This note covers the area of flat shapes, the arc length and area of a sector, and the volume and surface area of prisms, cylinders, pyramids, cones and spheres, including compound solids and frustums. It also covers converting units of area and volume. Mensuration is used in design, packaging, construction and any problem that asks how much material or space is needed.
Key ideas
Area of flat shapes.
- Rectangle: \(A = lw\). Triangle: \(A = \tfrac12 bh\), where \(h\) is the perpendicular height.
- Parallelogram: \(A = bh\), again using the perpendicular height, not the slanted side.
- Trapezium: \(A = \tfrac12(a+b)h\), where \(a\) and \(b\) are the parallel sides.
- Circle: \(A = \pi r^2\) and circumference \(C = 2\pi r = \pi d\).
- For a compound shape, split it into simple shapes, or subtract a cut-out from a larger shape.
Arcs and sectors. A sector with angle \(\theta\) (in degrees) is the fraction \(\dfrac{\theta}{360}\) of the whole circle. \[ \text{arc length} = \frac{\theta}{360} \times 2\pi r, \qquad \text{sector area} = \frac{\theta}{360} \times \pi r^2 \] The perimeter of a sector is the arc length plus two radii.
Volume of solids.
- Prism (any constant cross-section): \(V = \text{area of cross-section} \times \text{length}\).
- Cylinder: \(V = \pi r^2 h\).
- Pyramid: \(V = \tfrac13 \times \text{base area} \times \text{perpendicular height}\).
- Cone: \(V = \tfrac13 \pi r^2 h\), using the vertical height \(h\).
- Sphere: \(V = \tfrac43 \pi r^3\). A hemisphere has half of this.
Surface area.
- Prism: add the areas of all faces.
- Cylinder: curved surface \(2\pi r h\), plus \(2\pi r^2\) for two circular ends (if closed).
- Cone: curved surface \(\pi r l\), where \(l\) is the slant height, plus \(\pi r^2\) for the base. Find \(l\) from \(l^2 = r^2 + h^2\).
- Sphere: \(4\pi r^2\). A solid hemisphere has curved surface \(2\pi r^2\) and a flat face of \(\pi r^2\).
- Pyramid: area of the base plus the areas of the triangular faces.
Frustum (Higher). A frustum is what remains when a small cone is cut from the top of a larger cone by a plane parallel to the base. Its volume is the volume of the large cone minus the volume of the small cone. The two cones are similar, so use ratios of lengths to find the missing radius or height.
Units. Length units convert by a factor, area units by the factor squared, and volume units by the factor cubed. For example, \(1\text{ m} = 100\text{ cm}\), so \(1\text{ m}^2 = 10\,000\text{ cm}^2\) and \(1\text{ m}^3 = 1\,000\,000\text{ cm}^3\). Also, \(1\text{ litre} = 1000\text{ cm}^3\).
Answer form. If the question says "give your answer in terms of \(\pi\)", leave \(\pi\) in the answer, for example \(96\pi\) cm³. Otherwise use the \(\pi\) button on your calculator and give 3 significant figures.
Definitions and formulae
| Term or formula | Meaning / when to use |
|---|---|
| \(\tfrac12(a+b)h\) | Area of a trapezium. |
| \(\pi r^2,\ 2\pi r\) | Area and circumference of a circle. |
| \(\dfrac{\theta}{360}\times 2\pi r\) | Arc length for a sector of angle \(\theta\). |
| \(\dfrac{\theta}{360}\times \pi r^2\) | Area of a sector. |
| \(V = \pi r^2 h\) | Volume of a cylinder. |
| \(V = \tfrac13 \pi r^2 h\) | Volume of a cone. |
| \(V = \tfrac13 Ah\) | Volume of a pyramid with base area \(A\). |
| \(V = \tfrac43\pi r^3,\ S = 4\pi r^2\) | Volume and surface area of a sphere. |
| \(S = \pi r l\) | Curved surface area of a cone, with slant height \(l\). |
| \(S = 2\pi r h\) | Curved surface area of a cylinder. |
Worked examples
Example 1 A sector of a circle has radius 9 cm and angle \(140^\circ\). Find its arc length, its area, and its perimeter.
The sector is \(\dfrac{140}{360}\) of the circle.
Arc length \(= \dfrac{140}{360} \times 2\pi \times 9 = \dfrac{140}{360} \times 56.55 = 22.0\) cm (3 s.f.).
Area \(= \dfrac{140}{360} \times \pi \times 9^2 = \dfrac{140}{360} \times 254.47 = 99.0\) cm² (3 s.f.).
Perimeter \(=\) arc length \(+\) two radii \(= 21.99 + 18 = 40.0\) cm (3 s.f.).
Answer: arc 22.0 cm, area 99.0 cm², perimeter 40.0 cm
Example 2 A conical funnel has base radius 7 cm and vertical height 24 cm. Find (a) its slant height, (b) its volume, (c) its total surface area including the base.
(a) \(l^2 = 7^2 + 24^2 = 49 + 576 = 625\), so \(l = 25\) cm.
(b) \(V = \tfrac13 \pi r^2 h = \tfrac13 \times \pi \times 49 \times 24 = 392\pi = 1230\) cm³ (3 s.f.).
(c) Curved surface \(= \pi r l = \pi \times 7 \times 25 = 175\pi\).
Base \(= \pi r^2 = 49\pi\).
Total \(= 175\pi + 49\pi = 224\pi = 704\) cm² (3 s.f.).
Answer: (a) 25 cm (b) \(392\pi \approx 1230\) cm³ (c) \(224\pi \approx 704\) cm²
Example 3 A storage tank is a cylinder of radius 30 cm and height 100 cm with a solid hemisphere on top, so the whole solid is closed. Find (a) the volume in litres, (b) the total outside surface area in m².
(a) Cylinder: \(\pi \times 30^2 \times 100 = 90\,000\pi\) cm³.
Hemisphere: \(\tfrac12 \times \tfrac43 \pi \times 30^3 = \tfrac23 \pi \times 27\,000 = 18\,000\pi\) cm³.
Total \(= 108\,000\pi = 339\,292\) cm³. Divide by 1000 to convert to litres: 339 litres (3 s.f.).
(b) Curved surface of the cylinder: \(2\pi \times 30 \times 100 = 6000\pi\).
Flat circular base: \(\pi \times 30^2 = 900\pi\).
Curved surface of the hemisphere: \(2\pi \times 30^2 = 1800\pi\).
Total \(= 8700\pi = 27\,332\) cm². Divide by \(10\,000\) to convert to m²: 2.73 m² (3 s.f.).
The top of the cylinder is joined to the hemisphere, so it is not part of the outside surface.
Answer: (a) 339 litres (b) 2.73 m²
Common errors
- Common error: using the slant height in a volume formula, or the vertical height in the curved surface area of a cone.
- Common error: using the diameter instead of the radius in \(\pi r^2\) and \(\tfrac43\pi r^3\).
- Common error: converting area and volume units with the length factor. \(1\text{ m}^2\) is \(10\,000\text{ cm}^2\), not 100 cm².
- Common error: leaving out the flat face (or including a face that is hidden inside a compound solid) when finding surface area.
- Common error: using the slanted side of a parallelogram or trapezium as the height.
Exam tip: For compound solids, list each part with its formula on a separate line. Write the surface pieces you are including, and add them at the end.
Check yourself
- Find the area of a trapezium with parallel sides 9 cm and 15 cm, and a distance of 6 cm between them.
Answer
\(\tfrac12 (9 + 15) \times 6 = 12 \times 6 = 72\) cm². - Find the volume and surface area of a sphere of radius 3.5 cm.
Answer
\(V = \tfrac43 \pi (3.5)^3 = 180\) cm³ (3 s.f.). \(S = 4\pi (3.5)^2 = 154\) cm² (3 s.f.). - A rectangular tabletop has an area of 2.5 m². Give its area in cm².
Answer
\(2.5 \times 10\,000 = 25\,000\) cm². - A cone of height 12 cm and base radius 9 cm has a small cone of height 4 cm cut from its top (parallel to the base). Find the volume of the frustum that remains, in terms of \(\pi\) (Higher).
Answer
The small cone is similar with scale factor \(\tfrac{4}{12} = \tfrac13\), so its radius is 3 cm. Large cone: \(\tfrac13 \pi (81)(12) = 324\pi\). Small cone: \(\tfrac13 \pi (9)(4) = 12\pi\). Frustum \(= 312\pi\) cm³, which is about 980 cm³.