Surface Areas and Volumes · Answer guide

Lesson 1 · Count only exposed surfaces

Build a surface-area sum after identifying hidden joins.

1. A solid consists of a cylinder of radius 2 and height 5, with a hemisphere of the same radius on top. Its flat bottom is exposed. Find total exposed area.

   Answer: 32π. Exposed area = cylinder curve 2πrh + hemisphere curve 2πr² + bottom πr² = 32π. The joined circle is hidden.

2. A solid consists of a cylinder of radius 3 and height 4, with a hemisphere of the same radius on top. Its flat bottom is exposed. Find total exposed area.

   Answer: 51π. Exposed area = cylinder curve 2πrh + hemisphere curve 2πr² + bottom πr² = 51π. The joined circle is hidden.

3. A solid consists of a cylinder of radius 4 and height 6, with a hemisphere of the same radius on top. Its flat bottom is exposed. Find total exposed area.

   Answer: 96π. Exposed area = cylinder curve 2πrh + hemisphere curve 2πr² + bottom πr² = 96π. The joined circle is hidden.

Discuss: Sketch a cylinder with both ends capped by hemispheres. List the surfaces that remain exposed and derive an area expression.

Teaching note: Adding the total surface areas of the separate pieces counts the joined circles twice. Decide what is exposed first. Open responses need discussion; they are not automatically marked.

Lesson 2 · Add volumes of solid parts

Use each part’s formula and one consistent unit.

1. A solid cylinder (radius 3, height 6) is joined to a hemisphere of radius 3 with no overlap. What is the combined volume?

   Answer: 72π. Volume = πr²h + (2/3)πr³ = 54π + 18π = 72π.

2. A solid cylinder (radius 6, height 3) is joined to a hemisphere of radius 6 with no overlap. What is the combined volume?

   Answer: 252π. Volume = πr²h + (2/3)πr³ = 108π + 144π = 252π.

3. A solid cylinder (radius 3, height 9) is joined to a hemisphere of radius 3 with no overlap. What is the combined volume?

   Answer: 99π. Volume = πr²h + (2/3)πr³ = 81π + 18π = 99π.

Discuss: Draw a composite object, label radius versus diameter and convert every dimension to centimetres before calculating volume.

Teaching note: Surface-area rules and volume rules differ. A joined circular face has area but no thickness, so it does not remove volume. Open responses need discussion; they are not automatically marked.

Lesson 3 · Use volume conservation in an extension

Distinguish conserved volume from surface area during remoulding.

1. A sphere of radius 4 cm is remoulded into spheres of radius 2 cm with no loss. How many small spheres can be made?

   Answer: 8. Number = large volume/small volume = [(4/3)π×4³]/[(4/3)π×2³] = (4/2)³ = 8.

2. A sphere of radius 6 cm is remoulded into spheres of radius 2 cm with no loss. How many small spheres can be made?

   Answer: 27. Number = large volume/small volume = [(4/3)π×6³]/[(4/3)π×2³] = (6/2)³ = 27.

3. A sphere of radius 9 cm is remoulded into spheres of radius 3 cm with no loss. How many small spheres can be made?

   Answer: 27. Number = large volume/small volume = [(4/3)π×9³]/[(4/3)π×3³] = (9/3)³ = 27.

Discuss: Compare total area before and after splitting one radius-4 sphere into radius-2 spheres. Explain why the area grows although volume stays equal.

Teaching note: Dividing radii gives a length ratio, not a number of equal-volume pieces. Cube the radius ratio for similar spheres. Open responses need discussion; they are not automatically marked.

Final check · Set 1

Try one set after the lessons. Use the other set after reviewing any errors.

1. A solid consists of a cylinder of radius 2 and height 9, with a hemisphere of the same radius on top. Its flat bottom is exposed. Find total exposed area.

   48π. Exposed area = cylinder curve 2πrh + hemisphere curve 2πr² + bottom πr² = 48π. The joined circle is hidden.

2. A solid consists of a cylinder of radius 6 and height 7, with a hemisphere of the same radius on top. Its flat bottom is exposed. Find total exposed area.

   192π. Exposed area = cylinder curve 2πrh + hemisphere curve 2πr² + bottom πr² = 192π. The joined circle is hidden.

3. A solid cylinder (radius 9, height 3) is joined to a hemisphere of radius 9 with no overlap. What is the combined volume?

   729π. Volume = πr²h + (2/3)πr³ = 243π + 486π = 729π.

4. A solid cylinder (radius 3, height 12) is joined to a hemisphere of radius 3 with no overlap. What is the combined volume?

   126π. Volume = πr²h + (2/3)πr³ = 108π + 18π = 126π.

5. A sphere of radius 10 cm is remoulded into spheres of radius 2 cm with no loss. How many small spheres can be made?

   125. Number = large volume/small volume = [(4/3)π×10³]/[(4/3)π×2³] = (10/2)³ = 125.

6. A sphere of radius 12 cm is remoulded into spheres of radius 4 cm with no loss. How many small spheres can be made?

   27. Number = large volume/small volume = [(4/3)π×12³]/[(4/3)π×4³] = (12/4)³ = 27.

Final check · Set 2

Try one set after the lessons. Use the other set after reviewing any errors.

1. A solid consists of a cylinder of radius 3 and height 10, with a hemisphere of the same radius on top. Its flat bottom is exposed. Find total exposed area.

   87π. Exposed area = cylinder curve 2πrh + hemisphere curve 2πr² + bottom πr² = 87π. The joined circle is hidden.

2. A solid consists of a cylinder of radius 4 and height 12, with a hemisphere of the same radius on top. Its flat bottom is exposed. Find total exposed area.

   144π. Exposed area = cylinder curve 2πrh + hemisphere curve 2πr² + bottom πr² = 144π. The joined circle is hidden.

3. A solid cylinder (radius 6, height 9) is joined to a hemisphere of radius 6 with no overlap. What is the combined volume?

   468π. Volume = πr²h + (2/3)πr³ = 324π + 144π = 468π.

4. A solid cylinder (radius 9, height 6) is joined to a hemisphere of radius 9 with no overlap. What is the combined volume?

   972π. Volume = πr²h + (2/3)πr³ = 486π + 486π = 972π.

5. A sphere of radius 15 cm is remoulded into spheres of radius 3 cm with no loss. How many small spheres can be made?

   125. Number = large volume/small volume = [(4/3)π×15³]/[(4/3)π×3³] = (15/3)³ = 125.

6. A sphere of radius 6 cm is remoulded into spheres of radius 3 cm with no loss. How many small spheres can be made?

   8. Number = large volume/small volume = [(4/3)π×6³]/[(4/3)π×3³] = (6/3)³ = 8.