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Edexcel GCSE Maths revision notes

Geometry and measures

Section G
25 specification points

Notes and three levels of exam-style practice for each registered specification point in this section.

Checked against Edexcel 1MA1 section G

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.

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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

Notes
Worked answers & exam appearances →
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • 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 9090^\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 PSQRPS\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 PSQRPS\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.

Tier 1 · Easy

ORIGINAL

1

In triangle PQRPQR, which vertex is opposite the side PRPR?

(1)

(Total for Question 1 is 1 mark)

Tier 2 · Standard

ORIGINAL

1

Lines ABAB and CDCD are parallel. A line through EE meets ABAB at FF at a right angle. State the relationship between EFEF and CDCD.

(1)

(Total for Question 1 is 1 mark)

Tier 3 · Hard

ORIGINAL

1

Draw quadrilateral ABCDABCD with ABCDAB\parallel CD. Draw diagonal ACAC. Mark a point EE on ACAC, then draw the line through EE perpendicular to ABAB, meeting ABAB at FF and CDCD at GG.

(3)

(Total for Question 1 is 3 marks)

Your progress and exam materials
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

Notes
Worked answers & exam appearances →
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • 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.

Tier 1 · Easy

ORIGINAL

1

Describe how to construct the perpendicular bisector of a line segment ABAB using a ruler and compasses.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1

Two straight paths meet at OO. A lamp must be equally distant from the two paths and no more than 66 m from OO. Describe the locus of possible positions inside the angle between the paths.

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

1

Points AA and BB are 88 cm apart. A point PP must satisfy PA=PBPA=PB and PA5PA\leq5 cm. Describe and construct the complete locus of PP.

(4)

(Total for Question 1 is 4 marks)

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

Notes
Worked answers & exam appearances →
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Use angle facts as a chain of justified steps. Angles at a point total 360360^\circ, angles on a straight line total 180180^\circ, and vertically opposite angles are equal.
  • When a transversal crosses parallel lines, corresponding and alternate angles are equal; interior angles on the same side total 180180^\circ.
  • A triangle totals 180180^\circ.
  • Drawing diagonals from one vertex divides an nn-gon into n2n-2 triangles, so its interior-angle sum is (n2)×180(n-2)\times180^\circ.
  • In exam working, write the relevant equality or subtraction and name the angle fact, including the parallel lines that make it valid.
Alternate angles are equal when a transversal crosses parallel lines.
Worked example

A regular polygon has each exterior angle equal to 2424^\circ. Find its number of sides and each interior angle.

  1. 1.Exterior angles of any polygon total 360360^\circ, so n=360÷24=15n=360\div24=15.
  2. 2.An interior and exterior angle on a straight line total 180180^\circ.
  3. 3.Interior angle =18024=156=180^\circ-24^\circ=156^\circ.

Answer: The polygon has 1515 sides and each interior angle is 156156^\circ.

Common mistakes

  • Don't claim alternate or corresponding angles are equal without establishing that the two lines are parallel.
  • Don't use n×180n\times180^\circ instead of (n2)×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.

Tier 1 · Easy

ORIGINAL

1

Two adjacent angles on a straight line are 6363^\circ and xx^\circ. Work out xx.

(1)

(Total for Question 1 is 1 mark)

Tier 2 · Standard

ORIGINAL

1

Two parallel lines are crossed by a transversal. A pair of alternate angles are (4x+7)(4x+7)^\circ and (7x38)(7x-38)^\circ. Find xx and the size of these angles.

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

1

A polygon has 1313 sides. Twelve of its interior angles are each 155155^\circ. Work out the remaining interior angle.

(4)

(Total for Question 1 is 4 marks)

G4

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

Notes
Worked answers & exam appearances →
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • 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.

Tier 1 · Easy

ORIGINAL

1

Name the quadrilateral that has exactly one pair of parallel sides.

(1)

(Total for Question 1 is 1 mark)

Evidence from answers you checked

Checked automatically against the model answer once you submit.

Tier 2 · Standard

ORIGINAL

1

One interior angle of a rhombus is 6868^\circ. Work out the other three interior angles.

(2)

(Total for Question 1 is 2 marks)

Tier 3 · Hard

ORIGINAL

1

The diagonals of quadrilateral WXYZWXYZ bisect each other and are equal in length. Explain why WXYZWXYZ must be a rectangle, and why the information does not prove that it is a square.

(3)

(Total for Question 1 is 3 marks)

G5

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

Notes
Worked answers & exam appearances →
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • 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 ABCDEF\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.

Tier 1 · Easy

ORIGINAL

1

Triangle ABCABC has side lengths 55 cm, 77 cm and 99 cm. Triangle PQRPQR has side lengths 55 cm, 77 cm and 99 cm. State the congruence criterion.

(1)

(Total for Question 1 is 1 mark)

Evidence from answers you checked

Checked automatically against the model answer once you submit.

Tier 2 · Standard

ORIGINAL

1

Triangles ABCABC and DEFDEF are right-angled at BB and EE. Also AC=DF=13AC=DF=13 cm and AB=DE=5AB=DE=5 cm. Explain why the triangles are congruent.

(2)

(Total for Question 1 is 2 marks)

Tier 3 · Hard

ORIGINAL

1

In quadrilateral ABCDABCD, diagonal ACAC is drawn. Given that AB=ADAB=AD and BAC=CAD\angle BAC=\angle CAD, prove that BC=CDBC=CD.

(3)

(Total for Question 1 is 3 marks)

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

Notes
Worked answers & exam appearances →
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • 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=16925=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.

Tier 1 · Easy

ORIGINAL

1

Triangle ABCABC is isosceles with AB=ACAB=AC. Angle BB is 4747^\circ. Work out angle AA.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1

A right-angled triangle has perpendicular sides 99 cm and 1212 cm. Work out the length of its hypotenuse.

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

1

In quadrilateral ABCDABCD, ABCDAB\parallel CD and AB=CDAB=CD. Diagonal ACAC is drawn. Prove that BC=ADBC=AD.

(4)

(Total for Question 1 is 4 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)

Notes
Worked answers & exam appearances →
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • 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 PP'.

  1. 1.CP=(52,13)=(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+6,34)=(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.

Tier 1 · Easy

ORIGINAL

1

Point A(3,4)A(-3,4) is translated by the vector (52)\begin{pmatrix}5\\-2\end{pmatrix}. Find the coordinates of its image.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1

Point P(4,1)P(4,-1) is rotated 9090^\circ anticlockwise about the origin. Find the coordinates of its image PP'.

(2)

(Total for Question 1 is 2 marks)

Tier 3 · Hard

ORIGINAL

1

Triangle ABCABC has vertices A(6,3)A(6,3), B(0,5)B(0,5) and C(2,1)C(-2,-1). It is enlarged by scale factor 12-\frac12 about centre (2,1)(2,-1). Find the three image coordinates.

(4)

(Total for Question 1 is 4 marks)

G8

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

Notes
Worked answers & exam appearances →
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • 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 180180^\circ about the origin.

Answer: A rotation of 180180^\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.

Tier 1 · Easy

ORIGINAL

1

A shape is rotated and then translated. State two properties of the shape that must remain invariant.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1

A shape is translated first by (32)\begin{pmatrix}3\\-2\end{pmatrix} and then by (57)\begin{pmatrix}-5\\7\end{pmatrix}. Describe the single equivalent transformation.

(2)

(Total for Question 1 is 2 marks)

Tier 3 · Hard

ORIGINAL

1

A shape is reflected in the xx-axis and then reflected in the line y=xy=x. Describe the single equivalent transformation and state whether orientation is preserved.

(4)

(Total for Question 1 is 4 marks)

G9

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

Notes
Worked answers & exam appearances →
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • 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.

Tier 1 · Easy

ORIGINAL

1

A circle has radius 6.56.5 cm. Write down its diameter.

(1)

(Total for Question 1 is 1 mark)

Tier 2 · Standard

ORIGINAL

1

A straight segment joins two points on a circle but does not pass through its centre. Another straight line meets the circle at exactly one point. Give the geometric name of each.

(2)

(Total for Question 1 is 2 marks)

Tier 3 · Hard

ORIGINAL

1

A circle is centred at OO. Distinct points AA and BB are on its circumference, and ABAB is not a diameter. Describe precisely the boundaries of the minor sector AOBAOB and the minor segment cut off by ABAB.

(3)

(Total for Question 1 is 3 marks)

G10

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

Notes
Worked answers & exam appearances →
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • 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 9090^\circ; and opposite angles in a cyclic quadrilateral total 180180^\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 180180^\circ.
  2. 2.ADC=18073=107\angle ADC=180^\circ-73^\circ=107^\circ.
  3. 3.BAD=18041=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.

Tier 1 · Easy

ORIGINAL

1

ABAB is a diameter of a circle and CC is another point on the circumference. Find ACB\angle ACB.

(1)

(Total for Question 1 is 1 mark)

Tier 2 · Standard

ORIGINAL

1

In a circle with centre OO, the angle AOBAOB is 124124^\circ. Point CC lies on the major arc ABAB. Work out ACB\angle ACB.

(2)

(Total for Question 1 is 2 marks)

Tier 3 · Hard

ORIGINAL

1

From a point XX outside a circle with centre WW, tangents touch the circle at YY and ZZ. Prove that XY=XZXY=XZ. Given that YXW=34\angle YXW=34^\circ, work out YWZ\angle YWZ.

(5)

(Total for Question 1 is 5 marks)

G11

Solve geometrical problems on coordinate axes

Notes
Worked answers & exam appearances →
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • 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 (x2x1)2+(y2y1)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 82=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.

Tier 1 · Easy

ORIGINAL

1

Find the midpoint of the line segment joining (5,2)(-5,2) to (7,8)(7,8).

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1

Points A(1,4)A(-1,4) and B(5,4)B(5,-4) are joined. Find the exact length ABAB.

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

1

Triangle ABCABC has vertices A(2,1)A(-2,1), B(4,3)B(4,3) and C(2,9)C(2,9). Show that the triangle is right-angled and isosceles, then work out its area.

(5)

(Total for Question 1 is 5 marks)

G12

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

Notes
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Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • 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.

Tier 1 · Easy

ORIGINAL

1

Write down the number of faces and the number of vertices of a triangular prism.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1

A solid has one circular plane face, one curved surface, one circular edge and one vertex. Name the solid.

(2)

(Total for Question 1 is 2 marks)

Tier 3 · Hard

ORIGINAL

1

A prism has a regular 99-sided polygon as each end. Work out its numbers of faces, edges and vertices.

(3)

(Total for Question 1 is 3 marks)

G13

Construct and interpret plans and elevations of 3D shapes

Notes
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Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • 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.

Tier 1 · Easy

ORIGINAL

1

A cuboid is 88 cm long, 55 cm wide and 33 cm high. State the shape and dimensions of its plan.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1

A numbered plan of unit-cube stacks has two rows. From north to south, the rows are (2,1)(2,1) and (3,0)(3,0), with entries ordered west to east. Give the visible column heights in the front elevation viewed from the south and in the side elevation viewed from the east.

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

1

A solid uses vertical stacks of unit cubes on all four cells of a 22 by 22 plan. Its front elevation viewed from the south has heights (3,2)(3,2) from west to east. Its side elevation viewed from the east has heights (2,3)(2,3) from north to south. Find the least possible number of cubes and give one numbered plan that achieves it.

(4)

(Total for Question 1 is 4 marks)

G14

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

Notes
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Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • 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=10000 cm21\text{ m}^2=10\,000\text{ cm}^2 and 1 m3=1000000 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=10000100^2=10\,000.
  2. 2.Multiply the area: 2.4×10000=240002.4\times10\,000=24\,000.
  3. 3.Attach square centimetres because an area was converted.

Answer: 24000 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.

Tier 1 · Easy

ORIGINAL

1

Convert 3.75 kg3.75\text{ kg} to grams.

(1)

(Total for Question 1 is 1 mark)

Tier 2 · Standard

ORIGINAL

1

A rectangular water tank is 0.8 m0.8\text{ m} long, 0.5 m0.5\text{ m} wide and 0.6 m0.6\text{ m} high. It is 65%65\% full. Work out the volume of water in litres.

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

1

A floor measures 4.8 m4.8\text{ m} by 3.6 m3.6\text{ m}. Square tiles have side length 30 cm30\text{ cm} and are sold in boxes of 1212. A decorator buys at least 8%8\% more tiles than the exact number needed. Each box costs £14.7514.75. Work out the total cost.

(5)

(Total for Question 1 is 5 marks)

G15

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

Notes
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Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • 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 052052^\circ.
  • Draw a north line at the point where the journey starts.
  • Reverse bearings differ by 180180^\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 1:250001:25\,000, two locations are 6.46.4 cm apart. Find the real distance in kilometres.

  1. 1.Multiply by the scale factor: 6.4×25000=1600006.4\times25\,000=160\,000 cm.
  2. 2.Convert centimetres to metres: 160000÷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.

Tier 1 · Easy

ORIGINAL

1

A ray from point PP makes an angle of 6868^\circ clockwise from north. Write its bearing from PP.

(1)

(Total for Question 1 is 1 mark)

Tier 2 · Standard

ORIGINAL

1

On a map with north at the top, point BB is 6 cm6\text{ cm} east and 8 cm8\text{ cm} north of point AA. The scale is 1:200001:20\,000. Work out the real straight-line distance from AA to BB and the bearing of BB from AA. Give the bearing to the nearest degree.

(4)

(Total for Question 1 is 4 marks)

Tier 3 · Hard

ORIGINAL

1

On a scale drawing, a road of actual length 3.15 km3.15\text{ km} is represented by a line 8.4 cm8.4\text{ cm} long. Find the scale in the form 1:n1:n. Another road is 11.2 cm11.2\text{ cm} long on the same drawing. Work out its actual length in kilometres.

(4)

(Total for Question 1 is 4 marks)

G16

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

Notes
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Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Use A=12bhA=\frac12bh for a triangle, A=bhA=bh for a parallelogram and A=12(a+b)hA=\frac12(a+b)h for a trapezium, where hh is perpendicular to the base or parallel sides.
  • A right prism has a constant cross-section, so V=cross-sectional area×lengthV=\text{cross-sectional area}\times\text{length}.
  • A cylinder is a right prism with circular cross-section, giving V=πr2hV=\pi r^2h.
  • Mark the cross-section first, calculate its area with square units, then multiply by the prism length to obtain cubic units.
  • Examiners expect correct substitution and units; a sloping side is not a perpendicular height unless a right angle establishes it.
A trapezium uses the perpendicular height between its parallel sides.
Worked example

A triangular prism is 1010 cm long. Its triangular cross-section has base 66 cm and perpendicular height 44 cm. Find the volume.

  1. 1.Cross-sectional area =12×6×4=12 cm2=\frac12\times6\times4=12\text{ cm}^2.
  2. 2.Multiply by prism length: V=12×10V=12\times10.
  3. 3.Use cubic units for volume.

Answer: 120 cm3120\text{ cm}^3.

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.

Tier 1 · Easy

ORIGINAL

1

A trapezium has parallel sides of lengths 7 cm7\text{ cm} and 13 cm13\text{ cm} and perpendicular height 6 cm6\text{ cm}. Work out its area.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1

A cylinder has radius 4 cm4\text{ cm} and height 9 cm9\text{ cm}. Work out its volume in terms of π\pi.

(2)

(Total for Question 1 is 2 marks)

Tier 3 · Hard

ORIGINAL

1

A right triangular prism is 10 cm10\text{ cm} long. Its triangular cross-section has base x cmx\text{ cm} and perpendicular height (x+2) cm(x+2)\text{ cm}. The volume is 600 cm3600\text{ cm}^3. Find xx.

(4)

(Total for Question 1 is 4 marks)

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

Notes
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Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • For a circle, use C=2πr=πdC=2\pi r=\pi d and A=πr2A=\pi r^2. Composite areas add included regions and subtract cut-outs; composite perimeters include only the outside boundary.
  • A cone or pyramid has volume V=13×base area×perpendicular heightV=\frac13\times\text{base area}\times\text{perpendicular height}. A cone's curved surface area is πrl\pi rl, where ll is slant height.
  • A sphere has surface area 4πr24\pi r^2 and volume 43πr3\frac43\pi r^3.
  • For a composite solid, divide it into recognised solids and exclude internal joined faces from surface area.
  • Examiners expect radius, perpendicular height and slant height to be distinguished correctly.
A cone's perpendicular height hh and slant height ll have different roles.
Worked example

A solid cone has radius 33 cm, perpendicular height 44 cm and slant height 55 cm. Find its total surface area and volume in terms of π\pi.

  1. 1.Curved area =πrl=π(3)(5)=15π cm2=\pi rl=\pi(3)(5)=15\pi\text{ cm}^2; base area =πr2=9π cm2=\pi r^2=9\pi\text{ cm}^2.
  2. 2.Total surface area =15π+9π=24π cm2=15\pi+9\pi=24\pi\text{ cm}^2.
  3. 3.Volume =13πr2h=13π(32)(4)=12π cm3=\frac13\pi r^2h=\frac13\pi(3^2)(4)=12\pi\text{ cm}^3.

Answer: Total surface area =24π cm2=24\pi\text{ cm}^2; volume =12π cm3=12\pi\text{ cm}^3.

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.

Tier 1 · Easy

ORIGINAL

1

A circle has diameter 13 cm13\text{ cm}. Write its circumference in terms of π\pi.

(1)

(Total for Question 1 is 1 mark)

Tier 2 · Standard

ORIGINAL

1

A semicircle of diameter 10 cm10\text{ cm} is attached to one of the shorter sides of an 18 cm18\text{ cm} by 10 cm10\text{ cm} rectangle. The shared diameter is inside the shape. Work out the area and perimeter of the composite shape in terms of π\pi.

(4)

(Total for Question 1 is 4 marks)

Tier 3 · Hard

ORIGINAL

1

A solid cone has radius 6 cm6\text{ cm}, perpendicular height 8 cm8\text{ cm} and slant height 10 cm10\text{ cm}. Work out its total surface area, including the circular base, and its volume. Give both answers in terms of π\pi.

(5)

(Total for Question 1 is 5 marks)

G18

Calculate arc lengths, angles and areas of sectors of circles

Notes
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Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • A sector or arc occupies the same fraction of a circle as its central angle occupies of 360360^\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 120120^\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.

Tier 1 · Easy

ORIGINAL

1

Work out the arc length of a quarter-circle with radius 8 cm8\text{ cm}. Give the answer in terms of π\pi.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1

A sector has radius 9 cm9\text{ cm} and arc length 7π cm7\pi\text{ cm}. Work out the angle of the sector and its area in terms of π\pi.

(4)

(Total for Question 1 is 4 marks)

Tier 3 · Hard

ORIGINAL

1

A shape is an annular sector with angle 144144^\circ, outer radius 12 cm12\text{ cm} and inner radius 7 cm7\text{ cm}. Work out its area and perimeter in terms of π\pi.

(5)

(Total for Question 1 is 5 marks)

G19

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

Notes
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Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • 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.

Tier 1 · Easy

ORIGINAL

1

Triangle ABCABC has AB=5 cmAB=5\text{ cm}, AC=7 cmAC=7\text{ cm} and BAC=42\angle BAC=42^\circ. Triangle PQRPQR has PQ=5 cmPQ=5\text{ cm}, PR=7 cmPR=7\text{ cm} and QPR=42\angle QPR=42^\circ. State why the triangles are congruent.

(1)

(Total for Question 1 is 1 mark)

Tier 2 · Standard

ORIGINAL

1

Two similar shapes have corresponding lengths in the ratio smaller:larger =3:5=3:5. A side on the smaller shape is 12 cm12\text{ cm}. Another side on the larger shape is 35 cm35\text{ cm}. Work out the corresponding missing lengths.

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

1

Triangle ABCABC has side lengths 8 cm8\text{ cm}, 11 cm11\text{ cm} and 13 cm13\text{ cm}. Similar triangle DEFDEF has longest side 19.5 cm19.5\text{ cm}. Work out the other two side lengths of triangle DEFDEF and its perimeter.

(4)

(Total for Question 1 is 4 marks)

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

Notes
Worked answers & exam appearances →
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • In a right-angled triangle, Pythagoras gives a2+b2=c2a^2+b^2=c^2, with cc opposite the right angle. Relative to angle θ\theta, sinθ=oppositehypotenuse\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}, cosθ=adjacenthypotenuse\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}} and tanθ=oppositeadjacent\tan\theta=\frac{\text{opposite}}{\text{adjacent}}.
  • Label the sides first, choose the relationship containing the known and required values, then rearrange. Use an inverse trigonometric function for an angle.
  • Foundation questions use right-angled triangles in two dimensions.
  • Higher tier: identify an appropriate right-angled cross-section in three dimensions or use the later general-triangle rules.
  • Examiners expect a diagram, substitution, correct units, sensible rounding and a clear statement of what the result represents.
Another way to see this:
Opposite, adjacent and hypotenuse are labelled relative to the chosen angle.
Worked example

A ladder is 88 m long and its foot is 33 m from a vertical wall. Find the height reached and the angle with the ground, both to 11 decimal place.

  1. 1.The ladder is the hypotenuse, so h=8232=55=7.416h=\sqrt{8^2-3^2}=\sqrt{55}=7.416\ldots m.
  2. 2.For ground angle θ\theta, cosθ=3/8\cos\theta=3/8.
  3. 3.θ=cos1(3/8)=67.975\theta=\cos^{-1}(3/8)=67.975\ldots^\circ; round only at the end.

Answer: Height =7.4=7.4 m; angle =68.0=68.0^\circ.

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.

Tier 1 · Easy

ORIGINAL

1

A right-angled triangle has perpendicular sides 6 cm6\text{ cm} and 8 cm8\text{ cm}. Work out the hypotenuse.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1

A straight ladder of length 7.2 m7.2\text{ m} rests against a vertical wall. Its foot is 2.1 m2.1\text{ m} from the wall. Work out the height reached by the ladder and the angle it makes with the ground. Give each answer to 11 decimal place.

(4)

(Total for Question 1 is 4 marks)

Tier 3 · Hard

ORIGINAL

1

Isosceles triangle ABCABC has AB=AC=13 cmAB=AC=13\text{ cm} and BC=10 cmBC=10\text{ cm}. Work out the perpendicular height from AA to BCBC and angle ABCABC. Give the angle to 11 decimal place.

(4)

(Total for Question 1 is 4 marks)

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°

Notes
Worked answers & exam appearances →
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Know exact sine and cosine values for 00^\circ, 3030^\circ, 4545^\circ, 6060^\circ and 9090^\circ, and exact tangent values through 6060^\circ. The sequences are sinθ:0,12,22,32,1\sin\theta:0,\frac12,\frac{\sqrt2}{2},\frac{\sqrt3}{2},1 and cosθ:1,32,22,12,0\cos\theta:1,\frac{\sqrt3}{2},\frac{\sqrt2}{2},\frac12,0.
  • Also tan0=0\tan0^\circ=0, tan30=13\tan30^\circ=\frac1{\sqrt3}, tan45=1\tan45^\circ=1 and tan60=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 3sin602cos453\sin60^\circ-2\cos45^\circ.

  1. 1.Use sin60=32\sin60^\circ=\frac{\sqrt3}{2} and cos45=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: 3322\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 sin30=12\sin30^\circ=\frac12 with sin60=32\sin60^\circ=\frac{\sqrt3}{2}.

Exam tip

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

Tier 1 · Easy

ORIGINAL

1

Work out the exact value of sin30+cos60\sin30^\circ+\cos60^\circ.

(1)

(Total for Question 1 is 1 mark)

Evidence from answers you checked

Checked automatically against the model answer once you submit.

Tier 2 · Standard

ORIGINAL

1

Work out the exact value of 2sin45cos45+cos602\sin45^\circ\cos45^\circ+\cos60^\circ.

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

1

The hypotenuse of a right-angled triangle is 12 cm12\text{ cm} and one acute angle is 3030^\circ. Work out its exact area.

(4)

(Total for Question 1 is 4 marks)

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]

Notes
Worked answers & exam appearances →
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Higher tier only. The sine rule asinA=bsinB=csinC\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+c22bccosAa^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 4747^\circ. Find the third side to 11 decimal place.

  1. 1.Use the cosine rule with the required side opposite 4747^\circ: c2=82+1122(8)(11)cos47c^2=8^2+11^2-2(8)(11)\cos47^\circ.
  2. 2.c2=64.968c^2=64.968\ldots, so c=64.968=8.060c=\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.

Tier 1 · Easy

ORIGINAL

1

In triangle ABCABC, side a=7 cma=7\text{ cm}, angle A=30A=30^\circ and angle B=90B=90^\circ. Work out side bb.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1

Two sides of a triangle are 7 cm7\text{ cm} and 10 cm10\text{ cm}, and the included angle is 6060^\circ. Work out the exact length of the third side.

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

1

In triangle ABCABC, A=35A=35^\circ, a=8 cma=8\text{ cm} and b=11 cmb=11\text{ cm}. Find all possible values of angles BB and CC. Give each angle to 11 decimal place.

(5)

(Total for Question 1 is 5 marks)

G23

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

Notes
Worked answers & exam appearances →
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • Higher tier only. Use A=12absinCA=\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 bsinCb\sin C.
  • Rearrange it to find a missing side or use inverse sine for an angle.
  • Because sinC=sin(180C)\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 bsinCb\sin C.
Worked example

Two sides of a triangle are 99 cm and 1212 cm, with included angle 4040^\circ. Find the area to 11 decimal place.

  1. 1.Substitute the two sides and included angle: A=12(9)(12)sin40A=\frac12(9)(12)\sin40^\circ.
  2. 2.A=54sin40=34.710A=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 12absinC\frac12ab\sin C.

Tier 1 · Easy

ORIGINAL

1

Two sides of a triangle are 10 cm10\text{ cm} and 7 cm7\text{ cm}, and their included angle is 3030^\circ. Work out the area.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1

A triangle has area 48 cm248\text{ cm}^2. Two sides have lengths 12 cm12\text{ cm} and x cmx\text{ cm}, and their included angle is 3030^\circ. Work out xx.

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

1

Two sides of a triangle are 10 cm10\text{ cm} and 12 cm12\text{ cm}. Its area is 303 cm230\sqrt3\text{ cm}^2. Find both possible values of the included angle.

(4)

(Total for Question 1 is 4 marks)

G24

Describe translations as 2D vectors

Notes
Worked answers & exam appearances →
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • 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 =15=4=1-5=-4.
  3. 3.Write horizontal above vertical in a column vector.

Answer: (64)\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.

Tier 1 · Easy

ORIGINAL

1

Point A=(3,4)A=(-3,4) is translated to A=(2,1)A'=(2,-1). Describe the translation as a vector.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1

Triangle ABCABC has vertices A=(1,2)A=(1,-2), B=(5,1)B=(5,1) and C=(2,4)C=(2,4). It is translated by (43)\binom{-4}{3}. Write the coordinates of the three image vertices.

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

1

A translation maps P=(6,4)P=(-6,4) to P=(1,2)P'=(1,-2). The same translation maps Q=(3,k)Q=(3,k) to Q=(10,5)Q'=(10,5). Find kk and describe the translation as a vector.

(4)

(Total for Question 1 is 4 marks)

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

Notes
Worked answers & exam appearances →
Evidence from your answers: none yet
Your confidence:

A self-report of how sure you feel. It does not measure mastery. Evidence from your answers reaches secure after the latest Tier 2/3 attempt is correct, with three correct distinct drills across at least two dates and two practice sources.

Explanation

  • 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(21)(54)3\binom{2}{-1}-\binom{5}{4}.

  1. 1.Multiply first: 3(21)=(63)3\binom{2}{-1}=\binom{6}{-3}.
  2. 2.Subtract corresponding components: (6534)\binom{6-5}{-3-4}.
  3. 3.Simplify both components.

Answer: (17)\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.

Tier 1 · Easy

ORIGINAL

1

Work out (35)+(72)\binom{3}{-5}+\binom{-7}{2}.

(1)

(Total for Question 1 is 1 mark)

Tier 2 · Standard

ORIGINAL

1

Points AA and BB have position vectors a=(21)\mathbf{a}=\binom{2}{-1} and b=(85)\mathbf{b}=\binom{8}{5}. Point MM is the midpoint of ABAB. Find the position vector of MM and the vector AM\overrightarrow{AM}.

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

1

A walker starts at P=(3,2)P=(-3,2). The walker makes successive displacements (43)\binom{4}{3}, (15)\binom{-1}{5} and 2(12)-2\binom{1}{-2}. Work out the resultant displacement, the walker's finishing coordinates and the vector that would take the walker directly back to PP.

(4)

(Total for Question 1 is 4 marks)

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