Add up the lengths of every side. For a compound shape with missing side lengths, use the fact that opposite/parallel sides in a rectilinear figure must add up consistently to find any missing lengths before summing.

Only the area of a triangle is provided in Cambridge's List of Formulas — the rest you must recall:

where for the trapezium, $a$ and $b$ are the two parallel sides and $h$ is the perpendicular distance between them.

Every one of these formulas relies on perpendicular height, not a slanted side length. Using a slanted side by mistake (rather than the shortest, perpendicular distance to the opposite side/base) is the most common source of wrong answers on this topic — always check that the height you're using meets the base at a right angle.

Identify the pair of parallel sides first (these become the "base(s)" in the formula), then find the perpendicular distance between them — often this height is given directly, or it needs a right-angled triangle (formed with a dropped perpendicular) solved first using Pythagoras or trigonometry.

Rearrange the relevant formula to make the unknown length the subject, then substitute the known area and other dimension. For a parallelogram with a known area and base, rearranging $A = b \times h$ gives $h = A \div b$ — this "rearrange first, then substitute" approach works for any of the four area formulas and is a common way Extended questions test the same formulas in reverse.

Area and perimeter of these basic shapes are the building blocks for the whole of E5: a sector (E5.3) is just a fraction of a circle's area, a prism's volume (E5.4) is its cross-sectional area multiplied by length, and a compound shape (E5.5) is nothing more than several of these basic areas added or subtracted together. Being fluent with rectangles, triangles, parallelograms and trapezia here means every later mensuration question becomes a case of "which shapes is this built from?" rather than learning a new formula each time.

A trapezium has parallel sides of $7\text{ cm}$ and $11\text{ cm}$, and a perpendicular height of $5\text{ cm}$ between them. Find its area.

1. $A = \tfrac{1}{2}(a+b)h = \tfrac12(7+11)(5)$
2. $A = \tfrac12(18)(5) = \tfrac12(90) = 45\text{ cm}^2$

A parallelogram has an area of $54\text{ cm}^2$ and a base of $9\text{ cm}$. Find its perpendicular height.

1. Rearrange $A = b \times h$ to make $h$ the subject: $h = A \div b$.
2. $h = 54 \div 9 = 6\text{ cm}$