Mensuration — Cylinder, Cone, and Sphere
Introduction
Mensuration deals with measurement of geometric figures. In ICSE Class 10, you learn to compute the surface areas and volumes of three-dimensional solids — cylinders, cones, spheres, and hemispheres. Problems involving combinations of solids are also important.
Cylinder
Right Circular Cylinder
A cylinder has two parallel circular bases and a curved lateral surface.
| Quantity | Formula | Units |
|---|---|---|
| Curved surface area (CSA) | 2πrh | Square units |
| Total surface area (TSA) | 2πr(h + r) | Square units |
| Volume | πr²h | Cubic units |
Where r = radius of base, h = height.
Cone
Right Circular Cone
A cone has a circular base and a vertex directly above/below the centre of the base.
| Quantity | Formula | Units |
|---|---|---|
| Slant height (l) | l = √(r² + h²) | Linear units |
| Curved surface area (CSA) | πrl | Square units |
| Total surface area (TSA) | πr(l + r) | Square units |
| Volume | ¹/₃πr²h | Cubic units |
Where r = radius of base, h = vertical height, l = slant height.
Sphere and Hemisphere
Sphere
| Quantity | Formula |
|---|---|
| Surface area | 4πr² |
| Volume | ⁴/₃πr³ |
Hemisphere
| Quantity | Formula |
|---|---|
| Curved surface area (CSA) | 2πr² |
| Total surface area (TSA) | 3πr² |
| Volume | ²/₃πr³ |
Combination of Solids
ICSE problems frequently involve solids formed by combining two or more shapes. Common combinations:
- A cone on top of a cylinder (e.g., a tent with a cylindrical base).
- A hemisphere on top of a cylinder (e.g., a storage tank).
- A cone on top of a hemisphere (e.g., an ice cream cone).
Approach: Compute the surface area or volume of each part separately and add them.
Worked Examples
Example 1: Cylinder Volume and Surface Area
A cylindrical tank has a radius of 7 m and height of 10 m. Find (a) volume, (b) curved surface area, (c) total surface area.
Solution: (a) Volume = πr²h = ²²/₇ × 7² × 10 = 22 × 7 × 10 = 1540 m³
(b) CSA = 2πrh = 2 × ²²/₇ × 7 × 10 = 2 × 22 × 10 = 440 m²
(c) TSA = 2πr(h + r) = 2 × ²²/₇ × 7 × (10 + 7) = 44 × 17 = 748 m²
Example 2: Cone Slant Height and Volume
A cone has base radius 6 cm and height 8 cm. Find (a) slant height, (b) volume.
Solution: (a) l = √(r² + h²) = √(36 + 64) = √100 = 10 cm
(b) Volume = ¹/₃πr²h = ¹/₃ × 3.14 × 36 × 8 = 3.14 × 12 × 8 = 301.44 cm³
Example 3: Sphere and Hemisphere
(a) Find the surface area of a sphere of radius 7 cm. (b) Find the volume of a hemisphere of radius 7 cm.
Solution: (a) Surface area = 4πr² = 4 × ²²/₇ × 49 = 616 cm²
(b) Volume of hemisphere = ²/₃πr³ = ²/₃ × ²²/₇ × 343 = (2 × 22 × 49) / 3 = 2156 / 3 = 718.67 cm³
Example 4: Combination of Solids
A tent is in the shape of a cylinder surmounted by a cone. The cylindrical part is 6 m in diameter and 4 m high. The cone has the same base diameter and a slant height of 5 m. Find the total surface area of the tent.
Solution: Radius r = 3 m Cylindrical part: CSA = 2πrh = 2 × π × 3 × 4 = 24π m² Conical part: CSA = πrl = π × 3 × 5 = 15π m² Total canvas needed = 24π + 15π = 39π = 39 × 3.14 = 122.46 m²
(Base is not included since the tent is open at the bottom.)
Example 5: Volume of Combined Solid
A solid consists of a cylinder of height 10 cm and radius 3.5 cm with a hemisphere of the same radius on top. Find the total volume.
Solution: Volume of cylinder = πr²h = ²²/₇ × 3.5² × 10 = ²²/₇ × 12.25 × 10 = 22 × 1.75 × 10 = 385 cm³ Volume of hemisphere = ²/₃πr³ = ²/₃ × ²²/₇ × 3.5³ = ²/₃ × ²²/₇ × 42.875 = (2 × 22 × 6.125) / 3 = 269.5 / 3 ≈ 89.83 cm³ Total volume = 385 + 89.83 = 474.83 cm³
Common Mistakes and Fixes
| Mistake | Fix |
|---|---|
| Confusing CSA and TSA | CSA excludes the base(s); TSA includes all surfaces |
| Using diameter instead of radius | Always halve the diameter to get the radius |
| Forgetting the ¹/₃ factor for cone/hemisphere volume | Cone: ¹/₃πr²h; Hemisphere: ²/₃πr³ |
| Not matching units | Convert all lengths to the same unit before calculating |
| Including the base of a tent in canvas area | The base is open — only CSA is needed |
ICSE Exam Focus
Mensuration carries 10–14 marks in ICSE exams — a major topic. Questions include:
- Direct formula application for area and volume.
- Finding missing dimensions (r, h, l) given area or volume.
- Combination of solids (surface area and volume).
- Cost-based word problems (painting, filling, making).
Marks Blueprint:
| Topic | Marks |
|---|---|
| Cylinder (area/volume) | 3 |
| Cone (area/volume) | 3 |
| Sphere/Hemisphere (area/volume) | 3 |
| Combination of solids | 4 |
| Cost/word problems | 3 |
Self-Test Questions
-
Find the volume and total surface area of a cylinder with radius 5 cm and height 14 cm.
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A cone has a slant height of 13 cm and base radius 5 cm. Find its height and volume.
-
The surface area of a sphere is 616 cm². Find its radius and volume.
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A toy is in the form of a cone mounted on a hemisphere of the same radius 3.5 cm. The total height of the toy is 15.5 cm. Find the volume of the toy.
-
How many cylindrical pencils of radius 3.5 mm and height 14 cm can be made from a cuboidal block of wood measuring 14 cm × 11 cm × 7 cm?
-
A hemispherical bowl of internal radius 9 cm is full of water. The water is emptied into a cylindrical bottle of radius 3 cm. Find the height of water in the bottle.
In ICSE, always write the formula before substituting values. Partial marks are awarded for correct formula expression, even if the arithmetic is wrong.
