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C5.1 Units of measure

A metric conversion uses the linear scale factor once for length, squared for area and cubed for volume; the numerical value grows when the unit becomes smaller.

Quantity	Core equivalence	Conversion factor
length	$1\text{ m}=100\text{ cm}=1000\text{ mm}$ ; $1\text{ km}=1000\text{ m}$	use the linear factor
area	$1\text{ m}^2=10\,000\text{ cm}^2$ ; $1\text{ km}^2=1\,000\,000\text{ m}^2$	square the linear factor
volume	$1\text{ m}^3=1\,000\,000\text{ cm}^3$	cube the linear factor
capacity	$1\text{ litre}=1000\text{ ml}=1000\text{ cm}^3$ ; $1\text{ m}^3=1000\text{ litres}$	use the stated equivalence
mass	$1\text{ kg}=1000\text{ g}$	factor $1000$

First identify the quantity type from the unit exponent. Write one exact equivalence, raise its linear factor to the same exponent, then multiply when converting to smaller units or divide when converting to larger units. Keep the target unit beside the result and check that the size change is sensible.

To convert $0.17\text{ m}^2$ to $\text{cm}^2$ , use $1\text{ m}^2=100^2\text{ cm}^2$ : $0.17\times10\,000=1700\text{ cm}^2$ . To convert $8500\text{ cm}^3$ to litres, use $1000\text{ cm}^3=1$ litre: $8500\div1000=8.5$ litres.

Do not use a length factor unchanged for area or volume: $1\text{ m}^2$ is not $100\text{ cm}^2$ , and $1\text{ m}^3$ is not $100\text{ cm}^3$ . Capacity equivalences connect cubic units to litres; they are not ordinary linear conversions.

C5.2 Area and perimeter

Perimeter is the total distance around a shape; area is the surface enclosed. Choose the calculation from the quantity asked for, not just from the shape's name.

Shape	Area	Perimeter
rectangle, length $l$ , width $w$	$A=lw$	$P=2l+2w$
triangle, base $b$ , perpendicular height $h$	$A=\frac12bh$	add its three side lengths
parallelogram, base $b$ , perpendicular height $h$	$A=bh$	add all sides, or $2(a+b)$
trapezium, parallel sides $a,b$ , perpendicular height $h$	$A=\frac12(a+b)h$	add its four side lengths

Mark the parallel sides and the perpendicular height before substituting. Write the formula, insert values with consistent units, calculate, then attach linear units to perimeter or square units to area. If a dimension is unknown, form the area or perimeter equation first and solve it instead of guessing from the diagram.

A trapezium has parallel sides $7$ cm and $11$ cm and perpendicular height $5$ cm. Its area is $\frac12(7+11)\times5=45\text{ cm}^2$ . If a triangle has area $27\text{ cm}^2$ and base $6$ cm, then $\frac12\times6\times h=27$ , so $h=9$ cm.

A sloping side is not a height unless it is perpendicular to the chosen base. Area alone does not determine perimeter, so every boundary length must be known or derived. Keep compound shapes, circular arcs and sectors for their later Topics.

C5.3 Circles, arcs and sectors

A circle is controlled by one length: its radius $r$ . The diameter crosses the centre from edge to edge, so $d=2r$ . Circumference measures the boundary and area measures the enclosed surface.

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

First decide whether the given length is a radius or diameter. Convert units before substituting, keep $\pi$ on the calculator until the end, and attach linear units to circumference or square units to area. For an inverse problem, set the formula equal to the known value: divide by $2\pi$ to recover a radius from circumference, or divide by $\pi$ and then take the positive square root to recover a radius from area.

A circle has diameter $12$ cm. Then $r=6$ cm, so $C=12\pi$ cm and $A=36\pi\text{ cm}^2$ . If instead its area is $150\text{ cm}^2$ , then $r=\sqrt{150/\pi}=6.91\ldots$ cm. Use the requested accuracy; leave the answer in terms of $\pi$ when asked.

Do not square the diameter in $A=\pi r^2$ : halve it first. Circumference cannot be reported in square units, and area cannot be reported in linear units. Compound regions made from several circles belong to a later Topic.

A sector with central angle $\theta$ is the fraction $\theta/360$ of a full circle. The same fraction scales the full circumference to an arc length and the full area to a sector area.

Quantity	Full circle	Sector with angle $\theta$	Units
boundary along the curve	$2\pi r$	$\dfrac{\theta}{360}\times2\pi r$	length units
enclosed surface	$\pi r^2$	$\dfrac{\theta}{360}\times\pi r^2$	square units

For Core, use sector angles that divide $360^\circ$ exactly. Write the fraction $\theta/360$ , choose circumference for an arc or circle area for a sector, then multiply. If the total perimeter of a sector is required, add the two radii to the arc length; those straight edges are not part of the arc.

For $r=9$ cm and $\theta=72^\circ$ , the sector is $72/360=1/5$ of the circle. Arc length $=\frac15(2\pi\times9)=\frac{18\pi}{5}$ cm, while sector area $=\frac15(\pi\times9^2)=\frac{81\pi}{5}\text{ cm}^2$ . Its total perimeter would be $18+\frac{18\pi}{5}$ cm.

Use the angle at the centre, not an angle on the circumference. Scaling the radius by $\theta/360$ is not valid: scale the completed circumference or area formula. This Core objective does not introduce unrestricted or major-sector calculations.

C5.4 Surface area and volume

Surface area totals the exposed two-dimensional faces of a solid; volume measures the three-dimensional space inside it. Choose the quantity first, then identify the dimensions the matching relationship needs.

Solid	Total surface area	Volume
cuboid $l\times w\times h$	$2(lw+lh+wh)$	$lwh$
prism, cross-section area $B$ , perimeter $P$ , length $L$	$2B+PL$	$BL$
cylinder, radius $r$ , height $h$	$2\pi r^2+2\pi rh$	$\pi r^2h$
sphere, radius $r$	$4\pi r^2$	$\frac43\pi r^3$
pyramid, base area $B$ , perpendicular height $h$	base plus every triangular face	$\frac13Bh$
cone, radius $r$ , perpendicular height $h$ , slant height $s$	$\pi r^2+\pi rs$	$\frac13\pi r^2h$

For a surface-area problem, list each exposed face or curved surface once; a closed cylinder and cone include their circular bases. For volume, a uniform prism—including a cylinder—is cross-section area multiplied by length, while a pyramid or cone has one third of the volume of a matching prism with the same base and perpendicular height. Keep every length in one unit before calculating, then use square units for surface area and cubic units for volume. For an unknown dimension, write the full equation and rearrange.

A closed cylinder has radius $3$ cm and height $8$ cm. Its total surface area is $2\pi(3)^2+2\pi(3)(8)=66\pi\text{ cm}^2$ , while its volume is $\pi(3)^2(8)=72\pi\text{ cm}^3$ . Leaving $\pi$ exact keeps both values accurate until a decimal is requested.

A cone's slant height $s$ belongs in curved surface area, but its perpendicular height $h$ belongs in volume. Do not include internal joins in surface area. Compound solids, removed pieces and fractions of solids belong to the next Topic.

C5.5 Compound shapes and parts of shapes

A compound shape is built from familiar pieces or from a familiar shape with a piece removed. Area combines whole regions; perimeter follows only the final outside boundary.

Quantity	Reliable construction	What not to count
area	add non-overlapping component areas, or enclosing area minus removed areas	overlaps twice
perimeter	trace the outside edge once and add its straight and curved lengths	shared or internal edges

First mark missing lengths using totals, equal sides, radii or diameters. Choose a decomposition whose pieces do not overlap. Write every component area before adding or subtracting; for a shaded region, identify clearly which area is removed. For perimeter, start at one point and travel once around the finished shape, recording each boundary segment in order. Use arc length—not circle area—for curved edges, and keep units consistent.

A $10$ cm by $6$ cm rectangle has a $4$ cm by $2$ cm corner removed. The remaining area is $10\times6-4\times2=52\text{ cm}^2$ . Tracing its six outside edges gives $10+4+4+2+6+6=32$ cm. The two cut edges must be included because they become exposed, but the removed corner is not part of the area.

Do not add the perimeters of separate pieces: their shared edges would be counted even though they are inside the final shape. Area uses square units; perimeter uses linear units. Three-dimensional solids are handled in the next card.

For a compound solid, volume combines the space occupied by its components, while surface area counts only surfaces exposed after the components are joined or cut.

Situation	Volume	Surface area
solids joined	add component volumes	add only exposed faces; omit both sides of each join
piece removed or empty space	whole volume minus removed volume	include any new cut faces that become exposed
fraction of a solid	multiply the full volume by the fraction	take the relevant curved fraction and include exposed cut faces

Draw up a component list before calculating. Keep one unit throughout, use the correct full-solid formula for each component, and preserve $\pi$ until the end. A join does not change total volume, but it hides matching surfaces. For an unknown length, build the combined or remaining volume equation first and then rearrange.

A cylinder of radius $3$ cm and height $5$ cm is joined to a hemisphere of the same radius. Volume $=\pi(3)^2(5)+\frac12\left(\frac43\pi(3)^3\right)=63\pi\text{ cm}^3$ . Its exposed surface area is cylinder curve $30\pi$ , one circular base $9\pi$ , and hemisphere curve $18\pi$ , giving $57\pi\text{ cm}^2$ . The circular join is internal, so it is not counted.

For an isolated hemisphere, the flat circular cut face may be exposed; at a cylinder–hemisphere join it is hidden. Never subtract a joined face from volume. Frustums and other Extended-only developments are outside this Core Topic.

5. Mensuration

Exam analysis

Most tested topics

C5.1 Units of measure

Convert metric units by dimension

C5.2 Area and perimeter

Calculate area and perimeter of basic shapes

C5.3 Circles, arcs and sectors

Calculate with circumference and circle area

Calculate arc length and sector area

C5.4 Surface area and volume

Calculate surface area and volume of solids

C5.5 Compound shapes and parts of shapes

Solve compound area and perimeter problems

Solve compound solid and partial-solid problems