G Geometry and measures — 1MA1 revision notes

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

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 $90^\circ$ .
In $\angle ABC$ , the middle letter $B$ is the vertex; in triangle $ABC$ , side $AC$ is opposite $B$ .
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 $PQR$ with $PQ=PR$ . Put $S$ on $QR$ , draw $PS\perp QR$ , and state which side is opposite $P$ .

1.Draw and label triangle $PQR$ , then mark $PQ$ and $PR$ with matching equality ticks.
2.Place $S$ on segment $QR$ , join $P$ to $S$ , and add a right-angle mark at $S$ .
3.The side not touching vertex $P$ is $QR$ , so $QR$ is opposite $P$ .

Answer: A correctly labelled diagram with $PQ=PR$ , $S$ on $QR$ and $PS\perp QR$ ; the opposite side is $QR$ .

Common mistakes

Don't read $\angle ABC$ as the angle at $A$ instead of the middle letter $B$ .
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

PQR

, which vertex is opposite the side

PR

?

(1)

(Total for Question 1 is 1 mark)

Tier 2 · Standard

ORIGINAL

1

Lines

AB

and

CD

are parallel. A line through

E

meets

AB

at

F

at a right angle. State the relationship between

EF

and

CD

.

(1)

(Total for Question 1 is 1 mark)

Tier 3 · Hard

ORIGINAL

1

Draw quadrilateral

ABCD

with

AB\parallel CD

. Draw diagonal

AC

. Mark a point

E

on

AC

, then draw the line through

E

perpendicular to

AB

, meeting

AB

at

F

and

CD

at

G

.

(3)

(Total for Question 1 is 3 marks)

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

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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

Ruler-and-compass constructions must show the arcs that create the result. For a perpendicular bisector of $AB$ , draw equal-radius arcs from $A$ and $B$ with radius greater than half of $AB$ , then join their intersections.
Every point on this line is equidistant from $A$ and $B$ .
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 $A$ and $B$ and within $4$ km of $A$ . Describe the complete locus of possible positions.

1.Construct the perpendicular bisector of $AB$ because equal distances from $A$ and $B$ are required.
2.Draw the circle centred at $A$ with radius $4$ km because the distance from $A$ is at most $4$ km.
3.Keep the part of the perpendicular bisector inside or on that circle.

Answer: The segment of the perpendicular bisector of $AB$ lying inside or on the circle centre $A$ , radius $4$ 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

AB

using a ruler and compasses.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1

Two straight paths meet at

O

. A lamp must be equally distant from the two paths and no more than

6

m from

O

. 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

A

and

B

are

8

cm apart. A point

P

must satisfy

PA=PB

and

PA\leq5

cm. Describe and construct the complete locus of

P

.

(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

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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.

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.

Explanation

Angles at a point total $360^\circ$ , angles on a straight line total $180^\circ$ , 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 $180^\circ$ .
Angles in a triangle add up to $180^\circ$ , and an exterior angle of a triangle equals the sum of the two interior opposite angles.
Splitting an $n$ -sided polygon into $n-2$ triangles gives an interior-angle sum of $(n-2)\times180^\circ$ ; the exterior angles of any polygon total $360^\circ$ , so a regular polygon has exterior angle $360^\circ\div n$ .
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^\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.The exterior angles of any polygon add up to $360^\circ$ , so the number of sides is $360\div24=15$ .
2.An interior angle and its exterior angle lie on a straight line, so they add up to $180^\circ$ .
3.Each interior angle is $180-24=156^\circ$ .

Answer: $15$ sides; each interior angle is $156^\circ$ .

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Additional guidance
1	for $360\div24$	M1	oe
	for $15$	A1	cao
	for $156$	B1	ft $180-24$

Common mistakes

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

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 $360^\circ$ does not show a shape is a trapezium.

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

63^\circ

and

x^\circ

. Work out

x

.

(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)^\circ

and

(7x-38)^\circ

. Find

x

and the size of these angles.

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

1

A polygon has

13

sides. Twelve of its interior angles are each

155^\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

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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

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 $ABCD$ bisect each other and meet at right angles. State the most specific guaranteed quadrilateral and justify it.

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

Answer: $ABCD$ 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

68^\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

WXYZ

bisect each other and are equal in length. Explain why

WXYZ

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

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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 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 $ABC$ and $DEF$ satisfy $AB=DE$ , $AC=DF$ and $\angle BAC=\angle EDF$ . Prove that $BC=EF$ .

1. $AB=DE$ and $AC=DF$ give two pairs of corresponding equal sides.
2. $\angle BAC=\angle EDF$ is the included angle between those pairs.
3.Therefore $\triangle ABC\cong\triangle DEF$ by SAS, so corresponding sides $BC$ and $EF$ are equal.

Answer: $BC=EF$ 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

ABC

has side lengths

5

cm,

7

cm and

9

cm. Triangle

PQR

has side lengths

5

cm,

7

cm and

9

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

ABC

and

DEF

are right-angled at

B

and

E

. Also

AC=DF=13

cm and

AB=DE=5

cm. Explain why the triangles are congruent.

(2)

(Total for Question 1 is 2 marks)

Tier 3 · Hard

ORIGINAL

1

In quadrilateral

ABCD

, diagonal

AC

is drawn. Given that

AB=AD

and

\angle BAC=\angle CAD

, prove that

BC=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

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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

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 $a^2+b^2=c^2$ , where $c$ 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 $13$ cm and base $10$ cm. Find its perpendicular height.

1.The perpendicular from the apex bisects the $10$ cm base, giving a right triangle with base $5$ cm.
2.Apply Pythagoras: $h^2+5^2=13^2$ .
3. $h^2=169-25=144$ , so $h=12$ cm, taking the positive root.

Answer: The perpendicular height is $12$ cm.

Common mistakes

Don't use the full $10$ cm base in the right triangle instead of the bisected length $5$ 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

ABC

is isosceles with

AB=AC

. Angle

B

is

47^\circ

. Work out angle

A

.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1

A right-angled triangle has perpendicular sides

9

cm and

12

cm. Work out the length of its hypotenuse.

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

1

In quadrilateral

ABCD

,

AB\parallel CD

and

AB=CD

. Diagonal

AC

is drawn. Prove that

BC=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

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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 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 $C$ , use $\overrightarrow{CP'}=k\overrightarrow{CP}$ .
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 $P(5,1)$ is enlarged by scale factor $2$ about centre $C(2,3)$ . Find $P'$ .

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

Answer: $P'=(8,-1)$ .

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)

is translated by the vector

\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)

is rotated

90^\circ

anticlockwise about the origin. Find the coordinates of its image

P'

.

(2)

(Total for Question 1 is 2 marks)

Tier 3 · Hard

ORIGINAL

1

Triangle

ABC

has vertices

A(6,3)

,

B(0,5)

and

C(-2,-1)

. It is enlarged by scale factor

-\frac12

about centre

(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

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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

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 $x$ -axis and then in the $y$ -axis. Describe the single equivalent transformation.

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

Answer: A rotation of $180^\circ$ 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.

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

\begin{pmatrix}3\\-2\end{pmatrix}

and then by

\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

x

-axis and then reflected in the line

y=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

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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 circle vocabulary precisely. A radius joins the centre to the circumference.
A diameter is a chord through the centre and has length $2r$ . 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 $18$ cm. A chord $AB$ does not pass through the centre. State the radius and name the two regions cut off by chord $AB$ .

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

Answer: The radius is $9$ cm; chord $AB$ 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

6.5

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

O

. Distinct points

A

and

B

are on its circumference, and

AB

is not a diameter. Describe precisely the boundaries of the minor sector

AOB

and the minor segment cut off by

AB

.

(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

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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

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 $90^\circ$ ; and opposite angles in a cyclic quadrilateral total $180^\circ$ .
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 $A,B,C,D$ lie on a circle. $\angle ABC=73^\circ$ and $\angle BCD=41^\circ$ . Find $\angle ADC$ and $\angle BAD$ .

1.Opposite angles in cyclic quadrilateral $ABCD$ total $180^\circ$ .
2. $\angle ADC=180^\circ-73^\circ=107^\circ$ .
3. $\angle BAD=180^\circ-41^\circ=139^\circ$ .

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

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

AB

is a diameter of a circle and

C

is another point on the circumference. Find

\angle ACB

.

(1)

(Total for Question 1 is 1 mark)

Tier 2 · Standard

ORIGINAL

1

In a circle with centre

O

, the angle

AOB

is

124^\circ

. Point

C

lies on the major arc

AB

. Work out

\angle ACB

.

(2)

(Total for Question 1 is 2 marks)

Tier 3 · Hard

ORIGINAL

1

From a point

X

outside a circle with centre

W

, tangents touch the circle at

Y

and

Z

. Prove that

XY=XZ

. Given that

\angle YXW=34^\circ

, work out

\angle YWZ

.

(5)

(Total for Question 1 is 5 marks)

G11

Solve geometrical problems on coordinate axes

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

Coordinate geometry turns a diagram into exact calculations. The midpoint of $(x_1,y_1)$ and $(x_2,y_2)$ is $\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)$ .
Horizontal and vertical coordinate changes form a right triangle, so distance is $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ .
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 $A(-3,2)$ and $B(5,8)$ are endpoints of a segment. Find its midpoint and exact length.

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

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

Common mistakes

Don't calculate the midpoint by subtracting coordinates instead of averaging each coordinate pair.
Don't write $5-(-3)=2$ , 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)

to

(7,8)

.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1

Points

A(-1,4)

and

B(5,-4)

are joined. Find the exact length

AB

.

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

1

Triangle

ABC

has vertices

A(-2,1)

,

B(4,3)

and

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

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 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 $n$ -sided prism has two congruent polygonal ends and $n$ rectangular side faces, so it has $n+2$ faces, $3n$ edges and $2n$ 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.For a prism with $n=6$ , faces $=n+2=8$ .
2.Edges $=3n=18$ .
3.Vertices $=2n=12$ .

Answer: $8$ faces, $18$ edges and $12$ 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 $n+2$ , $3n$ and $2n$ 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

9

-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

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 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 $7$ cm long, $4$ cm deep and $3$ cm high. State the dimensions of its plan, front elevation looking along the depth, and side elevation.

1.The plan shows length and depth, so it is $7$ cm by $4$ cm.
2.Looking along the depth, the front elevation shows length and height: $7$ cm by $3$ cm.
3.The side elevation shows depth and height: $4$ cm by $3$ cm.

Answer: Plan $7\times4$ cm; front elevation $7\times3$ cm; side elevation $4\times3$ 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

8

cm long,

5

cm wide and

3

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)

and

(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

2

by

2

plan. Its front elevation viewed from the south has heights

(3,2)

from west to east. Its side elevation viewed from the east has heights

(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

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

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 $1\text{ m}=100\text{ cm}$ , $1\text{ m}^2=10\,000\text{ cm}^2$ and $1\text{ m}^3=1\,000\,000\text{ cm}^3$ .
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 $2.4\text{ m}^2$ to $\text{cm}^2$ .

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

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

Common mistakes

Don't multiply a square-metre value by $100$ instead of $100^2$ .
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\text{ kg}

to grams.

(1)

(Total for Question 1 is 1 mark)

Tier 2 · Standard

ORIGINAL

1

A rectangular water tank is

0.8\text{ m}

long,

0.5\text{ m}

wide and

0.6\text{ m}

high. It is

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\text{ m}

by

3.6\text{ m}

. Square tiles have side length

30\text{ cm}

and are sold in boxes of

12

. A decorator buys at least

8\%

more tiles than the exact number needed. Each box costs £

14.75

. 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 $1:n$ , one unit on the drawing represents $n$ of the same units in reality, so multiply drawing lengths by $n$ and then convert units.
A bearing is the clockwise angle from north at the starting point and is written with three figures, such as $052^\circ$ .
Draw a north line at the point where the journey starts.
Reverse bearings differ by $180^\circ$ .
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 $1:25\,000$ , two locations are $6.4$ cm apart. Find the real distance in kilometres.

1.Multiply by the scale factor: $6.4\times25\,000=160\,000$ cm.
2.Convert centimetres to metres: $160\,000\div100=1600$ m.
3.Convert metres to kilometres: $1600\div1000=1.6$ km.

Answer: $1.6$ 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

P

makes an angle of

68^\circ

clockwise from north. Write its bearing from

P

.

(1)

(Total for Question 1 is 1 mark)

Tier 2 · Standard

ORIGINAL

1

On a map with north at the top, point

B

is

6\text{ cm}

east and

8\text{ cm}

north of point

A

. The scale is

1:20\,000

. Work out the real straight-line distance from

A

to

B

and the bearing of

B

from

A

. 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\text{ km}

is represented by a line

8.4\text{ cm}

long. Find the scale in the form

1:n

. Another road is

11.2\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

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.

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.

Explanation

Area formulae: triangle $\frac12bh$ , parallelogram $bh$ , trapezium $\frac12(a+b)h$ , where $h$ 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 $V=\text{area of cross-section}\times\text{length}$ ; a cylinder is the case with a circular cross-section, $V=\pi r^2h$ .
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 cm $^2$ or cm $^3$ with the number.

A trapezium uses the perpendicular height between its parallel sides.

Worked example

1

The diagram shows a prism

12

cm long. The cross-section of the prism is a trapezium with parallel sides

8

cm and

5

cm and perpendicular height

6

cm. Work out the volume of the prism.

(3)

(Total for Question 1 is 3 marks)

Working

1.Area of the trapezium cross-section: $\frac12(8+5)\times6=\frac12\times13\times6=39$ cm $^2$ .
2.Volume of a prism is cross-sectional area multiplied by length: $39\times12$ .
3. $39\times12=468$ , in cubic centimetres.

Answer: $468$ cm $^3$

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Additional guidance
1	for a method to find the area of the trapezium, eg $\frac12(8+5)\times6$ (= 39)	M1	oe
	(dep) for a complete method to find the volume, eg “39” $\times12$	M1
	for $468$	A1	cao

Common mistakes

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

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.

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\text{ cm}

and

13\text{ cm}

and perpendicular height

6\text{ cm}

. Work out its area.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1

A cylinder has radius

4\text{ cm}

and height

9\text{ cm}

. Work out its volume in terms of

\pi

.

(2)

(Total for Question 1 is 2 marks)

Tier 3 · Hard

ORIGINAL

1

A right triangular prism is

10\text{ cm}

long. Its triangular cross-section has base

x\text{ cm}

and perpendicular height

(x+2)\text{ cm}

. The volume is

600\text{ cm}^3

. Find

x

.

(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

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.

Definition

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

Explanation

Circle facts: $C=2\pi r=\pi d$ and $A=\pi r^2$ , with the radius half the diameter; using $\pi r$ or $\pi r^2$ 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 $\pi r$ plus nothing for the diameter joined to the rest.
Volumes: cone or pyramid $\frac13\times\text{base area}\times\text{height}$ , sphere $\frac43\pi r^3$ , with the formulae given on the paper.
Surface area of a cone is the curved part $\pi rl$ , $l$ the slant height, plus the base $\pi r^2$ ; a hemisphere's is $2\pi r^2$ plus its circular base.
Exact answers stay as multiples of $\pi$ .

A cone's perpendicular height

h

and slant height

l

have different roles.

Worked example

1

A solid cone has base radius

3

cm, perpendicular height

4

cm and slant height

5

cm. Curved surface area of a cone

=\pi rl

. Work out the total surface area of the cone. Give your answer in terms of

\pi

.

(3)

(Total for Question 1 is 3 marks)

Working

1.Curved surface area $=\pi rl=\pi\times3\times5=15\pi$ cm $^2$ .
2.Base area $=\pi r^2=\pi\times3^2=9\pi$ cm $^2$ .
3.Total surface area $=15\pi+9\pi=24\pi$ cm $^2$ .

Answer: $24\pi$ cm $^2$

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Additional guidance
1	for a process to find the curved surface area, eg $\pi\times3\times5$ (= $15\pi$ )	P1
	for adding the circular base, eg “ $15\pi$ ” $+\pi\times3^2$	P1	Slant height, not perpendicular height, in $\pi rl$
	for $24\pi$	A1	cao

Common mistakes

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

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 $\pi r^2$ or $\pi r$ instead of $2\pi r$ 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.

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\text{ cm}

. Write its circumference in terms of

\pi

.

(1)

(Total for Question 1 is 1 mark)

Tier 2 · Standard

ORIGINAL

1

A semicircle of diameter

10\text{ cm}

is attached to one of the shorter sides of an

18\text{ cm}

by

10\text{ cm}

rectangle. The shared diameter is inside the shape. Work out the area and perimeter of the composite shape in terms of

\pi

.

(4)

(Total for Question 1 is 4 marks)

Tier 3 · Hard

ORIGINAL

1

A solid cone has radius

6\text{ cm}

, perpendicular height

8\text{ cm}

and slant height

10\text{ cm}

. Work out its total surface area, including the circular base, and its volume. Give both answers in terms of

\pi

.

(5)

(Total for Question 1 is 5 marks)

G18

Calculate arc lengths, angles and areas of sectors of circles

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 sector or arc occupies the same fraction of a circle as its central angle occupies of $360^\circ$ . For angle $\theta$ , arc length is $\frac{\theta}{360}\times2\pi r$ and sector area is $\frac{\theta}{360}\times\pi r^2$ .
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 $\pi$ unless a decimal accuracy is specified.

A sector's arc and area are the fraction

\theta/360

of the full circle.

Worked example

A sector has radius $9$ cm and angle $120^\circ$ . Find its arc length and area in terms of $\pi$ .

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

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

Common mistakes

Don't use $\theta/180$ instead of $\theta/360$ 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 $\theta/360$ 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\text{ cm}

. Give the answer in terms of

\pi

.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1

A sector has radius

9\text{ cm}

and arc length

7\pi\text{ cm}

. Work out the angle of the sector and its area in terms of

\pi

.

(4)

(Total for Question 1 is 4 marks)

Tier 3 · Hard

ORIGINAL

1

A shape is an annular sector with angle

144^\circ

, outer radius

12\text{ cm}

and inner radius

7\text{ cm}

. Work out its area and perimeter in terms of

\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

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 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 $k$ , the area scale factor is $k^2$ and the volume scale factor is $k^3$ .
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 $8$ cm and $12$ cm. Another side of the smaller triangle is $14$ cm. Find the corresponding larger side.

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

Answer: $21$ cm.

Common mistakes

Don't pair sides that do not lie opposite corresponding equal angles.
Don't use $k$ instead of $k^2$ or $k^3$ 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

ABC

has

AB=5\text{ cm}

,

AC=7\text{ cm}

and

\angle BAC=42^\circ

. Triangle

PQR

has

PQ=5\text{ cm}

,

PR=7\text{ cm}

and

\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

. A side on the smaller shape is

12\text{ cm}

. Another side on the larger shape is

35\text{ cm}

. Work out the corresponding missing lengths.

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

1

Triangle

ABC

has side lengths

8\text{ cm}

,

11\text{ cm}

and

13\text{ cm}

. Similar triangle

DEF

has longest side

19.5\text{ cm}

. Work out the other two side lengths of triangle

DEF

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.

Definition

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

Explanation

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 $x^2=8.5^2-4^2$ gives $x=7.5$ .
A trigonometric ratio links one angle and two sides; write the statement with the values in it, such as $\cos53^\circ=\dfrac{x}{14.5}$ , 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 $\sqrt{a^2+b^2+c^2}$ .
Non-right-angled triangles use the sine and cosine rules.

Another way to see this:StandardAlgebraic

Opposite, adjacent and hypotenuse are labelled relative to the chosen angle.

Worked example

1

A ladder of length

8

m rests against a vertical wall. The foot of the ladder is

3

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

1

decimal place.

(2)

(Total for Question 1 is 2 marks)

Working

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

Answer: $68.0^\circ$

Mark scheme

Mark scheme for question 1
Question	Answer	Mark	Additional guidance
1	for a correct trig statement, eg $\cos\theta=\dfrac{3}{8}$ or $\theta=\cos^{-1}(3\div8)$	M1	Can use a combination of skills, eg $\sqrt{55}$ then $\tan$ , but must have only one unknown
	for an answer in the range $67.9$ to $68$	A1	If an answer in the range is seen in working and then rounded incorrectly award full marks

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.

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.

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\text{ cm}

and

8\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\text{ m}

rests against a vertical wall. Its foot is

2.1\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

1

decimal place.

(4)

(Total for Question 1 is 4 marks)

Tier 3 · Hard

ORIGINAL

1

Isosceles triangle

ABC

has

AB=AC=13\text{ cm}

and

BC=10\text{ cm}

. Work out the perpendicular height from

A

to

BC

and angle

ABC

. Give the angle to

1

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 $0^\circ$ , $30^\circ$ , $45^\circ$ , $60^\circ$ and $90^\circ$ , and exact tangent values through $60^\circ$ . The sequences are $\sin\theta:0,\frac12,\frac{\sqrt2}{2},\frac{\sqrt3}{2},1$ and $\cos\theta:1,\frac{\sqrt3}{2},\frac{\sqrt2}{2},\frac12,0$ .
Also $\tan0^\circ=0$ , $\tan30^\circ=\frac1{\sqrt3}$ , $\tan45^\circ=1$ and $\tan60^\circ=\sqrt3$ . The $1:\sqrt3:2$ and $1:1:\sqrt2$ 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:\sqrt3:2

and

1:1:\sqrt2

triangles generate the exact trigonometric values.

Worked example

Work out the exact value of $3\sin60^\circ-2\cos45^\circ$ .

1.Use $\sin60^\circ=\frac{\sqrt3}{2}$ and $\cos45^\circ=\frac{\sqrt2}{2}$ .
2.Substitute: $3\left(\frac{\sqrt3}{2}\right)-2\left(\frac{\sqrt2}{2}\right)$ .
3.Simplify without decimals.

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

Common mistakes

Don't enter the values into a calculator and give decimals when the question asks for exact form.
Don't swap $\sin30^\circ=\frac12$ with $\sin60^\circ=\frac{\sqrt3}{2}$ .

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^\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

2\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\text{ cm}

and one acute angle is

30^\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 $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$ pairs each side with its opposite angle.
Use it when a complete opposite side-angle pair is known. The cosine rule $a^2=b^2+c^2-2bc\cos A$ 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 $8$ cm and $11$ cm, with included angle $47^\circ$ . Find the third side to $1$ decimal place.

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

Answer: $8.1$ cm.

Common mistakes

Don't pair side $a$ with an angle other than its opposite angle $A$ 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 $a$ opposite $A$ before choosing a rule; this prevents most substitution errors.

Tier 1 · Easy

ORIGINAL

1

In triangle

ABC

, side

a=7\text{ cm}

, angle

A=30^\circ

and angle

B=90^\circ

. Work out side

b

.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1

Two sides of a triangle are

7\text{ cm}

and

10\text{ cm}

, and the included angle is

60^\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

ABC

,

A=35^\circ

,

a=8\text{ cm}

and

b=11\text{ cm}

. Find all possible values of angles

B

and

C

. Give each angle to

1

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 $A=\frac12ab\sin C$ when two sides $a$ and $b$ and their included angle $C$ are known.
The formula comes from $\frac12\times\text{base}\times\text{perpendicular height}$ , with height $b\sin C$ .
Rearrange it to find a missing side or use inverse sine for an angle.
Because $\sin C=\sin(180^\circ-C)$ , 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

C

produces perpendicular height

b\sin C

.

Worked example

Two sides of a triangle are $9$ cm and $12$ cm, with included angle $40^\circ$ . Find the area to $1$ decimal place.

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

Answer: $34.7\text{ cm}^2$ .

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 $\frac12ab\sin C$ .

Tier 1 · Easy

ORIGINAL

1

Two sides of a triangle are

10\text{ cm}

and

7\text{ cm}

, and their included angle is

30^\circ

. Work out the area.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1

A triangle has area

48\text{ cm}^2

. Two sides have lengths

12\text{ cm}

and

x\text{ cm}

, and their included angle is

30^\circ

. Work out

x

.

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

1

Two sides of a triangle are

10\text{ cm}

and

12\text{ cm}

. Its area is

30\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 $\binom{x}{y}$ : 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 $A(-2,5)$ is translated to $A'(4,1)$ . Find the translation vector.

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

Answer: $\binom{6}{-4}$ .

Common mistakes

Don't subtract object minus image and obtain the reverse vector.
Don't write $(4,1)$ , 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)

is translated to

A'=(2,-1)

. Describe the translation as a vector.

(2)

(Total for Question 1 is 2 marks)

Tier 2 · Standard

ORIGINAL

1

Triangle

ABC

has vertices

A=(1,-2)

,

B=(5,1)

and

C=(2,4)

. It is translated by

\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)

to

P'=(1,-2)

. The same translation maps

Q=(3,k)

to

Q'=(10,5)

. Find

k

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 $\mathbf a$ . Add column vectors component by component.
Subtracting a vector adds its reverse, while multiplying by scalar $k$ multiplies every component and reverses direction when $k<0$ .
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:

\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC}

.

Worked example

Work out $3\binom{2}{-1}-\binom{5}{4}$ .

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

Answer: $\binom{1}{-7}$ .

Common mistakes

Don't subtract only the top component and add the bottom components.
Don't reverse $\overrightarrow{AB}$ and $\overrightarrow{BA}$ 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

\binom{3}{-5}+\binom{-7}{2}

.

(1)

(Total for Question 1 is 1 mark)

Tier 2 · Standard

ORIGINAL

1

Points

A

and

B

have position vectors

\mathbf{a}=\binom{2}{-1}

and

\mathbf{b}=\binom{8}{5}

. Point

M

is the midpoint of

AB

. Find the position vector of

M

and the vector

\overrightarrow{AM}

.

(3)

(Total for Question 1 is 3 marks)

Tier 3 · Hard

ORIGINAL

1

A walker starts at

P=(-3,2)

. The walker makes successive displacements

\binom{4}{3}

,

\binom{-1}{5}

and

-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

P

.

(4)

(Total for Question 1 is 4 marks)