Area, Perimeter and Volume

These formulae must be known:

Shape	Formula	Notes
Rectangle	$A=lw$
Triangle	$A=\frac{1}{2}bh$	$h$ is perpendicular height
Parallelogram	$A=bh$	$h$ is perpendicular height, not slant side
Trapezium	$A=\frac{1}{2}(a+b)h$	$a$ and $b$ are parallel sides, $h$ perpendicular height
Circle	$A=\pi r^2$

Worked example — find the area of a trapezium with parallel sides 8 cm and 12 cm and perpendicular height 5 cm.

$A=\frac{1}{2}(8+12)\times 5=\frac{1}{2}\times 20\times 5=50$ cm² ✓

Composite shapes — split into simpler shapes, find each area, add (or subtract if a piece is cut out).

Worked example — an L-shape: 10 cm × 6 cm rectangle with a 4 cm × 3 cm corner removed. Area $=60-12=48$ cm².

A prism has a constant cross-section throughout its length.

$V=A_{\text{cross-section}}\times l$

Prism	Formula
Cuboid	$V=lwh$
Cylinder	$V=\pi r^{2}h$
Triangular prism	$V=\frac{1}{2}bh\times l$

Surface area of a cylinder:

$SA=2\pi r^2+2\pi rh=2\pi r(r+h)$

Worked example — cylinder with radius 4 cm, height 9 cm:

$V=\pi \times 16\times 9=144\pi \approx 452.4$ cm³

$SA=2\pi (4)(4+9)=2\pi \times 52=104\pi \approx 326.7$ cm² ✓

Circle formulae:

$C=2\pi r=\pi dA=\pi r^2$

Perimeter of composite shapes involving circles: include straight edges plus arc lengths; do not use the full circumference unless the question involves a complete circle.

Worked example — find the perimeter and area of a semicircle of radius 6 cm.

Perimeter $=\pi r+2r=6\pi +12\approx 30.85$ cm ✓

Area $=\frac{1}{2}\pi r^2=18\pi \approx 56.55$ cm² ✓

Bearings (G15): measured clockwise from north, written as three digits (e.g., $045°$, $270°$). To find the bearing from $B$ back to $A$ given the bearing from $A$ to $B$: add or subtract $180°$.

These formulae are given on the exam formula sheet but must be applied correctly:

Solid	Volume	Surface area
Sphere	$V=\frac{4}{3}\pi r^3$	$SA=4\pi r^2$
Cone	$V=\frac{1}{3}\pi r^{2}h$	$SA=\pi rl+\pi r^2$ ($l$ = slant height)
Pyramid	$V=\frac{1}{3}\times \text{base area}\times h$	sum of base + triangular faces

Worked example — cone:

Radius 5 cm, perpendicular height 12 cm. Slant height: $l=\sqrt{5^2+12^2}=\sqrt{169}=13$ cm.

$V=\frac{1}{3}\pi \times 25\times 12=100\pi \approx 314.2$ cm³

$SA=\pi \times 5\times 13+\pi \times 25=65\pi +25\pi =90\pi \approx 282.7$ cm² ✓

Composite solids — add volumes of component parts (e.g., a hemisphere placed on top of a cylinder).

Unit conversion for area and volume:

$1{\text{ m}}^2=10000{\text{ cm}}^2$ (multiply by $100^2$)

$1{\text{ m}}^3=1000000{\text{ cm}}^3$ (multiply by $100^3$)

Scale drawings (G15): a scale of $1:n$ means every 1 unit on the drawing represents $n$ units in reality.

Worked example — find the real area of a room drawn at scale $1:50$, where the drawing shows $6\text{ cm}\times 8\text{ cm}$.

Real dimensions: $6\times 50=300$ cm and $8\times 50=400$ cm.

Real area $=300\times 400=120000$ cm² $=12$ m² ✓

Note: the area scale factor is $50^2=2500$; drawing area $=48$ cm²; real area $=48\times 2500=120000$ cm² ✓

1. Parallelogram — using the slant side as height

Area of parallelogram $=b\times h$, where $h$ is the perpendicular height (the vertical drop between the parallel sides). Using the slant side length instead gives a larger, incorrect answer.

2. Cone surface area — forgetting the base

Total surface area $=\pi rl+\pi r^2$. Omitting $\pi r^2$ (the circular base) is a common error. Check whether the question asks for total surface area or curved surface area only.

3. Unit conversion for area — multiplying by the length scale factor

$1{\text{ m}}^2=10000{\text{ cm}}^2$ (not 100). The length conversion (×100) must be squared because area is two-dimensional.

4. Composite shapes — double-counting shared boundaries

When finding the surface area of a composite solid (e.g., a cone placed on a cylinder), the circular contact face between the two solids is internal and must not be counted.

Mistake	Correction
"Volume of sphere radius 3: $\frac{4}{3}\pi \times 3=4\pi$"	$V=\frac{4}{3}\pi \times 3^3=36\pi$ — cube the radius, not multiply
"Trapezium with parallel sides 5 and 7, height 4: area = $(5+7)\times 4=48$"	$A=\frac{1}{2}(5+7)\times 4=24$ — include the $\frac{1}{2}$
"Cylinder SA: $2\pi rh$ only"	Total SA $=2\pi r^2+2\pi rh$ — include both circular ends