths to the same unit before substitution.
Should I round during the working?
No. Keep exact values or enough decimal places during working and round only in the final answer, according to the accuracy asked in the question.