What this chapter is about
This chapter extends your knowledge of finding surface areas and volumes of basic solids (cuboid, cylinder, cone, sphere) to more complex situations. You will learn to calculate these measurements for combinations of solids — objects formed by joining two or more basic shapes together, such as a toy shaped like a cone attached to a hemisphere, or a storage tank made of a cylinder with hemispherical ends.
You will also study what happens when a solid is converted from one shape to another. When a metalsmith melts a sphere and recasts it into a wire (cylinder), the volume remains the same. This principle helps solve practical problems about moulding, melting, and reshaping objects.
Finally, you will learn to find the volume of a frustum of a cone — the shape left when you slice off the top of a cone with a cut parallel to its base. This shape appears in buckets, lampshades, and drinking glasses. After this chapter, you should be able to handle real-world measurement problems involving composite objects.
Key ideas
- The surface area of a combination of solids equals the sum of the curved/lateral surface areas of each part, minus any surfaces that are hidden or joined internally.
- When two solids are joined, the total volume equals the sum of their individual volumes.
- When a solid is melted and recast into another shape, the volume remains unchanged — this is the principle of conservation of volume.
- A frustum of a cone is the portion that remains after cutting a cone by a plane parallel to its base and removing the smaller cone formed above.
- The slant height of a frustum can be found using the Pythagorean theorem: l = √[h² + (R − r)²], where h is the vertical height, R is the larger radius, and r is the smaller radius.
- Always identify which surfaces are exposed (painted, covered, or visible) before calculating surface area.
Formulas and facts to remember
Basic solids (for reference):
- Curved surface area of cylinder = 2πrh
- Total surface area of cylinder = 2πr(r + h)
- Volume of cylinder = πr²h
- Curved surface area of cone = πrl (where l is slant height)
- Volume of cone = (1/3)πr²h
- Surface area of sphere = 4πr²
- Volume of sphere = (4/3)πr³
- Curved surface area of hemisphere = 2πr²
- Volume of hemisphere = (2/3)πr³
Frustum of a cone:
- Volume of frustum = (1/3)πh(R² + r² + Rr), where R and r are the radii of the two circular ends and h is the height.
- Curved surface area of frustum = π(R + r)l, where l = √[h² + (R − r)²] is the slant height.
- Total surface area of frustum = π(R + r)l + πR² + πr² (includes both circular ends).
Conversion principle:
- Volume before reshaping = Volume after reshaping.
Worked examples
Example 1: A toy shaped like a cone mounted on a hemisphere
A wooden toy consists of a cone of height 8 cm mounted on a hemisphere. Both have the same base radius of 3 cm. Find the total surface area of the toy. (Use π = 22/7)
Solution: The toy has two exposed surfaces: the curved surface of the cone and the curved surface of the hemisphere. The circular base of the cone sits on the hemisphere and is not visible.
Slant height of cone, l = √(h² + r²) = √(8² + 3²) = √(64 + 9) = √73 cm
Curved surface area of cone = πrl = (22/7) × 3 × √73 = (66/7) × √73 cm²
Curved surface area of hemisphere = 2πr² = 2 × (22/7) × 3² = 2 × (22/7) × 9 = 396/7 cm²
Total surface area = (66√73)/7 + 396/7 = (66 × 8.54 + 396)/7 ≈ (563.64 + 396)/7 ≈ 137.1 cm²
Example 2: Melting and recasting
A solid metal sphere of radius 6 cm is melted and recast into small cylinders, each of radius 1 cm and height 2 cm. How many cylinders can be made?
Solution: Volume of sphere = (4/3)πr³ = (4/3) × π × 6³ = (4/3) × π × 216 = 288π cm³
Volume of one small cylinder = πr²h = π × 1² × 2 = 2π cm³
Number of cylinders = Volume of sphere ÷ Volume of one cylinder = 288π ÷ 2π = 144
Therefore, 144 small cylinders can be made.
Example 3: Volume of a bucket (frustum)
A bucket is in the shape of a frustum of a cone. Its top and bottom radii are 20 cm and 10 cm respectively, and its depth is 30 cm. Find the capacity of the bucket in litres. (Use π = 3.14)
Solution: Here R = 20 cm, r = 10 cm, h = 30 cm
Volume of frustum = (1/3)πh(R² + r² + Rr) = (1/3) × 3.14 × 30 × (20² + 10² + 20 × 10) = (1/3) × 3.14 × 30 × (400 + 100 + 200) = (1/3) × 3.14 × 30 × 700 = 3.14 × 10 × 700 = 21980 cm³
Since 1 litre = 1000 cm³, capacity = 21980 ÷ 1000 = 21.98 litres
Common mistakes
- Adding total surface areas of each solid instead of only the exposed surfaces → Identify and exclude the joined or hidden surfaces before adding.
- Using the wrong formula for slant height of frustum; writing l = √(h² + R² + r²) → The correct formula is l = √[h² + (R − r)²].
- Forgetting to convert units when capacity is asked in litres → Remember 1 litre = 1000 cm³; always check the unit asked.
- Using curved surface area formula when total surface area is needed → Read carefully whether the problem asks for curved, lateral, or total surface area.
- Assuming the number of objects must be a whole number and rounding up → When recasting, round down because you cannot make a fraction of an object from leftover material.
Quick revision
- For combined solids: add volumes; add only exposed surface areas.
- Melting and recasting problems: equate volumes before and after.
- Frustum volume: (1/3)πh(R² + r² + Rr); slant height: l = √[h² + (R − r)²].
- Always draw a rough figure and label dimensions before calculating.
- Convert cm³ to litres by dividing by 1000.