Combination of Solids
A combination of solids (or composite solid) is a three-dimensional shape formed by joining two or more basic solids — such as cylinders, cones, hemispheres, and cuboids. This topic is covered in Class 10 CBSE Chapter 13 (Surface Areas and Volumes).
Real-life objects are rarely perfect cubes or spheres. An ice cream cone with a hemisphere on top, a capsule (cylinder + two hemispheres), or a tent (cylinder + cone) are all combinations of standard solids.
To find the total surface area or volume of a combined solid, we break it into its component parts, calculate individually, and then add or subtract as appropriate — making sure not to count any surfaces that are joined together (and hence not exposed).
What is Combination of Solids - Surface Area, Volume & Solved Examples?
Definition: A combined solid is formed by attaching two or more standard 3D shapes (cylinder, cone, hemisphere, sphere, cuboid, frustum) together.
Key Principles:
- Volume of combined solid = Sum of volumes of individual parts
- Surface area of combined solid = Sum of surface areas of individual parts − Areas of joined surfaces (internal surfaces that are not exposed)
Standard Formulas Used:
| Solid | CSA | TSA | Volume |
|---|---|---|---|
| Cylinder | 2πrh | 2πr(h + r) | πr²h |
| Cone | πrl | πr(l + r) | ⅓πr²h |
| Hemisphere | 2πr² | 3πr² | ⅔πr³ |
| Sphere | 4πr² | 4πr² | ⁴⁄₃πr³ |
| Cuboid | 2h(l+b) | 2(lb+bh+hl) | lbh |
Combination of Solids Formula
Surface Area of Combined Solids:
TSA = Sum of visible surface areas of all parts
Important: Subtract the areas where two solids are joined (these surfaces are hidden).
Volume of Combined Solids:
V = Sum of volumes of all individual parts
Common Combinations:
- Cylinder + Cone: TSA = CSA of cylinder + CSA of cone + base area of cylinder (if bottom is flat)
- Cylinder + Hemisphere: TSA = CSA of cylinder + CSA of hemisphere + one base area (if bottom is flat)
- Cylinder + 2 Hemispheres (capsule): TSA = CSA of cylinder + 2 × CSA of hemisphere
- Cone + Hemisphere: TSA = CSA of cone + CSA of hemisphere (no flat surfaces)
Derivation and Proof
Why we subtract joined surfaces:
- When two solids are joined, the surfaces at the junction are no longer visible (they are internal).
- For example, a hemisphere placed on a cylinder: the flat circular face of the hemisphere sits on the top circular face of the cylinder.
- Both these circular faces (each = πr²) become internal and are not part of the outer surface.
- Therefore: TSA = CSA of cylinder + CSA of hemisphere + bottom base of cylinder (not the top).
Why volumes simply add:
- Volume measures the total space occupied, regardless of how shapes are joined.
- No volume is "lost" at the junction — the matter fills both shapes entirely.
- Therefore: Volume of combined solid = V₁ + V₂ + ... (simple addition).
Types and Properties
Common Types of Combined Solids:
- Toy/Rocket: Cone on top of a cylinder — TSA = CSA(cone) + CSA(cylinder) + base area of cylinder
- Capsule: Two hemispheres on either end of a cylinder — TSA = CSA(cylinder) + 2 × CSA(hemisphere)
- Ice cream cone: Hemisphere on top of a cone — TSA = CSA(cone) + CSA(hemisphere)
- Tent: Cone on top of a cylinder — TSA = CSA(cone) + CSA(cylinder)
- Water tank: Hemisphere on top of a cylinder — volume and surface area calculations
- Pencil: Cone on top of a cylinder
- Gulab jamun: Cylinder with two hemispheres at ends
Solved Examples
Example 1: Tent — Cylinder with Conical Top
Problem: A tent is in the shape of a cylinder with a conical top. The radius of the base is 7 m, height of cylindrical part is 3 m, and slant height of cone is 5 m. Find the total surface area of the tent (excluding the base). Use π = 22/7.
Solution:
Given: r = 7 m, h(cyl) = 3 m, l(cone) = 5 m
CSA of cylinder:
- = 2πrh = 2 × (22/7) × 7 × 3 = 132 m²
CSA of cone:
- = πrl = (22/7) × 7 × 5 = 110 m²
Total surface area (no base):
- = 132 + 110 = 242 m²
Answer: TSA of tent = 242 m²
Example 2: Capsule — Cylinder with Two Hemispheres
Problem: A medicine capsule is in the shape of a cylinder with two hemispheres at each end. The total length is 14 mm and the diameter is 4 mm. Find the total surface area.
Solution:
Given: Diameter = 4 mm → r = 2 mm
Length of cylindrical part: 14 − 2(2) = 14 − 4 = 10 mm
CSA of cylinder:
- = 2πrh = 2π(2)(10) = 40π mm²
CSA of 2 hemispheres:
- = 2 × 2πr² = 4π(4) = 16π mm²
TSA:
- = 40π + 16π = 56π
- = 56 × 22/7 = 176 mm²
Answer: TSA = 176 mm²
Example 3: Toy — Hemisphere with Cone
Problem: A toy is made by mounting a cone on a hemisphere. The radius is 3.5 cm and total height is 15.5 cm. Find the TSA of the toy.
Solution:
Given: r = 3.5 cm, total height = 15.5 cm
Height of cone: 15.5 − 3.5 = 12 cm
Slant height of cone:
- l = √(h² + r²) = √(144 + 12.25) = √156.25 = 12.5 cm
CSA of cone:
- = πrl = (22/7) × 3.5 × 12.5 = 137.5 cm²
CSA of hemisphere:
- = 2πr² = 2 × (22/7) × (3.5)² = 77 cm²
TSA:
- = 137.5 + 77 = 214.5 cm²
Answer: TSA of toy = 214.5 cm²
Example 4: Volume of Ice Cream Cone with Hemisphere
Problem: An ice cream cone with a hemispherical top has a radius of 3 cm. The height of the conical part is 12 cm. Find the volume of ice cream.
Solution:
Given: r = 3 cm, h(cone) = 12 cm
Volume of cone:
- = ⅓πr²h = ⅓ × π × 9 × 12 = 36π cm³
Volume of hemisphere:
- = ⅔πr³ = ⅔ × π × 27 = 18π cm³
Total volume:
- = 36π + 18π = 54π
- = 54 × 22/7 = 169.71 cm³
Answer: Volume of ice cream = 54π cm³ ≈ 169.71 cm³
Example 5: Wooden Toy — Cylinder + Cone + Hemisphere
Problem: A wooden toy is in the form of a cone mounted on a cylinder, with a hemisphere at the bottom. The radius is 5 cm, cylinder height is 10 cm, and cone height is 6 cm. Find the total volume.
Solution:
Given: r = 5 cm, h(cyl) = 10 cm, h(cone) = 6 cm
Volume of cylinder:
- = πr²h = π × 25 × 10 = 250π cm³
Volume of cone:
- = ⅓πr²h = ⅓ × π × 25 × 6 = 50π cm³
Volume of hemisphere:
- = ⅔πr³ = ⅔ × π × 125 = 250π/3 cm³
Total volume:
- = 250π + 50π + 250π/3
- = (750π + 150π + 250π)/3
- = 1150π/3
- = 1150 × 22/(7 × 3) = 1204.76 cm³
Answer: Total volume = 1150π/3 cm³ ≈ 1204.76 cm³
Example 6: Gulab Jamun Shape
Problem: A gulab jamun has the shape of a cylinder with two hemispherical ends. The diameter is 2.8 cm and the total length is 5 cm. Find the volume of each gulab jamun.
Solution:
Given: d = 2.8 cm → r = 1.4 cm, total length = 5 cm
Length of cylindrical part: 5 − 2(1.4) = 5 − 2.8 = 2.2 cm
Volume of cylinder:
- = πr²h = (22/7) × (1.4)² × 2.2 = (22/7) × 1.96 × 2.2 = 13.55 cm³
Volume of 2 hemispheres = Volume of 1 sphere:
- = ⁴⁄₃πr³ = (4/3) × (22/7) × (1.4)³ = (4/3) × (22/7) × 2.744 = 11.50 cm³
Total volume:
- = 13.55 + 11.50 = 25.05 cm³
Answer: Volume = 25.05 cm³
Example 7: Water Tank — Cylinder with Hemispherical Dome
Problem: A water tank consists of a cylinder with a hemispherical top. The internal radius is 1.5 m and the total internal height is 5 m. Find the volume of water the tank can hold.
Solution:
Given: r = 1.5 m, total height = 5 m
Height of cylindrical part: 5 − 1.5 = 3.5 m
Volume of cylinder:
- = πr²h = π × 2.25 × 3.5 = 7.875π m³
Volume of hemisphere:
- = ⅔πr³ = ⅔ × π × 3.375 = 2.25π m³
Total volume:
- = 7.875π + 2.25π = 10.125π
- = 10.125 × 3.1416 = 31.81 m³
Answer: Capacity = 10.125π m³ ≈ 31.81 m³
Example 8: Surface Area of a Rocket
Problem: A rocket is in the shape of a cylinder with a cone at the top. The radius is 2.5 cm, the cylinder height is 21 cm, and the cone height is 6 cm. Find the TSA (excluding the base).
Solution:
Given: r = 2.5 cm, h(cyl) = 21 cm, h(cone) = 6 cm
Slant height of cone:
- l = √(6² + 2.5²) = √(36 + 6.25) = √42.25 = 6.5 cm
CSA of cylinder:
- = 2πrh = 2 × (22/7) × 2.5 × 21 = 330 cm²
CSA of cone:
- = πrl = (22/7) × 2.5 × 6.5 = 51.07 cm²
Base area of cylinder:
- = πr² = (22/7) × 6.25 = 19.64 cm²
TSA:
- = 330 + 51.07 + 19.64 = 400.71 cm²
Answer: TSA = 400.71 cm²
Real-World Applications
Real-life applications:
- Architecture: Domes on buildings (hemisphere + cylinder), silos (cylinder + cone), and water tanks are all combined solids.
- Manufacturing: Capsule pills, pencils, bolts, and rocket models are designed as combined shapes.
- Packaging: Calculating material required for containers and packaging of irregular shapes.
- Civil engineering: Estimating concrete required for pillars with conical tops, reservoirs, and tanks.
- Toys: Rattles, spinning tops, and building blocks are combined solid shapes.
Key Points to Remember
- Volume of a combined solid = sum of volumes of individual parts.
- Surface area of a combined solid = sum of visible areas; subtract areas of joined surfaces.
- When a hemisphere sits on a cylinder, the two circular faces at the junction are internal (not counted in TSA).
- Always identify which surfaces are exposed and which are hidden.
- For a cylinder + cone (same radius): TSA = CSA(cyl) + CSA(cone) + base area of cylinder (bottom only).
- For a capsule (cyl + 2 hemispheres): TSA = CSA(cyl) + 2 × CSA(hemisphere). No flat surfaces are exposed.
- Slant height of cone: l = √(h² + r²).
- Use consistent units throughout the calculation.
- Remember: CSA = curved surface area only (no base); TSA = total surface area (including base if exposed).
Practice Problems
- A solid is made by placing a cone on a hemisphere, each of radius 7 cm. If the height of the cone is 24 cm, find the TSA and volume.
- A boiler consists of a cylinder of length 2 m with two hemispherical ends, each of radius 1 m. Find the TSA and volume.
- A toy is in the shape of a right circular cylinder with a hemisphere at one end and a cone at the other, all of the same radius 5 cm. Heights: cylinder 13 cm, cone 12 cm. Find the surface area of the toy.
- A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, with a cone surmounted on it of height 60 cm. Find the mass if 1 cm³ of iron weighs 8 g.
- A wooden article was made by scooping out a hemisphere from each end of a solid cylinder. If the cylinder height is 10 cm and base radius 3.5 cm, find the TSA.
- A tent is made of a cylinder of height 3 m and a cone of slant height 5 m. If the diameter of the base is 8 m, find the total canvas used.
Frequently Asked Questions
Q1. How do you find the surface area of a combined solid?
Add the visible (exposed) surface areas of all individual parts. Subtract the areas of surfaces that are joined together (internal surfaces). For example, a cone on a cylinder: TSA = CSA of cone + CSA of cylinder + base of cylinder (the top of the cylinder and bottom of the cone are hidden).
Q2. Why do we subtract areas at the junction?
At the junction, both solids share a common face that is no longer visible from outside. For example, when a hemisphere sits on a cylinder, the circular base of the hemisphere and the circular top of the cylinder are hidden inside.
Q3. Do volumes get subtracted when combining solids?
No. When solids are joined together, volumes are simply added. Volume is subtracted only when one solid is scooped out of another (e.g., a cone carved out from a cylinder).
Q4. What is the difference between CSA and TSA?
CSA (Curved Surface Area) includes only the curved surface, not the flat bases. TSA (Total Surface Area) includes all surfaces — curved and flat. In combined solids, we typically use CSA of parts and add only the flat surfaces that are exposed.
Q5. How do you find the slant height of a cone?
Slant height l = √(h² + r²), where h is the height and r is the radius. This is derived from the Pythagoras theorem applied to the right triangle formed by h, r, and l.
Q6. What is a frustum?
A frustum is formed when a smaller cone is cut off from the top of a larger cone by a plane parallel to the base. It is essentially a cone with its tip removed, resulting in two circular faces of different radii.
Q7. How do you find the volume of a capsule?
A capsule = cylinder + 2 hemispheres. Volume = πr²h + 2(⅔πr³) = πr²h + ⁴⁄₃πr³, where h is the cylindrical part length and r is the common radius.
Q8. What value of π should be used in board exams?
Use π = 22/7 or π = 3.14 as specified in the problem. If not specified, use 22/7 for fractions involving 7 in the denominator, and 3.14 otherwise.
Related Topics
- Conversion of Solids
- Frustum of a Cone
- Surface Area of Cone
- Surface Area of Cylinder
- 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
- Curved Surface Area of Frustum
