Perimeter, Area and Circles

Perimeter and area are among the most heavily tested ideas in OCR GCSE Mathematics (J560), because they appear in straightforward recall questions and inside longer problem-solving contexts about gardens, fields, packaging and floor plans. This lesson covers perimeter and area of the standard 2D shapes, the circumference and area of a circle, composite shapes, and — at Higher tier — arcs and sectors. The recurring exam skill is choosing the right formula, substituting carefully, and giving the answer with correct units. Recalling and using a formula is AO1; deciding how to split a composite shape is AO2; and applying area and perimeter to a real context is AO3.

Key Vocabulary

Term	Meaning
Perimeter	The total distance around the outside of a 2D shape
Area	The space inside a shape, in square units (cm$^{2}$2, m$^{2}$2)
Circumference	The perimeter of a circle
Arc	A portion of the circumference of a circle
Sector	The region between two radii and an arc
Composite shape	A shape made from two or more simpler shapes
Radius / Diameter	Centre to edge / right across through the centre ($d = 2r$d=2r)

Perimeter and Area of 2D Shapes

Perimeter is a length, so it is measured in centimetres or metres; area is a two-dimensional amount of surface, measured in square units such as cm$^{2}$2 or m$^{2}$2. Keeping that distinction clear stops the single most common mistake in the topic, which is finding one quantity when the question actually asks for the other. A quick way to stay on track is to read the units demanded by the question: if the mark scheme wants cm$^{2}$2 you are after area, and if it wants cm you are after perimeter. The standard area formulae are not printed on the J560 formula sheet, so you must commit them to memory and be able to apply them without prompting.

A rectangle has area $A = l \times w$A=l×w and perimeter $P = 2(l + w)$P=2(l+w). A triangle has area $A = \dfrac{1}{2} \times b \times h$A=21​×b×h, where it is essential that $h$h is the perpendicular height — the straight-line distance from the base to the opposite vertex, not the length of a sloping side. A parallelogram has area $A = b \times h$A=b×h, again using the perpendicular height between the two parallel sides. The area of a trapezium, $A = \dfrac{1}{2}(a + b)h$A=21​(a+b)h where $a$a and $b$b are the two parallel sides, is provided on the OCR formula sheet, but you should still be completely fluent with it because it appears so often. In every case, the units of the answer are the square of the length units used, so a rectangle measured in metres has its area in square metres.

Shape	Area	Perimeter
Rectangle	$A = l \times w$A=l×w	$P = 2(l + w)$P=2(l+w)
Triangle	$A = \dfrac{1}{2}bh$A=21​bh	sum of the three sides
Parallelogram	$A = bh$A=bh	sum of the four sides
Trapezium	$A = \dfrac{1}{2}(a + b)h$A=21​(a+b)h	sum of the four sides

Worked Example 1

A rectangle has length $12$12 cm and width $5$5 cm. Work out its perimeter and area.

Perimeter $= 2(12 + 5) = 2 \times 17 = 34$=2(12+5)=2×17=34 cm. Area $= 12 \times 5 = 60$=12×5=60 cm$^{2}$2.

Answer: perimeter $34$34 cm, area $60$60 cm$^{2}$2.

Worked Example 2

A triangle has base $8$8 cm and perpendicular height $6$6 cm. Work out its area.

Area $= \dfrac{1}{2} \times 8 \times 6 = 24$=21​×8×6=24 cm$^{2}$2.

Answer: $24$24 cm$^{2}$2.

Common error: using a slanted side as the height. The height must be perpendicular to the chosen base.

Worked Example 3

A trapezium has parallel sides $6$6 cm and $10$10 cm and perpendicular height $4$4 cm. Work out its area.

Area $= \dfrac{1}{2}(6 + 10) \times 4 = \dfrac{1}{2} \times 16 \times 4 = 32$=21​(6+10)×4=21​×16×4=32 cm$^{2}$2.

Answer: $32$32 cm$^{2}$2.

Worked Example 4

A triangle has area $48$48 cm$^{2}$2 and base $16$16 cm. Work out its perpendicular height.

Rearrange $A = \dfrac{1}{2}bh$A=21​bh: $48 = \dfrac{1}{2} \times 16 \times h = 8h$48=21​×16×h=8h, so $h = 6$h=6 cm.

Answer: $6$6 cm.

Working Backwards from a Given Area

Many higher-scoring questions give you the area and ask for a missing length, which means rearranging the formula rather than using it directly. The technique is to substitute the known area, then solve the resulting equation for the unknown. For a triangle you divide by $\dfrac{1}{2}$21​ and by the base; for a rectangle you divide the area by the known side; for a circle you divide by $\pi$π and then take a square root. Treating the area formula as an equation to be solved, rather than a one-way recipe, is exactly the algebra-into-geometry crossover that OCR likes to test, and it rewards candidates who set the working out line by line.

Worked Example 4b

A rectangle has area $96$96 cm$^{2}$2 and length $12$12 cm. Work out its width and hence its perimeter.

Width $= \dfrac{96}{12} = 8$=1296​=8 cm. Perimeter $= 2(12 + 8) = 40$=2(12+8)=40 cm.

Answer: width $8$8 cm, perimeter $40$40 cm.

Circumference and Area of a Circle

Every circle calculation comes from two formulae, both built on the special number $\pi$π (pi), which is approximately $3.14159$3.14159 and represents how many times the diameter fits around the circumference. The circumference — the distance all the way round — is $C = \pi d = 2\pi r$C=πd=2πr, and the area enclosed is $A = \pi r^{2}$A=πr2. These two are worth saying out loud until they are automatic: circumference uses the diameter (or twice the radius) to the first power, while area uses the radius squared. The figure shows the radius and diameter that the formulae depend on. The commonest slip by far is mixing up radius and diameter, because the diameter is twice the radius; so before you substitute anything, pause and identify exactly which length the question has given you. If you are told the diameter but the formula needs the radius, halve it first.

Worked Example 5

A circle has radius $5$5 cm. Work out its circumference to $1$1 decimal place.

$C = 2\pi r = 2 \times \pi \times 5 = 10\pi \approx 31.4$C=2πr=2×π×5=10π≈31.4 cm.

Answer: $31.4$31.4 cm.

Worked Example 6

A circle has diameter $14$14 cm. Work out its area to $1$1 decimal place.

First the radius: $r = 14 \div 2 = 7$r=14÷2=7 cm. Then $A = \pi r^{2} = \pi \times 7^{2} = 49\pi \approx 153.9$A=πr2=π×72=49π≈153.9 cm$^{2}$2.

Answer: $153.9$153.9 cm$^{2}$2.

Common error: substituting the diameter into $A = \pi r^{2}$A=πr2 without halving it first, which inflates the area fourfold.

Worked Example 7

The circumference of a circle is $40$40 cm. Work out its radius to $2$2 decimal places.

$C = 2\pi r$C=2πr, so $40 = 2\pi r$40=2πr and $r = \dfrac{40}{2\pi} = \dfrac{20}{\pi} \approx 6.37$r=2π40​=π20​≈6.37 cm.

Answer: $6.37$6.37 cm.

Worked Example 8

A circle has area $49\pi$49π cm$^{2}$2. Work out its circumference, giving your answer in terms of $\pi$π.

$\pi r^{2} = 49\pi$πr2=49π, so $r^{2} = 49$r2=49 and $r = 7$r=7 cm. Then $C = 2\pi r = 14\pi$C=2πr=14π cm.

Answer: $14\pi$14π cm.

Arcs and Sectors [H]

This section is Higher tier. A sector is a "pizza slice" of a circle bounded by two radii and an arc, and an arc is the curved part of the circumference between the ends of those radii. The key insight is that a sector is simply a fraction of the whole circle, and that fraction is decided by the centre angle: an angle of $\theta$θ degrees represents $\dfrac{\theta}{360}$360θ​ of a full turn, so the sector takes up exactly $\dfrac{\theta}{360}$360θ​ of the circle. Everything else follows from this one idea. Take that fraction of the whole circumference to get the arc length, $\dfrac{\theta}{360} \times 2\pi r$360θ​×2πr, and take the same fraction of the whole area to get the sector area, $\dfrac{\theta}{360} \times \pi r^{2}$360θ​×πr2. Because both formulae are just "fraction of the circle", you do not need to memorise them as separate facts if you remember the fraction. The figure shows a sector with its centre angle marked.