Mini-Lesson

Geometry & Measures

11 missions. Guess, learn, then check. Your place is saved.

Mission 1Why can tiles be squares or hexagons but never regular pentagons?
Mission 2Two lookouts see one fire. How do two bearings pinpoint it?
Mission 3Why do runners in outer lanes start further ahead?
Mission 4Roll a sheet into a tall tube or a wide one. Which holds more?
Mission 5Will a 1.2 m umbrella fit flat in a 1 m by 0.8 m box?
Mission 6How can you find a tree's height with a protractor and a tape?
Mission 7Two bearings, 10 km apart: how far away is the lighthouse?
Mission 8A picture was spun round one nail. Can you find where the nail is?
Mission 9A giant ape 10 times a gorilla's height: why would its legs snap?
Mission 10Draw a triangle on a diameter. Why is the top corner always 90°?
Mission 11A drone flies 3 east, 4 north, then 6 east, 8 north. Straight line?

The specification points and sources are listed under the lesson: see Sources.

Mission 1

Why can tiles be squares or hexagons but never regular pentagons?

Make a guess: Why can tiles be squares or hexagons but never regular pentagons?

Geometry · angles and polygons

Angle facts and the reasons to write

Name angles with three letters: angle $A BC$ ABC is the angle at $B$ B.
Every angle you find needs a reason written in words. Examiners do not accept "Z angles" or "F angles": use the proper names.

Fact	Reason to write
Angles round a point add to 360°	angles at a point sum to 360°
Angles on a straight line add to 180°	angles on a straight line sum to 180°
Opposite angles at a crossing are equal	vertically opposite angles are equal
Parallel lines: the "Z" pair are equal	alternate angles are equal
Parallel lines: the "F" pair are equal	corresponding angles are equal
Parallel lines: the "C" pair add to 180°	co-interior angles sum to 180°

Worked example Geometry · angles and polygons

Angles on parallel lines, with reasons

Question

In the diagram,

A B

is parallel to

C D

. Angle

EGB = 72°

. Work out angle

x

(angle

G H C

). Give a reason for each step.

1
Use the straight line $EF$ EF: angle $BG H = 180° - 72° = 108°$ BGH = 180° − 72° = 108° (angles on a straight line sum to 180°)
2
Angles $BG H$ BGH and $G H C$ GHC are a "Z" pair between the parallel lines: angle $G H C = 108°$ GHC = 108° (alternate angles are equal)
3
Check another way: angle $A G H = 72°$ AGH = 72° (vertically opposite angle $EGB$ EGB), and $72° + 108° = 180°$ 72° + 108° = 180° for the co-interior pair

Answer

x = 108°

⚠️ Watch out

writing only numbers, or "Z angles", loses the reasoning marks even when the angle is right.

Quick check

?Angles

p

and

q

lie between two parallel lines, on the same side of a line crossing them.

p = 74°

. What is

q

?

Geometry · angles and polygons

Triangles and quadrilaterals

Angles in a triangle sum to 180°. Proof: draw a line through one corner parallel to the opposite side; the two alternate angles and the corner angle make a straight line.
An exterior angle of a triangle equals the sum of the two interior opposite angles.
Isosceles: two equal sides and two equal base angles. Equilateral: three equal sides, every angle 60°.
Angles in a quadrilateral sum to 360°: a diagonal splits it into two triangles.

Quadrilateral	Key properties	Lines of symmetry	Rotational order
Square	4 equal sides, 4 right angles, diagonals equal and cross at 90°	4	4
Rectangle	4 right angles, opposite sides equal, diagonals equal	2	2
Rhombus	4 equal sides, opposite angles equal, diagonals cross at 90°	2	2
Parallelogram	opposite sides parallel and equal, opposite angles equal	0	2
Kite	two pairs of equal adjacent sides, one pair of equal opposite angles, diagonals cross at 90°	1	1
Trapezium	one pair of parallel sides	0 (1 if isosceles)	1

Sort it

?Is each statement always true, or only sometimes true?

Tap a statement, then tap the group it belongs in. Then press Check.

Geometry · angles and polygons

Angles in polygons

From one corner, an $n$ n-sided polygon splits into $n - 2$ n − 2 triangles, so the interior angle sum is

(n - 2) \times 180°

The exterior angles of any polygon sum to 360° (walk round it and you turn once).
In a regular polygon all angles are equal: exterior angle $= 360° \div n$ = 360° ÷ n, and interior angle $= 180° -$ = 180° − exterior angle (they sit on a straight line).

Worked example Geometry · angles and polygons

Number of sides from an interior angle

Question

Each interior angle of a regular polygon is 156°. How many sides does it have?

1
Interior and exterior angles sit on a straight line: exterior angle $= 180° - 156° = 24°$ = 180° − 156° = 24°
2
The exterior angles sum to 360°: $n = 360° \div 24° = 15$ n = 360° ÷ 24° = 15
3
Check: $(15 - 2) \times 180° = 2340°$ (15 − 2) × 180° = 2340° and $2340° \div 15 = 156°$ 2340° ÷ 15 = 156°

Answer

15 sides

⚠️ Watch out

dividing 360 by the interior angle (

360 \div 156 \approx 2.31

) gives nonsense. Use the exterior angle.

Calculate

?At one point on a patterned floor, a square tile, a regular hexagon tile and one more tile meet with no gaps and no overlaps. Work out the angle of the third tile at that point.

degrees

Calculate

?The third tile in screen 8 is a regular polygon with interior angle 150°. How many sides does it have?

Mission check

?Why can identical regular pentagons never tile a floor?

Mission 2

Two lookouts see one fire. How do two bearings pinpoint it?

Make a guess: Two lookouts see one fire. How do two bearings pinpoint it?

Geometry · bearings and constructions

Three-figure bearings

A bearing is an angle measured clockwise from north, written with three figures: 065°, 008°, 300°.
"The bearing of $B$ B from $A$ A": stand at $A$ A, face north, turn clockwise until you face $B$ B.
Compass points as bearings: N 000°, NE 045°, E 090°, SE 135°, S 180°, SW 225°, W 270°, NW 315°.
The back bearing (the bearing of $A$ A from $B$ B) differs by exactly 180°, because the north lines at $A$ A and $B$ B are parallel.

Worked example Geometry · bearings and constructions

Finding a back bearing

Question

The bearing of

B

from

A

is 065°. The bearing of

Q

from

P

is 300°. Find the bearing of

A

from

B

, and the bearing of

P

from

Q

.

1
065° is less than 180°, so add 180°: $65° + 180° = 245°$ 65° + 180° = 245°
2
300° is more than 180°, so subtract 180°: $300° - 180° = 120°$ 300° − 180° = 120°
3
Write each answer with three figures

Answer

A

from

B

is 245°;

P

from

Q

is 120°

⚠️ Watch out

"the bearing of

A

from

B

" means you stand at

B

. Reading the question the wrong way round gives the original bearing back.

Calculate

?The bearing of

B

from

A

is 138°. Work out the bearing of

A

from

B

.

degrees

Geometry · bearings and constructions

Scale drawings: fixing a point from two bearings

A scale: "1 cm represents 1 km", or 1 : 100,000.
Two bearings fix a point: draw a north line at each known point, measure each bearing clockwise, draw the lines. The point is where they cross.
Measure the drawn distance; convert with the scale.
Triangle from three sides: draw one side; from its ends draw arcs of the other two lengths; join to the crossing. Leave the arcs.

Quick check

?On a map with scale 1 : 20,000, two villages are 6.5 cm apart. How far apart are they in real life?

Calculate

?A ship sails from port

P

to buoy

Q

on a bearing of 070°. At

Q

it turns and sails to lighthouse

R

on a bearing of 160°. Work out the size of angle

PQR

.

degrees

Geometry · bearings and constructions

Ruler-and-compass constructions

Perpendicular bisector of $A B$ AB: equal arcs (radius over half $A B$ AB) from $A$ A and $B$ B cross on both sides; join the crossings. Its points are equidistant from $A$ A and $B$ B.
Angle bisector: an arc from the vertex cuts both arms; equal arcs from those points cross; join the vertex to the crossing.
Perpendicular from $P$ P to a line: an arc from $P$ P cuts the line twice; bisect that segment. The perpendicular distance is the shortest.
Leave every arc visible.

Geometry · bearings and constructions

Loci

A locus is the set of all points that obey a rule.
Fixed distance from a point: a circle. "Within 5 m" means inside it.
Fixed distance from a line segment: two parallel lines joined by semicircles (a racetrack shape).
Equidistant from two points: the perpendicular bisector. "Closer to $A$ A than $B$ B" is the side containing $A$ A.
Equidistant from two lines that meet: the angle bisector.
Shade the region that obeys every condition, and label it.

Sort it

?Which locus describes each set of points?

Tap a statement, then tap the group it belongs in. Then press Check.

Mission check

?Tower

B

is 8 km due east of tower

A

. A fire is on a bearing of 045° from

A

and 315° from

B

. Where is the fire?

Mission 3

Why do runners in outer lanes start further ahead?

Make a guess: Why do runners in outer lanes start further ahead?

Measures · perimeter and area

Areas of triangles, parallelograms and trapezia

Perimeter is the total distance round the edge (cm). Area is the space inside (cm²).
Always use the perpendicular height, measured at 90° to the base, never a slanted side.
Split compound shapes (including kites and rhombuses) into rectangles and triangles, then add or subtract.

Shape	Area
Rectangle	$length \times width$ length × width
Triangle	$\frac{1}{2} \times b \times h$ 1/2 × b × h
Parallelogram	$b \times h$ b × h
Trapezium	$\frac{1}{2} (a + b) h$ 1/2(a + b)h, where $a$ a and $b$ b are the parallel sides

Worked example Measures · perimeter and area

Area of a trapezium

Question

A trapezium has parallel sides 7 cm and 13 cm. Its perpendicular height is 6 cm and each slanted side is 6.7 cm. Find its area.

1
Pick out the parallel sides and the perpendicular height: $a = 7$ a = 7, $b = 13$ b = 13, $h = 6$ h = 6
2
Add the parallel sides: $7 + 13 = 20$ 7 + 13 = 20
3
Multiply by the height and halve: $\frac{1}{2} \times 20 \times 6 = 60$ 1/2 × 20 × 6 = 60

Answer

60 cm²

⚠️ Watch out

using the slanted side 6.7 cm as the height gives 67 cm², which is wrong. The 6.7 cm is extra information.

Quick check

?A parallelogram has base 9 cm and slanted sides 5 cm. Its perpendicular height is 4 cm. What is its area?

Measures · perimeter and area

Circles

Circumference $C = π d = 2 π r$ C = π d = 2π r. Area $A = π r^{2}$ A = π r². Learn both.
Halve a diameter before using $π r^{2}$ π r²: squaring the diameter is the commonest slip.
Exact answers are left in terms of $π$ π: for $r = 3$ r = 3, $A = 9 π$ A = 9π cm².
A semicircle's perimeter is half the circumference plus the diameter.

Worked example Measures · perimeter and area

A track-shaped perimeter and area

Question

A shape is a 10 cm by 6 cm rectangle with a semicircle on each 6 cm end. Find its perimeter and its area, exactly and to 1 decimal place.

1
The two semicircles make one circle of diameter 6 cm: curved length $= π \times 6 = 6 π$ = π × 6 = 6π
2
Add the two straight sides: perimeter $= 20 + 6 π \approx 38.8$ = 20 + 6π ≈ 38.8 cm
3
The two semicircles make one circle of radius 3 cm: area $= π \times 3^{2} = 9 π$ = π × 3² = 9π
4
Add the rectangle: area $= 60 + 9 π \approx 88.3$ = 60 + 9π ≈ 88.3 cm²

Answer

perimeter

20 + 6 π \approx 38.8

cm; area

60 + 9 π \approx 88.3

cm²

⚠️ Watch out

the 6 cm sides are inside the shape, not on its edge, so they are not part of the perimeter.

Calculate

?Find the perimeter of a semicircle with diameter 12 cm. Give your answer to 1 decimal place.

cm

Calculate

?A rectangular lawn is 12 m by 8 m. A semicircular flower bed with diameter 8 m is cut out of one end. Grass seed comes in boxes that each cover 15 m². How many boxes are needed to seed the lawn?

Measures · perimeter and area

Arcs and sectors

A sector is a "pizza slice" cut by two radii; the arc is its curved edge.
A sector with angle $θ$ θ is the fraction $\frac{θ}{360}$ θ/360 of the whole circle:

arc length = \frac{θ}{360} \times 2 π r sector area = \frac{θ}{360} \times π r^{2}

The perimeter of a sector is the arc plus two radii.

Calculate

?A sector has radius 9 cm and angle 40°. Find its area, to 1 decimal place.

cm²

Mission check

?Each lane on a track is 1.2 m wide, so lane 2's bends have a radius 1.2 m bigger than lane 1's. The straights are the same. About how much further is one lap in lane 2?

Mission 4

Roll a sheet into a tall tube or a wide one. Which holds more?

Make a guess: Roll a sheet into a tall tube or a wide one. Which holds more?

Measures · volume and surface area

Prisms and cylinders

A prism has the same cross-section all the way through. A cuboid has 6 faces, 12 edges and 8 vertices.
Volume of any prism:

V = area of cross-section \times length

Cuboid: $V = l \times w \times h$ V = l × w × h. Cylinder: the cross-section is a circle, so $V = π r^{2} h$ V = π r² h.
Units: cm³ or m³. 1 litre $= 1000$ = 1000 cm³, and 1 m³ $= 1000$ = 1000 litres.

Worked example Measures · volume and surface area

Volume of a triangular prism

Question

A prism has a right-angled triangle as its cross-section, with shorter sides 6 cm and 8 cm. The prism is 15 cm long. Find its volume.

1
Area of the cross-section: $\frac{1}{2} \times 6 \times 8 = 24$ 1/2 × 6 × 8 = 24 cm²
2
Multiply by the length: $24 \times 15 = 360$ 24 × 15 = 360

Answer

360 cm³

⚠️ Watch out

forgetting to halve for the triangle gives

6 \times 8 \times 15 = 720

cm³, double the true volume.

Calculate

?A cylindrical water tank has radius 0.5 m and height 1.2 m. How many litres does it hold when full? Give your answer to the nearest litre.

litres

Measures · volume and surface area

Surface area, nets, plans and elevations

Surface area is the total area of the faces. A net shows them unfolded flat.
Cuboid: $S A = 2 (lw + l h + w h)$ SA = 2(lw + lh + wh).
A cylinder's net is two circles and a rectangle whose width is the circumference $2 π r$ 2π r. Curved surface $= 2 π r h$ = 2π r h; total $= 2 π r^{2} + 2 π r h$ = 2π r² + 2π r h.
The plan is the view from above; the front and side elevations are the views from the front and side. Draw them to scale on squared paper.

Quick check

?A cuboid measures 5 cm by 4 cm by 3 cm. What is its total surface area?

Measures · volume and surface area

Pyramids, cones and spheres

Pyramid: $V = \frac{1}{3} \times base area \times h$ V = 1/3 × base area × h.
Cone: $V = \frac{1}{3} π r^{2} h$ V = 1/3π r² h, a third of the cylinder with the same base and height. Curved surface $= π r l$ = π r l, where the slant height $l$ l comes from Pythagoras: $l^{2} = r^{2} + h^{2}$ l² = r² + h².
Sphere: $V = \frac{4}{3} π r^{3}$ V = 4/3π r³ and $S A = 4 π r^{2}$ SA = 4π r². Half a sphere is a hemisphere.
A frustum is a cone with its top cut off parallel to the base: volume $=$ = large cone $-$ − small cone.
Cone and sphere formulae are given in the exam (OCR also gives the pyramid formula); learn the prism and cylinder ones.

Worked example Measures · volume and surface area

Volume and curved surface of a cone

Question

A cone has base radius 3 cm and perpendicular height 4 cm. Find its volume and its curved surface area, in terms of

π

and to 1 decimal place.

1
Volume: $\frac{1}{3} \times π \times 3^{2} \times 4 = 12 π \approx 37.7$ 1/3 × π × 3² × 4 = 12π ≈ 37.7 cm³
2
Compare: a cylinder with the same base and height holds $π \times 3^{2} \times 4 = 36 π$ π × 3² × 4 = 36π, three times as much
3
Slant height by Pythagoras: $l = 3^{2} + 4^{2} = 5$ l = √(3² + 4²) = 5 cm
4
Curved surface: $π r l = π \times 3 \times 5 = 15 π \approx 47.1$ π r l = π × 3 × 5 = 15π ≈ 47.1 cm²

Answer

volume

12 π \approx 37.7

cm³; curved surface area

15 π \approx 47.1

cm²

⚠️ Watch out

π r l

needs the slant height. Using the perpendicular height gives

π \times 3 \times 4 = 12 π

cm², which is too small.

Calculate

?A solid is a cylinder of radius 3 cm and height 10 cm with half a sphere (a hemisphere) of radius 3 cm on top. Find its volume, to 3 significant figures.

cm³

Mission check

?One 20 cm by 30 cm sheet is rolled into a tube two ways: tall (30 cm high) or short (20 cm high). Which holds more?

Mission 5

Will a 1.2 m umbrella fit flat in a 1 m by 0.8 m box?

Make a guess: Will a 1.2 m umbrella fit flat in a 1 m by 0.8 m box?

Geometry · Pythagoras' theorem

The hypotenuse and the theorem

In a right-angled triangle, the side opposite the right angle is the hypotenuse. It is always the longest side.
Pythagoras' theorem: if the shorter sides are $a$ a and $b$ b and the hypotenuse is $c$ c, then

a^{2} + b^{2} = c^{2}

The theorem is about the squares of the sides, not the sides themselves.
It works the other way too: sides 6, 8 and 10 give $36 + 64 = 100$ 36 + 64 = 100, so that triangle is right-angled. Sides 6, 8 and 11 do not.

Worked example Geometry · Pythagoras' theorem

Finding the hypotenuse

Question

A right-angled triangle has shorter sides 8 cm and 5 cm. Find the length of the hypotenuse to 3 significant figures.

1
Square the two shorter sides and add: $8^{2} + 5^{2} = 64 + 25 = 89$ 8² + 5² = 64 + 25 = 89
2
That is $c^{2}$ c², so square-root: $c = 89 = 9.433 \dots$ c = √89 = 9.433…
3
Round to 3 significant figures and check it is longer than 8 cm.

Answer

9.43 cm

⚠️ Watch out

stopping at 89 forgets the square root. A "hypotenuse" of 89 cm next to sides of 8 cm and 5 cm is impossible.

Calculate

?A right-angled triangle has shorter sides 9.5 cm and 6.2 cm. Find the length of the hypotenuse. Give your answer to 1 decimal place.

cm

Geometry · Pythagoras' theorem

Finding a shorter side

Rearrange the theorem: $a^{2} = c^{2} - b^{2}$ a² = c² − b².
Square, subtract, square-root. Subtract because the hypotenuse is the biggest square.
Check: your answer must be shorter than the hypotenuse.
First spot the hypotenuse: it is opposite the right angle, wherever the triangle is turned.

Worked example Geometry · Pythagoras' theorem

A ladder against a wall

Question

A 5 m ladder leans against a vertical wall. Its foot is 1.4 m from the wall on level ground. How high up the wall does it reach?

1
The ladder is the hypotenuse (opposite the right angle between wall and ground): $c = 5$ c = 5
2
Square and subtract: $h^{2} = 5^{2} - 1. 4^{2} = 25 - 1.96 = 23.04$ h² = 5² − 1.4² = 25 − 1.96 = 23.04
3
Square-root: $h = 23.04 = 4.8$ h = √23.04 = 4.8

Answer

4.8 m

⚠️ Watch out

adding the squares gives

26.96 \approx 5.19

m, which is longer than the ladder itself.

Quick check

?A right-angled triangle has hypotenuse 13.5 cm and one shorter side 8.1 cm. What is the length of the third side,

x

?

Calculate

?A rectangular field is 120 m long and 50 m wide. Sam walks from one corner to the opposite corner along the edges. Ali walks straight across the diagonal. How much shorter is Ali's walk?

m

Geometry · Pythagoras' theorem

Pythagoras in 3-D (Higher)

To find the longest diagonal of a cuboid, use Pythagoras twice.
First the diagonal across the base: $d^{2} = a^{2} + b^{2}$ d² = a² + b². Then the base diagonal and the height make a right angle: $g^{2} = d^{2} + c^{2}$ g² = d² + c².
Together: $g^{2} = a^{2} + b^{2} + c^{2}$ g² = a² + b² + c², so $g = a^{2} + b^{2} + c^{2}$ g = √(a² + b² + c²).
A box 1 m by 0.8 m by 0.5 m deep: $g = 1 + 0.64 + 0.25 = 1.89 \approx 1.37$ g = √(1 + 0.64 + 0.25) = √1.89 ≈ 1.37 m.

Calculate

?A box is a cuboid 60 cm long, 40 cm wide and 30 cm high. Find the length of the longest straight rod that fits inside it. Give your answer to 1 decimal place.

cm

Mission check

?Will a 1.2 m umbrella fit flat in a box 1 m by 0.8 m?

Mission 6

How can you find a tree's height with a protractor and a tape?

Make a guess: How can you find a tree's height with a protractor and a tape?

Geometry · trigonometry

Naming the sides and the three ratios

Mark the angle $θ$ θ you know or want. The hypotenuse is opposite the right angle, the opposite faces $θ$ θ, the adjacent is next to $θ$ θ (and not the hypotenuse).
$sin θ = \frac{opp}{hyp} cos θ = \frac{adj}{hyp} tan θ = \frac{opp}{adj}$ sin θ = opp/hyp cos θ = adj/hyp tan θ = opp/adj
Memory aid: SOH CAH TOA. Pick the ratio that uses the side you have and the side you want.
Each ratio is fixed for a given angle because all right-angled triangles with that angle are similar.

Worked example Geometry · trigonometry

Finding a side

Question

In a right-angled triangle the hypotenuse is 12 cm and one angle is 35°. Find the side opposite the 35° angle, to 3 significant figures.

1
Label from 35°: hyp = 12, opp = $x$ x. These are O and H, so use sine.
2
Write the ratio: $sin 3 5^{\circ} = \frac{x}{12}$ sin 35° = x/12
3
Multiply both sides by 12: $x = 12 sin 3 5^{\circ} = 6.883 \dots$ x = 12sin 35° = 6.883…

Answer

6.88 cm

⚠️ Watch out

check your calculator is in degrees. In radians

12 sin 35

gives

- 5.14

, which is impossible for a length.

Calculate

?A right-angled triangle has an angle of 28°. The side adjacent to it is 9 cm. Find the side opposite the 28° angle. Give your answer to 3 significant figures.

cm

Worked example Geometry · trigonometry

Finding an angle

Question

A right-angled triangle has opposite side 5 cm and adjacent side 8 cm from angle

θ

. Find

θ

to 1 decimal place.

1
O and A, so use tangent: $tan θ = \frac{5}{8} = 0.625$ tan θ = 5/8 = 0.625
2
Use the inverse function to undo tan: $θ = tan^{- 1} (0.625)$ θ = tan⁻¹(0.625)
3
Calculate: $θ = 32.005 \dots^{\circ}$ θ = 32.005…°

Answer

θ = 32. 0^{\circ}

⚠️ Watch out

tan θ = \frac{8}{5}

gives 58.0°, the other acute angle. Opposite goes on top.

Quick check

?In a right-angled triangle one angle is 40°. The side opposite it is 6 cm. The hypotenuse is

x

. Which calculation gives

x

?

Geometry · trigonometry

Angles of elevation and depression

The angle of elevation is measured up from the horizontal to the line of sight.
The angle of depression is measured down from the horizontal.
Looking up from A to B and down from B to A gives equal angles (alternate angles between parallel horizontals).
Measure from where your eye is: add the eye height back on at the end.

Calculate

?Priya stands on level ground 25 m from the base of a tree. Her eyes are 1.6 m above the ground. The angle of elevation of the top of the tree from her eyes is 32°. Find the height of the tree, to 1 decimal place.

m

Geometry · trigonometry

Exact values

Learn these (from half an equilateral triangle of side 2 and an isosceles right-angled triangle with shorter sides 1):

$θ$ θ	0°	30°	45°	60°	90°
$sin θ$ sin θ	0	$\frac{1}{2}$ 1/2	$\frac{2}{2}$ (√2)/2	$\frac{3}{2}$ (√3)/2	1
$cos θ$ cos θ	1	$\frac{3}{2}$ (√3)/2	$\frac{2}{2}$ (√2)/2	$\frac{1}{2}$ 1/2	0
$tan θ$ tan θ	0	$\frac{3}{3}$ (√3)/3	1	$3$ √3	—

$\frac{3}{3} = \frac{1}{3}$ (√3)/3 = 1/(√3) and $\frac{2}{2} = \frac{1}{2}$ (√2)/2 = 1/(√2). $tan 9 0^{\circ}$ tan 90° is undefined.
Non-calculator use: hypotenuse 10 and angle 30° give opposite $10 \times \frac{1}{2} = 5$ 10 × 1/2 = 5.

Worked example Geometry · trigonometry

The angle between a line and a plane (Higher)

Question

A cuboid ABCDEFGH has base ABCD with AB = 8 cm, BC = 6 cm, and height CG = 5 cm. Find the angle between the diagonal AG and the base ABCD, to 1 decimal place.

1
Drop from G to the base: it lands on C, so the angle is $∠ G A C$ ∠ GAC in triangle GAC (right angle at C).
2
Find AC across the base with Pythagoras: $A C = 8^{2} + 6^{2} = 10$ AC = √(8² + 6²) = 10
3
In triangle GAC, opp = CG = 5 and adj = AC = 10: $tan θ = \frac{5}{10}$ tan θ = 5/10
4
Inverse: $θ = tan^{- 1} (0.5) = 26.56 \dots^{\circ}$ θ = tan⁻¹(0.5) = 26.56…°

Answer

26.6°

⚠️ Watch out

using the edge AB = 8 instead of AC gives 32.0°. The angle with a plane uses the line's projection onto it.

Mission check

?Sol stands on level ground 20 m from a tree, eyes 1.5 m above the ground, and measures a 40° angle up to the top. How tall is the tree?

Mission 7

Two bearings, 10 km apart: how far away is the lighthouse?

Make a guess: Two bearings, 10 km apart: how far away is the lighthouse?

Geometry · non-right-angled triangles

The sine rule (Higher)

Label angles $A$ A, $B$ B, $C$ C and the opposite sides $a$ a, $b$ b, $c$ c.
Sine rule: each side divided by the sine of its opposite angle gives the same value.

\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C} or \frac{sin A}{a} = \frac{sin B}{b} = \frac{sin C}{c}

Use it when you know a side and its opposite angle; use the second form for an angle.
These formulas are given on the formula sheet.

Worked example Geometry · non-right-angled triangles

A bearings problem with the sine rule (Higher)

Question

Boat B is 12 km due east of boat A. A buoy C is on a bearing of 050° from A and 300° from B. Find the distance BC, to 3 significant figures.

1
Angle at A: from east (090°) back to 050° is $9 0^{\circ} - 5 0^{\circ} = 4 0^{\circ}$ 90° − 50° = 40°
2
Angle at B: from west (270°) round to 300° is $3 0^{\circ}$ 30°, so $C = 18 0^{\circ} - 4 0^{\circ} - 3 0^{\circ} = 11 0^{\circ}$ C = 180° − 40° − 30° = 110°
3
Pair AB = 12 with its opposite angle $C = 11 0^{\circ}$ C = 110°; BC is opposite $A = 4 0^{\circ}$ A = 40°: $\frac{BC}{sin 4 0 ^{\circ}} = \frac{12}{sin 11 0 ^{\circ}}$ BC/(sin 40°) = 12/(sin 110°)
4
Rearrange: $BC = \frac{12 sin 4 0 ^{\circ}}{sin 11 0 ^{\circ}} = 8.208 \dots$ BC = (12sin 40°)/(sin 110°) = 8.208…

Answer

BC = 8.21 km

⚠️ Watch out

using

sin 3 0^{\circ}

on top pairs BC with the wrong angle and gives 6.39 km, which is AC.

Calculate

?In triangle ABC,

a = 9

cm,

b = 7

cm and angle

A = 7 2^{\circ}

. Find angle

B

, to 1 decimal place.

degrees

Geometry · non-right-angled triangles

The cosine rule (Higher)

Use the cosine rule when you know two sides and the angle between them, or all three sides.

a^{2} = b^{2} + c^{2} - 2 b c cos A cos A = \frac{b ^{2} + c ^{2} - a ^{2}}{2 b c}

$a$ a is the side opposite the angle $A$ A; $b$ b and $c$ c are the sides that make angle $A$ A.
If $A = 9 0^{\circ}$ A = 90°, $cos A = 0$ cos A = 0 and it becomes Pythagoras.
For an obtuse angle $cos A$ cos A is negative, so the calculator gives the obtuse angle directly.

Worked example Geometry · non-right-angled triangles

Finding an angle from three sides (Higher)

Question

A triangle has sides 5 cm, 7 cm and 10 cm. Find its largest angle, to 1 decimal place.

1
The largest angle is opposite the longest side, so $a = 10$ a = 10, $b = 5$ b = 5, $c = 7$ c = 7
2
Substitute: $cos A = \frac{5 ^{2} + 7 ^{2} - 1 0 ^{2}}{2 \times 5 \times 7} = \frac{- 26}{70}$ cos A = (5² + 7² − 10²)/(2 × 5 × 7) = (−26)/70
3
Evaluate: $cos A = - 0.3714 \dots$ cos A = −0.3714…
4
Inverse cosine: $A = 111.80 \dots^{\circ}$ A = 111.80…°

Answer

111. 8^{\circ}

⚠️ Watch out

the negative cosine is correct, not a mistake: it tells you the angle is obtuse.

Calculate

?In triangle ABC,

b = 8

cm,

c = 11

cm and the angle between them is

A = 3 8^{\circ}

. Find side

a

, to 3 significant figures.

cm

Sort it

?Which rule do you start with?

Tap a statement, then tap the group it belongs in. Then press Check.

Geometry · non-right-angled triangles

Area of any triangle (Higher)

Two sides and the angle between them give the area:

Area = \frac{1}{2} ab sin C

The angle must be the included angle, between sides $a$ a and $b$ b.
Example: sides 6 cm and 9 cm with 30° between them: $\frac{1}{2} \times 6 \times 9 \times sin 3 0^{\circ} = 13.5 cm^{2}$ 1/2 × 6 × 9 × sin 30° = 13.5 cm².
If you know all three sides, find an angle with the cosine rule first.

Calculate

?A triangular garden has sides 8 m, 11 m and 13 m. Find its area, to 3 significant figures.

m²

Mission check

?A ship at A sees a lighthouse L on a bearing of 040°. It sails 10 km due north to B, where L is on a bearing of 120°. How far is the ship at B from the lighthouse?

Mission 8

A picture was spun round one nail. Can you find where the nail is?

Make a guess: A picture was spun round one nail. Can you find where the nail is?

Geometry · transformations

Describing a transformation fully

A full description names the transformation and gives every fact needed to repeat it exactly.

Transformation	Give	Image is
Reflection	the mirror line as an equation, e.g. $x = 3$ x = 3 or $y = - 1$ y = −1	congruent
Rotation	centre, angle, direction	congruent
Translation	a column vector	congruent
Enlargement	centre, scale factor	similar

Reflections, rotations and translations keep lengths and angles. Enlargements keep angles but change lengths.
Give one transformation only. Two moves for a "single transformation" question score nothing.
Slanted mirror lines such as $y = x$ y = x and $y = - x$ y = −x: Foundation and Higher.

Worked example Geometry · transformations

Describing a rotation

Question

Describe fully the single transformation that maps triangle A, with vertices

(1, 1)

,

(3, 1)

,

(1, 2)

, onto triangle B, with vertices

(- 1, 1)

,

(- 1, 3)

,

(- 2, 1)

.

1
Same size and shape, but turned round: it is a rotation.
2
Find the angle: side $(1, 1)$ (1, 1) to $(3, 1)$ (3, 1) points right; its image $(- 1, 1)$ (−1, 1) to $(- 1, 3)$ (−1, 3) points up. That is a quarter-turn anticlockwise, 90°.
3
Find the centre: try $(0, 0)$ (0, 0) with tracing paper. Each vertex and its image are the same distance from it.
4
Write all three facts with the word "rotation".

Answer

Rotation, 90° anticlockwise, centre

(0, 0)

.

⚠️ Watch out

leaving out the centre, the angle or the direction each loses a mark, and "turn" is not accepted for "rotation".

Quick check

?Triangle P has vertices

(1, 1)

,

(3, 1)

and

(1, 4)

. Triangle Q has vertices

(7, 1)

,

(5, 1)

and

(7, 4)

. Which is the single transformation that maps P onto Q?

Geometry · transformations

Translations and enlargements

Translation by $(a b)$ ( a ; b ): the top number moves right (+) or left (−); the bottom number moves up (+) or down (−).
Enlargement: every distance from the centre is multiplied by the scale factor.
A scale factor between 0 and 1 makes the image smaller. Scale factor $\frac{1}{2}$ 1/2, centre $(0, 0)$ (0, 0): $(4, 6) \to (2, 3)$ (4, 6) → (2, 3).
Fractional scale factors: Foundation and Higher.
Higher: a negative scale factor puts the image on the other side of the centre, upside down. Scale factor $- 2$ −2, centre $(0, 0)$ (0, 0): $(1, 3) \to (- 2, - 6)$ (1, 3) → (−2, −6).

Worked example Geometry · transformations

Enlargement from a centre

Question

Enlarge triangle T, with vertices

(2, 3)

,

(3, 3)

,

(2, 5)

, by scale factor 2, centre

(1, 2)

.

1
Count from the centre to each vertex: $(11)$ ( 1 ; 1 ), $(21)$ ( 2 ; 1 ), $(13)$ ( 1 ; 3 )
2
Multiply each by 2: $(22)$ ( 2 ; 2 ), $(42)$ ( 4 ; 2 ), $(26)$ ( 2 ; 6 )
3
Count out from the centre again: $(3, 4)$ (3, 4), $(5, 4)$ (5, 4), $(3, 8)$ (3, 8)
4
Check: the base was 1 and is now 2; the height was 2 and is now 4.

Answer

The image has vertices

(3, 4)

,

(5, 4)

and

(3, 8)

.

⚠️ Watch out

doubling the coordinates themselves, e.g.

(2, 3) \to (4, 6)

, is an enlargement about the origin, not about

(1, 2)

.

Calculate

?Triangle T has a vertex at

P (1, 2)

. T is enlarged by scale factor 3, centre

(- 1, 0)

. Find the

x

-coordinate of the image of P.

Geometry · transformations

Finding the centre of a rotation

The centre is the one point that does not move: an invariant point.
Every other point stays the same distance from the centre: $CP = C P^{'}$ CP = CP'.
So the centre lies on the perpendicular bisector of $P P^{'}$ PP'. Two such bisectors cross at the centre.
On a grid: pin tracing paper at a guess and turn it. If the shape lands on the image, that is the centre.

Calculate

?The point

P (5, 3)

is rotated 90° clockwise about

(2, 1)

. Find the

y

-coordinate of its image

P^{'}

.

Worked example Geometry · transformations

Two reflections make one move (Higher)

Question

Triangle T has vertices

(2, 1)

,

(3, 1)

,

(2, 3)

. Reflect T in

x = 1

to give

T_{1}

, then reflect

T_{1}

in

x = 4

to give

T_{2}

. Describe fully the single transformation that maps T onto

T_{2}

.

1
Reflect in $x = 1$ x = 1: $T_{1}$ T₁ is $(0, 1)$ (0, 1), $(- 1, 1)$ (−1, 1), $(0, 3)$ (0, 3)
2
Reflect in $x = 4$ x = 4: $T_{2}$ T₂ is $(8, 1)$ (8, 1), $(9, 1)$ (9, 1), $(8, 3)$ (8, 3)
3
Compare T with $T_{2}$ T₂: every vertex has moved 6 right and 0 up, and $T_{2}$ T₂ is the same way round as T.
4
Invariance: each reflection fixes the points on its own mirror, but the combined move fixes no point at all.

Answer

Translation by

(60)

.

⚠️ Watch out

the shift is twice the gap between the mirrors (

2 \times 3 = 6

), and doing the reflections in the other order gives

(- 6 0)

.

Mission check

?The nail problem: a picture turns 90° clockwise about a nail. Corner

A (5, 6)

moves to

A^{'} (7, 0)

and corner

B (1, 6)

moves to

B^{'} (7, 4)

. Where is the nail?

Mission 9

A giant ape 10 times a gorilla's height: why would its legs snap?

Make a guess: A giant ape 10 times a gorilla's height: why would its legs snap?

Geometry · congruence and similarity

Congruent or similar?

Congruent shapes are the same shape and size: one fits exactly on the other (after turning or flipping).
Similar shapes are the same shape: corresponding angles are equal and every length is multiplied by the same scale factor $k$ k.
$k = \frac{length on the new shape}{corresponding length on the original}$ k = (length on the new shape)/(corresponding length on the original)
A 4 by 6 rectangle and a 6 by 9 rectangle are similar ( $k = 1.5$ k = 1.5); a 4 by 6 and a 6 by 8 are not.
Triangles with equal angles are similar. Give a reason for each equal angle.

Geometry · congruence and similarity

Proving triangles congruent

Two triangles are congruent if they match by one of these conditions:
SSS: three sides equal. SAS: two sides and the angle between them.
ASA: two angles and a corresponding side. RHS: right angle, hypotenuse and one other side.
AAA proves only similarity; two sides and a non-included angle is not a condition.
In a proof, list the three equal facts with a reason for each, then name the condition.

Sort it

?Each pair of triangles has the facts shown equal. Is that enough to prove them congruent?

Tap a statement, then tap the group it belongs in. Then press Check.

Worked example Geometry · congruence and similarity

A missing length in similar triangles

Question

In triangle ABC, D is on AB and E is on AC with DE parallel to BC. AD = 4 cm, DB = 6 cm and DE = 6 cm. Find BC.

1
Show similarity: angle A is shared; $∠ A D E = ∠ A BC$ ∠ ADE = ∠ ABC (corresponding angles), so the triangles are similar.
2
Use whole sides from A: $A B = 4 + 6 = 10$ AB = 4 + 6 = 10 cm
3
Scale factor from ADE to ABC: $k = \frac{A B}{A D} = \frac{10}{4} = 2.5$ k = AB/AD = 10/4 = 2.5
4
Multiply: $BC = 6 \times 2.5 = 15$ BC = 6 × 2.5 = 15

Answer

BC = 15 cm

⚠️ Watch out

using

\frac{D B}{A D} = 1.5

gives 9 cm. DB is not a side of either triangle.

Calculate

?In triangle PQR, S is on PQ and T is on PR with ST parallel to QR. PS = 5 cm, ST = 4 cm and QR = 10 cm. Find the length SQ.

cm

Geometry · congruence and similarity

Area and volume scale factors (Higher)

If lengths are multiplied by $k$ k, then:
areas (including surface areas) are multiplied by $k^{2}$ k²
volumes (and so masses of the same material) are multiplied by $k^{3}$ k³
Double every length of a cube: area of each face $\times 4$ × 4, volume $\times 8$ × 8.
Given an area or volume factor, work back to $k$ k: area factor 9 means $k = 3$ k = 3; volume factor 8 means $k = 2$ k = 2.

Worked example Geometry · congruence and similarity

Two similar bottles (Higher)

Question

Two bottles are mathematically similar. The small one is 12 cm tall and holds 400 ml. The large one is 18 cm tall. How much does the large one hold?

1
Length scale factor: $k = \frac{18}{12} = 1.5$ k = 18/12 = 1.5
2
Capacity is a volume, so use $k^{3} = 1. 5^{3} = 3.375$ k³ = 1.5³ = 3.375
3
Multiply: $400 \times 3.375 = 1350$ 400 × 3.375 = 1350

Answer

1,350 ml

⚠️ Watch out

400 \times 1.5 = 600

ml uses the length factor and

400 \times 2.25 = 900

ml the area factor.

Calculate

?Two solid cylinders are mathematically similar. Their surface areas are 80 cm² and 180 cm². The smaller one has volume 120 cm³. Find the volume of the larger one.

cm³

Quick check

?A statue is similar to a model and 3 times as tall. Painting the model's surface takes 0.5 litres. How much paint does the statue need?

Mission check

?A giant ape is similar to a gorilla but 10 times as tall. Why would its legs snap?

Mission 10

Draw a triangle on a diameter. Why is the top corner always 90°?

Make a guess: Draw a triangle on a diameter. Why is the top corner always 90°?

Geometry · circle theorems

The parts of a circle

Radius: centre to the circle. Diameter: a chord through the centre, twice the radius.
Chord: a straight line joining two points on the circle. Tangent: a straight line touching the circle at one point.
Arc: part of the circumference. Sector: the slice between two radii and an arc. Segment: the region between a chord and an arc.
Higher: a quadrilateral with all four corners on a circle is a cyclic quadrilateral.

Geometry · circle theorems

Angle in a semicircle (Higher)

AB is a diameter and C is any point on the circle. Then angle $A CB = 9 0^{\circ}$ ACB = 90°: the angle in a semicircle is a right angle.
Why: join C to the centre O. $O A = OB = OC$ OA = OB = OC (radii), so triangles $A OC$ AOC and $BOC$ BOC are isosceles.
Call the base angles $a, a$ a, a and $b, b$ b, b. Triangle $A BC$ ABC gives $2 a + 2 b = 18 0^{\circ}$ 2a + 2b = 180°, so $a + b = 9 0^{\circ}$ a + b = 90°.
Exam reason: "angle in a semicircle is 90°".

Quick check

?AB is a diameter of a circle and C is a point on the circle. Angle

C A B = 3 5^{\circ}

. Find angle

A BC

.

Geometry · circle theorems

Tangents and chords (Higher)

A tangent is perpendicular to the radius at the point of contact.
Two tangents from the same outside point are equal in length (by RHS congruence), so they make a kite with the two radii.
The perpendicular from the centre to a chord bisects the chord. The converse holds too: the line from the centre to the midpoint of a chord is perpendicular to it.

Calculate

?Exam-style (Higher). PA and PB are tangents from a point P to a circle, centre O, touching it at A and B. Angle

A PB = 5 0^{\circ}

. Find angle

O A B

.

degrees

Geometry · circle theorems

Four angle theorems (Higher)

The angle at the centre is twice the angle at the circumference standing on the same arc.
Angles in the same segment are equal (they stand on the same chord, on the same side).
Opposite angles of a cyclic quadrilateral add up to 180°.
Alternate segment theorem: the angle between a tangent and a chord equals the angle in the alternate segment.
In an exam, quote the theorem in words for every step. "Circle theorem" alone earns no reason mark.

Worked example Geometry · circle theorems

Chasing angles with reasons (Higher)

Question

A, B, C and D lie on a circle, centre O. A is on the major arc BD and C is on the minor arc BD. Angle

BO D = 14 0^{\circ}

. Find angle

B A D

and angle

BC D

, giving reasons.

1
Angle $B A D$ BAD stands on arc BD, like $BO D$ BOD: $B A D = 140 \div 2 = 7 0^{\circ}$ BAD = 140 ÷ 2 = 70° (angle at the centre is twice the angle at the circumference)
2
ABCD is a cyclic quadrilateral: $BC D = 180 - 70 = 11 0^{\circ}$ BCD = 180 − 70 = 110° (opposite angles of a cyclic quadrilateral add up to 180°)

Answer

Angle

B A D = 7 0^{\circ}

and angle

BC D = 11 0^{\circ}

.

⚠️ Watch out

halving 140° for angle

BC D

too. C is on the other side of BD, so it does not stand on the same arc as

BO D

.

Calculate

?Exam-style (Higher). The tangent to a circle at A makes an angle of 64° with the chord AB. C is a point on the circle in the alternate segment, and angle

A BC = 7 1^{\circ}

. Find angle

B A C

.

degrees

Worked example Geometry · circle theorems

Proving the angle at the centre theorem (Higher)

Question

Prove that the angle subtended by an arc at the centre is twice the angle at the circumference.

1
Let A, B, C lie on the circle, centre O. Join CO and extend it to D.
2
$O A = OC$ OA = OC (radii), so triangle $A OC$ AOC is isosceles: angle $O A C$ OAC = angle $OC A = x$ OCA = x.
3
Exterior angle of triangle $A OC$ AOC: angle $A O D = x + x = 2 x$ AOD = x + x = 2x.
4
In the same way, $OB = OC$ OB = OC gives angle $OCB = y$ OCB = y and angle $BO D = 2 y$ BOD = 2y.
5
So angle $A OB = 2 x + 2 y = 2 (x + y) = 2 \times$ AOB = 2x + 2y = 2(x + y) = 2 × angle $A CB$ ACB.

Answer

Angle

A OB = 2 \times

angle

A CB

, as required.

⚠️ Watch out

every equal-angle claim needs its reason ("radii", "base angles of an isosceles triangle"); a proof with bare numbers scores nothing.

Mission check

?AB is a diameter and C is any point on the circle. Why is angle

A CB

always 90°?

Mission 11

A drone flies 3 east, 4 north, then 6 east, 8 north. Straight line?

Make a guess: A drone flies 3 east, 4 north, then 6 east, 8 north. Straight line?

Geometry · vectors

Column vectors

A vector has a size and a direction. The column vector $(x y)$ ( x ; y ) means $x$ x across (right is +) and $y$ y up (up is +).
Vectors are printed in bold, $a$ a, or written $A B$ AB→ (from A to B). By hand, underline the letter.
A translation is described by a column vector: $(3 - 4)$ ( 3 ; −4 ) moves $(1, 2)$ (1, 2) to $(4, - 2)$ (4, −2).
Equal vectors have the same size and direction, wherever they are drawn.
$B A = - A B$ BA→ = −AB→: same size, opposite direction.

Geometry · vectors

Adding, subtracting and scaling (Foundation and Higher)

Add or subtract matching components: $(2 - 1) + (- 3 4) = (- 1 3)$ ( 2 ; −1 ) + ( −3 ; 4 ) = ( −1 ; 3 )
On a diagram, $a + b$ a + b means "follow a, then b": draw b starting where a ends. The resultant joins the start to the finish.
$a - b = a + (- b)$ a − b = a + (−b): follow a, then b backwards.
Multiplying by a scalar (a number) multiplies both components: $3 (2 - 1) = (6 - 3)$ 3( 2 ; −1 ) = ( 6 ; −3 )

Worked example Geometry · vectors

Combining column vectors (Foundation and Higher)

Question

a = (2 - 1)

and

b = (- 3 4)

. Find

3 a - 2 b

as a column vector.

1
Scale a: $3 a = (6 - 3)$ 3a = ( 6 ; −3 )
2
Scale b: $2 b = (- 6 8)$ 2b = ( −6 ; 8 )
3
Subtract the matching components: $(6 - (- 6) - 3 - 8)$ ( 6 − (−6) ; −3 − 8 )

Answer

3 a - 2 b = (12 - 11)

⚠️ Watch out

6 - (- 6) = 12

. Adding

2 b

by mistake gives

(05)

.

Calculate

?

a = (4 - 2)

and

b = (- 1 3)

. Find the bottom component of

2 a - 3 b

.

Geometry · vectors

Parallel vectors (Foundation and Higher)

$k a$ ka is parallel to $a$ a and $k$ k times as long. If $k > 0$ k > 0 it points the same way; if $k < 0$ k < 0, the opposite way.
Test: is one vector a number times the other? $(- 6 3) = - 3 (2 - 1)$ ( −6 ; 3 ) = −3( 2 ; −1 ), so these are parallel (opposite directions).
$(23)$ ( 2 ; 3 ) and $(32)$ ( 3 ; 2 ) are not parallel: no single number turns 2 into 3 and 3 into 2.
Two journeys line up exactly when the second is a positive multiple of the first.

Sort it

?Which vectors are parallel to

(2 - 3)

?

Tap a statement, then tap the group it belongs in. Then press Check.

Geometry · vectors

Vectors in geometry (Higher)

Follow a route along known vectors: $A B = A O + OB = - a + b$ AB→ = AO→ + OB→ = −a + b
Midpoint M of AB: $OM = a + \frac{1}{2} (b - a) = \frac{1}{2} a + \frac{1}{2} b$ OM→ = a + 1/2(b − a) = 1/2a + 1/2b
Show lines are parallel: one vector is a multiple of the other.
Show X, Y, Z are collinear: $Y Z = k X Y$ YZ→ = kXY→ and they share point Y. Example: $X Y = 2 a + b$ XY→ = 2a + b, $Y Z = 6 a + 3 b = 3 X Y$ YZ→ = 6a + 3b = 3XY→.

Calculate

?Exam-style (Higher). OAB is a triangle with

O A = a

and

OB = b

. P lies on AB with

A P : PB = 1 : 3

. Write

OP = m a + n b

. Find

m

as a decimal.

Mission check

?A drone flies

(34)

and then

(68)

. Did it fly in a straight line?

Recap

Big ideas

Every angle needs a reason: angles at a point (360°), on a straight line (180°), vertically opposite, alternate, corresponding and co-interior angles.
Triangle 180°, quadrilateral 360°, any polygon $(n - 2) \times 180°$ (n−2) × 180°; exterior angles of a polygon always sum to 360°.
For a regular polygon, find the exterior angle $360° \div n$ 360° ÷ n first, then interior $= 180° -$ = 180° − exterior.
Bearings are measured clockwise from north and written with three figures; a back bearing differs by 180°.
Two bearing lines from two known points cross at exactly one place: that is how you fix a position on a scale drawing.
Loci: fixed distance from a point is a circle; equidistant from two points is the perpendicular bisector; equidistant from two lines is the angle bisector.
Area: rectangle $lw$ lw, triangle $\frac{1}{2} bh$ 1/2bh, parallelogram $bh$ bh, trapezium $\frac{1}{2} (a + b) h$ 1/2(a+b)h, always with the perpendicular height.
Circle: $C = 2 π r = π d$ C = 2π r = π d, $A = π r^{2}$ A = π r²; give exact answers in terms of $π$ π when asked; split compound shapes into parts.
Sector $= \frac{θ}{360}$ = θ/360 of the circle for both arc length and area.
Prism volume $=$ = cross-section area $\times$ × length; cylinder $V = π r^{2} h$ V = π r² h; 1 litre $= 1000$ = 1000 cm³.
Surface area is the sum of the faces: use a net; a cylinder's curved surface unrolls to a $2 π r$ 2π r by $h$ h rectangle.
Pyramid and cone volumes are a third of the matching prism or cylinder; sphere $V = \frac{4}{3} π r^{3}$ V = 4/3π r³, $S A = 4 π r^{2}$ SA = 4π r².
The hypotenuse is opposite the right angle and is the longest side: $a^{2} + b^{2} = c^{2}$ a² + b² = c².
Hypotenuse: square, add, square-root. Shorter side: square, subtract, square-root.
Higher: in a cuboid use Pythagoras twice, $g = a^{2} + b^{2} + c^{2}$ g = √(a² + b² + c²).
Label hyp, opp and adj from the angle; choose the ratio with SOH CAH TOA.
Unknown side: write the ratio and rearrange. Unknown angle: use $sin^{- 1}$ sin⁻¹, $cos^{- 1}$ cos⁻¹ or $tan^{- 1}$ tan⁻¹.
Know the exact values for 0°, 30°, 45°, 60°, 90°; in 3-D, find a right-angled triangle inside the solid (Higher/Extended).
Label each side with the same letter as its opposite angle.
Sine rule for a side–angle pair; cosine rule for two sides with the included angle, or three sides.
Area $= \frac{1}{2} ab sin C$ = 1/2 absin C with the angle between the two sides.
Describe fully: reflection (mirror line equation), rotation (centre, angle, direction), translation (column vector), enlargement (centre, scale factor).
The centre of a rotation is its invariant point: it is equally far from every point and that point's image.
Two reflections in parallel mirrors make a translation of twice the gap between them.
Similar: equal angles, every length $\times k$ × k. Use whole corresponding sides to find $k$ k.
Congruent triangles: prove with SSS, SAS, ASA or RHS, giving a reason for each fact.
Lengths $\times k$ × k, areas $\times k^{2}$ × k², volumes $\times k^{3}$ × k³.
Tangent ⟂ radius; two tangents from a point are equal; the perpendicular from the centre bisects a chord.
Angle in a semicircle is 90°; angle at the centre is twice the angle at the circumference; angles in the same segment are equal; opposite angles of a cyclic quadrilateral sum to 180°; alternate segment theorem.
Prove theorems from equal radii and isosceles triangles, giving a reason for every step.
Column vector $(x y)$ ( x ; y ): $x$ x right, $y$ y up. Add, subtract and scale component by component.
$k a$ ka is parallel to $a$ a; a positive $k$ k keeps the direction.
Higher: route along known vectors; collinear means parallel and sharing a point.

🏆

Mini-lesson complete!

⭐⭐⭐

You've worked through Geometry & Measures for AQA GCSE Maths (8300). 🎉

Mission 1Why can tiles be squares or hexagons but never regular pentagons?
Mission 2Two lookouts see one fire. How do two bearings pinpoint it?
Mission 3Why do runners in outer lanes start further ahead?
Mission 4Roll a sheet into a tall tube or a wide one. Which holds more?
Mission 5Will a 1.2 m umbrella fit flat in a 1 m by 0.8 m box?
Mission 6How can you find a tree's height with a protractor and a tape?
Mission 7Two bearings, 10 km apart: how far away is the lighthouse?
Mission 8A picture was spun round one nail. Can you find where the nail is?
Mission 9A giant ape 10 times a gorilla's height: why would its legs snap?
Mission 10Draw a triangle on a diameter. Why is the top corner always 90°?
Mission 11A drone flies 3 east, 4 north, then 6 east, 8 north. Straight line?

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Specification points and sources for this lesson are listed at the bottom of the page.