Calculate perimeter and area of 2-D shapes including circles, sectors and compound shapes, and the surface area and volume of prisms, cylinders, pyramids, cones and spheres, with appropriate units.

What this dot point is asking

This is the mensuration core of WJEC geometry: finding perimeter and area of two-dimensional shapes, and surface area and volume of three-dimensional solids. You need the area formulae for rectangles, triangles, parallelograms and trapezia, the circle formulae for circumference and area, sector arc length and area, and the volume and surface-area formulae for prisms, cylinders, pyramids, cones and spheres. Compound shapes, where you split a figure into known parts, are a recurring exam style. Working in the right units, and converting between them, secures the final mark.

Area of 2-D shapes

The standard plane shapes each have a formula.

For a compound shape, split it into rectangles and triangles, find each area, then add (or subtract a cut-out). Perimeter is the total distance around the outside, so add every boundary length, taking care with the curved parts of any circular section.

Circles and sectors

The circle formulae use the radius $r$.

The circumference is $C = 2\pi r$ (or $\pi d$ using the diameter) and the area is $A = \pi r^2$. A sector is a "slice" with angle $\theta$ at the centre, taking the fraction $\tfrac{\theta}{360^\circ}$ of the whole circle: its arc length is $\tfrac{\theta}{360^\circ}\times 2\pi r$ and its area is $\tfrac{\theta}{360^\circ}\times \pi r^2$. A semicircle is the special case $\theta = 180^\circ$ (half) and a quarter circle is $\theta = 90^\circ$.

Volume of solids

Volume measures the space inside a 3-D solid, in cubic units.

So a triangular prism with cross-section area $20$ cm$^2$ and length $9$ cm has volume $20 \times 9 = 180$ cm$^3$. The cone and pyramid both carry the $\tfrac{1}{3}$ factor, which is easy to forget.

Surface area

Surface area is the total area of all the faces, found by adding them.

For a cone, the curved surface area is $\pi r l$ (where $l$ is the slant height) and you add the base $\pi r^2$ for a solid cone; for a sphere the surface area is $4\pi r^2$.

Units and conversions

Choosing and converting units correctly is part of the marks.

Area uses square units, so $1$ m$^2 = 100^2 = 10\,000$ cm$^2$, and volume uses cubic units, so $1$ m$^3 = 100^3 = 1\,000\,000$ cm$^3$. Capacity links to volume: $1$ cm$^3 = 1$ ml and $1$ litre $= 1000$ cm$^3$. Always convert all lengths to the same unit before substituting into a formula, and give the answer with the matching square or cubic unit.

Why this matters

Mensuration is one of the most heavily assessed geometry strands, appearing in pure calculation, in real-life contexts (volumes of containers, areas of land) and combined with Pythagoras and trigonometry on harder problems. The marks reward selecting the right formula, substituting accurately and, crucially, stating units. Compound shapes and "in terms of $\pi$" answers are WJEC favourites, and reverse problems (find the radius given the volume) test the same formulae rearranged, linking this topic back to algebra.