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Geometry

Syllabus topic 4 · angles, circles, similarity and constructions

What the check suggests about this topic

On the Foundations Check, angles, shape and Pythagoras came back secure to step 5 and broke at step 6. Step 6 is the Extended rung. So the ordinary angle facts are in place — what is not yet in place is the Extended layer: circle theorems, similar-shape scale factors, and multi-step angle chases where one fact feeds the next.

Two other things from the check land directly on this topic:

What this guide covers

CodeSub-topicExpected difficulty for you
E4.1Geometrical termsLow — vocabulary, but Cambridge marks the words
E4.2Geometrical constructionsLow — method, not calculation
E4.3Scale drawings and bearingsLow — three-figure form catches people out
E4.4SimilarityMedium — the k² and k³ step is the dropped mark
E4.5SymmetryLow
E4.6AnglesMedium — rebuilt here from the bottom
E4.7Circle theorems IMedium — new territory above your break
E4.8Circle theorems IILow, once the first five are secure

Paper 2 is non-calculator: 2 hours, 100 marks, 50% of your grade. Every number in this guide is chosen so it can be done by hand, and surds are left as surds.

The thing that is already working

Number sense and the four operations: solid on all seven rungs. No gap anywhere, right up to Extended. Geometry is unusually kind to that strength — most circle-theorem questions are one clean subtraction or one halving, and you can already do those without thinking.

Which means the marks in this topic are not about arithmetic. They are about naming the reason. Cambridge awards the angle and the reason separately, and the reason is the mark most often thrown away by people who got the number right.

How this guide works

Every idea appears four times, with less help each time.

  1. Worked in full — every step, with a reason on each line.
  2. Last step is yours — work out the last line before you press the button.
  3. Last two are yours — the same, harder.
  4. All yours — type an answer and check it.

Section 9 is a mixed set: fifteen questions, unlabelled and out of order, because in an exam nobody tells you which theorem applies.

E4.1 · low1 · Geometrical terms — the words that carry marks▼
▶  Watch: E4.1 Geometrical terms
Short hand-worked explanations for exactly this sub-topic. Opens on YouTube in a new tab. Watch one, then come straight back and try the questions below — watching without testing yourself feels like learning but is not.

Geometry has more vocabulary than any other maths topic, and Cambridge tests the words directly. “Name the type of angle”, “state which two triangles are congruent”, “write down the mathematical name of the quadrilateral” — these are one-mark questions where the mark is the word.

Angles, by size

acute less than 90° right exactly 90° obtuse between 90° and 180° reflex between 180° and 360°
The mistake: calling a 217° angle “obtuse”. Obtuse stops at 180°. Anything past a straight line and short of a full turn is reflex. Also note that 90° is called right, never acute.

Lines and points

WordWhat it means
PointA position with no size. Labelled with a capital letter.
LineStraight, and goes on for ever in both directions.
Line segmentThe bit of a line between two points. AB usually means this.
RayStarts at a point and goes on for ever one way.
PlaneA flat surface going on for ever. A page is a piece of one.
ParallelSame direction, never meet. Marked with matching arrowheads.
PerpendicularMeeting at 90°.
Perpendicular bisectorThe line that cuts a line segment exactly in half and crosses it at 90°. Every point on it is the same distance from both ends of the segment.
VertexA corner. Plural: vertices.

Parts of a circle — learn these now, sections 7 and 8 assume them

O radius diameter chord tangent sector segment arc
WordWhat it is
RadiusCentre to the edge. Every radius in one circle is the same length — that fact does a lot of work later.
DiameterRight across through the centre. Twice the radius.
ChordA straight line joining two points on the circle. A chord through the centre is a diameter, the longest chord.
TangentA line that touches the circle at exactly one point.
ArcPart of the curved edge.
SemicircleHalf a circle, cut off by a diameter.
Minor and major arcTwo points on a circle split it into two arcs: the shorter one is the minor arc, the longer one the major arc. Minor and major sectors and segments are named the same way.
SectorA slice, bounded by two radii and an arc.
SegmentCut off by a chord. The smaller piece is the minor segment.
CircumferenceThe whole way round the edge.

Triangles and quadrilaterals

ShapeProperties Cambridge expects you to state
Equilateral triangleThree equal sides, three angles of 60°.
Isosceles triangleTwo equal sides; the angles opposite them are equal.
Scalene triangleAll sides different, all angles different.
SquareFour equal sides, four right angles, diagonals equal and bisect at 90°.
RectangleOpposite sides equal, four right angles, diagonals equal.
ParallelogramTwo pairs of parallel sides, opposite angles equal, no line of symmetry.
RhombusA parallelogram with all four sides equal; diagonals cross at 90°.
TrapeziumExactly one pair of parallel sides.
KiteTwo pairs of adjacent equal sides; one line of symmetry; diagonals cross at 90°.

Polygons and solids

polygonsidesone interior angle if regular
pentagon5108°
hexagon6120°
octagon8135°
decagon10144°

Regular means all the sides are equal and all the angles are equal. Anything else is irregular: a rectangle that is not a square has equal angles but unequal sides, so it is an irregular quadrilateral. One interior angle of a regular polygon is 180° − (360° ÷ number of sides).

wordwhat it means
facea flat surface of a solid
edgewhere two faces meet
vertex (vertices)a corner
prismthe same cross-section all the way along: cube, cuboid, triangular prism, cylinder
pyramida flat base and sloping triangular faces meeting at one point, the apex
hemispherehalf a sphere
frustuma cone or pyramid with its top cut off parallel to the base
Counting a prism: if the cross-section has n sides, the prism has n + 2 faces, 3n edges and 2n vertices. A triangular prism (n = 3) has 5 faces, 9 edges and 6 vertices.
All yours
What is the name of a polygon with 10 sides?
Work out one interior angle of a regular octagon, in degrees.
How many edges does a triangular prism have?
How many vertices does a square-based pyramid have?
A cone has its top cut off parallel to its base. What is the solid that is left called?

Congruent or similar — the one pair of words that must not blur

Congruent = identical. Same shape, same size. One would fit exactly on the other.
Similar = same shape, different size. All angles equal, all sides in the same ratio.

0580 does not ask you to prove that two shapes are congruent, so there are no congruence rules to learn. What it does test is using the two words correctly. A reflection, rotation or translation always gives a congruent image; an enlargement gives a similar one, with every length multiplied by the scale factor.

Reading the diagram: tick marks on sides mean those sides are equal, matching arcs at corners mean those angles are equal, and arrowheads on lines mean parallel. Those marks are given information — use them, and do not assume anything that is not marked.
Worked in full
An isosceles triangle has an angle of 34° between its two equal sides. Work out the other two angles.
1
The two equal sides are given
So the two angles opposite them are equal. Call each one b.
2
34 + b + b = 180
The angles in a triangle add to 180°.
3
2b = 180 − 34
Take the known angle off both sides. Do this subtraction slowly.
4
2b = 146
180 − 34: take off 30 to get 150, then take off 4 to get 146.
5
b = 73
Halve it. Both remaining angles are 73°.
6
Check: 34 + 73 + 73 = 180
It does. A check like this costs five seconds and catches every sign slip.
Last step is yours
A triangle has two equal sides. One of the base angles is 52°. Work out the apex angle.
1
The other base angle is also 52°
Base angles of an isosceles triangle are equal.
2
52 + 52 = 104
Add the two known angles first, then subtract once. One subtraction, one chance to slip.
3
apex = 180 − 104 = 76°
Angles in a triangle add to 180°.
Last two are yours
A right-angled triangle is also isosceles. Work out its two equal angles.
1
One angle is 90°
That is what right-angled means.
2
The other two are equal
Isosceles: the angles opposite the two equal sides are equal.
3
180 − 90 = 90
What is left for the other two angles between them.
4
90 ÷ 2 = 45° each
Share it equally. Every right-angled isosceles triangle is 90, 45, 45.
All yours
An isosceles triangle has an apex angle of 96°. Work out one of the base angles, in degrees.

Four more, on the words

Two shapes have exactly the same shape but different sizes. Are they congruent or similar?
An angle measures 217°. Name it.
Name the quadrilateral with two pairs of parallel sides and all four sides equal, but no right angles.
A straight line joining two points on a circle, not passing through the centre, is called a what?
E4.2 · low2 · Geometrical constructions▼
▶  Watch: E4.2 Geometrical constructions
Short hand-worked explanations for exactly this sub-topic. Opens on YouTube in a new tab. Watch one, then come straight back and try the questions below — watching without testing yourself feels like learning but is not.

E4.2 is three skills and only three: measuring and drawing lines and angles, constructing a triangle from its three sides with a ruler and a pair of compasses only, and drawing and using nets of solids. Many GCSE books and videos also teach bisector constructions and loci. The 0580 syllabus asks for neither, so they are not taught here.

Construction questions are marked on the method you can see. The arcs are the evidence. A perfect triangle with no arcs showing scores less than a slightly wobbly one with the arcs left in.

Four things must be on the desk: a sharp pencil, a ruler marked in mm, a protractor, and a pair of compasses that does not slip. Every straight edge is drawn with the ruler, never freehand.

Measuring and drawing lines and angles

Lines. Put the 0 mark of the ruler on one end of the line, not the end of the ruler — most rulers have a few millimetres of plastic before the 0. Read the other end to the nearest millimetre and write the length in centimetres to one decimal place (8.7 cm) or in millimetres (87 mm). 1 cm = 10 mm.

Angles. A protractor has two scales, one running each way round, so every line through its centre crosses two numbers that add up to 180. Only one of them is your angle.

180 0 170 10 160 20 150 30 140 40 130 50 120 60 110 70 100 80 90 90 80 100 70 110 60 120 50 130 40 140 30 150 20 160 10 170 0 180 50° 50 130 The left arm lies on the 0 of the OUTER scale, so read the outer scale: the angle is 50°. The inner scale says 130 on the same line. It counts from the other end of the 0° line: that is the trap. Check: the angle is clearly acute, so it must be less than 90°. Every line through the centre crosses two numbers that add up to 180. Only one of them is the angle.
  1. Decide first whether the angle is acute (less than 90°) or obtuse (between 90° and 180°).
  2. Put the centre cross of the protractor exactly on the vertex.
  3. Turn the protractor until its 0° line lies along one arm.
  4. Read the scale that shows 0 on that arm, and count round to the other arm.
  5. Check the reading against step 1. An acute angle cannot read 130.
The mistake: reading the wrong scale. Where the arm crosses the protractor, two numbers are printed, 50 and 130, and both look like an answer. The decision in step 1 is what tells you which one it is.

Reflex angles. A reflex angle is between 180° and 360°, and a protractor stops at 180°. The two angles between the same pair of arms add up to 360°, so you work with the other one.

110° 250° Q P R Asked for: reflex angle PQR = 250°. A 180° protractor cannot reach it, so use the other angle at Q: 360 − 250 = 110. Draw 110° from QP in the usual way. The 250° is the angle on the outside: mark it with an arc the long way round. The two angles between the same two arms always add up to 360°.
To draw a reflex angle of 250°: work out 360 − 250 = 110, draw a 110° angle in the usual way, then mark the angle on the outside with an arc that goes the long way round. To measure a reflex angle: measure the angle on the inside and take it away from 360.
Non-calculator: take away from 360 in two hops: 360 − 125 = 260 − 25 = 235. Then check by adding back: 235 + 125 = 360.

Accuracy. Measure and draw lengths to the nearest millimetre and angles to the nearest degree. Mark schemes allow only a small error either way, so draw thin, sharp lines and read each scale with your eye directly above it.

Constructing a triangle from three sides

Construct triangle ABC with AB = 7 cm, BC = 5 cm, CA = 4 cm. Leave every arc showing. A B C C′ 7 cm 4 cm 5 cm Arcs: radius 4 cm, centre A, and radius 5 cm, centre B. They cross above AB at C and again below AB at C′. Either point gives the same triangle: one is the mirror image of the other, so they are congruent. Choose one; here it is C. Draw AB first, then the arcs. Join C to A and B with a ruler only once C is chosen. Never rub out an arc.
  1. Draw the longest side first, accurately, with a ruler. Label the ends.
  2. Open the compasses to the second length. Point on one end, draw an arc.
  3. Open the compasses to the third length. Point on the other end, draw an arc that crosses the first.
  4. Arcs that are long enough cross at two points, one on each side of the line you drew first. Either point gives the same triangle (the two triangles are mirror images, so they are congruent): choose one. It is the third vertex.
  5. Join the third vertex to both ends with a ruler. Leave every arc on the page.
The mistake: rubbing the arcs out to make the diagram look tidy. The arcs are the working. Erasing them is the same as rubbing out the algebra and leaving only the answer.
When the arcs will not meet: the two shorter sides must add up to more than the longest side. Sides of 3 cm, 4 cm and 8 cm make no triangle: 3 + 4 = 7, which is less than 8, so the two arcs never reach each other. Check that before blaming the compasses.

A rhombus from two triangles

This is the syllabus's own example of a construction. All four sides of a rhombus are equal, so one diagonal splits it into two triangles whose three sides you already know. Draw the diagonal first, then do the triangle construction on both sides of it. This time you keep both crossing points: each one is the third vertex of a triangle.

A C B D 8 cm 5 cm 5 cm 5 cm 5 cm Rhombus ABCD: sides 5 cm, AC = 8 cm. 1. Draw the diagonal AC with a ruler. 2. Arcs of radius 5 cm from A: above and below. 3. Arcs of radius 5 cm from C, crossing both. 4. Above AC is B, below AC is D. Keep both. 5. Join AB, BC, CD and DA with a ruler. Two triangles, ABC and ADC, back to back on their shared side AC. Same construction as the triangle, done on both sides of the diagonal: both crossing points are used.
Worked in full
Construct triangle ABC with AB = 7 cm, BC = 5 cm and CA = 4 cm. Describe each step.
1
Draw AB = 7 cm with a ruler
Start with the longest side. It gives the other arcs room to cross cleanly.
2
Compasses set to 4 cm, point on A, draw an arc
Every point on that arc is 4 cm from A, so C is somewhere on it.
3
Compasses set to 5 cm, point on B, draw an arc
Every point on this one is 5 cm from B. C is on this one too.
4
The arcs cross twice, once on each side of AB
Both crossings are 4 cm from A and 5 cm from B. They give mirror-image triangles, which are congruent, so either will do. Choose one and call it C.
5
Join AC and BC with a ruler, leave the arcs
The arcs are the marks. Do not tidy them away.
Last step is yours
Construct rhombus ABCD with sides of 5 cm and diagonal AC = 8 cm. Then measure the other diagonal, BD.
1
Draw AC = 8 cm with a ruler
The diagonal is the side that the two triangles share.
2
Compasses set to 5 cm, point on A, draw one arc above AC and one below
B and D are both 5 cm from A.
3
Compasses still at 5 cm, point on C, cross both arcs
The crossing above AC is B and the crossing below is D. A rhombus uses both.
4
Join AB, BC, CD and DA with a ruler, leave the arcs
Two triangles, ABC and ADC, back to back on AC.
5
BD = 6 cm
Check by calculation: the diagonals of a rhombus cut each other in half at right angles. Half of AC is 4 and the side is 5, so half of BD is 3 (a 3, 4, 5 triangle) and BD = 6 cm.
Last two are yours
Construct triangle PQR with PR = 10 cm, PQ = 8 cm and QR = 6 cm. Then measure angle PQR.
1
Draw PR = 10 cm with a ruler
The longest side first.
2
Compasses set to 8 cm, point on P, draw an arc
Q is 8 cm from P, so it is somewhere on this arc.
3
Compasses set to 6 cm, point on R, draw an arc across the first
Q is also 6 cm from R.
4
The arcs cross at two points, one each side of PR: choose either as Q
Mirror images, so the same triangle. Join PQ and QR with a ruler and leave the arcs.
5
Angle PQR = 90°
Protractor centre on Q, 0 line along QP. Check: 6² + 8² = 36 + 64 = 100 = 10², so the angle opposite the 10 cm side is a right angle. A measurement that agrees with a calculation can be trusted.
All yours
A rhombus is constructed from two triangles on a diagonal of 24 cm, with every side 13 cm. How long is the other diagonal, in cm?

Three more

The second arm of an angle crosses 35 on one scale of a protractor and 145 on the other. The angle is obtuse. How many degrees is it?
You need to draw a reflex angle of 235° with a 180° protractor. What size of angle do you actually measure and draw?
A line is drawn 8.7 cm long. Give its length in millimetres.

Nets

A net is a flat shape that folds up into a solid. Every face of the solid appears on it exactly once, so the area of the net is the surface area of the solid. The net folds along the edges where two of its faces meet; every other edge on its outline is glued to another edge of the outline.

Six squares joined edge to edge can be arranged in 35 different shapes (turning a shape round or over does not make a new one). Exactly 11 of them fold into a cube, and these are all of them:

1–4–1: four in a row, with one square on each side of it (6 nets) 2–3–1 (3 nets) 2–2–2 (1) 3–3 (1) 1 2 3 4 5 6 7 8 9 10 11

Three of the other 24, and why each one fails:

Contains a 2 by 2 block: four squares round one point, but only three faces meet at a cube corner. Five in a row: a row of four already goes all the way round, so the fifth lands on the first. Two squares on the same side of the row of four: both fold over the same open end.
Three quick ways to rule a shape out. A cube net never contains a 2 by 2 block of squares: four faces would meet at one corner, and only three faces meet at a corner of a cube. It never has five squares in a row: a row of four already goes all the way round. And when it has four in a row, the other two squares sit one on each side of that row. Passing all three is not enough on its own: fold the shape in your head, or find it among the 11 above.

Reading a net: opposite faces and edges that meet

A B C D E F Same colour = opposite faces. C, D, E is a line of three: C and E. B, C, F is a line of three: B and F. A, B, C, D is a Z-shape: A and D. Same colour = edges that meet. Three squares round a point, with a 90° gap: the two edges beside the gap fold together, so the dots at their far ends meet.

Opposite faces. In a straight line of three squares, the two end squares are opposite: the middle one folds up between them. The two end squares of a Z-shape of four are opposite too. In every cube net these two rules find at least two of the three pairs, and the two faces left over make the third pair. Two squares that share an edge on the net are never opposite.

Edges that meet. Where three squares sit round one point with a 90° gap, the two edges either side of the gap fold onto each other, so their far ends meet as well. That is how to answer a question asking which point meets which.

Nets of cuboids, prisms and pyramids

A cuboid is three pairs of equal rectangles. A triangular prism is two equal triangles and three rectangles, one rectangle for each side of the triangle. A square-based pyramid is a square with a triangle on each edge. Before adding anything up, count the faces against the solid: 6, 5 and 5.

Net 1: cuboid 5 3 2 3 5 3 Net 2: triangular prism 10 3 4 5 Net 3: square-based pyramid 6 5 Each net is drawn to scale. Lengths are in cm.
Drawing a net accurately: draw rectangles with a ruler, and their right angles with a protractor. When you know all three sides of a triangle, draw it with the compass construction above. Every pair of edges that will be glued together must be the same length, or the net will not fold.
The mistake: using the height of a pyramid's triangle, measured on the net, as the height of the pyramid. On the net that length runs up the sloping face: it is the slant height. The pyramid's own height goes straight up from the centre of the base, and it is shorter.
Non-calculator: for a cuboid, work out the three different rectangles once and double the total. 2 × (15 + 10 + 6) is quicker, and safer, than adding six separate areas.
Worked in full
Net 1 above folds into a cuboid. Work out its surface area and its volume.
1
Three different rectangles: 5 × 3, 5 × 2 and 3 × 2
Read the lengths off the net. Each rectangle appears twice.
2
Their areas: 15, 10 and 6 cm²
One of each pair.
3
Surface area = 2 × (15 + 10 + 6) = 2 × 31 = 62 cm²
Double, because every face has an equal partner opposite it.
4
The cuboid is 5 cm by 3 cm by 2 cm
The three different edge lengths on the net are its length, width and height.
5
Volume = 5 × 3 × 2 = 30 cm³
Area is in cm², volume in cm³. Check the units before you finish.
Last step is yours
Net 2 folds into a triangular prism. Each triangle has sides of 3 cm, 4 cm and 5 cm, with a right angle between the 3 cm and 4 cm sides, and the prism is 10 cm long. Work out its surface area.
1
Two triangles: 2 × (½ × 3 × 4) = 12 cm²
The 3 cm and 4 cm sides meet at the right angle, so they are the base and the height.
2
Three rectangles, all 10 cm long: 10 × (3 + 4 + 5) = 120 cm²
One rectangle for each side of the triangle. Adding the widths first saves two multiplications.
3
Surface area = 12 + 120 = 132 cm²
Five faces, all counted.
Last two are yours
Net 3 folds into a square-based pyramid. The square has sides of 6 cm, and each triangle is 5 cm high on the net. Work out the surface area and the volume of the pyramid.
1
Base: 6 × 6 = 36 cm²
The square.
2
Each triangle: ½ × 6 × 5 = 15 cm², so four make 60 cm²
The 5 cm on the net is the slant height of each face.
3
Surface area = 36 + 60 = 96 cm²
Base plus the four triangles.
4
Height² = 5² − 3² = 16, so the height is 4 cm
The top is directly above the centre of the base, 3 cm in from each edge. The 5 cm slant height is the hypotenuse, not the height.
5
Volume = ⅓ × 36 × 4 = 48 cm³
The volume of a pyramid is one third of the base area × the height.
All yours
The net of a cube has a total area of 150 cm². Work out the volume of the cube, in cm³.

Four more

How many different nets does a cube have? (A net turned round or over counts as the same net.)
3 2 1 ?
On a dice, opposite faces add up to 7. The net above shows three of the numbers. What number goes on the face with the question mark?
P Q R S T
The net above is folded to make a cube. Which lettered point meets point P?
A closed box is a cuboid 10 cm by 6 cm by 4 cm. Work out the total area of its net, in cm².
E4.3 · low3 · Scale drawings and bearings▼
▶  Watch: E4.3 Scale drawings
Short hand-worked explanations for exactly this sub-topic. Opens on YouTube in a new tab. Watch one, then come straight back and try the questions below — watching without testing yourself feels like learning but is not.

Two skills sit in this sub-topic and they are usually examined together: reading a scale, and stating a bearing. Neither is difficult. Both are heavily penalised when written in the wrong form.

Scale

A scale of 1 : 25 000 means one unit on the map is 25 000 of the same unit in real life. The units must match before you multiply.

real distance = map distance × scale   |   map distance = real distance ÷ scale
Non-calculator: multiplying by 25 000 by hand: multiply by 25, then move the digits three places for the thousand. 6 × 25 = 150, so 6 × 25 000 = 150 000. Never reach for a calculator for a power of ten.
The mistake: answering in centimetres when the question asked for kilometres. After the multiplication you are in centimetres. Divide by 100 for metres, then by 1000 for kilometres — so divide by 100 000 in total.

Bearings — three rules, no exceptions

N A B 060° clockwise from north N 240° the bearing of A from B The two north lines are parallel, so the two bearings are co-interior: they differ by exactly 180°.
  1. Measured from north.
  2. Measured clockwise.
  3. Written with three figures always. Forty degrees is 040°, not 40°.
DirectionBearing
North000° or 360°
North-east045°
East090°
South-east135°
South180°
South-west225°
West270°
North-west315°
Back bearings: the bearing of A from B differs from the bearing of B from A by exactly 180°, because the two north lines are parallel and the two angles are co-interior. If the bearing you have is less than 180, add 180. If it is 180 or more, subtract 180. That choice keeps you inside 0 to 360.
The mistake: adding 180 when you should have subtracted, and producing a bearing over 360°. A bearing above 360 is always wrong. Look at the number first, then decide which way to go — that decision, not the arithmetic, is where the mark goes.
Worked in full
A map has scale 1 : 25 000. Two towns are 6 cm apart on the map. How far apart are they in real life, in km?
1
6 × 25 000
Map distance times the scale gives the real distance in the same unit, centimetres.
2
6 × 25 = 150, so the answer is 150 000 cm
Multiply by 25 first, then put the three zeros back.
3
150 000 ÷ 100 = 1500 m
100 cm in a metre.
4
1500 ÷ 1000 = 1.5 km
1000 m in a kilometre. The answer is 1.5 km.
5
Check it is sensible
A quarter of a kilometre per centimetre, six centimetres, so about one and a half kilometres. It fits.
Last step is yours
A map has scale 1 : 50 000. Two points are 3.4 cm apart on the map. Find the real distance in km.
1
3.4 × 50 000 = 170 000 cm
3.4 × 5 = 17, so 3.4 × 50 000 = 170 000.
2
170 000 ÷ 100 = 1700 m
Centimetres to metres.
3
1700 ÷ 1000 = 1.7 km
Metres to kilometres.
Last two are yours
The bearing of Q from P is 305°. Work out the bearing of P from Q.
1
305 is 180 or more
So subtract 180 rather than add it. Decide this before doing any arithmetic.
2
305 − 180 = 125
Take 180 off. Check the result is between 0 and 360 — it is.
3
Written as a bearing: 125°
Already three figures, so no leading zero needed here.
All yours
The bearing of Y from X is 072°. Work out the bearing of X from Y. Give three figures, digits only.

Five more

Write a bearing of forty degrees in correct three-figure form. Digits only, no degree sign.
Write the bearing for north-east. Digits only.
Write the bearing for due west. Digits only.
A model is made to a scale of 1 : 20. The real car is 4.6 m long. How long is the model, in cm?
On a map with scale 1 : 100 000, two places are 8 cm apart. Find the real distance in km.
E4.4 · medium4 · Similarity, and the k² / k³ trap▼
▶  Watch: E4.4 Similarity
Short hand-worked explanations for exactly this sub-topic. Opens on YouTube in a new tab. Watch one, then come straight back and try the questions below — watching without testing yourself feels like learning but is not.

Two shapes are similar when one is an enlargement of the other: every angle equal, every pair of corresponding sides in the same ratio. That ratio is the scale factor, written k.

6 cm 8 cm 10 cm triangle ABC 9 cm ? 15 cm triangle PQR — same shape, bigger ×1.5
k = (a length on the new shape) ÷ (the matching length on the old shape)

Get k the right way up by asking whether the answer should be bigger or smaller. Enlarging means k is greater than 1. If your k comes out less than 1 and the shape is clearly getting bigger, you have divided the wrong way round.

The part that is actually worth marks: k² and k³

length 1 area 1 length ×2 area ×4 = 2² length ×3 area ×9 = 3², volume ×27 = 3³ Multiply every length by k, and area goes ×k², volume goes ×k³.
What you are scalingMultiply byWhy
Lengths, perimeters, radii, heightskOne dimension
Areas, surface areask²Two dimensions, each stretched by k
Volumes, capacities, masses of the same materialk³Three dimensions
The mistake: doubling every length and then doubling the area too. Doubling the lengths quadruples the area and gives eight times the volume. This is the single most commonly dropped mark in the whole of E4.4, and it is dropped by people who understand similarity perfectly well.
Going backwards: if you are given the areas and want the lengths, square-root the area ratio. If you are given the volumes, cube-root them. Areas 9 : 25 means lengths 3 : 5. Volumes 8 : 27 means lengths 2 : 3.
Non-calculator: cube roots of 8, 27, 64, 125 and 1000 should come out of memory, and square roots up to 15² = 225. Similar-solids questions are built out of exactly those numbers, so on Paper 2 they are pure recall.
Worked in full
Triangles ABC and PQR are similar. AB = 6 cm, BC = 8 cm and PQ = 9 cm, where PQ corresponds to AB. Find QR.
1
Identify the pair that is fully known
AB = 6 and PQ = 9 both given, and they correspond. That pair gives k.
2
k = 9 ÷ 6 = 1.5
New over old. PQR is the bigger triangle, so k above 1 is what you expect.
3
QR corresponds to BC
So QR = BC × k.
4
QR = 8 × 1.5 = 12 cm
8 and a half of 8, which is 8 + 4 = 12.
5
Sense check
Every side of PQR should be half as big again as ABC. 6 to 9 and 8 to 12 — consistent.
Last step is yours
Two similar shapes have corresponding sides 5 cm and 12.5 cm. Another side of the small shape is 4 cm. Find the matching side of the large one.
1
k = 12.5 ÷ 5 = 2.5
New over old.
2
matching side = 4 × 2.5 = 10 cm
Multiply the known small length by k. 4 × 2.5 is 4 × 2 plus half of 4.
Last two are yours
Two similar shapes have corresponding lengths 4 cm and 10 cm. The smaller has area 24 cm². Find the area of the larger.
1
k = 10 ÷ 4 = 2.5
Length scale factor first, always.
2
Area scales by k²
Not by k. This is the step the mark is on.
3
k² = 2.5 × 2.5 = 6.25
2.5 squared. Do it as 2.5 × 2 = 5 plus 2.5 × 0.5 = 1.25.
4
area = 24 × 6.25 = 150 cm²
24 × 6 = 144, and 24 × 0.25 = 6, so 150 cm².
All yours
Two similar solids have heights 3 cm and 6 cm. The smaller has volume 20 cm³. Find the volume of the larger, in cm³.

Five more

Two similar shapes have lengths in the ratio 2 : 5. Write the ratio of their areas in the form a:b.
Two similar solids have volumes 27 cm³ and 64 cm³. Write the ratio of their lengths in the form a:b.
Two similar cones have surface areas 9 cm² and 25 cm². The smaller has height 6 cm. Find the height of the larger, in cm.
A photograph is enlarged by scale factor 3. By what factor does its area increase?
Two similar bottles have heights 8 cm and 12 cm. The small one holds 40 ml. How much does the large one hold, in ml?

Showing that two triangles are similar

Two triangles are similar if two pairs of angles are equal; the third pair must then be equal too, because each triangle adds to 180°. A “show that” question wants each pair named with its reason: common angle, corresponding angles, alternate angles or vertically opposite angles. Then match the sides that sit opposite equal angles to find k.

ABCDE4 cm6 cm5 cmBC = ?DE is parallel to BCangle A: commonangle ADE = angle ABC(corresponding angles)
Worked in full
In the diagram DE is parallel to BC, AD = 4 cm, DB = 6 cm and DE = 5 cm. (a) Show that triangles ADE and ABC are similar. (b) Find BC.
1
angle DAE = angle BAC
Reason: it is a common angle — the same angle belongs to both triangles.
2
angle ADE = angle ABC
Reason: corresponding angles, because DE is parallel to BC.
3
so the triangles are similar
Two pairs of equal angles is enough.
4
AB = 4 + 6 = 10 cm, so k = 10 ÷ 4 = 2.5
AD in the small triangle matches AB in the big one. Use the whole side AB, not DB.
5
BC = 5 × 2.5 = 12.5 cm
DE matches BC. Check: BC is bigger than DE, as it must be.
The mistake: using DB = 6 as the matching side and getting k = 6 ÷ 4 = 1.5. The big triangle is ABC, so its side is the whole of AB.
All yours
DE is parallel to BC, with D on AB and E on AC. AD = 3 cm, DB = 9 cm and DE = 4 cm. Find BC, in cm.
D lies on AC and angle ABD = angle ACB. AB = 8 cm and AD = 4 cm. Triangles ABD and ACB are similar. Find AC, in cm.
E4.5 · low5 · Symmetry▼
▶  Watch: E4.5 Symmetry
Short hand-worked explanations for exactly this sub-topic. Opens on YouTube in a new tab. Watch one, then come straight back and try the questions below — watching without testing yourself feels like learning but is not.

Two kinds of symmetry, and they are counted differently. Line symmetry counts folds. Rotational symmetry counts how many times a shape fits onto itself in one full turn.

square 4 lines, order 4 rhombus 2 lines, order 2 parallelogram 0 lines, order 2 A line of symmetry is a fold. Rotational order is how many times it fits itself in one full turn.
The mistake: saying a parallelogram has two lines of symmetry because it “looks symmetrical”. Fold a parallelogram down the middle either way and the halves do not land on each other. It has no lines of symmetry — but it does have rotational symmetry of order 2. Those two facts sit together and are easy to mix up.

The table to know cold

ShapeLines of symmetryOrder of rotational symmetry
Square44
Rectangle (not square)22
Parallelogram02
Rhombus22
Kite11
Trapezium (isosceles)11
Equilateral triangle33
Isosceles triangle11
Scalene triangle01
Regular polygon with n sidesnn
Circleinfinitely manyinfinite
Order 1 means none: every shape returns to itself after a full 360° turn, so the smallest possible order is 1. Never write order 0. A shape with no rotational symmetry has order 1.

The angle you turn through is 360 ÷ order. Order 5 means a turn of 72°, and a shape that looks the same after 72° has order 5.

Symmetry of solids

For a three-dimensional shape you count planes of symmetry — flat cuts that leave two mirror-image halves — and the order of rotational symmetry about an axis.

SolidPlanes of symmetryRotational symmetry
Cuboid, all three edges different3order 2 about each of three axes
Cube9order 4 about each face axis
Cylinderinfinitely many, plus one across the middleinfinite about its main axis
Square-based pyramid4order 4 about the vertical axis
Sphereinfinitely manyinfinite about any axis through the centre
Coneinfinitely many (every plane through its axis)infinite about its axis
Prism whose cross-section is a regular n-sided polygon (not a cube)n + 1: n through the main axis and 1 across the middleorder n about the main axis
Worked in full
State the number of lines of symmetry and the order of rotational symmetry of a regular octagon.
1
Regular means all sides and all angles equal
That is what makes the two counts equal.
2
An octagon has 8 sides
So there are 8 mirror lines: four through opposite vertices and four through opposite edge midpoints.
3
Lines of symmetry = 8
One for each side, for a regular polygon.
4
Order of rotational symmetry = 8
It fits onto itself every 360 ÷ 8 = 45°.
5
Check
Regular polygon with n sides: n lines, order n. Always.
Last step is yours
State the number of lines of symmetry and the order of rotational symmetry of a rhombus.
1
A rhombus has four equal sides but no right angles
Its two diagonals are the only fold lines.
2
Lines of symmetry = 2
Along each diagonal. Folding along a side-to-side line does not work.
3
Order of rotational symmetry = 2
A half turn maps it onto itself, and so does the full turn.
Last two are yours
State the number of lines of symmetry and the order of rotational symmetry of a parallelogram that is not a rhombus or a rectangle.
1
Try folding it along a diagonal
The halves do not match. The slant goes the wrong way on one of them.
2
Lines of symmetry = 0
None at all, despite how balanced it looks.
3
Now rotate it half a turn
It lands exactly on itself.
4
Order of rotational symmetry = 2
The half turn and the full turn. Order 2, never order 0.
All yours
Work out the order of rotational symmetry of a regular polygon with 12 sides.

Five more

How many lines of symmetry does a kite have?
How many planes of symmetry does a cuboid with three different edge lengths have?
A shape looks exactly the same after a rotation of 72°. What is its order of rotational symmetry?
How many lines of symmetry does a parallelogram that is not a rhombus or a rectangle have?
What is the smallest possible order of rotational symmetry for any shape?
All yours
Symmetry of solids
How many planes of symmetry does a prism with a regular pentagon as its cross-section have?
Write down the order of rotational symmetry of a regular hexagonal prism about its main axis.
How many planes of symmetry does a cone have? Type infinite if there is no limit.
E4.6 · medium6 · Angles — always with a reason▼
▶  Watch: E4.6 Angles
Short hand-worked explanations for exactly this sub-topic. Opens on YouTube in a new tab. Watch one, then come straight back and try the questions below — watching without testing yourself feels like learning but is not.

This is the sub-topic your check broke on, at step 6, so it is rebuilt here from the bottom rather than assumed. The facts themselves are step 3 and 4 material. What is Extended is chaining them — using one angle to unlock the next — and writing the reason every time.

Naming an angle with three letters

Angle PRQ (also written ∠PRQ or PR̂Q) is the angle at R: the middle letter is always the vertex, and the other two letters are points on its two arms. Angle PRQ and angle QRP are the same angle. Exam questions name angles this way, so read the middle letter first.

PQRangle PRQthe middle letter, R,is the corner
All yours
In triangle PQR, at which vertex is angle PQR? Type the letter.
Write a three-letter name for the angle at B between the lines BA and BC.

The reasons, word for word

Cambridge awards the angle and the reason on separate marks. The wording below is what mark schemes accept. Learn the phrases, not the idea — the idea you already have.

The factThe words that earn the mark
Straight lineangles on a straight line add to 180°
Full turnangles at a point add to 360°
Crossing linesvertically opposite angles are equal
Parallel lines, F shapecorresponding angles are equal
Parallel lines, Z shapealternate angles are equal
Parallel lines, C shapeco-interior angles add to 180°
Triangleangles in a triangle add to 180°
Triangle, side extendedthe exterior angle of a triangle equals the sum of the two opposite interior angles
Isosceles trianglebase angles of an isosceles triangle are equal
Quadrilateralangles in a quadrilateral add to 360°
Any polygonexterior angles of a polygon add to 360°
“because it looks like it” — 0 marks. “because they are Z angles” — risky. “alternate angles are equal” — the mark.

Parallel lines

a b c d e f g h b and f are corresponding (F shape) — equal d and e are alternate (Z shape) — equal d and f are co-interior (C shape) — add to 180° a and d are vertically opposite — equal The arrows mean the two lines are parallel. Without arrows you may not assume it.
The mistake: writing “co-interior angles are equal”. They are the one pair that is not equal — they add to 180°. This is exactly the kind of slightly-off rule that produces a confident wrong answer, and the check found five of those. If in doubt, look at the picture: two co-interior angles are obviously different sizes unless both happen to be 90°.
How to spot them without guessing: trace the shape the two angles make with the transversal. An F (corresponding) and a Z (alternate) both give equal angles. A C or U (co-interior) gives angles that add to 180°. The letters can be back to front or upside down and still count.
Non-calculator: subtracting from 180 is where sign slips live. Do it in two easy hops rather than one hard one: 180 − 47 is 180 − 40 = 140, then 140 − 7 = 133. Same for 360: 360 − 145 is 360 − 100 = 260, then 260 − 45 = 215. Two small subtractions beat one borrow.

Triangles and polygons

exterior Walk once round the outside and you turn through 360° in total — always, whatever the polygon. Regular polygon: one exterior angle = 360 ÷ n. Interior angle = 180 − exterior. interior sum = (n − 2) × 180
RuleFormulaWorth knowing
Angles in a triangle180°Everything else is built on this
Angles in a quadrilateral360°Two triangles stuck together
Interior angle sum, n sides(n − 2) × 180°n − 2 is the number of triangles it splits into
Exterior angle sum, any polygon360°Does not depend on n at all
Regular polygon, one exterior angle360 ÷ nUsually the fastest route in
Regular polygon, one interior angle180 − (360 ÷ n)Interior and exterior are on a straight line
Choose the exterior angle route: for a regular polygon, going through the exterior angle is almost always fewer steps than the interior sum. One interior angle of a regular 12-gon: exterior is 360 ÷ 12 = 30, so interior is 180 − 30 = 150. Doing it the other way needs (12 − 2) × 180 ÷ 12, which is three operations and a bigger multiplication.
Worked in full
Two parallel lines are crossed by a transversal. One angle is 63°. Work out the co-interior angle on the same side, and give a reason at every step.
1
The 63° angle and the angle directly below it on the other parallel line are corresponding
Reason: corresponding angles are equal. So that angle is also 63°.
2
That 63° angle and the one next to it lie on a straight line
Reason: angles on a straight line add to 180°.
3
180 − 63
Do it in hops: 180 − 60 = 120, then 120 − 3 = 117.
4
The co-interior angle is 117°
Reason: co-interior angles add to 180°. Either route gets there; the reason is what is marked.
5
Check: 63 + 117 = 180
It does. Any co-interior pair must total 180°.
Last step is yours
Two parallel lines are crossed by a transversal. One co-interior angle is 118°. Find the other.
1
Name the relationship first
They are co-interior: on the same side of the transversal, between the two parallel lines.
2
Co-interior angles add to 180°
So the other angle is 180 − 118.
3
180 − 118 = 62°
180 − 100 = 80, then 80 − 18 = 62. Reason: co-interior angles add to 180°.
Last two are yours
In a triangle, two of the interior angles are 47° and 68°. One side is extended. Find the exterior angle at the third vertex, and the third interior angle.
1
The exterior angle equals the sum of the two opposite interior angles
That is the named rule. It saves a step compared with finding the third angle first.
2
47 + 68
40 + 60 = 100, then 7 + 8 = 15.
3
exterior angle = 115°
Reason: the exterior angle of a triangle equals the sum of the two opposite interior angles.
4
third interior angle = 180 − 115 = 65°
Reason: angles on a straight line add to 180°. Check: 47 + 68 + 65 = 180.
All yours
A regular polygon has 15 sides. Work out the size of one exterior angle, in degrees.

Eight more — state the reason to yourself before you type the number

Two angles lie on a straight line. One is 47°. Find the other, in degrees.
Three angles meet at a point. Two of them are 90° and 145°. Find the third, in degrees.
Work out the sum of the interior angles of a decagon (10 sides), in degrees.
Work out one interior angle of a regular hexagon, in degrees.
A regular polygon has an exterior angle of 20°. How many sides does it have?
Two co-interior angles between parallel lines: one is 73°. Find the other, in degrees.
Two alternate angles between parallel lines: one is 56°. Find the other, in degrees.
A quadrilateral has angles 85°, 100° and 62°. Find the fourth angle, in degrees.
E4.7 · medium7 · Circle theorems I — the five that do most of the work▼
▶  Watch: E4.7 Circle theorems I
Short hand-worked explanations for exactly this sub-topic. Opens on YouTube in a new tab. Watch one, then come straight back and try the questions below — watching without testing yourself feels like learning but is not.

Five theorems. They are the reason circle questions look impossible and then take two lines. Every one of them is a licence to write down an angle you were not given, and each has a set phrase that earns the reason mark.

The five, with the wording

TheoremThe words that earn the mark
Angle in a semicirclethe angle in a semicircle is 90°
Tangent and radiusa tangent meets a radius at 90°
Angle at the centrethe angle at the centre is twice the angle at the circumference
Same segmentangles in the same segment are equal
Cyclic quadrilateralopposite angles of a cyclic quadrilateral add to 180°

One more that is not a theorem but does half the work in most questions: two radii always make an isosceles triangle, because all radii of a circle are equal. Reason: base angles of an isosceles triangle are equal.

O A B C AB is a diameter, so angle ACB = 90° reason: the angle in a semicircle is 90° O 120° 60° A B C angle AOB = 2 × angle ACB reason: the angle at the centre is twice the angle at the circumference
Reading a circle diagram: look for the centre first. If the centre is marked, the angle-at-the-centre theorem or an isosceles triangle is probably involved. If a line goes straight through the centre and both ends touch the circle, that is a diameter and you have a 90° angle for free. If four points on the circle are joined into a quadrilateral, opposite angles add to 180°.
A B C D angle ACB = angle ADB reason: angles in the same segment are equal P R S Q x 180 − x angle P + angle R = 180°, and angle Q + angle S = 180° reason: opposite angles of a cyclic quadrilateral add to 180°
The mistake: halving when you should double. Centre to circumference means halve; circumference to centre means double. Before you touch the arithmetic, decide which of the two angles is at the centre. The angle at the centre is always the larger of the two, so if your answer at the centre is smaller than the one at the circumference, you have gone the wrong way.
O T angle OTP = 90° reason: a tangent meets a radius at 90° O P A B PA = PB, and OP bisects angle APB reason: tangents from an external point are equal
Non-calculator: all five theorems reduce to halving, doubling, or subtracting from 180. Nothing here needs more than mental arithmetic, which is why circle theorems are good Paper 2 marks — the difficulty is entirely in spotting which theorem applies.
Worked in full
A, B and C are on a circle with centre O. The angle AOB at the centre is 130°. Find angle ACB, and give reasons.
1
Identify which angle is at the centre
AOB, because O is the centre. ACB is at the circumference, standing on the same arc AB.
2
The angle at the centre is twice the angle at the circumference
That is the reason. Both angles must stand on the same arc for it to apply.
3
angle ACB = 130 ÷ 2
Going from centre to circumference, so halve.
4
angle ACB = 65°
Reason: the angle at the centre is twice the angle at the circumference.
5
Check the direction
65 is smaller than 130. The circumference angle should be the smaller one, so this is the right way round.
Last step is yours
AB is a diameter of a circle and C lies on the circle. Angle ABC = 34°. Find angle BAC.
1
AB is a diameter, so angle ACB = 90°
Reason: the angle in a semicircle is 90°.
2
The three angles of triangle ABC add to 180°
So 90 + 34 + angle BAC = 180.
3
90 + 34 = 124
Add the two known angles first, then subtract once.
4
angle BAC = 180 − 124 = 56°
Reason: angles in a triangle add to 180°.
Last two are yours
PQRS is a cyclic quadrilateral. Angle P = 84° and angle Q = 97°. Find angles R and S.
1
P and R are opposite, and Q and S are opposite
Pair them off before doing any arithmetic. Getting the pairing wrong is the usual error here.
2
Opposite angles of a cyclic quadrilateral add to 180°
That is the reason for both parts.
3
angle R = 180 − 84 = 96°
180 − 80 = 100, then 100 − 4 = 96.
4
angle S = 180 − 97 = 83°
180 − 90 = 90, then 90 − 7 = 83. Check: 84 + 97 + 96 + 83 = 360.
All yours
The angle at the centre of a circle standing on arc AB is 146°. Work out the angle at the circumference standing on the same arc, in degrees.

Six more

AB is a diameter and C is a point on the circle. Write down angle ACB, in degrees.
Two angles stand on the same chord and lie in the same segment. One is 38°. Find the other, in degrees.
A cyclic quadrilateral has one angle of 115°. Find the angle opposite it, in degrees.
OT is a radius and a tangent touches the circle at T. Write down the angle between them, in degrees.
An angle at the circumference is 41°. Find the angle at the centre standing on the same arc, in degrees.
O is the centre and A and B are on the circle. Angle OAB = 25°. Find angle AOB, in degrees.
E4.8 · low8 · Circle theorems II — chords, tangents and the alternate segment▼
▶  Watch: E4.8 Circle theorems II
Short hand-worked explanations for exactly this sub-topic. Opens on YouTube in a new tab. Watch one, then come straight back and try the questions below — watching without testing yourself feels like learning but is not.

Four more results. These are lower difficulty than section 7, but only once the first five are secure — they tend to appear as the second or third step of a longer question rather than on their own.

The perpendicular from the centre bisects a chord

O 6 8 8 10 The perpendicular from the centre bisects the chord so a right-angled triangle appears: 6² + 8² = 10². Pythagoras finishes it. d d Equal chords are the same distance from the centre and the reverse is true: same distance means same length.

Drop a perpendicular from the centre onto a chord and it lands exactly on the midpoint. That manufactures a right-angled triangle whose hypotenuse is a radius, and Pythagoras finishes the job.

(half the chord)² + (distance from centre)² = radius²
The mistake: using the whole chord in Pythagoras instead of half of it. The right-angled triangle has the half-chord as one of its legs. Using 16 instead of 8 gives an impossible answer — a leg longer than the hypotenuse — which is at least easy to spot if you check.
Non-calculator: these questions are built out of Pythagorean triples: 3-4-5, 5-12-13, 6-8-10, 8-15-17, 9-12-15, 7-24-25. When they are not, leave the answer as a surd. √12 simplifies to 2√3, because 12 = 4 × 3 and √4 = 2. Always pull out the largest square factor.

Equal chords are equidistant from the centre

Two chords of the same length in one circle sit the same distance from the centre, and the converse holds: same distance means same length. It follows directly from the Pythagoras picture above — same radius and same distance forces the same half-chord.

Tangents from an external point are equal

From a point outside a circle you can draw exactly two tangents, and they have the same length. The four points make a kite: two radii equal, two tangents equal, two right angles where each tangent meets its radius. The line from the external point to the centre bisects the angle between the tangents.

tangent length² + radius² = (distance from centre to external point)²

That is Pythagoras again, using the right angle between tangent and radius.

The alternate segment theorem

T A C x x The alternate segment theorem The angle between the tangent and the chord equals the angle in the alternate segment — that is, the angle standing on the same chord but on the other side of it. tangent at T
How to find the alternate segment: the chord splits the circle into two pieces. The angle between the tangent and the chord is equal to the angle standing on that chord in the other piece — the one the angle is not leaning into. Trace the chord with a finger and look across it.
Worked in full
A circle has radius 10 cm. A chord of length 16 cm is drawn. Find the distance from the centre to the chord.
1
Draw the perpendicular from the centre to the chord
Reason: the perpendicular from the centre bisects the chord.
2
Half the chord is 16 ÷ 2 = 8 cm
This is the leg of the right-angled triangle, not 16.
3
8² + d² = 10²
Pythagoras. The radius is the hypotenuse, because it runs from the centre to the circle.
4
64 + d² = 100
Squares first. Both come straight from memory.
5
d² = 100 − 64 = 36
Subtract carefully. It is the hypotenuse squared minus the leg squared, not the other way round.
6
d = 6 cm
√36 = 6. Note 6, 8, 10 — a scaled 3, 4, 5 triangle.
Last step is yours
A circle has radius 13 cm and a chord of length 24 cm. Find the distance from the centre to the chord.
1
Half the chord is 24 ÷ 2 = 12 cm
The perpendicular from the centre bisects the chord.
2
12² + d² = 13²
Radius is the hypotenuse.
3
144 + d² = 169
From the squares table.
4
d² = 169 − 144 = 25, so d = 5 cm
√25 = 5. This is the 5, 12, 13 triple.
Last two are yours
A circle has radius 4 cm and a chord of length 4 cm. Find the distance from the centre to the chord, in surd form.
1
Half the chord is 2 cm
Half of 4.
2
2² + d² = 4²
Pythagoras, radius as hypotenuse.
3
d² = 16 − 4 = 12
Not a perfect square, so the answer will be a surd. That is fine, and expected on Paper 2.
4
d = √12 = 2√3 cm
12 = 4 × 3, and √4 = 2, so √12 = 2√3. Leave it like that.
All yours
A circle has radius 5 cm and a chord of length 6 cm. Work out the distance from the centre to the chord, in cm.

Six more

Tangents PA and PB are drawn from an external point P. PA = 9 cm. Write down PB, in cm.
The angle between a tangent and a chord is 54°. Find the angle in the alternate segment, in degrees.
Two chords in the same circle are both 10 cm long. One is 4 cm from the centre. How far is the other from the centre, in cm?
A circle has radius 2 cm and a chord of length 2 cm. Find the distance from the centre to the chord in surd form. Type a root as sqrt, for example 2sqrt5.
A chord of length 12 cm is 8 cm from the centre of a circle. Find the radius, in cm.
A tangent from point P touches a circle at T. OT = 5 cm and OP = 13 cm. Find PT, in cm.
mixed9 · Mixed set — no labels, no order▼

Fifteen questions from everything above, shuffled and unlabelled. Deciding which fact applies is the part ordinary revision quietly does for you, and it is the part the exam tests. Before you calculate anything, say the reason out loud. If you cannot name a reason, you are guessing.

nothing answered yet
1. A regular polygon has an interior angle of 156°. How many sides does it have?
2. The bearing of B from A is 236°. Find the bearing of A from B. Three figures, digits only.
3. Two similar solids have surface areas 16 cm² and 81 cm². Write the ratio of their volumes in the form a:b.
4. AB is a diameter of a circle, C is on the circle, and angle CAB = 29°. Find angle ABC, in degrees.
5. Two co-interior angles between parallel lines are 4x and 5x. Find the value of x.
6. A circle has radius 17 cm and a chord of length 16 cm. Find the distance from the centre to the chord, in cm.
7. How many lines of symmetry does a regular pentagon have?
8. A map has scale 1 : 200 000. Two towns are 4.5 cm apart on the map. Find the real distance in km.
9. A cyclic quadrilateral has angles 3x, 90°, 2x and 110°, where 3x and 2x are opposite each other. Find x.
10. The exterior angle of a triangle is 128°. One of the two opposite interior angles is 51°. Find the other, in degrees.
11. A tangent touches a circle at P. What is the angle between the tangent and the radius drawn to P, in degrees?
12. The angle at the centre standing on an arc is 5 times the angle at the circumference standing on it? True or false. Answer true or false.
13. Two similar triangles have areas 20 cm² and 45 cm². The smaller has a side of 8 cm. Find the matching side of the larger, in cm.
14. A tangent touches a circle at T. The angle between the tangent and a chord TA is 67°. Find the angle subtended by TA in the alternate segment, in degrees.
15. An isosceles triangle has two base angles of 4x and an apex angle of 2x. Find the value of x.

What to take away

Six things, and between them they cover most of what an examiner can ask in topic 4:

  1. Say the reason before the number. Cambridge marks them separately, and the reason is the mark most often thrown away by people who got the angle right.
  2. Co-interior angles add to 180°. They are the only parallel-line pair that is not equal, and they are the most common confident wrong answer.
  3. Subtract from 180 in two hops. 180 − 47 is 180 − 40 = 140, then − 7 = 133. Your sign errors live in that one line, not in the geometry.
  4. Length k, area k², volume k³. Find k from lengths first, then square or cube it. Going backwards, square-root the areas and cube-root the volumes.
  5. Centre to circumference, halve. Circumference to centre, double. The angle at the centre is always the bigger one, so you have a check built in.
  6. A perpendicular from the centre bisects the chord, which makes a right-angled triangle with the radius as hypotenuse. Use half the chord, never the whole one, and leave surds as surds.

Come back to section 9 in a week without reading anything above it. A high score on a cold run means this topic is set, and the next one can start.