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G Geometry and measures — notes on one page

Every Edexcel GCSE Maths 1MA1 specification point in G, read straight through: 25 points with definitions, explanations, mark-scheme wording and worked examples. Practice questions are in the printable question pack.

Checked against Edexcel 1MA1 section G. Review basis: the qualification registry sourced from the Pearson Edexcel Level 1/Level 2 GCSE (9-1) in Mathematics (1MA1) specification; registry verification recorded 9 July 2026.

How this checking works

G1 · Use conventional terms/notations: points, lines, vertices, edges, planes, parallel and perpendicular lines, right angles, polygons; standard triangle labelling; draw diagrams from written description

  • Geometry uses agreed words and symbols so a diagram communicates facts precisely. A point names a position; a line continues, while a line segment has endpoints.
  • Vertices are corners and edges join vertices. Parallel lines never meet, whereas perpendicular lines meet at 90∘90^\circ.
  • In ∠ABC\angle ABC, the middle letter BB is the vertex; in triangle ABCABC, side ACAC is opposite BB.
  • Draw a written description one instruction at a time, label every required point and add conventional arrow, equality and right-angle marks.
  • The examiner awards stated or marked relationships, never what a sketch merely appears to show.
Conventional marks show two parallel lines and a perpendicular transversal.

Worked example

Draw triangle PQRPQR with PQ=PRPQ=PR. Put SS on QRQR, draw PS⊥QRPS\perp QR, and state which side is opposite PP.

  1. 1.Draw and label triangle PQRPQR, then mark PQPQ and PRPR with matching equality ticks.
  2. 2.Place SS on segment QRQR, join PP to SS, and add a right-angle mark at SS.
  3. 3.The side not touching vertex PP is QRQR, so QRQR is opposite PP.

Answer: A correctly labelled diagram with PQ=PRPQ=PR, SS on QRQR and PS⊥QRPS\perp QR; the opposite side is QRQR.

Common mistakes

  • Don't read ∠ABC\angle ABC as the angle at AA instead of the middle letter BB.
  • Don't use a relationship because the sketch looks parallel, equal or perpendicular although it is not stated or marked.

Exam tip. In a construction or drawing question, keep every requested label and conventional mark visible because each can carry an independent mark.

G2 · Use standard ruler and compass constructions (perpendicular bisector, perpendicular from/at a point, angle bisector); construct figures, solve loci problems; perpendicular distance is shortest

  • Ruler-and-compass constructions must show the arcs that create the result. For a perpendicular bisector of ABAB, draw equal-radius arcs from AA and BB with radius greater than half of ABAB, then join their intersections.
  • Every point on this line is equidistant from AA and BB.
  • An angle bisector is built from an arc centred at the vertex and equal arcs from the two cut points.
  • Translate loci language into boundaries: a fixed distance from a point gives a circle; a fixed distance from a line gives two parallels.
  • Perpendicular distance is the shortest distance to a line.
Equal-radius arcs from both endpoints locate the perpendicular bisector.

Worked example

A park must be equally distant from towns AA and BB and within 44 km of AA. Describe the complete locus of possible positions.

  1. 1.Construct the perpendicular bisector of ABAB because equal distances from AA and BB are required.
  2. 2.Draw the circle centred at AA with radius 44 km because the distance from AA is at most 44 km.
  3. 3.Keep the part of the perpendicular bisector inside or on that circle.

Answer: The segment of the perpendicular bisector of ABAB lying inside or on the circle centre AA, radius 44 km.

Common mistakes

  • Don't use a measured midpoint or protractor line instead of leaving the intersecting compass arcs visible.
  • Don't draw only the boundary circle for 'within' and omit the permitted interior region.

Exam tip. For a loci question, draw each condition separately, then shade or state only their intersection.

G3 · Apply angles at a point, on a straight line, vertically opposite angles; use alternate and corresponding angles on parallel lines; derive and use the angle sum of a triangle and of any polygon

Definition. Specification G3: apply the properties of angles at a point, angles at a point on a straight line and vertically opposite angles; understand and use alternate and corresponding angles on parallel lines; derive and use the sum of angles in a triangle, and use it to deduce the angle sum in any polygon and the properties of regular polygons.

  • Angles at a point total 360∘360^\circ, angles on a straight line total 180∘180^\circ, and vertically opposite angles are equal. Where a line crosses two parallel lines, corresponding angles are equal, alternate angles are equal, and co-interior (allied) angles add up to 180∘180^\circ.
  • Angles in a triangle add up to 180∘180^\circ, and an exterior angle of a triangle equals the sum of the two interior opposite angles.
  • Splitting an nn-sided polygon into n−2n-2 triangles gives an interior-angle sum of (n−2)×180∘(n-2)\times180^\circ; the exterior angles of any polygon total 360∘360^\circ, so a regular polygon has exterior angle 360∘÷n360^\circ\div n.
  • In a reasoning question each step needs a calculation and the reason for it, with the key words in the reason and the reason tied to the parallel lines or triangle actually used.
  • Angles found should be marked on the diagram or named.
Alternate angles are equal when a transversal crosses parallel lines.

Worked example

1

Each exterior angle of a regular polygon is 24∘24^\circ. Work out the number of sides of the polygon and the size of each interior angle.

(3)

(Total for Question 1 is 3 marks)

Working

  1. 1.The exterior angles of any polygon add up to 360∘360^\circ, so the number of sides is 360÷24=15360\div24=15.
  2. 2.An interior angle and its exterior angle lie on a straight line, so they add up to 180∘180^\circ.
  3. 3.Each interior angle is 180−24=156∘180-24=156^\circ.

Answer: 1515 sides; each interior angle is 156∘156^\circ.

Mark scheme

Mark scheme for question 1
QuestionAnswerMarkAdditional guidance
1for 360÷24360\div24M1oe
for 1515A1cao
for 156156B1ft 180−24180-24

Mark-scheme wording to use

  • “Corresponding angles are equal … Allied angles / Co-interior angles add up to 180 … Angles on a straight line add up to 180 … Angles in a triangle add up to 180” — Pearson 1MA1, June 2020 Paper 1H, Q6
  • “When reasons are given the key words underlined must be present. Reasons need to be linked to their method; any reasons not linked, do not credit. There should be no incorrect reasons given” — Pearson 1MA1, June 2020 Paper 1H, Q6
  • “M1 for finding an exterior angle eg 360 ÷ 6 (= 60) or 360 ÷ 5 (= 72) or an interior angle eg 180 × 4 ÷ 6 (= 120) or 180 × 3 ÷ 5 (= 108) … M1 for a complete method eg 360 – “120” – “108” or “60” + ”72” A1 cao Answer only award no marks” — Pearson 1MA1, June 2022 Paper 1H, Q5

What the examiner wants. The correct name for the fact used. The November 2024 Paper 1H report says the reasoning mark was for an appropriate parallel-lines reason, usually “alternate angles”, and that “corresponding angles” was a common wrong reason for the same angle; the June 2024 Paper 2H report warns that showing the angles sum to 360∘360^\circ does not show a shape is a trapezium.

Common mistakes

  • Don't claim alternate or corresponding angles are equal without establishing that the two lines are parallel.
  • Don't use n×180∘n\times180^\circ instead of (n−2)×180∘(n-2)\times180^\circ for an interior-angle sum.

Exam tip. A 'give a reason' angle question needs the named fact, such as 'alternate angles, parallel lines', not just the arithmetic.

G4 · Derive and apply properties and definitions of special quadrilaterals (square, rectangle, parallelogram, trapezium, kite, rhombus), triangles and other plane figures using appropriate language

  • Classify a plane shape using properties that must hold, not how the drawing looks. A parallelogram has two pairs of parallel opposite sides; its opposite sides and angles are equal, and its diagonals bisect each other.
  • A rectangle is a parallelogram with four right angles and equal diagonals. A rhombus has four equal sides and perpendicular diagonals, while a square is both a rectangle and a rhombus.
  • A kite has two pairs of adjacent equal sides.
  • An isosceles triangle has equal base angles.
  • Examiners expect a property to be connected to a conclusion, especially in 'explain' and proof questions.
A parallelogram and its diagonals, which bisect each other.

Worked example

The diagonals of quadrilateral ABCDABCD bisect each other and meet at right angles. State the most specific guaranteed quadrilateral and justify it.

  1. 1.Diagonals that bisect each other establish that ABCDABCD is a parallelogram.
  2. 2.A parallelogram whose diagonals are perpendicular is a rhombus.
  3. 3.No right angle or equal diagonals are given, so a square is not guaranteed.

Answer: ABCDABCD must be a rhombus, but need not be a square.

Common mistakes

  • Don't state that every rhombus has four right angles because the diagram resembles a square.
  • Don't use 'the diagonals are equal' as a property of every parallelogram instead of rectangles and squares.

Exam tip. For a classification question, give the most specific shape forced by the information and state the defining property.

G5 · Use the basic congruence criteria for triangles (SSS, SAS, ASA, RHS)

  • Congruent triangles have exactly the same size and shape, so corresponding sides and angles are equal. Prove congruence using SSS, SAS, ASA or RHS.
  • SAS needs the angle included between the two known sides. RHS applies only to right-angled triangles and needs equal hypotenuses plus one equal shorter side.
  • Mark or state the three matching facts, then write triangle names in corresponding order.
  • Only after proving congruence may you use corresponding parts to establish another equality.
  • AAA proves similarity rather than congruence, and SSA does not determine a unique triangle, so neither is a valid general congruence test.
Matching two sides and the included angle establishes SAS congruence.

Worked example

Triangles ABCABC and DEFDEF satisfy AB=DEAB=DE, AC=DFAC=DF and ∠BAC=∠EDF\angle BAC=\angle EDF. Prove that BC=EFBC=EF.

  1. 1.AB=DEAB=DE and AC=DFAC=DF give two pairs of corresponding equal sides.
  2. 2.∠BAC=∠EDF\angle BAC=\angle EDF is the included angle between those pairs.
  3. 3.Therefore △ABC≅△DEF\triangle ABC\cong\triangle DEF by SAS, so corresponding sides BCBC and EFEF are equal.

Answer: BC=EFBC=EF by corresponding sides of congruent triangles.

Common mistakes

  • Don't use SAS with an angle that is not between the two stated equal sides.
  • Don't write AAA as a congruence proof even though the triangles could have different sizes.

Exam tip. In a proof, state the three matching facts before naming the congruence criterion and the corresponding conclusion.

G6 · Apply angle facts, congruence, similarity and quadrilateral properties to derive results about angles and sides, incl. Pythagoras' theorem and isosceles base angles, and obtain simple proofs

  • Combine established geometry facts to derive a new result. Equal sides in an isosceles triangle face equal angles, and equal angles face equal sides.
  • In a right-angled triangle, Pythagoras' theorem gives a2+b2=c2a^2+b^2=c^2, where cc is always the hypotenuse opposite the right angle.
  • Angle facts, quadrilateral properties, similarity and congruence can then form a proof chain.
  • Each line must follow from given information or a named result; a diagram's appearance is not evidence.
  • For a 'prove' question, show the intermediate equality or congruence and finish with the exact conclusion requested.
Equal sides in an isosceles triangle face equal base angles.

Worked example

An isosceles triangle has equal sides 1313 cm and base 1010 cm. Find its perpendicular height.

  1. 1.The perpendicular from the apex bisects the 1010 cm base, giving a right triangle with base 55 cm.
  2. 2.Apply Pythagoras: h2+52=132h^2+5^2=13^2.
  3. 3.h2=169−25=144h^2=169-25=144, so h=12h=12 cm, taking the positive root.

Answer: The perpendicular height is 1212 cm.

Common mistakes

  • Don't use the full 1010 cm base in the right triangle instead of the bisected length 55 cm.
  • Don't treat a sloping side as the hypotenuse without checking which side is opposite the right angle.

Exam tip. A simple proof needs a reason beside each claim; a correct final statement without the linked reasons does not earn all proof marks.

G7 · Identify, describe and construct congruent and similar shapes, incl. on coordinate axes, by rotation, reflection, translation and enlargement (including fractional and negative scale factors)

  • Describe a transformation completely: a translation needs a column vector; a rotation needs centre, angle and direction; a reflection needs its mirror line; an enlargement needs centre and scale factor. Translation, rotation and reflection preserve every length and angle, so object and image are congruent.
  • Enlargement preserves angles and multiplies every length from the centre by the scale factor. For centre CC, use CP′→=kCP→\overrightarrow{CP'}=k\overrightarrow{CP}.
  • A fractional factor shrinks the shape.
  • At Higher tier, a negative factor also puts every image point on the opposite side of the centre.
  • Examiners require every defining detail, not just the transformation name.
Corresponding points lie the same perpendicular distance from a reflection line.

Worked example

Point P(5,1)P(5,1) is enlarged by scale factor 22 about centre C(2,3)C(2,3). Find P′P'.

  1. 1.CP→=(5−2,1−3)=(3,−2)\overrightarrow{CP}=(5-2,1-3)=(3,-2).
  2. 2.Multiply by 22: CP′→=(6,−4)\overrightarrow{CP'}=(6,-4).
  3. 3.Add the centre: P′=(2+6,3−4)=(8,−1)P'=(2+6,3-4)=(8,-1).

Answer: P′=(8,−1)P'=(8,-1).

Common mistakes

  • Don't give 'rotation' without its centre, angle and direction, or 'reflection' without the mirror line.
  • Don't multiply the original coordinates by the scale factor even though the enlargement centre is not the origin.

Exam tip. For an enlargement, draw straight rays from the centre through corresponding points to check the centre and scale factor.

G8 · Describe the changes and invariance achieved by combinations of rotations, reflections and translations [Higher only]

  • Higher tier only. In a combination, apply transformations in the stated order because the image from one step becomes the object for the next.
  • Two translations combine by adding their vectors. Two reflections in parallel lines produce a translation; two reflections in intersecting lines produce a rotation, with angle twice the angle between the mirror lines.
  • Rotations, reflections and translations preserve lengths, angles, area and parallelism. One reflection reverses orientation; two reflections restore it.
  • Coordinate rules are a reliable way to identify a single equivalent transformation.
  • Reversing the order can change the result, so transformations do not generally commute.
Successive reflections in parallel lines give one translation.

Worked example

A point is reflected in the xx-axis and then in the yy-axis. Describe the single equivalent transformation.

  1. 1.Reflection in the xx-axis maps (x,y)(x,y) to (x,−y)(x,-y).
  2. 2.Reflection in the yy-axis then maps (x,−y)(x,-y) to (−x,−y)(-x,-y).
  3. 3.The rule (x,y)↦(−x,−y)(x,y)\mapsto(-x,-y) is a rotation of 180∘180^\circ about the origin.

Answer: A rotation of 180∘180^\circ about the origin.

Common mistakes

  • Don't apply the second transformation to the original shape instead of the first image.
  • Don't say that two reflections always make a translation, ignoring whether the mirror lines intersect.

Exam tip. To identify a combined transformation, track one general point and then state every parameter of the single result.

G9 · Identify and apply circle definitions and properties, including: centre, radius, chord, diameter, circumference, tangent, arc, sector and segment

  • Use circle vocabulary precisely. A radius joins the centre to the circumference.
  • A diameter is a chord through the centre and has length 2r2r. A chord joins two points on the circumference, while a tangent meets the circle at exactly one point.
  • An arc is part of the circumference. A sector is bounded by two radii and an arc; a segment is bounded by a chord and an arc.
  • The minor region uses the shorter arc and the major region uses the longer arc.
  • In a labelled-diagram question, identify the defining boundaries rather than relying on a region's appearance.
Key circle parts: centre, radius, diameter, chord and tangent.

Worked example

A circle has diameter 1818 cm. A chord ABAB does not pass through the centre. State the radius and name the two regions cut off by chord ABAB.

  1. 1.Use d=2rd=2r, so r=18÷2=9r=18\div2=9 cm.
  2. 2.A chord and each corresponding arc bound a segment.
  3. 3.The shorter-arc region is the minor segment and the longer-arc region is the major segment.

Answer: The radius is 99 cm; chord ABAB forms a minor segment and a major segment.

Common mistakes

  • Don't call any chord a diameter even though it does not pass through the centre.
  • Don't confuse a sector, bounded by two radii and an arc, with a segment, bounded by a chord and an arc.

Exam tip. For a circle-definition mark, name the boundary pieces explicitly: radii, chord or arc.

G10 · Apply and prove the standard circle theorems concerning angles, radii, tangents and chords, and use them to prove related results [Higher only]

  • Higher tier only.
  • Standard circle theorems connect angles, radii, tangents and chords.
  • The angle at the centre is twice the angle at the circumference standing on the same arc; angles in the same segment are equal; an angle in a semicircle is 90∘90^\circ; and opposite angles in a cyclic quadrilateral total 180∘180^\circ.
  • A radius is perpendicular to a tangent at the contact point, tangents from one external point are equal, and the alternate segment theorem equates the tangent-chord angle with the angle in the opposite segment.
  • A proof must identify the relevant common arc or chord and name each theorem used.
The angle at the centre is twice the angle at the circumference on the same arc.

Worked example

Points A,B,C,DA,B,C,D lie on a circle. ∠ABC=73∘\angle ABC=73^\circ and ∠BCD=41∘\angle BCD=41^\circ. Find ∠ADC\angle ADC and ∠BAD\angle BAD.

  1. 1.Opposite angles in cyclic quadrilateral ABCDABCD total 180∘180^\circ.
  2. 2.∠ADC=180∘−73∘=107∘\angle ADC=180^\circ-73^\circ=107^\circ.
  3. 3.∠BAD=180∘−41∘=139∘\angle BAD=180^\circ-41^\circ=139^\circ.

Answer: ∠ADC=107∘\angle ADC=107^\circ and ∠BAD=139∘\angle BAD=139^\circ.

Common mistakes

  • Don't halve a central angle without checking that both angles stand on the same arc.
  • Don't use 'angles in the same segment' for angles on opposite sides of a chord.

Exam tip. A circle-theorem 'give a reason' mark needs the theorem's name or an unambiguous full statement.

G11 · Solve geometrical problems on coordinate axes

  • Coordinate geometry turns a diagram into exact calculations. The midpoint of (x1,y1)(x_1,y_1) and (x2,y2)(x_2,y_2) is (x1+x22,y1+y22)\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right).
  • Horizontal and vertical coordinate changes form a right triangle, so distance is (x2−x1)2+(y2−y1)2\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}.
  • Coordinate differences can prove equal side lengths and identify horizontal or vertical lines; areas can be found from base and perpendicular height or by enclosing and subtracting simpler shapes.
  • Keep brackets when subtracting negative coordinates and show the substitution.
  • The axes and calculated differences, not the sketch's appearance, supply the evidence the examiner needs.
Coordinate differences form a right triangle for gradient and distance.

Worked example

Points A(−3,2)A(-3,2) and B(5,8)B(5,8) are endpoints of a segment. Find its midpoint and exact length.

  1. 1.Midpoint =(−3+52,2+82)=(1,5)=\left(\frac{-3+5}{2},\frac{2+8}{2}\right)=(1,5).
  2. 2.Coordinate changes are 5−(−3)=85-(-3)=8 and 8−2=68-2=6.
  3. 3.AB=82+62=100=10AB=\sqrt{8^2+6^2}=\sqrt{100}=10.

Answer: The midpoint is (1,5)(1,5) and AB=10AB=10.

Common mistakes

  • Don't calculate the midpoint by subtracting coordinates instead of averaging each coordinate pair.
  • Don't write 5−(−3)=25-(-3)=2, losing the second negative sign.

Exam tip. In a 'show that' coordinate proof, display the gradients or squared lengths that establish the claimed property.

G12 · Identify properties of the faces, surfaces, edges and vertices of: cubes, cuboids, prisms, cylinders, pyramids, cones and spheres

  • Describe a solid by its flat faces, curved surfaces, edges and vertices. A face is flat; an edge is where surfaces meet; a vertex is where edges meet.
  • An nn-sided prism has two congruent polygonal ends and nn rectangular side faces, so it has n+2n+2 faces, 3n3n edges and 2n2n vertices.
  • A pyramid has one polygonal base and triangular faces meeting at an apex.
  • A cylinder has two circular faces and one curved surface but no vertices; a cone has one circular face, one curved surface and one vertex; a sphere has only one curved surface.
  • Count hidden features once.
A prism has congruent polygonal ends joined by rectangular faces.

Worked example

A prism has hexagonal ends. Work out its numbers of faces, edges and vertices.

  1. 1.For a prism with n=6n=6, faces =n+2=8=n+2=8.
  2. 2.Edges =3n=18=3n=18.
  3. 3.Vertices =2n=12=2n=12.

Answer: 88 faces, 1818 edges and 1212 vertices.

Common mistakes

  • Don't count a cylinder's curved surface as a flat face or give the cylinder vertices.
  • Don't count only visible edges and omit hidden edges at the back of a solid.

Exam tip. For a prism, identify the number of sides on one end first, then use n+2n+2, 3n3n and 2n2n as a check.

G13 · Construct and interpret plans and elevations of 3D shapes

  • A plan is the view from directly above. A front or side elevation is an orthographic view from the named horizontal direction, drawn without perspective.
  • Corresponding widths align between the plan and front elevation; depths align between the plan and side elevation. For stacks of unit cubes, a numbered plan records the height in each occupied cell.
  • An elevation shows, for each viewing column, the greatest height visible along the line of sight.
  • Use true lengths in the viewing plane and include changes of outline, but do not add hidden internal edges unless requested.
  • Examiners compare dimensions and visible column heights, not artistic realism.
A numbered plan projects to the greatest visible height in each elevation column.

Worked example

A cuboid measures 77 cm long, 44 cm deep and 33 cm high. State the dimensions of its plan, front elevation looking along the depth, and side elevation.

  1. 1.The plan shows length and depth, so it is 77 cm by 44 cm.
  2. 2.Looking along the depth, the front elevation shows length and height: 77 cm by 33 cm.
  3. 3.The side elevation shows depth and height: 44 cm by 33 cm.

Answer: Plan 7×47\times4 cm; front elevation 7×37\times3 cm; side elevation 4×34\times3 cm.

Common mistakes

  • Don't include the height in the plan even though the plan is viewed from above.
  • Don't add all stack heights along a line of sight instead of taking the greatest visible height.

Exam tip. Before drawing an elevation, write the two dimensions visible from that direction and project matching corners with straight construction lines.

G14 · Use standard units of measure and related concepts (length, area, volume/capacity, mass, time, money, etc.)

  • Choose standard units that match the quantity: length uses mm, cm, m or km; area uses square units; volume uses cubic units; capacity uses ml or litres; mass uses g or kg; and time uses seconds, minutes or hours.
  • Convert all measurements to compatible units before calculating.
  • A linear conversion factor must be squared for area and cubed for volume: because 1 m=100 cm1\text{ m}=100\text{ cm}, 1 m2=10 000 cm21\text{ m}^2=10\,000\text{ cm}^2 and 1 m3=1 000 000 cm31\text{ m}^3=1\,000\,000\text{ cm}^3.
  • Examiners expect a numerical value with the correct unit and sensible rounding for the context, especially money and time.
Linear conversion factors are squared for area and cubed for volume.

Worked example

Convert 2.4 m22.4\text{ m}^2 to cm2\text{cm}^2.

  1. 1.1 m=100 cm1\text{ m}=100\text{ cm}, so the area factor is 1002=10 000100^2=10\,000.
  2. 2.Multiply the area: 2.4×10 000=24 0002.4\times10\,000=24\,000.
  3. 3.Attach square centimetres because an area was converted.

Answer: 24 000 cm224\,000\text{ cm}^2.

Common mistakes

  • Don't multiply a square-metre value by 100100 instead of 1002100^2.
  • Don't carry a length unit such as cm into a final answer that is an area or volume.

Exam tip. Write the unit conversion before the arithmetic; this makes the required power and final unit visible for method marks.

G15 · Measure line segments and angles in geometric figures, including interpreting maps and scale drawings and use of bearings

  • Measure only from an accurate scale drawing, using a ruler or protractor to a precision justified by the scale. For scale 1:n1:n, one unit on the drawing represents nn of the same units in reality, so multiply drawing lengths by nn and then convert units.
  • A bearing is the clockwise angle from north at the starting point and is written with three figures, such as 052∘052^\circ.
  • Draw a north line at the point where the journey starts.
  • Reverse bearings differ by 180∘180^\circ.
  • Examiners expect the scale working, the correct direction of measurement and a three-figure bearing.
A bearing is measured clockwise from north at the starting point.

Worked example

On a map with scale 1:25 0001:25\,000, two locations are 6.46.4 cm apart. Find the real distance in kilometres.

  1. 1.Multiply by the scale factor: 6.4×25 000=160 0006.4\times25\,000=160\,000 cm.
  2. 2.Convert centimetres to metres: 160 000÷100=1600160\,000\div100=1600 m.
  3. 3.Convert metres to kilometres: 1600÷1000=1.61600\div1000=1.6 km.

Answer: 1.61.6 km.

Common mistakes

  • Don't measure a bearing anticlockwise or from an east-west line.
  • Don't multiply by the scale factor but leave the real distance in centimetres when kilometres are requested.

Exam tip. For a bearing mark, draw the north line at the departure point and write the answer using exactly three figures.

G16 · Know and apply formulae to calculate: area of triangles, parallelograms, trapezia; volume of cuboids and other right prisms (including cylinders)

Definition. Specification G16: know and apply formulae to calculate the area of triangles, parallelograms and trapezia, and the volume of cuboids and other right prisms, including cylinders.

  • Area formulae: triangle 12bh\frac12bh, parallelogram bhbh, trapezium 12(a+b)h\frac12(a+b)h, where hh is the perpendicular height between the base and the parallel side, never a sloping edge. A right prism has the same cross-section all the way along, so V=area of cross-section×lengthV=\text{area of cross-section}\times\text{length}; a cylinder is the case with a circular cross-section, V=πr2hV=\pi r^2h.
  • Identify and draw the cross-section first, find its area in square units, then multiply by the length to get cubic units.
  • Compound shapes are split into rectangles and triangles, or a larger shape with a piece subtracted; each area is a process mark.
  • Convert to consistent units before dividing one area by another, for example tiles into a wall.
  • Units are often a separate mark, so write cm2^2 or cm3^3 with the number.
A trapezium uses the perpendicular height between its parallel sides.

Worked example

1

The diagram shows a prism 1212 cm long. The cross-section of the prism is a trapezium with parallel sides 88 cm and 55 cm and perpendicular height 66 cm. Work out the volume of the prism.

(3)

(Total for Question 1 is 3 marks)

Working

  1. 1.Area of the trapezium cross-section: 12(8+5)×6=12×13×6=39\frac12(8+5)\times6=\frac12\times13\times6=39 cm2^2.
  2. 2.Volume of a prism is cross-sectional area multiplied by length: 39×1239\times12.
  3. 3.39×12=46839\times12=468, in cubic centimetres.

Answer: 468468 cm3^3

Mark scheme

Mark scheme for question 1
QuestionAnswerMarkAdditional guidance
1for a method to find the area of the trapezium, eg 12(8+5)×6\frac12(8+5)\times6 (= 39)M1oe
(dep) for a complete method to find the volume, eg “39” ×12\times12M1
for 468468A1cao

Mark-scheme wording to use

  • “for a complete method to find the volume eg 4 x 10 x 15 … (indep) cm3” — Pearson 1MA1, June 2020 Paper 2F, Q11
  • “for a process to find the area of one shape, eg 10 × 8 (= 80) … for a complete process to find the total area, eg “80” – “12” (= 68)” — Pearson 1MA1, June 2024 Paper 1H, Q3(a)
  • “M1 a full method to find the area of the trapezium M1 a full method to calculate both areas in consistent units M1 for the area of the trapezium ÷ area of a tile (with consistent units) M1 (dep on previous M) for complete method to find the number of packs required” — Pearson 1MA1, Specimen 1 Paper 3H, Q5(a)

What the examiner wants. The cross-section identified before any multiplying. The June 2024 Paper 3H report says many could not find the volume of a cylinder, with errors ranging from finding the surface area to multiplying the radius by the height, while the Paper 1H report notes that splitting the compound floor into two rectangles and showing each area earned the process marks even when the arithmetic slipped.

Common mistakes

  • Don't use the sloping side of a triangle or trapezium as hh without a perpendicular mark.
  • Don't stop after finding the cross-sectional area and fail to multiply by the prism length.

Exam tip. For a prism, write 'cross-sectional area × length' before substituting so both stages of the method are clear.

G17 · Know circumference of a circle = 2πr = πd and area = πr²; calculate perimeters of 2D shapes incl. circles, areas of circles and composite shapes, surface area and volume of spheres, pyramids, cones

Definition. Specification G17: know the formulae circumference of a circle =2πr=πd=2\pi r=\pi d and area of a circle =πr2=\pi r^2; calculate perimeters of 2D shapes, including circles; areas of circles and composite shapes; surface area and volume of spheres, pyramids, cones and composite solids.

  • Circle facts: C=2πr=πdC=2\pi r=\pi d and A=πr2A=\pi r^2, with the radius half the diameter; using πr\pi r or πr2\pi r^2 for a circumference scores nothing. A composite area is a sum of rectangles, triangles, trapezia and parts of circles, with a cut-out subtracted; show each area separately because each earns a process mark, and keep unrounded values until the end.
  • Perimeter counts only the outside edge, so a semicircular end contributes πr\pi r plus nothing for the diameter joined to the rest.
  • Volumes: cone or pyramid 13×base area×height\frac13\times\text{base area}\times\text{height}, sphere 43πr3\frac43\pi r^3, with the formulae given on the paper.
  • Surface area of a cone is the curved part πrl\pi rl, ll the slant height, plus the base πr2\pi r^2; a hemisphere's is 2πr22\pi r^2 plus its circular base.
  • Exact answers stay as multiples of π\pi.
A cone's perpendicular height hh and slant height ll have different roles.

Worked example

1

A solid cone has base radius 33 cm, perpendicular height 44 cm and slant height 55 cm. Curved surface area of a cone =πrl=\pi rl. Work out the total surface area of the cone. Give your answer in terms of π\pi.

(3)

(Total for Question 1 is 3 marks)

Working

  1. 1.Curved surface area =πrl=π×3×5=15π=\pi rl=\pi\times3\times5=15\pi cm2^2.
  2. 2.Base area =πr2=π×32=9π=\pi r^2=\pi\times3^2=9\pi cm2^2.
  3. 3.Total surface area =15π+9π=24π=15\pi+9\pi=24\pi cm2^2.

Answer: 24π24\pi cm2^2

Mark scheme

Mark scheme for question 1
QuestionAnswerMarkAdditional guidance
1for a process to find the curved surface area, eg π×3×5\pi\times3\times5 (= 15π15\pi)P1
for adding the circular base, eg “15π15\pi” +π×32+\pi\times3^2P1Slant height, not perpendicular height, in πrl\pi rl
for 24π24\piA1cao

Mark-scheme wording to use

  • “for correctly finding the area of at least three sections, eg 3 of 11 × 7 (= 77), or 9 × 7 (= 63), or 1/2 × 11 × 9 (= 49.5), or 1/4 × π × 7^2 (= 38.4845..)” — Pearson 1MA1, November 2022 Paper 2H, Q4
  • “P1 starts process by using 250/3 π and 1/2 × 4/3 πr^3 to find radius P1 starts process using ½ curved surface area eg (4 × π × “5”^2 ) ÷ 2 P1 complete process shown eg (4 × π × “5”^2 ) ÷ 2 + ( π × “5”^2 ) A1 for 75π” — Pearson 1MA1, Specimen 1 Paper 1H, Q18
  • “P1 Process to find volume of cone or hemisphere P1 Process to total volume of solid P1 Process to find volume of cylinder P1 Equates 2 volumes A1 Reaches h = x/2” — Pearson 1MA1, Specimen 2 Paper 1H, Q11

What the examiner wants. The right formula at the first step. The June 2024 Paper 1H report says many with some knowledge of sectors went wrong immediately by using πr2\pi r^2 or πr\pi r instead of 2πr2\pi r and gained no marks, while those who made a correct start usually finished; the Paper 3H report records volume-of-cylinder attempts that found the surface area or multiplied radius by height instead.

Common mistakes

  • Don't use the slant height ll in the cone-volume formula instead of the perpendicular height hh.
  • Don't add the circular base twice or omit it when total surface area is requested.

Exam tip. Underline 'curved' or 'total' surface area, then list the exposed faces before calculating.

G18 · Calculate arc lengths, angles and areas of sectors of circles

  • A sector or arc occupies the same fraction of a circle as its central angle occupies of 360∘360^\circ. For angle θ\theta, arc length is θ360×2πr\frac{\theta}{360}\times2\pi r and sector area is θ360×πr2\frac{\theta}{360}\times\pi r^2.
  • Rearrange these relationships when the angle or radius is unknown.
  • An arc length contains only the curved boundary; a sector perimeter also contains the two radii.
  • For an annular sector, subtract the inner sector area from the outer one and include both arcs in the perimeter.
  • Examiners expect exact answers in terms of π\pi unless a decimal accuracy is specified.
A sector's arc and area are the fraction θ/360\theta/360 of the full circle.

Worked example

A sector has radius 99 cm and angle 120∘120^\circ. Find its arc length and area in terms of π\pi.

  1. 1.Arc length =120360×2π(9)=6π=\frac{120}{360}\times2\pi(9)=6\pi cm.
  2. 2.Sector area =120360×π(92)=27π cm2=\frac{120}{360}\times\pi(9^2)=27\pi\text{ cm}^2.
  3. 3.Keep the exact π\pi forms because no decimal accuracy is requested.

Answer: Arc length =6π=6\pi cm; area =27π cm2=27\pi\text{ cm}^2.

Common mistakes

  • Don't use θ/180\theta/180 instead of θ/360\theta/360 for the sector fraction.
  • Don't add the two radii when the question asks for arc length rather than sector perimeter.

Exam tip. Write the fraction θ/360\theta/360 first; it earns the method whether the question asks for an arc, area or rearranged angle.

G19 · Apply the concepts of congruence and similarity, including the relationships between lengths, areas and volumes in similar figures

  • Congruent figures have the same size and shape. Similar figures have equal corresponding angles and corresponding lengths in one constant ratio.
  • Match sides by their positions opposite equal angles, then calculate a scale factor as image length divided by original length.
  • Multiply every original length by the same factor, or divide to reverse the enlargement.
  • Higher tier: if the length scale factor is kk, the area scale factor is k2k^2 and the volume scale factor is k3k^3.
  • Examiners expect corresponding quantities to be paired consistently; a ratio written in the wrong direction gives every later value incorrectly.
Similar triangles keep equal angles while all corresponding lengths share one scale factor.

Worked example

Two similar triangles have corresponding sides 88 cm and 1212 cm. Another side of the smaller triangle is 1414 cm. Find the corresponding larger side.

  1. 1.Scale factor from smaller to larger =12÷8=1.5=12\div8=1.5.
  2. 2.Multiply the corresponding smaller length: 14×1.5=2114\times1.5=21.
  3. 3.Attach centimetres because a length has been scaled.

Answer: 2121 cm.

Common mistakes

  • Don't pair sides that do not lie opposite corresponding equal angles.
  • Don't use kk instead of k2k^2 or k3k^3 for a related area or volume (Higher tier).

Exam tip. Write the scale-factor direction in words, such as 'small to large', before multiplying or dividing.

G20 · Know Pythagoras' theorem a² + b² = c² and the trigonometric ratios sin, cos and tan; apply them to find angles and lengths in right-angled and, where possible, general triangles in 2D and 3D figures

Definition. Specification G20: know the formulae for Pythagoras' theorem a2+b2=c2a^2+b^2=c^2 and the trigonometric ratios sin⁡θ=oppositehypotenuse\sin\theta=\dfrac{\text{opposite}}{\text{hypotenuse}}, cos⁡θ=adjacenthypotenuse\cos\theta=\dfrac{\text{adjacent}}{\text{hypotenuse}} and tan⁡θ=oppositeadjacent\tan\theta=\dfrac{\text{opposite}}{\text{adjacent}}; apply them to find angles and lengths in right-angled triangles and, where possible, general triangles in two- and three-dimensional figures.

  • Label the triangle first: the hypotenuse is opposite the right angle, and opposite and adjacent are named from the angle in use. Pythagoras finds a side from two others: add the squares for the hypotenuse, subtract for a shorter side, so x2=8.52−42x^2=8.5^2-4^2 gives x=7.5x=7.5.
  • A trigonometric ratio links one angle and two sides; write the statement with the values in it, such as cos⁡53∘=x14.5\cos53^\circ=\dfrac{x}{14.5}, then rearrange, and use the inverse function for an angle. Round only at the end and give at least three significant figures unless told otherwise.
  • To test whether a triangle is right-angled, compare the sum of the two smaller squares with the largest square and state the conclusion.
  • Higher tier: in three dimensions find a right-angled triangle inside the solid, often needing Pythagoras twice; the diagonal of a cuboid is a2+b2+c2\sqrt{a^2+b^2+c^2}.
  • Non-right-angled triangles use the sine and cosine rules.
Opposite, adjacent and hypotenuse are labelled relative to the chosen angle.

Worked example

1

A ladder of length 88 m rests against a vertical wall. The foot of the ladder is 33 m from the base of the wall on horizontal ground. Work out the angle the ladder makes with the ground. Give your answer correct to 11 decimal place.

(2)

(Total for Question 1 is 2 marks)

Working

  1. 1.Relative to the angle θ\theta at the ground, the 33 m side is adjacent and the 88 m ladder is the hypotenuse.
  2. 2.Choose the ratio with those two sides: cos⁡θ=38\cos\theta=\dfrac{3}{8}.
  3. 3.Use the inverse function: θ=cos⁡−1(38)=67.975…∘\theta=\cos^{-1}\left(\dfrac{3}{8}\right)=67.975\ldots^\circ.
  4. 4.Round at the end: 68.0∘68.0^\circ.

Answer: 68.0∘68.0^\circ

Mark scheme

Mark scheme for question 1
QuestionAnswerMarkAdditional guidance
1for a correct trig statement, eg cos⁡θ=38\cos\theta=\dfrac{3}{8} or θ=cos⁡−1(3÷8)\theta=\cos^{-1}(3\div8)M1Can use a combination of skills, eg 55\sqrt{55} then tan⁡\tan, but must have only one unknown
for an answer in the range 67.967.9 to 6868A1If an answer in the range is seen in working and then rounded incorrectly award full marks

Mark-scheme wording to use

  • “for a correct trig statement, eg 14.5 × cos 53 or cos 53 = x ÷ 14.5 … answer in the range 8.726 to 8.73” — Pearson 1MA1, November 2022 Paper 2H, Q5
  • “M1 for correct use of Pythagoras, eg.8.5^2 – 4^2 (= 56.25) or 4^2 + x^2 = 8.5^2 Must have values substituted … A1 for 7.5 or 7 1/2 or 15/2” — Pearson 1MA1, June 2022 Paper 3H, Q1
  • “M1 Derive AC=9 cm and identify as hypotenuse M1 4^2 + 7^2 A1 for using eg AC = √(4^2 + 7^2) or 65 and 81 C1 for concluding explanation that ABC is not a right-angled triangle with evidence.” — Pearson 1MA1, Specimen 2 Paper 1H, Q4

What the examiner wants. The right version of the formula for the side being found. The June 2024 Paper 2H report calls a plain Pythagoras question a positive start for most, but notes a proportion who mixed up the formula and worked as if finding the hypotenuse; the November 2024 Paper 2H report adds that many forgot to halve the base when using Pythagoras for the height of an isosceles triangle.

Common mistakes

  • Don't label the side opposite the right angle as adjacent instead of hypotenuse.
  • Don't use ordinary sine or cosine when an inverse function is needed to find an angle.

Exam tip. For a multi-step triangle question, keep unrounded calculator values for later steps and round only the final answers.

G21 · Know the exact values of sin θ and cos θ for θ = 0°, 30°, 45°, 60° and 90°; know the exact value of tan θ for θ = 0°, 30°, 45° and 60°

  • Know exact sine and cosine values for 0∘0^\circ, 30∘30^\circ, 45∘45^\circ, 60∘60^\circ and 90∘90^\circ, and exact tangent values through 60∘60^\circ. The sequences are sin⁡θ:0,12,22,32,1\sin\theta:0,\frac12,\frac{\sqrt2}{2},\frac{\sqrt3}{2},1 and cos⁡θ:1,32,22,12,0\cos\theta:1,\frac{\sqrt3}{2},\frac{\sqrt2}{2},\frac12,0.
  • Also tan⁡0∘=0\tan0^\circ=0, tan⁡30∘=13\tan30^\circ=\frac1{\sqrt3}, tan⁡45∘=1\tan45^\circ=1 and tan⁡60∘=3\tan60^\circ=\sqrt3. The 1:3:21:\sqrt3:2 and 1:1:21:1:\sqrt2 triangles explain these values by pairing opposite, adjacent and hypotenuse sides with the marked angle.
  • If a value is forgotten, reconstruct it from the appropriate special triangle rather than using a decimal.
  • Use exact values throughout multi-step area or algebra calculations.
  • Examiners require fractions and surds.
The 1:3:21:\sqrt3:2 and 1:1:21:1:\sqrt2 triangles generate the exact trigonometric values.

Worked example

Work out the exact value of 3sin⁡60∘−2cos⁡45∘3\sin60^\circ-2\cos45^\circ.

  1. 1.Use sin⁡60∘=32\sin60^\circ=\frac{\sqrt3}{2} and cos⁡45∘=22\cos45^\circ=\frac{\sqrt2}{2}.
  2. 2.Substitute: 3(32)−2(22)3\left(\frac{\sqrt3}{2}\right)-2\left(\frac{\sqrt2}{2}\right).
  3. 3.Simplify without decimals.

Answer: 332−2\frac{3\sqrt3}{2}-\sqrt2.

Common mistakes

  • Don't enter the values into a calculator and give decimals when the question asks for exact form.
  • Don't swap sin⁡30∘=12\sin30^\circ=\frac12 with sin⁡60∘=32\sin60^\circ=\frac{\sqrt3}{2}.

Exam tip. Write each exact trig value before substitution; the unsimplified exact form usually secures the method.

G22 · Know and apply the sine rule a/sin A = b/sin B = c/sin C, and cosine rule a² = b² + c² - 2bc cos A, to find unknown lengths and angles [Higher only]

  • Higher tier only. The sine rule asin⁡A=bsin⁡B=csin⁡C\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C} pairs each side with its opposite angle.
  • Use it when a complete opposite side-angle pair is known. The cosine rule a2=b2+c2−2bccos⁡Aa^2=b^2+c^2-2bc\cos A finds a side from two sides and their included angle, or an angle from three sides.
  • Label the triangle before substituting so the lowercase side is opposite its matching capital angle.
  • When an inverse sine gives an angle, check the supplementary value against the triangle angle sum and other information.
  • Examiners expect an unrounded substitution followed by the requested final accuracy.
In the sine and cosine rules, each lowercase side is opposite its matching capital angle.

Worked example

Two sides of a triangle are 88 cm and 1111 cm, with included angle 47∘47^\circ. Find the third side to 11 decimal place.

  1. 1.Use the cosine rule with the required side opposite 47∘47^\circ: c2=82+112−2(8)(11)cos⁡47∘c^2=8^2+11^2-2(8)(11)\cos47^\circ.
  2. 2.c2=64.968…c^2=64.968\ldots, so c=64.968…=8.060…c=\sqrt{64.968\ldots}=8.060\ldots.
  3. 3.Round the final length to 11 decimal place.

Answer: 8.18.1 cm.

Common mistakes

  • Don't pair side aa with an angle other than its opposite angle AA in the sine rule.
  • Don't use the cosine rule with an angle that is not included between the two substituted sides.

Exam tip. Sketch and label aa opposite AA before choosing a rule; this prevents most substitution errors.

G23 · Know and apply Area = ½ ab sin C to calculate the area, sides or angles of any triangle [Higher only]

  • Higher tier only. Use A=12absin⁡CA=\frac12ab\sin C when two sides aa and bb and their included angle CC are known.
  • The formula comes from 12×base×perpendicular height\frac12\times\text{base}\times\text{perpendicular height}, with height bsin⁡Cb\sin C.
  • Rearrange it to find a missing side or use inverse sine for an angle.
  • Because sin⁡C=sin⁡(180∘−C)\sin C=\sin(180^\circ-C), an angle calculation may have both an acute and an obtuse solution; test both against the stated triangle.
  • Examiners expect the angle between the two substituted sides, correct square units for area, and all valid angle solutions unless the diagram or context excludes one.
The included angle CC produces perpendicular height bsin⁡Cb\sin C.

Worked example

Two sides of a triangle are 99 cm and 1212 cm, with included angle 40∘40^\circ. Find the area to 11 decimal place.

  1. 1.Substitute the two sides and included angle: A=12(9)(12)sin⁡40∘A=\frac12(9)(12)\sin40^\circ.
  2. 2.A=54sin⁡40∘=34.710…A=54\sin40^\circ=34.710\ldots.
  3. 3.Round the final area to 11 decimal place and use square units.

Answer: 34.7 cm234.7\text{ cm}^2.

Common mistakes

  • Don't substitute an angle that is not between the two chosen sides.
  • Don't find one inverse-sine angle and discard the supplementary solution without checking it.

Exam tip. On the diagram, circle the included angle between the two substituted sides before using 12absin⁡C\frac12ab\sin C.

G24 · Describe translations as 2D vectors

  • A translation moves every point by the same horizontal and vertical displacement, preserving lengths, angles and orientation. Describe it with a column vector (xy)\binom{x}{y}: a positive top component moves right and a negative one moves left; a positive bottom component moves up and a negative one moves down.
  • To recover the vector from object and image coordinates, calculate image minus object in each coordinate.
  • Apply the same vector to every vertex when constructing an image.
  • Examiners expect a vector, not a destination coordinate, and the object-to-image direction matters.
  • Matching one pair of corresponding points is enough if the shapes are genuinely translations.
A translation gives every corresponding point the same displacement vector.

Worked example

Point A(−2,5)A(-2,5) is translated to A′(4,1)A'(4,1). Find the translation vector.

  1. 1.Horizontal displacement =4−(−2)=6=4-(-2)=6.
  2. 2.Vertical displacement =1−5=−4=1-5=-4.
  3. 3.Write horizontal above vertical in a column vector.

Answer: (6−4)\binom{6}{-4}.

Common mistakes

  • Don't subtract object minus image and obtain the reverse vector.
  • Don't write (4,1)(4,1), the image coordinate, instead of the displacement vector.

Exam tip. Use image minus object for both coordinates, then check the signs against the visible direction of movement.

G25 · Apply addition and subtraction of vectors, multiplication of vectors by a scalar, and diagrammatic and column representations of vectors; use vectors to construct geometric arguments and proofs

  • A vector has magnitude and direction and may be shown as a directed segment, a column vector or a symbol such as a\mathbf a. Add column vectors component by component.
  • Subtracting a vector adds its reverse, while multiplying by scalar kk multiplies every component and reverses direction when k<0k<0.
  • For a route through successive points, add vectors in travel order; position-vector differences use destination minus start.
  • Higher tier: compare alternative routes, or show one direction vector is a scalar multiple of another, to prove points are collinear or lines parallel.
  • Examiners expect a connected vector equation and an explicit geometric conclusion.
Successive vectors add head-to-tail: AC→=AB→+BC→\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC}.

Worked example

Work out 3(2−1)−(54)3\binom{2}{-1}-\binom{5}{4}.

  1. 1.Multiply first: 3(2−1)=(6−3)3\binom{2}{-1}=\binom{6}{-3}.
  2. 2.Subtract corresponding components: (6−5−3−4)\binom{6-5}{-3-4}.
  3. 3.Simplify both components.

Answer: (1−7)\binom{1}{-7}.

Common mistakes

  • Don't subtract only the top component and add the bottom components.
  • Don't reverse AB→\overrightarrow{AB} and BA→\overrightarrow{BA} even though they have opposite directions.

Exam tip. In a vector proof, finish with words such as 'therefore parallel' or 'therefore collinear' after the scalar-multiple equation.