Areas of rectangles, triangles, parallelograms and trapezia, and the volume and surface area of prisms, cylinders, cones, spheres and pyramids.

What this dot point is asking

AQA wants you to find the area of standard 2D shapes and the volume and surface area of standard 3D solids, choosing the right formula and keeping units consistent. Some formulae (cone, sphere, pyramid) are given on the AQA formulae sheet; the basic ones are not, so you must know them. These questions often combine shapes, so compound figures and "find the missing dimension" problems are common.

Areas of 2D shapes

The height in every triangle and parallelogram formula is the perpendicular height, not a slanted side. For a compound shape (an L-shape, say) split it into rectangles, find each area, and add them. To find a missing dimension, rearrange: a triangle of area $24\,\text{cm}^2$ and base $8\,\text{cm}$ has height $h = \dfrac{2 \times 24}{8} = 6\,\text{cm}$.

Volume of prisms and cylinders

A prism has the same cross-section all along its length. Its volume is the area of that cross-section multiplied by the length.

A triangular prism with a cross-sectional triangle of area $15\,\text{cm}^2$ and length $12\,\text{cm}$ has volume $15 \times 12 = 180\,\text{cm}^3$. A cylinder of radius $3\,\text{cm}$ and height $10\,\text{cm}$ has volume $\pi \times 3^2 \times 10 = 90\pi \approx 283\,\text{cm}^3$.

Surface area

Surface area is the total area of every face. For a cylinder, the surface is two circles plus the curved part: $2\pi r^2 + 2\pi r h$. For a cuboid, add the areas of all six rectangular faces. To unwrap a curved surface, picture the cylinder's side as a rectangle of width $2\pi r$ (the circumference) and height $h$.

Spheres, cones and pyramids

The formulae sheet provides the sphere volume $\tfrac{4}{3}\pi r^3$ and surface area $4\pi r^2$, the cone volume $\tfrac{1}{3}\pi r^2 h$ and curved surface area $\pi r l$ (where $l$ is the slant height), and the pyramid volume $\tfrac{1}{3} \times \text{base area} \times \text{height}$. For a cone, the slant height $l$ is found from the radius and vertical height by Pythagoras: $l = \sqrt{r^2 + h^2}$.

Compound solids and working backwards

Higher-tier questions often combine solids, such as a cylinder topped with a hemisphere, or ask you to work backwards from a given volume. For a compound solid, find each part's volume or surface area separately, then add (or subtract for a hole), taking care not to double-count the joining face: when a hemisphere sits on a cylinder, the flat circular base of the hemisphere is hidden, so it is not part of the total surface area. To work backwards, rearrange the formula: a sphere of volume $36\pi\,\text{cm}^3$ has $\tfrac{4}{3}\pi r^3 = 36\pi$, so $r^3 = 27$ and $r = 3\,\text{cm}$.

Units and conversion factors

A reliable source of lost marks is unit conversion, especially for area and volume. Because area is a length squared, $1\,\text{m}^2 = 100^2 = 10\,000\,\text{cm}^2$, not $100\,\text{cm}^2$. Because volume is a length cubed, $1\,\text{m}^3 = 100^3 = 1\,000\,000\,\text{cm}^3$. Capacity links to volume through $1\,\text{cm}^3 = 1\,\text{ml}$ and $1\,\text{litre} = 1000\,\text{cm}^3$. Converting consistently, and at the right power, before or after the calculation keeps these answers correct.