Area of triangles, parallelograms and trapeziums; circumference and area of circles; volume and surface area of prisms and cylinders; and the volume and surface area of cones, spheres and pyramids (Higher tier).

What this dot point is asking

Edexcel expects you to find the area of 2D shapes, the circumference and area of circles, and the surface area and volume of 3D solids, from prisms and cylinders up to cones, spheres and pyramids at Higher tier. Some formulae are given on the formulae sheet, but you must know how to select and apply them, and which dimensions to use.

Area of 2D shapes

The core area formulae apply to the most common straight-sided shapes.

The perpendicular height matters: in a triangle or parallelogram, use the height at right angles to the base, not a slanting side. Compound shapes are split into rectangles and triangles, found separately, then added.

Circumference and area of a circle

A circle is defined by its radius $r$ (centre to edge) or diameter $d = 2r$.

So a circle of radius $5\,\text{cm}$ has circumference $2\pi \times 5 = 10\pi \approx 31.4\,\text{cm}$ and area $\pi \times 5^2 = 25\pi \approx 78.5\,\text{cm}^2$. Questions often ask for the answer "in terms of $\pi$", which means leaving $\pi$ in rather than rounding.

Volume and surface area of prisms and cylinders

A prism has the same cross-section all along its length. A cylinder is a prism with a circular cross-section.

Cones, spheres and pyramids (Higher)

Higher tier adds curved solids, whose formulae are on the formulae sheet but must be applied correctly.

The key formulae are: sphere volume $\tfrac{4}{3}\pi r^3$ and surface area $4\pi r^2$; cone volume $\tfrac{1}{3}\pi r^2 h$ and curved surface area $\pi r l$ (where $l$ is the slant height); pyramid volume $\tfrac{1}{3} \times \text{base area} \times \text{height}$.

Surface area and compound solids

Surface area is the total area of every face or curved surface, so the method is to find each surface separately and add them. For a closed cylinder, that is the two circular ends ($2 \times \pi r^2$) plus the curved surface, which "unrolls" into a rectangle of width $2\pi r$ (the circumference) and height $h$, giving $2\pi r h$. For a cone, the surface area is the base circle $\pi r^2$ plus the curved surface $\pi r l$. Compound solids, such as a cylinder topped by a hemisphere, are handled by adding the relevant volumes or surface areas, taking care not to double-count a hidden face where two solids join. Working in terms of $\pi$ until the final line keeps the arithmetic exact.

Try this

Q1. Work out the area of a trapezium with parallel sides $6\,\text{cm}$ and $10\,\text{cm}$ that are $4\,\text{cm}$ apart. [2 marks]

• Cue. $\tfrac{1}{2}(6 + 10) \times 4 = \tfrac{1}{2} \times 16 \times 4 = 32\,\text{cm}^2$.

Q2. A triangular prism has a cross-section of area $15\,\text{cm}^2$ and length $9\,\text{cm}$. Work out its volume. [2 marks]

• Cue. Volume $=$ cross-section $\times$ length $= 15 \times 9 = 135\,\text{cm}^3$.