Learning Objectives
- Calculate volumes and surface areas of pyramids, cones, spheres and hemispheres.
- Use slant height correctly when finding curved surface area of a cone.
- Solve problems involving compound solids, missing dimensions and frustums.
- Convert consistently between metric units of length, area, volume and capacity.
- Recognise planes and axes of symmetry for pyramids and cones as required by the syllabus.
Key Terms
- Pyramid
- A solid with a polygonal base and triangular faces meeting at an apex.
- Cone
- A solid with a circular base and curved surface meeting at an apex.
- Sphere
- A solid whose surface points are all the same distance from the centre.
- Hemisphere
- Half of a sphere.
- Slant height
- The distance along the curved surface of a cone from the rim to the apex.
- Frustum
- The part of a cone or pyramid left after the top is cut off parallel to the base.
- Plane of symmetry
- A plane that divides a solid into two mirror-image halves.
- Axis of symmetry
- A line about which a solid can rotate and appear unchanged.
What This Chapter Covers
- Volume and surface-area formulas
- Slant height and cone calculations
- Spheres, hemispheres and compound solids
- Frustums by subtraction
- Planes and axes of symmetry for relevant solids
Units And Dimensional Consistency
Lengths use units such as cm, areas use squared units such as cm², and volumes use cubed units such as cm³. Convert all dimensions to the same unit before substituting. Area conversion factors are squared and volume conversion factors are cubed.
| Conversion | Equivalent |
|---|---|
| Length | 1\text{ m}=100\text{ cm} |
| Area | 1\text{ m}^2=10\,000\text{ cm}^2 |
| Volume | 1\text{ m}^3=1\,000\,000\text{ cm}^3 |
| Capacity | 1\text{ litre}=1000\text{ cm}^3 and 1\text{ m}^3=1000\text{ litres} |
Pyramids
The perpendicular height is measured from the apex at right angles to the base plane. It is not generally the sloping edge. The volume is one third of the volume of a prism with the same base area and perpendicular height.
Worked Example: Square-Based Pyramid
Question: A square-based pyramid has base side 10 cm and perpendicular height 12 cm. Find its volume.
- Base area =10^2=100 cm².
- V=\frac13(100)(12).
Answer: 400\text{ cm}^3.
Surface Area Of A Pyramid
Add the base area and the areas of all triangular faces. The height of each triangular face is a slant height measured perpendicular to the corresponding base edge within that face. Different faces may have different slant heights in a non-regular pyramid.
Worked Example: Total Surface Area Of A Square Pyramid
Question: A square pyramid has base side 8 cm. Each triangular face has perpendicular slant height 5 cm. Find total surface area.
- Base area =8^2=64.
- Area of one triangle =\frac12(8)(5)=20.
- Four triangular faces have area 80.
Answer: 144\text{ cm}^2.
Cones
A cone has radius r, perpendicular height h and slant height l. For a right circular cone these form a right triangle, so l^2=r^2+h^2. The curved surface area uses l, while the volume uses h.
Worked Example: Cone Volume And Area
Question: A cone has radius 6 cm and perpendicular height 8 cm. Find its volume and total surface area.
- Slant height l=\sqrt{6^2+8^2}=10 cm.
- Volume =\frac13\pi(6^2)(8)=96\pi cm³.
- Total surface area =\pi(6)(10)+\pi(6^2)=96\pi cm².
Answer: Volume 96\pi\text{ cm}^3; total surface area 96\pi\text{ cm}^2.
Spheres And Hemispheres
A hemisphere has half the volume of a sphere. Its curved surface area is 2\pi r^2. If total surface area is required, add the circular base \pi r^2, giving 3\pi r^2.
Worked Example: Hemisphere
Question: A solid hemisphere has radius 7 cm. Find its volume and total surface area.
- Volume =\frac12\times\frac43\pi(7^3)=\frac{686}{3}\pi cm³.
- Curved area =2\pi(7^2)=98\pi cm².
- Base area =49\pi cm².
- Total area =147\pi cm².
Answer: Volume \frac{686}{3}\pi\text{ cm}^3; total surface area 147\pi\text{ cm}^2.
Finding A Missing Dimension
Substitute known values into the appropriate formula and rearrange. If the unknown is a radius, remember that it may be squared or cubed. Use a root only after isolating that power.
Worked Example: Radius From Volume
Question: A sphere has volume 288\pi cm³. Find its radius.
- \frac43\pi r^3=288\pi.
- Cancel \pi and multiply by 3/4: r^3=216.
- Take the cube root.
Answer: r=6 cm.
Compound Solids
Split a compound solid into standard solids. Add volumes for joined parts and subtract volumes for holes or removed parts. For surface area, include only exposed surfaces. A circular face hidden at a join is not part of the external surface.
Worked Example: Cylinder With Hemispherical End
Question: A capsule consists of a cylinder of radius 3 cm and length 10 cm with a hemisphere at each end. Find its volume.
- Cylinder volume =\pi(3^2)(10)=90\pi.
- Two hemispheres make one sphere: \frac43\pi(3^3)=36\pi.
- Add the volumes.
Answer: 126\pi\text{ cm}^3.
Frustums
A frustum can be treated as a large cone or pyramid with a smaller similar cone or pyramid removed. Determine the missing height using similarity if necessary, then subtract volumes. For surface area, subtract or calculate the appropriate curved areas and add exposed circular or polygonal ends.
Worked Example: Cone Frustum By Subtraction
Question: A large cone has radius 9 cm and height 12 cm. A similar small cone of radius 3 cm is removed from the top. Find the frustum volume.
- Linear scale factor small:large =3/9=1/3.
- Small cone height =12/3=4 cm.
- Large volume =\frac13\pi(9^2)(12)=324\pi.
- Small volume =\frac13\pi(3^2)(4)=12\pi.
- Subtract.
Answer: 312\pi\text{ cm}^3.
Symmetry Of Pyramids And Cones
The syllabus requires recognition of symmetry properties of prisms, cylinders, pyramids and cones. In this chapter the focus is on pyramids and cones. A right circular cone has infinitely many planes of symmetry through its central axis and one axis of rotational symmetry. A regular square-based pyramid has four vertical planes of symmetry and one rotational axis through the apex and centre of the base; its rotational order is 4.
| Solid | Planes of symmetry | Axis or order |
|---|---|---|
| Right circular cone | Infinitely many planes through the central axis | One axis; any rotation angle about it preserves the solid |
| Regular square-based pyramid | 4 vertical planes | One axis of rotational symmetry, order 4 |
| Regular triangular pyramid with equilateral base and apex above centre | 3 vertical planes | One rotational axis, order 3 |
Accuracy And Use Of Pi
If the question asks for an exact answer, leave \pi in the result. Otherwise use the calculator value of \pi and round at the end. Keep intermediate values unrounded, especially slant heights found with Pythagoras.
Examination Guidance
- Distinguish perpendicular height from slant height before selecting a formula.
- For total surface area, list every exposed face or surface before calculating.
- Keep answers in terms of \pi when requested.
- Use cubed units for volume and squared units for surface area.
- In compound solids, do not count internal joined faces as exposed area.
Common Mistakes
- Using slant height in the cone volume formula.
- Forgetting the circular base when total surface area of a cone is requested.
- Using 2\pi r^2 as total area of a hemisphere; this is only its curved area.
- Applying a length conversion factor directly to area or volume.
- Adding the volume of a removed section instead of subtracting it.
Knowledge Check And Practice
1. Find the volume of a pyramid with base area 54 cm² and height 10 cm.
2. Find the volume of a cone with radius 4 cm and height 9 cm.
3. A cone has radius 5 cm and height 12 cm. Find its slant height.
4. Find the total surface area of a sphere of radius 3 cm.
5. Find the volume of a hemisphere of radius 6 cm.
6. Convert 0.004\text{ m}^3 to cm³.
7. Why is the common circular face of two joined solids excluded from external surface area?
8. How many planes of symmetry does a right circular cone have?
