Mensuration: Perimeter, Area & Volume

Why mensuration matters

Mensuration is the branch of geometry that measures lengths, areas and volumes of shapes. On the 4MA1 Higher paper you will be asked to find the perimeter of a compound shape, the surface area of a cylinder, the volume of a cone, or how a volume changes when a shape is enlarged. The formulae are not all given to you, so learning them — and knowing when each applies — is essential marks.

Key terms Perimeter is the total distance around the outside of a 2D shape.

Area is the amount of surface a 2D shape covers, measured in square units (cm$^2$, m$^2$).

Volume is the space a 3D solid fills, measured in cubic units (cm$^3$, m$^3$).

Surface area is the total area of all the faces of a solid.

Area of 2D shapes

The four "polygon" formulae you must memorise:

Here $h$ is always the perpendicular height, not a slanted side.

Watch out For a triangle or parallelogram the height must be measured at a right angle to the base. If a question gives you a sloping side, it is a trap — only use it for perimeter, not area.

Worked example A trapezium has parallel sides $a = 6$ cm and $b = 10$ cm, with perpendicular height $h = 4$ cm.

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

Circles: circumference and area

For a circle of radius $r$ (diameter $d = 2r$):

$$C = 2\pi r = \pi d \qquad A = \pi r^2$$

Exam tip Keep answers in terms of $\pi$ until the final line, then round. If a question says "give your answer to 3 significant figures", use the $\pi$ button — never $3.14$, which loses accuracy.

Arc length and sector area

A sector is a "pizza slice" bounded by two radii and an arc, with angle $\theta$ at the centre. Take the fraction $\dfrac{\theta}{360}$ of the whole circle:

$$\text{Arc length} = \frac{\theta}{360}\times 2\pi r \qquad \text{Sector area} = \frac{\theta}{360}\times \pi r^2$$

Worked example A sector has radius $r = 9$ cm and angle $\theta = 80^\circ$.

Arc length $= \dfrac{80}{360}\times 2\pi \times 9 = \dfrac{2}{9}\times 18\pi = 4\pi \approx 12.6$ cm.

Sector area $= \dfrac{80}{360}\times \pi \times 9^2 = \dfrac{2}{9}\times 81\pi = 18\pi \approx 56.5$ cm$^2$.

The full perimeter of the sector also includes the two radii: $12.6 + 9 + 9 = 30.6$ cm.

Compound shapes

Break a compound shape into pieces you recognise, then add or subtract.

Worked example A running track end is a rectangle $20 \times 14$ m with a semicircle (radius $7$ m) on one end.

Area $= (20\times 14) + \tfrac{1}{2}\pi (7)^2 = 280 + 76.97 = 357$ m$^2$ (3 s.f.).

Surface area and volume of solids

For any prism (a solid with a constant cross-section):

$$\text{Volume} = \text{area of cross-section} \times \text{length}$$

The headline solid formulae:

In the cone, $l$ is the slant height and $h$ is the perpendicular height; they are linked by Pythagoras: $l^2 = r^2 + h^2$.

Exam tip The cone and sphere formulae are given on the 4MA1 formula sheet, but the cylinder and prism are not. Learn the cylinder cold — it appears almost every series.

Worked example A cone has base radius $r = 6$ cm and perpendicular height $h = 8$ cm.

Volume $= \tfrac{1}{3}\pi r^2 h = \tfrac{1}{3}\pi (6)^2 (8) = \tfrac{1}{3}\pi \times 288 = 96\pi \approx 302$ cm$^3$.

For surface area, find the slant height: $l = \sqrt{6^2 + 8^2} = \sqrt{100} = 10$ cm.

Curved surface $= \pi r l = \pi \times 6 \times 10 = 60\pi$. Total $= \pi r^2 + \pi r l = 36\pi + 60\pi = 96\pi \approx 302$ cm$^2$.

Worked example A sphere has radius $r = 5$ cm.

Volume $= \tfrac{4}{3}\pi (5)^3 = \tfrac{4}{3}\pi \times 125 = \tfrac{500}{3}\pi \approx 524$ cm$^3$.

Converting area and volume units

This is one of the most common slips on the paper. Lengths scale by the conversion factor; areas scale by its square and volumes by its cube.

Worked example Convert $3.5$ m$^2$ to cm$^2$: $3.5 \times 10\,000 = 35\,000$ cm$^2$.

Convert $2\,000\,000$ cm$^3$ to m$^3$: $2\,000\,000 \div 1\,000\,000 = 2$ m$^3$.

Similar shapes: the $k$, $k^2$, $k^3$ rule

Two shapes are similar if one is an enlargement of the other. If the length scale factor is $k$, then:

every length (and perimeter) is multiplied by $k$,
every area (and surface area) is multiplied by $k^2$,
every volume is multiplied by $k^3$.

Key terms The length scale factor $k$ is found by dividing a length on the larger shape by the matching length on the smaller. Work backwards: area scale factor $k^2 \Rightarrow k = \sqrt{k^2}$; volume scale factor $k^3 \Rightarrow k = \sqrt[3]{k^3}$.

Worked example Two similar bottles have heights $12$ cm and $18$ cm. The small bottle holds $400$ ml.

Length scale factor $k = \dfrac{18}{12} = 1.5$.

Volume scale factor $= k^3 = 1.5^3 = 3.375$.

Large bottle volume $= 400 \times 3.375 = 1350$ ml.

Worked example Two similar tins have surface areas $50$ cm$^2$ and $200$ cm$^2$. The small tin has radius $3$ cm.

Area scale factor $= \dfrac{200}{50} = 4$, so $k = \sqrt{4} = 2$.

Large tin radius $= 3 \times 2 = 6$ cm.

Real world Scale factors explain why a baby elephant cannot simply be a shrunken adult: doubling every length multiplies weight (volume) by $2^3 = 8$, but bone strength (cross-sectional area) only by $2^2 = 4$. Big animals need proportionally thicker legs.

Formulae summary

Exam tip Always write the units and check they match the dimension: a length answer in cm, area in cm$^2$, volume in cm$^3$. Mismatched units are an easy way to lose the final mark on an otherwise perfect solution.