G Geometry and measures — printable coverage pack

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

Use precise geometric language: a vertex is a corner, an edge joins vertices, parallel lines never meet, and perpendicular lines meet at a right angle.
Label an angle by placing its vertex in the middle, so $\angle ABC$ is the angle at $B$ ; in triangle $ABC$ , side $AB$ is opposite vertex $C$ .
Build a diagram one instruction at a time, marking equal lengths, parallel lines and right angles with conventional symbols rather than relying on appearance.
A common error is to assume a sketch is drawn to scale; only stated facts and marked relationships may be used.

Tier 1 · Easy

1. In triangle $PQR$ , which vertex is opposite the side $PR$ ?[1 mark]
Answer
- Vertex $Q$
Method: The side $PR$ joins vertices $P$ and $R$ . The remaining vertex is $Q$ , so $Q$ is opposite $PR$ .

Tier 2 · Standard

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 mark]
Answer
- $EF$ is perpendicular to $CD$ .
Method: Because $AB\parallel CD$ , a line perpendicular to $AB$ is also perpendicular to $CD$ . Therefore $EF\perp CD$ .

Tier 3 · Hard

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 marks]
Answer
- A correctly labelled diagram with $AB\parallel CD$ , $A,E,C$ collinear, $F,E,G$ collinear, and right-angle marks at $F$ and $G$ .
Method: Draw and arrow-mark the parallel sides $AB$ and $CD$ , then join $A$ to $C$ . Place $E$ on $AC$ . Through $E$ , draw one straight line meeting $AB$ at $F$ and $CD$ at $G$ , and mark both perpendicular intersections as right angles.

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

A perpendicular bisector crosses a segment at its midpoint at $90^\circ$ ; every point on it is equidistant from the segment's endpoints.
Construct an angle bisector using equal-radius arcs from the vertex and then intersecting arcs from the two arms; points on it are equidistant from the arms.
Translate locus conditions into standard boundaries: a fixed distance from a point gives a circle, while a fixed distance from a straight line gives two parallel lines.
Keep compass arcs visible and do not measure to replace a construction; the perpendicular distance to a line is the shortest distance.

Tier 1 · Easy

1. Describe how to construct the perpendicular bisector of a line segment $AB$ using a ruler and compasses.[2 marks]
Answer
- Draw equal-radius arcs from $A$ and $B$ that meet above and below $AB$ , then join the two arc intersections.
Method: Set the compass radius to more than half of $AB$ . Without changing it, draw arcs centred at $A$ and $B$ so that they intersect twice. A straight line through the intersections is the perpendicular bisector of $AB$ .

Tier 2 · Standard

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 marks]
Answer
- The part of the internal angle bisector from $O$ up to and including the point $6$ m from $O$ .
Method: Points equidistant from two intersecting lines lie on an angle bisector. Restricting the distance from $O$ to at most $6$ m keeps only the segment of the internal bisector inside the circle centred at $O$ with radius $6$ m.

Tier 3 · Hard

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 marks]
Answer
- The part of the perpendicular bisector of $AB$ lying inside or on both circles of radius $5$ cm centred at $A$ and $B$ ; it is the segment joining the circles' two intersection points.
Method: Construct the perpendicular bisector of $AB$ because $PA=PB$ . Draw circles of radius $5$ cm centred at $A$ and $B$ . The condition $PA\leq5$ keeps points inside both circles, so the required locus is the perpendicular-bisector segment between their two intersections.

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

Angles at a point sum to $360^\circ$ , angles on a straight line sum to $180^\circ$ , and vertically opposite angles are equal.
When a transversal crosses parallel lines, alternate and corresponding angles are equal; co-interior angles sum to $180^\circ$ .
Splitting an $n$ -sided polygon into $n-2$ triangles gives an interior-angle sum of $(n-2)\times180^\circ$ .
Do not quote alternate or corresponding angles unless the relevant lines are stated or shown to be parallel, and name the angle fact used.

Tier 1 · Easy

1. Two adjacent angles on a straight line are $63^\circ$ and $x^\circ$ . Work out $x$ .[1 mark]
Answer
- $117^\circ$
Method: Angles on a straight line total $180^\circ$ , so $x=180^\circ-63^\circ=117^\circ$ .

Tier 2 · Standard

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 marks]
Answer
- $x=15$
- $67^\circ$
Method: Alternate angles between parallel lines are equal, so $4x+7=7x-38$ . Hence $45=3x$ and $x=15$ . Substitution gives $4(15)+7=67^\circ$ .

Tier 3 · Hard

1. A polygon has $13$ sides. Twelve of its interior angles are each $155^\circ$ . Work out the remaining interior angle.[4 marks]
Answer
- $120^\circ$
Method: The interior-angle sum is $(13-2)\times180^\circ=1980^\circ$ . The twelve known angles total $12\times155^\circ=1860^\circ$ . The remaining angle is $1980^\circ-1860^\circ=120^\circ$ .

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

Classify a shape from properties that must always hold: for example, a parallelogram has two pairs of parallel opposite sides and a trapezium has one pair of parallel sides.
A rectangle has four right angles, a rhombus has four equal sides, and a square satisfies both definitions as well as being a parallelogram.
Use diagonal properties carefully: parallelogram diagonals bisect each other, rectangle diagonals are also equal, and rhombus diagonals are also perpendicular.
A common error is to use a property that is merely possible rather than guaranteed; one diagram cannot establish a definition.

Tier 1 · Easy

1. Name the quadrilateral that has exactly one pair of parallel sides.[1 mark]
Answer
- Trapezium
Method: A quadrilateral with exactly one pair of parallel sides is a trapezium.

Tier 2 · Standard

1. One interior angle of a rhombus is $68^\circ$ . Work out the other three interior angles.[2 marks]
Answer
- $112^\circ$ , $68^\circ$ , $112^\circ$
Method: Opposite angles in a rhombus are equal and adjacent angles sum to $180^\circ$ . The opposite angle is $68^\circ$ , and each adjacent angle is $180^\circ-68^\circ=112^\circ$ .

Tier 3 · Hard

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 marks]
Answer
- Diagonals that bisect each other make the quadrilateral a parallelogram; equal diagonals then make it a rectangle. A square would additionally require four equal sides or perpendicular diagonals, which has not been given.
Method: First use the converse parallelogram property: diagonals that bisect each other establish a parallelogram. In a parallelogram, equal diagonals establish a rectangle. Equal diagonals alone do not establish equal sides or perpendicular diagonals, so a non-square rectangle remains possible.

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

Congruent triangles are identical in size and shape, so all corresponding sides and angles are equal.
Use SSS for three side pairs, SAS for two side pairs and the included angle, ASA for two angle pairs and a side, and RHS for right triangles with equal hypotenuse and one other side.
Write vertices in corresponding order when stating congruence, then transfer only matching sides or angles.
AAA proves similarity, not congruence, and SSA is not a valid general congruence test.

Tier 1 · Easy

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 mark]
Answer
- SSS
Method: All three side lengths of one triangle match all three side lengths of the other, so the criterion is SSS.

Tier 2 · Standard

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 marks]
Answer
- They are congruent by RHS.
Method: Both triangles are right-angled, their hypotenuses $AC$ and $DF$ are equal, and one corresponding pair of shorter sides $AB$ and $DE$ is equal. Therefore the triangles are congruent by RHS.

Tier 3 · Hard

1. In quadrilateral $ABCD$ , diagonal $AC$ is drawn. Given that $AB=AD$ and $\angle BAC=\angle CAD$ , prove that $BC=CD$ .[3 marks]
Answer
- $BC=CD$
Method: Compare triangles $BAC$ and $DAC$ . We have $AB=AD$ , $AC$ is common, and the included angles $\angle BAC$ and $\angle CAD$ are equal. The triangles are congruent by SAS, so corresponding sides $BC$ and $DC$ are equal.

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

In an isosceles triangle, equal sides face equal angles; the converse also holds, so equal angles face equal sides.
For a right-angled triangle, use $a^2+b^2=c^2$ with $c$ opposite the right angle, and take the positive square root for a length.
A geometric proof should form a connected chain in which every equality follows from a stated angle fact, shape property, similarity or congruence result.
Do not use the fact being proved as a reason, and do not infer equal sides or angles just because they look equal in a sketch.

Tier 1 · Easy

1. Triangle $ABC$ is isosceles with $AB=AC$ . Angle $B$ is $47^\circ$ . Work out angle $A$ .[2 marks]
Answer
- $86^\circ$
Method: Equal sides face equal angles, so angle $C$ is also $47^\circ$ . The angles in a triangle sum to $180^\circ$ , giving $A=180^\circ-47^\circ-47^\circ=86^\circ$ .

Tier 2 · Standard

1. A right-angled triangle has perpendicular sides $9$ cm and $12$ cm. Work out the length of its hypotenuse.[3 marks]
Answer
- $15$ cm
Method: By Pythagoras' theorem, $c^2=9^2+12^2=81+144=225$ . Therefore $c=\sqrt{225}=15$ cm.

Tier 3 · Hard

1. In quadrilateral $ABCD$ , $AB\parallel CD$ and $AB=CD$ . Diagonal $AC$ is drawn. Prove that $BC=AD$ .[4 marks]
Answer
- $BC=AD$
Method: Because $AB\parallel CD$ , $\angle BAC=\angle DCA$ as alternate angles. Also $AB=CD$ and $AC$ is common. Thus triangles $BAC$ and $DCA$ are congruent by SAS. Corresponding sides $BC$ and $DA$ are therefore equal.

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

Describe a translation by a vector, a rotation by its centre, angle and direction, a reflection by its mirror line, and an enlargement by its centre and scale factor.
Translations, rotations and reflections preserve lengths and angles, so the image is congruent to the object; an enlargement preserves angles and scales every length by the same factor.
For an enlargement about centre $C$ , multiply the vector from $C$ to each point by the scale factor; a negative factor places the image on the opposite side of $C$ .
A common error is to omit part of a full transformation description or to multiply coordinates directly when the centre of enlargement is not the origin.

Tier 1 · Easy

1. Point $A(-3,4)$ is translated by the vector $\begin{pmatrix}5\\-2\end{pmatrix}$ . Find the coordinates of its image.[2 marks]
Answer
- $(2,2)$
Method: Add the vector components to the coordinates: $(-3+5,4-2)=(2,2)$ .

Tier 2 · Standard

1. Point $P(4,-1)$ is rotated $90^\circ$ anticlockwise about the origin. Find the coordinates of its image $P'$ .[2 marks]
Answer
- $P'(1,4)$
Method: A $90^\circ$ anticlockwise rotation about the origin maps $(x,y)$ to $(-y,x)$ . Therefore $(4,-1)$ maps to $(1,4)$ .

Tier 3 · Hard

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 marks]
Answer
- $A'(0,-3)$
- $B'(3,-4)$
- $C'(4,-1)$
Method: Subtract the centre, multiply by $-\frac12$ , then add the centre. For $A$ , $(4,4)\mapsto(-2,-2)$ , giving $(0,-3)$ . For $B$ , $(-2,6)\mapsto(1,-3)$ , giving $(3,-4)$ . For $C$ , $(-4,0)\mapsto(2,0)$ , giving $(4,-1)$ .

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

Track a combination in the stated order: the output of the first transformation is the input to the next.
Two translations combine by adding their vectors, while some pairs of reflections combine to form a rotation or a translation.
Rotations, reflections and translations preserve lengths, angle sizes, area and parallelism; reflections reverse orientation, whereas rotations and translations preserve it.
Do not reverse the order of a combination: transformations need not commute, so changing the order can change the final image.

Tier 1 · Easy

1. A shape is rotated and then translated. State two properties of the shape that must remain invariant.[2 marks]
Answer
- Any two of: side lengths, angle sizes, area, parallelism, or orientation remain unchanged.
Method: Both a rotation and a translation are rigid transformations. Each preserves lengths and angles, so it also preserves area and parallel lines; neither reverses orientation.

Tier 2 · Standard

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 marks]
Answer
- A translation by $\begin{pmatrix}-2\\5\end{pmatrix}$ .
Method: Add the translation vectors component by component: $\begin{pmatrix}3\\-2\end{pmatrix}+\begin{pmatrix}-5\\7\end{pmatrix}=\begin{pmatrix}-2\\5\end{pmatrix}$ .

Tier 3 · Hard

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 marks]
Answer
- A rotation of $90^\circ$ anticlockwise about the origin; orientation is preserved.
Method: The first reflection maps $(x,y)$ to $(x,-y)$ . Reflection in $y=x$ then swaps the coordinates, giving $(-y,x)$ . This is the coordinate rule for a $90^\circ$ anticlockwise rotation about the origin. A rotation preserves orientation.

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

A radius joins the centre to the circumference, a diameter is a chord through the centre, and every diameter is twice the radius.
A tangent touches a circle at one point, while a chord joins two points on the circumference; an arc is a portion of the circumference.
A sector is bounded by two radii and an arc, whereas a segment is bounded by a chord and an arc.
Do not call the curved boundary a circle: the circumference is the boundary, while the circle includes the region inside it.

Tier 1 · Easy

1. A circle has radius $6.5$ cm. Write down its diameter.[1 mark]
Answer
- $13$ cm
Method: The diameter is twice the radius, so $d=2\times6.5=13$ cm.

Tier 2 · Standard

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 marks]
Answer
- The segment is a chord; the line is a tangent.
Method: A segment whose endpoints lie on the circumference is a chord. A line meeting a circle at exactly one point is a tangent.

Tier 3 · Hard

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 marks]
Answer
- The minor sector is bounded by radii $OA$ and $OB$ and the minor arc $AB$ ; the minor segment is bounded by chord $AB$ and the minor arc $AB$ .
Method: A sector uses two radii plus their connecting arc. A segment uses a chord plus its corresponding arc. Selecting the shorter arc $AB$ gives the minor sector and minor segment.

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

The angle at the centre is twice the angle at the circumference standing on the same arc; angles in the same segment are equal, and an angle in a semicircle is $90^\circ$ .
Opposite angles in a cyclic quadrilateral sum to $180^\circ$ , and the alternate segment theorem links the angle between a tangent and chord to the angle in the opposite segment.
A radius is perpendicular to a tangent at the point of contact; tangents from the same external point are equal, and the perpendicular from the centre bisects a chord.
Always identify the same chord or arc before applying an angle theorem, and state the theorem as the reason rather than quoting an unsupported angle.

Tier 1 · Easy

1. $AB$ is a diameter of a circle and $C$ is another point on the circumference. Find $\angle ACB$ .[1 mark]
Answer
- $90^\circ$
Method: The angle in a semicircle is $90^\circ$ , so $\angle ACB=90^\circ$ .

Tier 2 · Standard

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 marks]
Answer
- $62^\circ$
Method: Both angles stand on the minor arc $AB$ . The angle at the centre is twice the angle at the circumference, so $\angle ACB=124^\circ\div2=62^\circ$ .

Tier 3 · Hard

1. From a point $P$ outside a circle with centre $O$ , tangents touch the circle at $A$ and $B$ . Prove that $PA=PB$ and that $OP$ bisects $\angle APB$ .[4 marks]
Answer
- $PA=PB$ and $\angle APO=\angle OPB$
Method: Radii meet tangents at right angles, so triangles $OAP$ and $OBP$ are right-angled. They have equal hypotenuse $OP$ and equal radii $OA=OB$ , so they are congruent by RHS. Corresponding parts then give $PA=PB$ and $\angle APO=\angle OPB$ , proving that $OP$ bisects $\angle APB$ .

G11 · Solve geometrical problems on coordinate axes

Use coordinate differences to find horizontal and vertical lengths, and use the midpoint formula by averaging the two $x$ -coordinates and the two $y$ -coordinates.
The distance between $(x_1,y_1)$ and $(x_2,y_2)$ is $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$ by Pythagoras' theorem.
Gradients can establish geometry: equal gradients show parallel lines, while gradients with product $-1$ show perpendicular non-vertical lines.
Keep signs when subtracting negative coordinates, and do not read a length directly from a sketch that is not drawn to scale.

Tier 1 · Easy

1. Find the midpoint of the line segment joining $(-5,2)$ to $(7,8)$ .[2 marks]
Answer
- $(1,5)$
Method: Average the coordinates: $x=(-5+7)/2=1$ and $y=(2+8)/2=5$ . The midpoint is $(1,5)$ .

Tier 2 · Standard

1. Points $A(-1,4)$ and $B(5,-4)$ are joined. Find the exact length $AB$ .[3 marks]
Answer
- $10$
Method: The coordinate changes are $5-(-1)=6$ and $-4-4=-8$ . Therefore $AB=\sqrt{6^2+(-8)^2}=\sqrt{36+64}=\sqrt{100}=10$ .

Tier 3 · Hard

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 marks]
Answer
- Right-angled at $B$
- $AB=BC=2\sqrt{10}$
- Area $=20$ square units
Method: The gradients are $m_{AB}=2/6=1/3$ and $m_{BC}=6/(-2)=-3$ , whose product is $-1$ , so the angle at $B$ is $90^\circ$ . Also $AB=\sqrt{6^2+2^2}=\sqrt{40}=2\sqrt{10}$ and $BC=\sqrt{(-2)^2+6^2}=\sqrt{40}=2\sqrt{10}$ , so the triangle is isosceles. Its perpendicular equal sides have area $\frac12(2\sqrt{10})^2=20$ square units.

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

A face is flat, a curved surface is not a face, an edge is where surfaces meet, and a vertex is a point where edges meet.
An $n$ -sided prism has two congruent polygonal ends and $n$ rectangular side faces, giving $n+2$ faces, $3n$ edges and $2n$ vertices.
A pyramid has one polygonal base and triangular faces meeting at one apex; a cone has one circular base, one curved surface and one vertex.
Count each feature once, especially hidden edges, and do not give a sphere or cylinder vertices where no edges meet.

Tier 1 · Easy

1. Write down the number of faces and the number of vertices of a triangular prism.[2 marks]
Answer
- $5$ faces
- $6$ vertices
Method: A triangular prism has two triangular faces and three rectangular faces, making $5$ faces. Its two triangular ends have $3+3=6$ vertices.

Tier 2 · Standard

1. A solid has one circular plane face, one curved surface, one circular edge and one vertex. Name the solid.[2 marks]
Answer
- Cone
Method: The circular base is the one plane face, its rim is the circular edge, the side is curved, and the surfaces meet at one apex. These are the properties of a cone.

Tier 3 · Hard

1. A prism has a regular $9$ -sided polygon as each end. Work out its numbers of faces, edges and vertices.[3 marks]
Answer
- $11$ faces
- $27$ edges
- $18$ vertices
Method: For an $n$ -sided prism, the counts are $n+2$ faces, $3n$ edges and $2n$ vertices. With $n=9$ , these are $11$ faces, $27$ edges and $18$ vertices.

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

A plan is the view from directly above; a front or side elevation is an orthographic view from the stated horizontal direction.
Project corresponding corners with parallel construction lines so widths align between the plan and front elevation and depths align between the plan and side elevation.
For stacks of cubes, an elevation shows the greatest height visible in each column along that viewing direction, while a numbered plan records each stack height.
Do not add perspective or show hidden internal edges unless asked; plans and elevations use true lengths in their viewing plane.

Tier 1 · Easy

1. A cuboid is $8$ cm long, $5$ cm wide and $3$ cm high. State the shape and dimensions of its plan.[2 marks]
Answer
- A rectangle $8$ cm by $5$ cm
Method: The plan is viewed from above, so it shows length and width but not height. Therefore it is an $8$ cm by $5$ cm rectangle.

Tier 2 · Standard

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 marks]
Answer
- Front elevation, west to east: $(3,1)$
- Side elevation, north to south: $(2,3)$
Method: From the south, take the greatest height in each west-east column: west gives $\max(2,3)=3$ and east gives $\max(1,0)=1$ . From the east, take the greatest height in each north-south row: north gives $\max(2,1)=2$ and south gives $\max(3,0)=3$ .

Tier 3 · Hard

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 marks]
Answer
- $7$ cubes; for example, from north to south the plan rows can be $(1,2)$ and $(3,1)$ , with entries west to east.
Method: The height $3$ required in both the west front column and south side row can be placed in the south-west cell. The height $2$ required in both the east front column and north side row can be placed in the north-east cell. The other two occupied cells need at least one cube each, giving $3+2+1+1=7$ . The plan rows $(1,2)$ and $(3,1)$ have exactly the stated elevations.

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

Standard units include millimetres, centimetres, metres and kilometres for length; square or cubic units for area and volume; litres for capacity; grams or kilograms for mass; and seconds, minutes or hours for time.
Convert every measurement to compatible units before calculating, remembering that area conversion factors are squared and volume conversion factors are cubed.
For example, because $1\text{ m}=100\text{ cm}$ , it follows that $1\text{ m}^2=10\,000\text{ cm}^2$ and $1\text{ m}^3=1\,000\,000\text{ cm}^3$ .
A common error is to use the linear conversion factor for an area or volume, such as multiplying square metres by $100$ instead of $10\,000$ to obtain square centimetres.

Tier 1 · Easy

1. Convert $3.75\text{ kg}$ to grams.[1 mark]
Answer
- $3750\text{ g}$
Method: There are $1000\text{ g}$ in $1\text{ kg}$ , so $3.75\times1000=3750$ . Therefore the mass is $3750\text{ g}$ .

Tier 2 · Standard

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 marks]
Answer
- $156$ litres
Method: The full volume is $0.8\times0.5\times0.6=0.24\text{ m}^3$ . The water occupies $0.65\times0.24=0.156\text{ m}^3$ . Since $1\text{ m}^3=1000$ litres, the volume is $0.156\times1000=156$ litres.

Tier 3 · Hard

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 marks]
Answer
- £ $265.50$
Method: Each tile has side $0.30\text{ m}$ , so its area is $0.30^2=0.09\text{ m}^2$ . The floor area is $4.8\times3.6=17.28\text{ m}^2$ , requiring exactly $17.28/0.09=192$ tiles. Including $8\%$ gives $192\times1.08=207.36$ , so at least $208$ tiles are needed. Since $208/12=17.\overline{3}$ , the decorator must buy $18$ boxes. The cost is $18\times14.75=265.50$ , so the total is £ $265.50$ .

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

Lengths and angles can be measured from an accurate drawing, while a map scale links a measured map distance to the corresponding real distance.
For a scale $1:n$ , multiply a map length by $n$ to obtain the real length in the same units; bearings are measured clockwise from north and written using three figures.
For example, a clockwise angle of $52^\circ$ from north is written as the bearing $052^\circ$ , and the reverse bearing is $232^\circ$ .
A common error is to measure a bearing anticlockwise or from an east-west line instead of clockwise from the north line at the starting point.

Tier 1 · Easy

1. A ray from point $P$ makes an angle of $68^\circ$ clockwise from north. Write its bearing from $P$ .[1 mark]
Answer
- $068^\circ$
Method: A bearing is the clockwise angle from north written with three figures. Therefore $68^\circ$ is written as $068^\circ$ .

Tier 2 · Standard

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 marks]
Answer
- $2\text{ km}$
- $037^\circ$
Method: The map distance is $\sqrt{6^2+8^2}=10\text{ cm}$ . The real distance is $10\times20\,000=200\,000\text{ cm}=2\text{ km}$ . If the bearing angle is $\theta$ , then $\tan\theta=6/8$ , so $\theta=36.9^\circ$ . Written as a three-figure bearing to the nearest degree, this is $037^\circ$ .

Tier 3 · Hard

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 marks]
Answer
- $1:37\,500$
- $4.2\text{ km}$
Method: Convert $3.15\text{ km}$ to $315\,000\text{ cm}$ . The scale factor is $315\,000/8.4=37\,500$ , so the scale is $1:37\,500$ . The second road represents $11.2\times37\,500=420\,000\text{ cm}=4.2\text{ km}$ .

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

Use $A=\frac12 bh$ for a triangle, $A=bh$ for a parallelogram and $A=\frac12(a+b)h$ for a trapezium, where each height is perpendicular to the relevant base or parallel sides.
A right prism has a constant cross-section, so its volume is the area of that cross-section multiplied by the prism length; for a cylinder this gives $V=\pi r^2h$ .
For example, a trapezium with parallel sides $5$ and $9$ and perpendicular height $4$ has area $\frac12(5+9)\times4=28$ square units.
A common error is to use a sloping side as the perpendicular height or to omit the factor of $\frac12$ in a triangle or trapezium calculation.

Tier 1 · Easy

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 marks]
Answer
- $60\text{ cm}^2$
Method: Use $A=\frac12(a+b)h$ . Then $A=\frac12(7+13)\times6=\frac12\times20\times6=60\text{ cm}^2$ .

Tier 2 · Standard

1. A cylinder has radius $4\text{ cm}$ and height $9\text{ cm}$ . Work out its volume in terms of $\pi$ .[2 marks]
Answer
- $144\pi\text{ cm}^3$
Method: The circular cross-section has area $\pi r^2=\pi(4^2)=16\pi\text{ cm}^2$ . Multiply by the height: $V=16\pi\times9=144\pi\text{ cm}^3$ .

Tier 3 · Hard

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 marks]
Answer
- $x=10$
Method: The cross-sectional area is $\frac12x(x+2)$ . Hence $10\times\frac12x(x+2)=600$ , so $x(x+2)=120$ . This gives $x^2+2x-120=0$ , which factorises to $(x+12)(x-10)=0$ . A length must be positive, so $x=10$ .

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

For a circle, circumference is $C=2\pi r=\pi d$ and area is $A=\pi r^2$ ; for composite shapes, add included regions or subtract cut-out regions.
For a pyramid or cone, $V=\frac13\times\text{base area}\times\text{perpendicular height}$ ; find surface area by adding the base and every exposed triangular face, or use $\pi rl$ for a cone's curved surface.
For example, a sphere of radius $r$ has surface area $4\pi r^2$ and volume $\frac43\pi r^3$ , so doubling $r$ multiplies these by $4$ and $8$ respectively.
A common error is to use the diameter where a formula requires the radius, or to include an internal shared edge or face in a perimeter or surface-area total.

Tier 1 · Easy

1. A circle has diameter $13\text{ cm}$ . Write its circumference in terms of $\pi$ .[1 mark]
Answer
- $13\pi\text{ cm}$
Method: Use $C=\pi d$ . With $d=13\text{ cm}$ , the circumference is $13\pi\text{ cm}$ .

Tier 2 · Standard

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 marks]
Answer
- Area $=180+\frac{25\pi}{2}\text{ cm}^2$
- Perimeter $=46+5\pi\text{ cm}$
Method: The rectangle area is $18\times10=180\text{ cm}^2$ . The semicircle has radius $5\text{ cm}$ , so its area is $\frac12\pi(5^2)=\frac{25\pi}{2}\text{ cm}^2$ . For the perimeter, include the two $18\text{ cm}$ sides, the unshared $10\text{ cm}$ side and the semicircular arc $\pi r=5\pi\text{ cm}$ . This gives area $180+\frac{25\pi}{2}\text{ cm}^2$ and perimeter $46+5\pi\text{ cm}$ .

Tier 3 · Hard

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 marks]
Answer
- Total surface area $=96\pi\text{ cm}^2$
- Volume $=96\pi\text{ cm}^3$
Method: The curved surface area is $\pi rl=\pi(6)(10)=60\pi\text{ cm}^2$ and the base area is $\pi r^2=36\pi\text{ cm}^2$ , giving $96\pi\text{ cm}^2$ in total. The volume is $\frac13\pi r^2h=\frac13\pi(6^2)(8)=96\pi\text{ cm}^3$ .

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

An arc or sector is the same fraction of a circle as its central angle is of $360^\circ$ .
Use arc length $=\frac{\theta}{360}\times2\pi r$ and sector area $=\frac{\theta}{360}\times\pi r^2$ , rearranging when the angle or radius is unknown.
For example, a $90^\circ$ sector is one quarter of a circle, so its arc length is $\frac14(2\pi r)$ and its area is $\frac14\pi r^2$ .
A common error is to calculate the full circumference or area without multiplying by $\frac{\theta}{360}$ , or to include radii when only arc length is requested.

Tier 1 · Easy

1. Work out the arc length of a quarter-circle with radius $8\text{ cm}$ . Give the answer in terms of $\pi$ .[2 marks]
Answer
- $4\pi\text{ cm}$
Method: A quarter-circle has angle $90^\circ$ . Its arc length is $\frac{90}{360}\times2\pi(8)=\frac14\times16\pi=4\pi\text{ cm}$ .

Tier 2 · Standard

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 marks]
Answer
- Angle $=140^\circ$
- Area $=\frac{63\pi}{2}\text{ cm}^2$
Method: Use $7\pi=\frac{\theta}{360}\times2\pi(9)$ . Cancelling $\pi$ and solving gives $\theta=140^\circ$ . The sector area is then $\frac{140}{360}\times\pi(9^2)=\frac{63\pi}{2}\text{ cm}^2$ .

Tier 3 · Hard

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 marks]
Answer
- Area $=38\pi\text{ cm}^2$
- Perimeter $=10+\frac{76\pi}{5}\text{ cm}$
Method: The area is $\frac{144}{360}\pi(12^2-7^2)=\frac25\pi(144-49)=38\pi\text{ cm}^2$ . The two arcs have total length $\frac25\times2\pi(12+7)=\frac{76\pi}{5}\text{ cm}$ . The two straight radial edges each have length $12-7=5\text{ cm}$ , so the perimeter is $10+\frac{76\pi}{5}\text{ cm}$ .

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

Congruent figures have the same size and shape, while similar figures have equal corresponding angles and corresponding lengths in a constant ratio.
Match corresponding sides in the same order, calculate one length scale factor, and multiply or divide every corresponding length by that factor.
At Higher tier only, a length scale factor $k$ gives area scale factor $k^2$ and volume scale factor $k^3$ ; for example, length ratio $2:5$ gives area ratio $4:25$ and volume ratio $8:125$ .
A common error is to compare non-corresponding sides or, at Higher tier, to use the linear scale factor directly for an area or volume.

Tier 1 · Easy

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 mark]
Answer
- They are congruent by SAS.
Method: The triangles have two pairs of equal corresponding sides, $AB=PQ$ and $AC=PR$ . The equal $42^\circ$ angle is included between those sides in each triangle, so the triangles are congruent by SAS.

Tier 2 · Standard

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 marks]
Answer
- $12\text{ cm}$ corresponds to $20\text{ cm}$
- $35\text{ cm}$ corresponds to $21\text{ cm}$
Method: The scale factor from smaller to larger is $5/3$ . Therefore $12\times\frac53=20\text{ cm}$ . To reverse the enlargement, multiply by $3/5$ , giving $35\times\frac35=21\text{ cm}$ .

Tier 3 · Hard

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 marks]
Answer
- Other sides $=12\text{ cm}$ and $16.5\text{ cm}$
- Perimeter $=48\text{ cm}$
Method: The longest sides correspond, so the scale factor is $19.5/13=1.5$ . The other sides are $8\times1.5=12\text{ cm}$ and $11\times1.5=16.5\text{ cm}$ . The perimeter is $12+16.5+19.5=48\text{ cm}$ .

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

In a right-angled triangle, $a^2+b^2=c^2$ with $c$ as the hypotenuse, and $\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$ , $\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$ , $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$ .
Label the triangle relative to the required angle, choose Pythagoras or the ratio containing the known and unknown sides, and use an inverse trigonometric function when finding an angle.
At Foundation tier, apply Pythagoras and trigonometry to right-angled triangles in two-dimensional figures, including those found by symmetry in isosceles triangles; at Higher tier only, extend to general (non-right) triangles and identify a right-angled cross-section before applying the same method in three dimensions.
A common error is to treat a non-right-angled triangle as right-angled, or to use an angle and sides that do not belong to the same right-angled cross-section.

Tier 1 · Easy

1. A right-angled triangle has perpendicular sides $6\text{ cm}$ and $8\text{ cm}$ . Work out the hypotenuse.[2 marks]
Answer
- $10\text{ cm}$
Method: By Pythagoras, $c^2=6^2+8^2=36+64=100$ . Since a length is positive, $c=\sqrt{100}=10\text{ cm}$ .

Tier 2 · Standard

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 marks]
Answer
- Height $=6.9\text{ m}$
- Angle $=73.0^\circ$
Method: The ladder is the hypotenuse, so the height is $\sqrt{7.2^2-2.1^2}=6.886\ldots\text{ m}=6.9\text{ m}$ . If the ground angle is $\theta$ , then $\cos\theta=2.1/7.2$ . Hence $\theta=\cos^{-1}(2.1/7.2)=73.0^\circ$ to $1$ decimal place.

Tier 3 · Hard

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 marks]
Answer
- Height $=12\text{ cm}$
- $\angle ABC=67.4^\circ$
Method: The perpendicular from $A$ bisects the $10\text{ cm}$ base, giving a right-angled triangle with hypotenuse $13\text{ cm}$ and base $5\text{ cm}$ . Its height is $\sqrt{13^2-5^2}=12\text{ cm}$ . For $\angle ABC=\theta$ , $\tan\theta=12/5$ , so $\theta=67.4^\circ$ to $1$ decimal place.

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

Know the exact sine and cosine values at $0^\circ$ , $30^\circ$ , $45^\circ$ , $60^\circ$ and $90^\circ$ , and the exact tangent values at $0^\circ$ , $30^\circ$ , $45^\circ$ and $60^\circ$ .
The complete sets are $\sin\theta:0,\frac12,\frac{\sqrt2}{2},\frac{\sqrt3}{2},1$ and $\cos\theta:1,\frac{\sqrt3}{2},\frac{\sqrt2}{2},\frac12,0$ for $\theta=0^\circ,30^\circ,45^\circ,60^\circ,90^\circ$ ; also $\tan0^\circ=0$ , $\tan30^\circ=\frac1{\sqrt3}$ , $\tan45^\circ=1$ , $\tan60^\circ=\sqrt3$ .
For example, the $1:\sqrt3:2$ triangle gives $\sin30^\circ=\frac12$ , $\cos30^\circ=\frac{\sqrt3}{2}$ and $\tan30^\circ=\frac1{\sqrt3}$ .
A common error is to give a decimal approximation when an exact value is required, or to interchange the opposite and adjacent sides for $30^\circ$ and $60^\circ$ .

Tier 1 · Easy

1. Work out the exact value of $\sin30^\circ+\cos60^\circ$ .[1 mark]
Answer
- $1$
Method: $\sin30^\circ=\frac12$ and $\cos60^\circ=\frac12$ . Therefore the sum is $\frac12+\frac12=1$ .

Tier 2 · Standard

1. Work out the exact value of $2\sin45^\circ\cos45^\circ+\cos60^\circ$ .[3 marks]
Answer
- $\frac32$
Method: $\sin45^\circ=\cos45^\circ=\frac{\sqrt2}{2}$ and $\cos60^\circ=\frac12$ . Therefore $2\sin45^\circ\cos45^\circ+\cos60^\circ=2(\frac{\sqrt2}{2})(\frac{\sqrt2}{2})+\frac12=1+\frac12=\frac32$ .

Tier 3 · Hard

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 marks]
Answer
- $18\sqrt3\text{ cm}^2$
Method: The side opposite $30^\circ$ is $12\sin30^\circ=12\times\frac12=6\text{ cm}$ . The adjacent side is $12\cos30^\circ=12\times\frac{\sqrt3}{2}=6\sqrt3\text{ cm}$ . Hence the area is $\frac12\times6\times6\sqrt3=18\sqrt3\text{ cm}^2$ .

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]

The sine rule $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$ pairs each side with its opposite angle, while the cosine rule connects three sides with one angle.
Use the sine rule when an opposite side-angle pair is known; use the cosine rule for two sides and their included angle or for all three sides.
When the sine rule gives $\sin B=k$ , check both $B=\sin^{-1}k$ and $B=180^\circ-\sin^{-1}k$ against the triangle angle sum.
A common error is to pair a side with a non-opposite angle or to ignore the second possible angle in an ambiguous sine-rule problem.

Tier 1 · Easy

1. In triangle $ABC$ , side $a=7\text{ cm}$ , angle $A=30^\circ$ and angle $B=90^\circ$ . Work out side $b$ .[2 marks]
Answer
- $14\text{ cm}$
Method: By the sine rule, $\frac{b}{\sin90^\circ}=\frac{7}{\sin30^\circ}$ . Hence $b=7\times\frac{1}{1/2}=14\text{ cm}$ .

Tier 2 · Standard

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 marks]
Answer
- $\sqrt{79}\text{ cm}$
Method: Using the cosine rule, $c^2=7^2+10^2-2(7)(10)\cos60^\circ=49+100-70=79$ . Therefore $c=\sqrt{79}\text{ cm}$ .

Tier 3 · Hard

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 marks]
Answer
- $B=52.1^\circ$ , $C=92.9^\circ$
- $B=127.9^\circ$ , $C=17.1^\circ$
Method: The sine rule gives $\sin B=\frac{11\sin35^\circ}{8}=0.7887\ldots$ . Therefore $B=52.1^\circ$ or $B=180^\circ-52.1^\circ=127.9^\circ$ . The corresponding third angles are $C=180^\circ-35^\circ-52.1^\circ=92.9^\circ$ and $C=180^\circ-35^\circ-127.9^\circ=17.1^\circ$ .

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

The area of a triangle is $A=\frac12ab\sin C$ , where $C$ is the included angle between sides $a$ and $b$ .
Substitute the two sides and their included angle to find area, or rearrange before using an inverse sine to find an angle.
If $\sin C=k$ , both $C=\sin^{-1}k$ and $C=180^\circ-\sin^{-1}k$ may give triangles with the same area.
A common error is to use an angle that is not between the two substituted sides, or to reject an obtuse solution without checking the information given.

Tier 1 · Easy

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 marks]
Answer
- $17.5\text{ cm}^2$
Method: $A=\frac12ab\sin C=\frac12(10)(7)\sin30^\circ=35\times\frac12=17.5\text{ cm}^2$ .

Tier 2 · Standard

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 marks]
Answer
- $x=16$
Method: Use $48=\frac12(12)(x)\sin30^\circ$ . Since $\sin30^\circ=\frac12$ , this becomes $48=3x$ . Therefore $x=16$ .

Tier 3 · Hard

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 marks]
Answer
- $60^\circ$ or $120^\circ$
Method: $30\sqrt3=\frac12(10)(12)\sin C=60\sin C$ , so $\sin C=\frac{\sqrt3}{2}$ . Between $0^\circ$ and $180^\circ$ , this occurs at $C=60^\circ$ and $C=120^\circ$ .

G24 · Describe translations as 2D vectors

A translation moves every point through the same horizontal and vertical displacement without changing the shape's size or orientation.
Describe a translation by the column vector $\binom{x}{y}$ , where positive $x$ is right, negative $x$ is left, positive $y$ is up and negative $y$ is down.
For example, moving $4$ units left and $3$ units up is the translation vector $\binom{-4}{3}$ .
A common error is to reverse the vector by subtracting the image coordinates from the object coordinates instead of calculating image minus object.

Tier 1 · Easy

1. Point $A=(-3,4)$ is translated to $A'=(2,-1)$ . Describe the translation as a vector.[2 marks]
Answer
- $\binom{5}{-5}$
Method: Calculate image minus object in each coordinate: $2-(-3)=5$ and $-1-4=-5$ . The translation vector is $\binom{5}{-5}$ .

Tier 2 · Standard

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 marks]
Answer
- $A'=(-3,1)$
- $B'=(1,4)$
- $C'=(-2,7)$
Method: Add $-4$ to every $x$ -coordinate and $3$ to every $y$ -coordinate. This gives $A'=(-3,1)$ , $B'=(1,4)$ and $C'=(-2,7)$ .

Tier 3 · Hard

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 marks]
Answer
- $k=11$
- $\binom{7}{-6}$
Method: From $P$ to $P'$ , the displacement is $1-(-6)=7$ horizontally and $-2-4=-6$ vertically, so the vector is $\binom{7}{-6}$ . Therefore the image of $Q=(3,k)$ has $y$ -coordinate $k-6$ . Since $k-6=5$ , $k=11$ .

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

Vectors have magnitude and direction; they can be represented by directed line segments, column vectors or algebraic symbols such as $\mathbf{a}$ and $\mathbf{b}$ .
Add vectors component by component, subtract to reverse a displacement, and multiply by a scalar to change magnitude and, for a negative scalar, direction.
For a route through successive points, add vectors in travel order; at Higher tier only, compare two vector routes or show one direction vector is a scalar multiple of another to construct a geometric proof.
A common error is to subtract vectors in the wrong order or to forget that multiplying by a negative scalar reverses the vector's direction.

Tier 1 · Easy

1. Work out $\binom{3}{-5}+\binom{-7}{2}$ .[1 mark]
Answer
- $\binom{-4}{-3}$
Method: Add corresponding components: $3+(-7)=-4$ and $-5+2=-3$ . Therefore the sum is $\binom{-4}{-3}$ .

Tier 2 · Standard

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 marks]
Answer
- $\overrightarrow{OM}=\binom{5}{2}$
- $\overrightarrow{AM}=\binom{3}{3}$
Method: The midpoint position vector is $\frac12(\mathbf{a}+\mathbf{b})=\frac12\binom{10}{4}=\binom{5}{2}$ . Then $\overrightarrow{AM}=\overrightarrow{OM}-\overrightarrow{OA}=\binom{5}{2}-\binom{2}{-1}=\binom{3}{3}$ .

Tier 3 · Hard

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 marks]
Answer
- Resultant $=\binom{1}{12}$
- Finish $=(-2,14)$
- Return vector $=\binom{-1}{-12}$
Method: First, $-2\binom{1}{-2}=\binom{-2}{4}$ . Adding all displacements gives $\binom{4}{3}+\binom{-1}{5}+\binom{-2}{4}=\binom{1}{12}$ . Add this to $P=(-3,2)$ to get the finish $(-2,14)$ . The return vector is the negative of the resultant, $\binom{-1}{-12}$ .