G Geometry and measures — revision question pack

25 specification points · notes, questions, answers and worked methods

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

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

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

1
In triangle $PQR$ , which vertex is opposite the side $PR$ ?
(1)
(Total for Question 1 is 1 mark)
2
Two line segments $QR$ and $RS$ meet at $R$ . Write down the angle at $R$ using three letters.
(1)
(Total for Question 2 is 1 mark)

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)
(Total for Question 1 is 1 mark)
2
Two plane faces of a solid meet along line segment $AB$ , and several edges meet at $A$ . Write down the geometric term for $AB$ and the geometric term for $A$ .
(2)
(Total for Question 2 is 2 marks)
3
Line segments $AB$ , $BC$ and $BD$ meet at $B$ . The segment $BD$ is at right angles to the segment $BC$ . Write down the angle between $BA$ and $BD$ using three letters, and write the right-angle relationship between $BD$ and $BC$ using symbols.
(2)
(Total for Question 3 is 2 marks)

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)
(Total for Question 1 is 3 marks)
2
Draw line segment $PQ$ of length $8$ cm. At $P$ , draw $PR\perp PQ$ with $PR=6$ cm, and join $R$ to $Q$ . Mark $U$ as the midpoint of $PR$ . Through $U$ , draw the line parallel to $RQ$ , meeting $PQ$ at $V$ . Mark all the stated relationships.
(3)
(Total for Question 2 is 3 marks)
3
A triangular-based pyramid has base $PQR$ and apex $S$ . Write down the edge where plane faces $SPQ$ and $SPR$ meet, the vertex that is not on plane $PQR$ , and the edge opposite vertex $Q$ in triangular face $QRS$ .
(3)
(Total for Question 3 is 3 marks)
4
Quadrilateral $PQRS$ has vertices in order. Which option satisfies all these conditions: $PQ\parallel RS$ , $QR\perp PQ$ , diagonal $PR$ is drawn, and $T$ lies on $PR$ ? A $P(0,0),Q(6,0),R(6,4),S(1,4),T(3,2)$ B $P(0,0),Q(6,0),R(6,4),S(1,4),T(3,3)$ C $P(0,0),Q(6,0),R(6,4),S(0,3),T(3,2)$ . Give three geometric facts that justify your choice.
(4)
(Total for Question 4 is 4 marks)
5
On a coordinate grid, plot $P(0,0)$ , $Q(8,0)$ , $R(6,6)$ and $S(2,6)$ , then join them in order. Draw diagonals $PR$ and $QS$ , labelling their intersection $E$ . Through $E$ , draw the line perpendicular to $SR$ , meeting $SR$ at $G$ . Write down the coordinates of $E$ and $G$ , then write the relationships between $PQ$ and $SR$ , and between $EG$ and $SR$ , using symbols.
(5)
(Total for Question 5 is 5 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

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

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)
2
Describe the locus of points exactly $4$ cm from a fixed point $O$ .
(1)
(Total for Question 2 is 1 mark)

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)
(Total for Question 1 is 3 marks)
2
A point $P$ is not on the straight line $l$ . Describe how to construct the perpendicular from $P$ to $l$ using a ruler and compasses.
(2)
(Total for Question 2 is 2 marks)
3
Point $A$ lies on a straight line $l$ . Describe how to construct the perpendicular to $l$ at $A$ using a ruler and compasses. Leave all construction arcs visible.
(2)
(Total for Question 3 is 2 marks)

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)
(Total for Question 1 is 4 marks)
2
Fixed points $A$ and $B$ lie on a straight line $l$ , with $AB=8$ cm. A point $P$ must be closer to $A$ than to $B$ and less than $3$ cm from $l$ . Describe the complete region in which $P$ can lie.
(4)
(Total for Question 2 is 4 marks)
3
Using a ruler and compasses only, draw a horizontal line segment $AC$ of length $8$ cm. Construct its perpendicular bisector. Mark $B$ on the bisector $3$ cm above $AC$ and $D$ on the bisector $3$ cm below $AC$ . Join $A$ , $B$ , $C$ , $D$ in order and write down the most specific name of the quadrilateral.
(4)
(Total for Question 3 is 4 marks)
4
Two perpendicular straight lines $l$ and $m$ cross at $O$ . A sensor must be exactly $3$ m from $l$ and exactly $4$ m from $m$ . Describe how to construct the complete locus of possible positions and state how many positions there are.
(4)
(Total for Question 4 is 4 marks)
5
Using a ruler and compasses only, draw $AB=AC=16$ cm with $AB\perp AC$ . Mark $D$ on $AB$ so that $AD=8$ cm and mark $E$ on $AC$ so that $AE=6$ cm. Construct the point $P$ inside triangle $ABC$ that is equidistant from segments $AB$ and $AC$ , and satisfies $PD=PE$ . Leave all construction arcs visible.
(5)
(Total for Question 5 is 5 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

Explanation

Use angle facts as a chain of justified steps. Angles at a point total $360^\circ$ , angles on a straight line total $180^\circ$ , and vertically opposite angles are equal.
When a transversal crosses parallel lines, corresponding and alternate angles are equal; interior angles on the same side total $180^\circ$ .
A triangle totals $180^\circ$ .
Drawing diagonals from one vertex divides an $n$ -gon into $n-2$ triangles, so its interior-angle sum is $(n-2)\times180^\circ$ .
In exam working, write the relevant equality or subtraction and name the angle fact, including the parallel lines that make it valid.

Alternate angles are equal when a transversal crosses parallel lines.

Worked example

A regular polygon has each exterior angle equal to $24^\circ$ . Find its number of sides and each interior angle.

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

Answer: The polygon has $15$ sides and each interior angle is $156^\circ$ .

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.

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

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)
2
Two straight lines cross. One angle at the intersection is $38^\circ$ . Write down the vertically opposite angle.
(1)
(Total for Question 2 is 1 mark)

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)
(Total for Question 1 is 3 marks)
2
An exterior angle of a triangle is $126^\circ$ . One of the two opposite interior angles is $49^\circ$ . Work out the other opposite interior angle.
(2)
(Total for Question 2 is 2 marks)
3
Three angles at a point are $(2x+10)^\circ$ , $(x+35)^\circ$ and $135^\circ$ . Work out $x$ and the size of the smallest angle.
(3)
(Total for Question 3 is 3 marks)

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)
(Total for Question 1 is 4 marks)
2
A straight line $DAE$ passes through vertex $A$ of triangle $ABC$ , with $DE\parallel BC$ . Given $\angle DAB=68^\circ$ and $\angle CAE=47^\circ$ , work out all three angles of triangle $ABC$ . For each stage of your working, give a reason.
(4)
(Total for Question 2 is 4 marks)
3
The five exterior angles of a convex pentagon are $54^\circ$ , $68^\circ$ , $70^\circ$ , $x^\circ$ and $2x^\circ$ . Work out $x$ and the largest interior angle of the pentagon.
(4)
(Total for Question 3 is 4 marks)
4
Rays $OA$ , $OB$ , $OC$ and $OD$ occur in that clockwise order. $AOC$ is a straight line. Angles $AOB$ and $BOC$ are $(3x+7)^\circ$ and $(5x-3)^\circ$ . The reflex angle $BOD$ is $270^\circ$ . Work out $x$ and the smaller angle $COD$ .
(4)
(Total for Question 4 is 4 marks)
5
The vertices of convex trapezium $ABCD$ are in order, with $AB\parallel CD$ . Angle $DAB=(4x+9)^\circ$ and angle $CDA=(7x-5)^\circ$ . Angle $ABC$ is $49^\circ$ larger than angle $DAB$ . Work out $x$ and all four interior angles of the trapezium.
(4)
(Total for Question 5 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

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

1
Name the quadrilateral that has exactly one pair of parallel sides.
(1)
(Total for Question 1 is 1 mark)
2
Write down one property of the diagonals of a rectangle.
(1)
(Total for Question 2 is 1 mark)

Tier 2 · Standard

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)
2
A parallelogram has adjacent side lengths $7$ cm and $11$ cm. Work out its perimeter.
(2)
(Total for Question 2 is 2 marks)
3
Quadrilateral $ABCD$ is a parallelogram. Also $AB=BC$ and $\angle ABC=90^\circ$ . Write down the most specific name of $ABCD$ and explain which two properties force this conclusion.
(3)
(Total for Question 3 is 3 marks)

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)
(Total for Question 1 is 3 marks)
2
In kite $ABCD$ , $AB=AD$ and $CB=CD$ . Angle $A$ is $74^\circ$ and angle $C$ is $126^\circ$ . Work out angles $B$ and $D$ .
(3)
(Total for Question 2 is 3 marks)
3
In parallelogram $ABCD$ , diagonal $AC$ bisects $\angle DAB$ . Prove that $ABCD$ is a rhombus. Include the geometric reason beside each equality.
(4)
(Total for Question 3 is 4 marks)
4
In rhombus $ABCD$ , diagonals $AC$ and $BD$ meet at $E$ , and $\angle BAC=31^\circ$ . Work out $\angle ABE$ and all four interior angles of the rhombus. Give a reason for each stage.
(4)
(Total for Question 4 is 4 marks)
5
The diagonals of convex quadrilateral $ABCD$ meet at $E$ at right angles. Points $A,E,C$ are in order and points $B,E,D$ are in order. Given $BE=DE=3$ cm, $AE=4$ cm and $CE=7$ cm, write down the most precise name for $ABCD$ and work out its perimeter. Explain why it is not a rhombus.
(4)
(Total for Question 5 is 4 marks)

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

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

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)
2
Triangles $ABC$ and $DEF$ have $AB=DE$ , $BC=EF$ and $\angle ABC=\angle DEF$ . A student calls this SSA because the angle was stated last. Write down the correct congruence criterion.
(1)
(Total for Question 2 is 1 mark)

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)
(Total for Question 1 is 2 marks)
2
Triangles $JKL$ and $MNP$ satisfy $\angle J=\angle M$ , $\angle K=\angle N$ and $JK=MN$ . Explain why the triangles are congruent.
(2)
(Total for Question 2 is 2 marks)
3
Triangles $ABC$ and $PQR$ satisfy $AB=PQ$ , $BC=QR$ and $AC=PR$ . Write down a congruence statement with the vertices in corresponding order, and write down the angle in triangle $PQR$ that corresponds to $\angle ACB$ .
(2)
(Total for Question 3 is 2 marks)

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)
(Total for Question 1 is 3 marks)
2
Triangle $ABC$ is isosceles with $AB=AC$ . Point $D$ is the midpoint of $BC$ , and $AD$ is drawn. Prove that $AD$ is perpendicular to $BC$ .
(4)
(Total for Question 2 is 4 marks)
3
Line segment $AB$ is $11$ cm long. Points $C$ and $D$ lie on opposite sides of $AB$ . Angles $ACB$ and $ADB$ are right angles, and $AC=AD=7$ cm. Prove that $BC=BD$ .
(3)
(Total for Question 3 is 3 marks)
4
The horizontal segment $AB$ is $8$ cm long. Point $C$ is above $AB$ and point $D$ is below $AB$ . Angles $CAB$ , $DAB$ , $CBA$ and $DBA$ are each $45^\circ$ . Prove that $AB$ is the perpendicular bisector of $CD$ . Give a reason for each stage.
(5)
(Total for Question 4 is 5 marks)
5
In quadrilateral $ABCD$ , the vertices are in order, $B$ and $D$ lie on opposite sides of $AC$ , $AB=CD$ and $BC=AD$ . Diagonal $AC$ is drawn. Prove that both pairs of opposite sides are parallel, and hence that $ABCD$ is a parallelogram.
(4)
(Total for Question 5 is 4 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

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

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)
2
A right-angled triangle has hypotenuse $10$ cm and one shorter side $6$ cm. Work out the other shorter side.
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

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)
2
Two triangles are similar. A side of length $8$ cm in the smaller triangle corresponds to a side of length $20$ cm in the larger triangle. Another side of the smaller triangle is $14$ cm. Work out the corresponding side of the larger triangle.
(2)
(Total for Question 2 is 2 marks)
3
A triangle has side lengths $9$ cm, $40$ cm and $41$ cm. Show that the triangle is right-angled, then work out its area.
(3)
(Total for Question 3 is 3 marks)

Tier 3 · Hard

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)
2
The diagonals of a rhombus are $10$ cm and $24$ cm. They bisect each other at right angles. Work out the perimeter of the rhombus.
(4)
(Total for Question 2 is 4 marks)
3
Quadrilateral $ABCD$ is a rectangle with $AB=12$ cm and $BC=5$ cm. Point $E$ lies on side $DC$ and $AE=BE$ . Prove that $E$ is the midpoint of $DC$ .
(4)
(Total for Question 3 is 4 marks)
4
Convex trapezium $ABCD$ has $AB\parallel CD$ , with the shorter side $CD$ directly above $AB$ . Given $AB=21$ cm, $CD=11$ cm and $AD=BC=13$ cm, work out the perpendicular height of the trapezium and the length of diagonal $AC$ .
(5)
(Total for Question 4 is 5 marks)
5
Triangle $ABC$ is right-angled at $A$ . The perpendicular from $A$ meets hypotenuse $BC$ at $D$ , where $BD=9$ cm and $DC=16$ cm. By using similar triangles, work out $AD$ , $AB$ and $AC$ .
(5)
(Total for Question 5 is 5 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)

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

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)
2
Point $B(-4,3)$ is reflected in the $y$ -axis. Write down the coordinates of its image.
(1)
(Total for Question 2 is 1 mark)

Tier 2 · Standard

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)
2
Point $C(5,-2)$ is enlarged by scale factor $3$ about the origin. Find the coordinates of its image.
(2)
(Total for Question 2 is 2 marks)
3
Triangle $ABC$ has vertices $A(-3,2)$ , $B(1,5)$ and $C(4,-2)$ . The triangle is reflected in the line $y=x$ . Find the coordinates of the three image vertices.
(2)
(Total for Question 3 is 2 marks)

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)
(Total for Question 1 is 4 marks)
2
Describe fully the single transformation that maps the triangle with vertices $(3,1)$ , $(6,1)$ and $(3,4)$ to the triangle with vertices $(1,5)$ , $(1,8)$ and $(-2,5)$ .
(3)
(Total for Question 2 is 3 marks)
3
An enlargement maps $A(2,1)$ to $A'(5,3)$ and $B(8,1)$ to $B'(8,3)$ . Describe the enlargement fully.
(4)
(Total for Question 3 is 4 marks)
4
A translation maps $A(-4,3)$ to $A'(2,-2)$ . Describe the translation fully. The same translation maps $B(1,4)$ to $B'$ , $C(-3,-2)$ to $C'$ , and $D$ to $D'(0,6)$ . Work out the coordinates of $B'$ , $C'$ and $D$ .
(4)
(Total for Question 4 is 4 marks)
5
Higher only: An enlargement with scale factor $-2$ maps $P(4,1)$ to $P'(-5,7)$ . Work out the centre of enlargement. The same enlargement maps $Q(-2,5)$ to $Q'$ . Work out the coordinates of $Q'$ .
(4)
(Total for Question 5 is 4 marks)

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

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

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)
2
Horizontal lines $m$ and $n$ are parallel and $3$ cm apart, with $n$ above $m$ . A shape is reflected first in line $m$ , then in line $n$ . Describe the single equivalent transformation.
(2)
(Total for Question 2 is 2 marks)

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)
(Total for Question 1 is 2 marks)
2
Point $P(2,-5)$ is reflected in the $y$ -axis and then rotated $90^\circ$ clockwise about the origin. Find the final image. Write down whether orientation is preserved by the combination.
(3)
(Total for Question 2 is 3 marks)
3
A shape of area $34$ cm $^2$ undergoes two rotations and three reflections. Write down whether its final image is congruent to the original, whether its orientation is preserved, and its final area.
(3)
(Total for Question 3 is 3 marks)

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)
(Total for Question 1 is 4 marks)
2
A shape is reflected first in line $l$ and then in line $m$ . The lines meet at $O$ , and the clockwise angle from $l$ to $m$ is $38^\circ$ . Describe the single equivalent transformation. Write down two properties that remain invariant.
(4)
(Total for Question 2 is 4 marks)
3
A shape is rotated $90^\circ$ clockwise about the origin and then rotated $90^\circ$ clockwise about $(4,0)$ . Describe the single equivalent transformation and state whether orientation is preserved.
(4)
(Total for Question 3 is 4 marks)
4
A line segment has endpoints $A(-1,2)$ and $B(3,5)$ . In route 1, reflect the segment in the line $x=2$ and then translate it by $\begin{pmatrix}3\\-4\end{pmatrix}$ . In route 2, apply the same translation first and then reflect in $x=2$ . Work out the final endpoints for both routes and explain how your answers show that the order matters.
(5)
(Total for Question 4 is 5 marks)
5
A shape is rotated $180^\circ$ about $(3,-2)$ and then reflected in the line $x=3$ . Describe the single equivalent transformation, state whether orientation is preserved, and describe all points that remain fixed.
(4)
(Total for Question 5 is 4 marks)

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

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

1
A circle has radius $6.5$ cm. Write down its diameter.
(1)
(Total for Question 1 is 1 mark)
2
Write down the name of the part of a circumference between two points on a circle.
(1)
(Total for Question 2 is 1 mark)

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)
(Total for Question 1 is 2 marks)
2
Draw a circle with centre $O$ . Mark points $A$ and $B$ on the circumference, draw chord $AB$ , and shade the minor segment cut off by $AB$ .
(2)
(Total for Question 2 is 2 marks)
3
Points $P$ and $Q$ lie on a circle and are not endpoints of a diameter. Write down the geometric name of the longer route from $P$ to $Q$ along the circumference, the straight segment $PQ$ , and the larger region between this route and $PQ$ .
(3)
(Total for Question 3 is 3 marks)

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)
(Total for Question 1 is 3 marks)
2
Distinct points $A$ and $B$ lie on a circle. The shorter arc $AB$ has length $14$ cm and the circumference is $50$ cm. Give the geometric name and length of the other arc from $A$ to $B$ . Work out what fraction of the circumference the shorter arc is. Give the fraction in its simplest form.
(3)
(Total for Question 2 is 3 marks)
3
A circle has centre $O$ , with $A$ , $B$ , $C$ and $D$ on its circumference. Segment $AB$ passes through $O$ , but segment $CD$ does not. Explain why the statement 'every diameter is a chord, but not every chord is a diameter' is correct.
(3)
(Total for Question 3 is 3 marks)
4
A circle with centre $O$ has radius $7$ cm. The perimeter of minor sector $AOB$ is $31$ cm. Chord $AB$ is drawn. Work out the lengths of the minor and major arcs $AB$ . Write down the names of the smaller and larger regions bounded by chord $AB$ and the circumference.
(5)
(Total for Question 4 is 5 marks)
5
The circumference of a circle is $84$ cm. Points $A$ and $B$ divide the circumference into two arcs whose lengths are in the ratio $2:5$ . Chord $AB$ is drawn. Work out both arc lengths, name the smaller region between $AB$ and the circumference, and name the larger region bounded by $OA$ , $OB$ and the circumference.
(4)
(Total for Question 5 is 4 marks)

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

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

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)
2
A tangent touches a circle at $T$ , radius $OT$ is drawn, and $L$ is another point on the tangent. Write down $\angle OTL$ .
(1)
(Total for Question 2 is 1 mark)

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)
(Total for Question 1 is 2 marks)
2
Opposite angles of a cyclic quadrilateral are $(3x+12)^\circ$ and $(5x-8)^\circ$ . Work out $x$ and the sizes of these two angles.
(3)
(Total for Question 2 is 3 marks)
3
A tangent $AT$ touches a circle at $A$ . Chord $AB$ is drawn, and point $C$ lies on the circle in the opposite segment. Points $T$ and $C$ are on opposite sides of $AB$ . Given $\angle TAB=43^\circ$ and $\angle ABC=72^\circ$ , work out $\angle ACB$ and $\angle BAC$ . Give a reason for each answer.
(3)
(Total for Question 3 is 3 marks)

Tier 3 · Hard

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)
2
A chord $AB$ of a circle is not a diameter. Its midpoint is $M$ and the centre is $O$ . Prove that $OM$ is perpendicular to $AB$ .
(4)
(Total for Question 2 is 4 marks)
3
The two tangents from a point $P$ meet a circle, centre $O$ , at the points $A$ and $B$ . The point $C$ lies on the major arc $AB$ . Given $\angle APB=74^\circ$ , work out $\angle AOB$ and $\angle ACB$ . Give a reason for each stage.
(4)
(Total for Question 3 is 4 marks)
4
A circle contains diameter $AB$ and another point $C$ on its circumference. The tangent at $C$ meets the extension of $AB$ beyond $B$ at $P$ . Given $\angle CAB=28^\circ$ , work out $\angle CPB$ . Give a reason for each stage.
(4)
(Total for Question 4 is 4 marks)
5
A circle has centre $O$ and radius $10$ cm. $AB$ is a horizontal diameter with $A$ to the left of $O$ . The tangent at $A$ is drawn. Point $E$ lies on $OB$ with $OE=6$ cm. The line through $E$ parallel to the tangent meets the circle at $C$ above $AB$ and $D$ below $AB$ . Prove that $E$ is the midpoint of $CD$ , then work out $CD$ and prove that triangle $ACD$ is isosceles.
(5)
(Total for Question 5 is 5 marks)

G11 · Solve geometrical problems on coordinate axes

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

1
Find the midpoint of the line segment joining $(-5,2)$ to $(7,8)$ .
(2)
(Total for Question 1 is 2 marks)
2
Points $P(-6,4)$ and $Q(3,4)$ are joined. Work out the length $PQ$ .
(1)
(Total for Question 2 is 1 mark)

Tier 2 · Standard

1
Points $A(-1,4)$ and $B(5,-4)$ are joined. Find the exact length $AB$ .
(3)
(Total for Question 1 is 3 marks)
2
The midpoint of $AB$ is $(2,1)$ . Point $A$ is $(-3,4)$ . Find the coordinates of $B$ .
(2)
(Total for Question 2 is 2 marks)
3
Triangle $ABC$ has vertices $A(-4,-2)$ , $B(8,-2)$ and $C(3,5)$ . Work out the area of the triangle.
(3)
(Total for Question 3 is 3 marks)

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)
(Total for Question 1 is 5 marks)
2
The points $A(1,2)$ , $B(7,2)$ , $C(9,7)$ and $D(3,7)$ are joined in order. Show that $ABCD$ is a parallelogram and work out its area.
(4)
(Total for Question 2 is 4 marks)
3
Point $P$ lies on the $y$ -axis and is equidistant from $A(-3,1)$ and $B(5,5)$ . Find the coordinates of $P$ .
(4)
(Total for Question 3 is 4 marks)
4
Points $A(-4,1)$ and $B(8,7)$ are endpoints of a diameter of a circle. Work out the centre and the exact radius of the circle. Point $C$ has coordinates $(5,-2)$ . Show that $C$ lies on the circle.
(4)
(Total for Question 4 is 4 marks)
5
Points $A(-4,0)$ , $B(1,6)$ and $C(6,0)$ are consecutive vertices of parallelogram $ABCD$ . Work out the coordinates of $D$ . Show that the parallelogram is a rhombus but not a square.
(5)
(Total for Question 5 is 5 marks)

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

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

1
Write down the number of faces and the number of vertices of a triangular prism.
(2)
(Total for Question 1 is 2 marks)
2
A cube and a cuboid both have $6$ faces. Write down one other number that is the same for both and say what it counts.
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

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)
2
Work out the numbers of faces, edges and vertices of a square-based pyramid.
(3)
(Total for Question 2 is 3 marks)
3
A solid has no vertices and two circular plane faces. Write down the name of the solid, its number of edges and its number of curved surfaces.
(3)
(Total for Question 3 is 3 marks)

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)
(Total for Question 1 is 3 marks)
2
A prism has $12$ faces. Work out the number of sides on each congruent end, name the prism, and work out its numbers of edges and vertices.
(4)
(Total for Question 2 is 4 marks)
3
One solid is a prism with $7$ -sided congruent ends. A second solid is a pyramid with a $7$ -sided base. Work out the total numbers of faces, edges and vertices across the two solids.
(4)
(Total for Question 3 is 4 marks)
4
A pyramid has a polygon with $n$ sides as its base. The pyramid has $5$ more edges than faces. Work out $n$ , name the pyramid, and write down its numbers of faces, edges and vertices.
(4)
(Total for Question 4 is 4 marks)
5
A triangular-based pyramid is joined to a triangular prism so that one triangular face of each solid matches exactly and is hidden inside the new solid. No other faces overlap, and no exterior faces become coplanar. Work out the numbers of exterior faces, distinct edges and distinct vertices of the new solid.
(5)
(Total for Question 5 is 5 marks)

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

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

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)
2
A vertical cylinder has radius $3$ cm and height $8$ cm. Write down the shape and dimensions of its plan.
(2)
(Total for Question 2 is 2 marks)

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)
(Total for Question 1 is 3 marks)
2
A triangular prism is $7$ cm long. Each end face is a right-angled triangle with a horizontal base $4$ cm and a vertical side $3$ cm on the left. Write down the shapes and dimensions of the plan and of the elevation seen when looking at an end face with the vertical side on the left.
(3)
(Total for Question 2 is 3 marks)
3
A right circular cone stands vertically with radius $4$ cm and perpendicular height $9$ cm. Write down the shapes and dimensions of its plan and its front elevation.
(3)
(Total for Question 3 is 3 marks)

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)
(Total for Question 1 is 4 marks)
2
A solid consists of a base cuboid $8$ cm long, $5$ cm deep and $2$ cm high, with a second cuboid $3$ cm long, $5$ cm deep and $4$ cm high. The second cuboid sits on top of the left-hand $3$ cm of the base. Describe the plan, the front elevation viewed along the depth, and the side elevation viewed from the right.
(4)
(Total for Question 2 is 4 marks)
3
A right circular cylinder has radius $3$ cm and length $10$ cm. Its axis is horizontal and runs from west to east. Write down the shapes and dimensions of the plan, the elevation viewed from the east, and the elevation viewed from the south.
(4)
(Total for Question 3 is 4 marks)
4
A pentagonal prism is $10$ cm long from west to east. Each end is shaped like a house: a horizontal base $6$ cm wide, vertical walls $3$ cm high and a roof apex centred $5$ cm above the base. Describe the plan, the east elevation and the south elevation, including the ridge and eave lines.
(5)
(Total for Question 4 is 5 marks)
5
A cuboid is $12$ cm wide, $8$ cm deep and $4$ cm high. A square-based pyramid is fixed centrally on its top face. The pyramid has a $6$ cm by $6$ cm base and a perpendicular height of $5$ cm. Describe the plan, front elevation and side elevation, including the projected apex, all visible joins and the total height.
(5)
(Total for Question 5 is 5 marks)

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

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

1
Convert $3.75\text{ kg}$ to grams.
(1)
(Total for Question 1 is 1 mark)
2
Change $3$ hours $18$ minutes into minutes.
(1)
(Total for Question 2 is 1 mark)

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)
(Total for Question 1 is 3 marks)
2
A roll contains $7.5\text{ m}$ of ribbon. Aisha cuts $18$ pieces, each $35\text{ cm}$ long. Work out the length of ribbon left. Give your answer in centimetres.
(2)
(Total for Question 2 is 2 marks)
3
A water butt contains $0.42\text{ m}^3$ of water. Then $135$ litres are used and $8$ buckets, each containing $12.5$ litres, are poured into the water butt. Work out the volume of water now in the water butt. Give your answer in litres.
(3)
(Total for Question 3 is 3 marks)

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)
(Total for Question 1 is 5 marks)
2
A bakery needs $2.6\text{ kg}$ of almonds. The almonds are sold in $375\text{ g}$ bags costing £ $2.15$ each. Work out the total cost of the bags needed.
(3)
(Total for Question 2 is 3 marks)
3
A square floor mat has area $0.81\text{ m}^2$ . Work out the perimeter of the mat in millimetres.
(3)
(Total for Question 3 is 3 marks)
4
A rectangular metal sheet measures $1.8\text{ m}$ by $75\text{ cm}$ . A rectangular opening measuring $320\text{ mm}$ by $450\text{ mm}$ is cut from the sheet. Work out the remaining area of the sheet in square centimetres.
(3)
(Total for Question 4 is 3 marks)
5
A flight leaves at $22{:}48$ . The flight lasts $7$ hours $35$ minutes. At the destination, clocks are $2$ hours ahead of the departure clocks. Work out the local arrival time. State whether the arrival is on the same day or the next day.
(3)
(Total for Question 5 is 3 marks)

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

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

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)
2
A ship travels in a south-east direction. Write down the bearing of the ship's direction.
(1)
(Total for Question 2 is 1 mark)

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)
(Total for Question 1 is 4 marks)
2
A route on a map consists of two straight sections of lengths $4.6\text{ cm}$ and $3.4\text{ cm}$ . The map scale is $1:50\,000$ . Work out the actual length of the route in kilometres.
(2)
(Total for Question 2 is 2 marks)
3
A rectangular playground measures $72\text{ m}$ by $45\text{ m}$ . A plan of the playground uses the scale $1 : 2500$ . Work out the dimensions of the playground on the plan. Give your answers in centimetres.
(3)
(Total for Question 3 is 3 marks)

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)
(Total for Question 1 is 4 marks)
2
A lighthouse is on a bearing of $046^\circ$ from a harbour. A boat leaves the lighthouse on a bearing of $166^\circ$ . Work out the smaller angle between the direction from the lighthouse to the harbour and the boat's direction.
(3)
(Total for Question 2 is 3 marks)
3
A rescue boat travels from $P$ to $Q$ on a bearing of $038^\circ$ . At $Q$ it turns $126^\circ$ clockwise and travels to $R$ . It then returns directly from $R$ to $Q$ . At $Q$ it turns $101^\circ$ anticlockwise from its return direction and begins a new course. Work out the bearing of $R$ from $Q$ , the bearing of $Q$ from $R$ and the bearing of the new course.
(4)
(Total for Question 3 is 4 marks)
4
A map has scale $1 : 40\,000$ . The map is enlarged to $125\%$ of its original size. On the enlarged copy, a path is $13.5\text{ cm}$ long. Work out the real length of the path in kilometres.
(4)
(Total for Question 4 is 4 marks)
5
Point $Q$ is due north of point $P$ . Point $R$ is on a bearing of $122^\circ$ from $P$ , and $\angle PRQ=37^\circ$ . Work out $\angle PQR$ and the bearing of $Q$ from $R$ .
(4)
(Total for Question 5 is 4 marks)

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

Explanation

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

A trapezium uses the perpendicular height between its parallel sides.

Worked example

A triangular prism is $10$ cm long. Its triangular cross-section has base $6$ cm and perpendicular height $4$ cm. Find the volume.

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

Answer: $120\text{ cm}^3$ .

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.

Exam tip

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

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)
(Total for Question 1 is 2 marks)
2
A parallelogram has a base of $12\text{ cm}$ and a perpendicular height of $7\text{ cm}$ . Work out the area of the parallelogram.
(1)
(Total for Question 2 is 1 mark)

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)
(Total for Question 1 is 2 marks)
2
A trapezium has area $76\text{ cm}^2$ , perpendicular height $8\text{ cm}$ and one parallel side of length $5\text{ cm}$ . Work out the length of the other parallel side.
(3)
(Total for Question 2 is 3 marks)
3
A cylinder has radius $7\text{ cm}$ and volume $588\pi\text{ cm}^3$ . Work out the height of the cylinder.
(3)
(Total for Question 3 is 3 marks)

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)
(Total for Question 1 is 4 marks)
2
A right prism has a rectangular cross-section measuring $12\text{ cm}$ by $8\text{ cm}$ . A rectangular hole measuring $6\text{ cm}$ by $3\text{ cm}$ runs through the full $20\text{ cm}$ length of the prism, with its $6\text{ cm}$ edge parallel to the $12\text{ cm}$ edge of the cross-section. Work out the volume of material in the prism.
(4)
(Total for Question 2 is 4 marks)
3
The length of a right prism is $25\text{ cm}$ . Perpendicular to this length, its constant cross-section starts as a rectangle with a horizontal side of $14\text{ cm}$ and a vertical side of $10\text{ cm}$ . A right-angled triangular piece is removed from the top-right corner: its $6\text{ cm}$ side lies along the top edge and its $4\text{ cm}$ side lies along the right edge. Work out the volume of material in the prism.
(4)
(Total for Question 3 is 4 marks)
4
Right prism $P$ is $14\text{ cm}$ long and has a triangular cross-section with base $12\text{ cm}$ and perpendicular height $9\text{ cm}$ . Right prism $Q$ is $13.5\text{ cm}$ long and has a parallelogram cross-section with base $7\text{ cm}$ and perpendicular height $8\text{ cm}$ . Show that the two prisms have the same volume.
(4)
(Total for Question 4 is 4 marks)
5
The radius of a cylinder is increased by $20\%$ and its height is reduced by $20\%$ . Sam says that these changes leave the volume unchanged. Is Sam correct? Work out the percentage change in the volume.
(4)
(Total for Question 5 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

Explanation

For a circle, use $C=2\pi r=\pi d$ and $A=\pi r^2$ . Composite areas add included regions and subtract cut-outs; composite perimeters include only the outside boundary.
A cone or pyramid has volume $V=\frac13\times\text{base area}\times\text{perpendicular height}$ . A cone's curved surface area is $\pi rl$ , where $l$ is slant height.
A sphere has surface area $4\pi r^2$ and volume $\frac43\pi r^3$ .
For a composite solid, divide it into recognised solids and exclude internal joined faces from surface area.
Examiners expect radius, perpendicular height and slant height to be distinguished correctly.

A cone's perpendicular height

h

and slant height

l

have different roles.

Worked example

A solid cone has radius $3$ cm, perpendicular height $4$ cm and slant height $5$ cm. Find its total surface area and volume in terms of $\pi$ .

1.Curved area $=\pi rl=\pi(3)(5)=15\pi\text{ cm}^2$ ; base area $=\pi r^2=9\pi\text{ cm}^2$ .
2.Total surface area $=15\pi+9\pi=24\pi\text{ cm}^2$ .
3.Volume $=\frac13\pi r^2h=\frac13\pi(3^2)(4)=12\pi\text{ cm}^3$ .

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

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.

Exam tip

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

Tier 1 · Easy

1
A circle has diameter $13\text{ cm}$ . Write its circumference in terms of $\pi$ .
(1)
(Total for Question 1 is 1 mark)
2
A circle has radius $6\text{ cm}$ . Work out its area in terms of $\pi$ .
(1)
(Total for Question 2 is 1 mark)

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)
(Total for Question 1 is 4 marks)
2
A square has side length $11\text{ cm}$ . A circular hole of radius $4\text{ cm}$ is cut from the square. Work out the area that remains in terms of $\pi$ .
(2)
(Total for Question 2 is 2 marks)
3
A circle has radius $8\text{ cm}$ . A square has the same perimeter as the circle. Work out the area of the square in terms of $\pi$ .
(3)
(Total for Question 3 is 3 marks)

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)
(Total for Question 1 is 5 marks)
2
The total surface area of a solid hemisphere, including its circular base, is $147\pi\text{ cm}^2$ . Work out the volume of the hemisphere in terms of $\pi$ .
(4)
(Total for Question 2 is 4 marks)
3
A circular pond has radius $7\text{ m}$ . A path surrounds the pond. The outer edge of the path is a circle with circumference $18\pi\text{ m}$ . Work out the area of the path in terms of $\pi$ .
(4)
(Total for Question 3 is 4 marks)
4
Higher only: A solid hemisphere of radius $5\text{ cm}$ is joined along its circular face to the circular base of a cone. The cone also has radius $5\text{ cm}$ and has slant height $13\text{ cm}$ . The joined circular faces are inside the solid. Work out the exterior surface area of the solid in terms of $\pi$ .
(4)
(Total for Question 4 is 4 marks)
5
Higher only: A pyramid has a trapezium as its base. The parallel sides of the trapezium are $8\text{ cm}$ and $14\text{ cm}$ , and the perpendicular distance between them is $6\text{ cm}$ . The volume of the pyramid is $330\text{ cm}^3$ . Work out the perpendicular height of the pyramid.
(4)
(Total for Question 5 is 4 marks)

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

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

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)
2
A semicircle has diameter $22\text{ cm}$ . Work out its perimeter in terms of $\pi$ .
(2)
(Total for Question 2 is 2 marks)

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)
(Total for Question 1 is 4 marks)
2
A semicircle has area $50\pi\text{ cm}^2$ . Work out the length of its curved edge in terms of $\pi$ .
(3)
(Total for Question 2 is 3 marks)
3
The minute hand of a clock is $9\text{ cm}$ long. Work out the distance travelled by the tip of the minute hand from $2{:}00$ to $2{:}25$ . Give your answer in terms of $\pi$ .
(3)
(Total for Question 3 is 3 marks)

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)
(Total for Question 1 is 5 marks)
2
A sector has angle $128^\circ$ and area $80\pi\text{ cm}^2$ . Work out the full perimeter of the sector, including the two radii. Give your answer in terms of $\pi$ .
(5)
(Total for Question 2 is 5 marks)
3
Sector A has radius $9\text{ cm}$ and angle $160^\circ$ . Sector B has radius $12\text{ cm}$ and the same area as sector A. Work out the angle of sector B and the difference between the arc lengths of the two sectors. Give the difference in terms of $\pi$ .
(4)
(Total for Question 3 is 4 marks)
4
Higher only: A sector has area $54\pi\text{ cm}^2$ and arc length $6\pi\text{ cm}$ . Work out the radius and the angle of the sector.
(5)
(Total for Question 4 is 5 marks)
5
Higher only: A clock face has radius $18\text{ cm}$ . At $4{:}20$ , the hour hand has moved continuously from the $4$ towards the $5$ . Work out the exact area of the smaller sector between the hour hand and the minute hand.
(4)
(Total for Question 5 is 4 marks)

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

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

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)
2
Two shapes are similar. The length scale factor from the smaller shape to the larger shape is $4$ . A side on the smaller shape is $3.5\text{ cm}$ . Work out the corresponding side on the larger shape.
(1)
(Total for Question 2 is 1 mark)

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)
(Total for Question 1 is 3 marks)
2
Two polygons are similar. Their perimeters are $24\text{ cm}$ and $36\text{ cm}$ . A side on the smaller polygon is $10\text{ cm}$ . Work out the corresponding side on the larger polygon.
(2)
(Total for Question 2 is 2 marks)
3
Two quadrilaterals are similar. A side of length $8\text{ cm}$ on the smaller quadrilateral corresponds to a side of length $12\text{ cm}$ on the larger quadrilateral. The difference between their perimeters is $26\text{ cm}$ . Work out both perimeters.
(3)
(Total for Question 3 is 3 marks)

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)
(Total for Question 1 is 4 marks)
2
In quadrilateral $PQRS$ , $PQ=PS$ and $RQ=RS$ . The diagonal $PR$ is drawn. Prove that triangle $PQR$ is congruent to triangle $PSR$ . Hence show that $\angle QPR=\angle SPR$ .
(3)
(Total for Question 2 is 3 marks)
3
Higher only: Solid $P$ and solid $Q$ are similar. For each solid, its volume is divided by its surface area. The result is $2.4\text{ cm}$ for solid $P$ and $4\text{ cm}$ for solid $Q$ . Work out the ratio of the surface area of $P$ to the surface area of $Q$ and the ratio of the volume of $P$ to the volume of $Q$ .
(4)
(Total for Question 3 is 4 marks)
4
Convex quadrilateral $ABCD$ has $AB$ parallel to $CD$ and $AB=CD$ . The diagonal $AC$ is drawn, with $B$ and $D$ on opposite sides of $AC$ . Prove that triangles $BAC$ and $DCA$ are congruent. Hence prove that $ABCD$ is a parallelogram.
(4)
(Total for Question 4 is 4 marks)
5
Higher only: A solid model is enlarged to make a mathematically similar solid. The volume increases by $237.5\%$ . Work out the percentage increase in the surface area.
(4)
(Total for Question 5 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

Explanation

In a right-angled triangle, Pythagoras gives $a^2+b^2=c^2$ , with $c$ opposite the right angle. Relative to angle $\theta$ , $\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$ , $\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$ and $\tan\theta=\frac{\text{opposite}}{\text{adjacent}}$ .
Label the sides first, choose the relationship containing the known and required values, then rearrange. Use an inverse trigonometric function for an angle.
Foundation questions use right-angled triangles in two dimensions.
Higher tier: identify an appropriate right-angled cross-section in three dimensions or use the later general-triangle rules.
Examiners expect a diagram, substitution, correct units, sensible rounding and a clear statement of what the result represents.

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

Worked example

A ladder is $8$ m long and its foot is $3$ m from a vertical wall. Find the height reached and the angle with the ground, both to $1$ decimal place.

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

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

Common mistakes

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

Exam tip

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

Tier 1 · Easy

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)
2
A right-angled triangle has hypotenuse $17\text{ cm}$ and one shorter side $15\text{ cm}$ . Work out the length of the other shorter side.
(2)
(Total for Question 2 is 2 marks)

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)
(Total for Question 1 is 4 marks)
2
In a right-angled triangle, the side opposite angle $x$ is $7\text{ cm}$ and the side adjacent to $x$ is $24\text{ cm}$ . Work out $x$ . Give the value of $x$ correct to $1$ decimal place.
(2)
(Total for Question 2 is 2 marks)
3
A kite is held by a straight taut string $25\text{ m}$ long. The string makes an angle of $38^\circ$ with the horizontal. The person's hand is $1.4\text{ m}$ above level ground. Work out the height of the kite above the ground, correct to $1$ decimal place.
(3)
(Total for Question 3 is 3 marks)

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)
(Total for Question 1 is 4 marks)
2
A cuboid measures $6\text{ cm}$ by $9\text{ cm}$ by $14\text{ cm}$ . Work out the length of its space diagonal. Give the diagonal length correct to $1$ decimal place.
(3)
(Total for Question 2 is 3 marks)
3
A rhombus has diagonals that meet at right angles and bisect each other. One diagonal is $10\text{ cm}$ long and each side of the rhombus is $13\text{ cm}$ long. Work out the length of the other diagonal and the area of the rhombus.
(4)
(Total for Question 3 is 4 marks)
4
A rectangle has area $240\text{ cm}^2$ and diagonal length $26\text{ cm}$ . Work out the perimeter of the rectangle.
(4)
(Total for Question 4 is 4 marks)
5
Rectangle $ABCD$ has $AB=16\text{ cm}$ and $BC=12\text{ cm}$ . Point $M$ is the midpoint of $CD$ . Work out $AM$ in exact form and work out $\angle MAB$ correct to $1$ decimal place.
(4)
(Total for Question 5 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°

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

1
Work out the exact value of $\sin30^\circ+\cos60^\circ$ .
(1)
(Total for Question 1 is 1 mark)
2
The angle $x$ is one of $0^\circ$ , $30^\circ$ , $45^\circ$ , $60^\circ$ or $90^\circ$ . Given that $\sin x=\cos x$ , write down $x$ .
(1)
(Total for Question 2 is 1 mark)

Tier 2 · Standard

1
Work out the exact value of $2\sin45^\circ\cos45^\circ+\cos60^\circ$ .
(3)
(Total for Question 1 is 3 marks)
2
The acute angle $x$ is one of $30^\circ$ , $45^\circ$ or $60^\circ$ . Given that $\sin x=\frac12$ , write down $x$ and the exact value of $\cos x$ .
(2)
(Total for Question 2 is 2 marks)
3
Solve $2x\sin30^\circ+3\cos60^\circ=\frac{7}{2}$ .
(3)
(Total for Question 3 is 3 marks)

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)
(Total for Question 1 is 4 marks)
2
A right-angled triangle has an acute angle of $60^\circ$ . The side between the $60^\circ$ angle and the right angle is $4\text{ cm}$ . Work out the exact area of the triangle.
(3)
(Total for Question 2 is 3 marks)
3
Hassan says $2\sin30^\circ+\cos60^\circ=\tan45^\circ$ . Is Hassan correct? You must show all your working.
(3)
(Total for Question 3 is 3 marks)
4
Solve $x^2\tan45^\circ-8x\sin30^\circ+3\cos0^\circ=0$ .
(4)
(Total for Question 4 is 4 marks)
5
Higher only: A right-angled triangle has an acute angle of $60^\circ$ and area $48\sqrt3\text{ cm}^2$ . Work out the exact perimeter of the triangle.
(4)
(Total for Question 5 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]

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

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)
2
A triangle has side lengths $5\text{ cm}$ , $13\text{ cm}$ and $15\text{ cm}$ . Work out the angle opposite the $15\text{ cm}$ side. Give the angle correct to $1$ decimal place.
(3)
(Total for Question 2 is 3 marks)

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)
(Total for Question 1 is 3 marks)
2
In triangle $ABC$ , $A=45^\circ$ , $a=7\sqrt2\text{ cm}$ and $b=7\text{ cm}$ . Work out angle $B$ .
(3)
(Total for Question 2 is 3 marks)
3
In triangle $ABC$ , $A=38^\circ$ , $B=79^\circ$ and the side opposite angle $A$ is $8.4\text{ cm}$ . Work out the length of the side opposite angle $B$ , correct to $1$ decimal place.
(3)
(Total for Question 3 is 3 marks)

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)
(Total for Question 1 is 5 marks)
2
Two sides of a triangle are $8\text{ cm}$ and $10\text{ cm}$ , and their included angle is $120^\circ$ . Work out the third side and the angle opposite the $10\text{ cm}$ side. Give both answers correct to $1$ decimal place.
(5)
(Total for Question 2 is 5 marks)
3
Convex quadrilateral $ABCD$ has diagonal $AC$ , with $B$ and $D$ on opposite sides of $AC$ . Given that $AB=7\text{ cm}$ , $BC=9\text{ cm}$ , $\angle ABC=58^\circ$ , $AD=6\text{ cm}$ and $CD=10\text{ cm}$ , work out $\angle ADC$ , correct to $1$ decimal place.
(5)
(Total for Question 3 is 5 marks)
4
A triangle has sides of lengths $(x+1)\text{ cm}$ , $(x+3)\text{ cm}$ and $2\sqrt{13}\text{ cm}$ . The angle between the sides of lengths $(x+1)\text{ cm}$ and $(x+3)\text{ cm}$ is $60^\circ$ . Work out $x$ .
(4)
(Total for Question 4 is 4 marks)
5
A triangle has side lengths $7\text{ cm}$ , $9\text{ cm}$ and $11\text{ cm}$ . Work out the difference between its largest angle and its smallest angle, correct to $1$ decimal place.
(4)
(Total for Question 5 is 4 marks)

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

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

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)
2
Triangle P has sides of $6\text{ cm}$ and $9\text{ cm}$ with an included angle of $30^\circ$ . Triangle Q has sides of $5\text{ cm}$ and $8\text{ cm}$ with an included angle of $90^\circ$ . Which triangle has the larger area? You must show your working.
(2)
(Total for Question 2 is 2 marks)

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)
(Total for Question 1 is 3 marks)
2
Two sides of a triangle are $7\text{ cm}$ and $11\text{ cm}$ . Work out the greatest possible area of the triangle and the included angle that gives this area.
(3)
(Total for Question 2 is 3 marks)
3
Convex quadrilateral $ABCD$ has diagonal $AC$ , with $B$ and $D$ on opposite sides of $AC$ . Given that $AB=7\text{ cm}$ , $AC=12\text{ cm}$ , $AD=9\text{ cm}$ , $\angle BAC=48^\circ$ and $\angle CAD=35^\circ$ , work out the area of the quadrilateral, correct to $1$ decimal place.
(4)
(Total for Question 3 is 4 marks)

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)
(Total for Question 1 is 4 marks)
2
Two triangles each have sides of lengths $8\text{ cm}$ and $13\text{ cm}$ . The included angle is $35^\circ$ in the first triangle and $145^\circ$ in the second triangle. Show that the triangles have equal areas.
(3)
(Total for Question 2 is 3 marks)
3
Two sides of a triangle are $7.5\text{ cm}$ and $20\text{ cm}$ . The area of the triangle is $60\text{ cm}^2$ and the included angle is acute. Work out the included angle and the length of the third side, giving each value correct to $1$ decimal place.
(5)
(Total for Question 3 is 5 marks)
4
Two sides of a hinged triangular frame have fixed lengths $8\text{ cm}$ and $15\text{ cm}$ . The included angle is increased from $30^\circ$ to $45^\circ$ . Work out the exact increase in the area of the triangle.
(4)
(Total for Question 4 is 4 marks)
5
A triangle has area $15\sqrt3\text{ cm}^2$ . Two of its sides have lengths $10\text{ cm}$ and $x\text{ cm}$ , and the angle between them is $60^\circ$ . Work out the value of $x$ .
(4)
(Total for Question 5 is 4 marks)

G24 · Describe translations as 2D vectors

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

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)
2
Point $P=(4,-1)$ is translated by $\binom{-3}{5}$ . Write down the coordinates of the image of $P$ .
(1)
(Total for Question 2 is 1 mark)

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)
(Total for Question 1 is 3 marks)
2
A translation by $\binom{6}{-4}$ maps point $Q$ to $Q'=(2,7)$ . Work out the coordinates of $Q$ .
(2)
(Total for Question 2 is 2 marks)
3
Triangle $T$ has vertices $(-5,1)$ , $(-2,1)$ and $(-4,4)$ . After a translation, its image has vertices $(1,-6)$ , $(4,-6)$ and $(2,-3)$ in the corresponding order. Describe the translation as a vector.
(3)
(Total for Question 3 is 3 marks)

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)
(Total for Question 1 is 4 marks)
2
A translation by vector $\mathbf{a}$ maps $P=(2,-1)$ to $(5,3)$ . A second translation by vector $\mathbf{b}$ maps $(5,3)$ to $(1,7)$ . Work out $\mathbf{a}$ , work out $\mathbf{b}$ , and write down the single vector that maps $P$ directly to $(1,7)$ .
(3)
(Total for Question 2 is 3 marks)
3
A mapping sends $A=(1,-2)$ to $A'=(-3,4)$ , $B=(5,0)$ to $B'=(1,6)$ and $C=(2,3)$ to $C'=(-2,8)$ . State whether this mapping can be a translation. Give a reason.
(3)
(Total for Question 3 is 3 marks)
4
Segment $AB$ has endpoints $A=(-4,7)$ and $B=(8,-1)$ . After a translation, the midpoint of its image $A'B'$ is $(9,3)$ . Work out the translation vector and the coordinates of $A'$ and $B'$ .
(4)
(Total for Question 4 is 4 marks)
5
Higher only: The line $y=2x+1$ is translated by $\binom{3}{k}$ . Its image passes through the point $(5,7)$ . Work out $k$ and write the equation of the image line.
(4)
(Total for Question 5 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

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

1
Work out $\binom{3}{-5}+\binom{-7}{2}$ .
(1)
(Total for Question 1 is 1 mark)
2
Work out $-3\binom{2}{-4}$ .
(1)
(Total for Question 2 is 1 mark)

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)
(Total for Question 1 is 3 marks)
2
In triangle $ABC$ , $\overrightarrow{AB}=\mathbf a$ and $\overrightarrow{BC}=\mathbf b$ . Write $\overrightarrow{CA}$ in terms of $\mathbf a$ and $\mathbf b$ .
(2)
(Total for Question 2 is 2 marks)
3
Given that $2\binom{x}{-3}+\binom{4}{y}=\binom{10}{-1}$ , find $x$ and $y$ .
(3)
(Total for Question 3 is 3 marks)

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)
(Total for Question 1 is 4 marks)
2
The vectors $\mathbf a$ and $\mathbf b$ are non-parallel. Points $A$ , $B$ and $C$ have position vectors $\overrightarrow{OA}=2\mathbf a+\mathbf b$ , $\overrightarrow{OB}=5\mathbf a-\mathbf b$ and $\overrightarrow{OC}=11\mathbf a-5\mathbf b$ . Prove that $A$ , $B$ and $C$ are collinear.
(4)
(Total for Question 2 is 4 marks)
3
Given that $p\binom{2}{1}+q\binom{-1}{3}=\binom{1}{11}$ , work out $p$ and $q$ .
(4)
(Total for Question 3 is 4 marks)
4
Higher only: The vertices of quadrilateral $ABCD$ , in order, are $A=(1,2)$ , $B=(6,4)$ , $C=(8,9)$ and $D=(3,7)$ . Use vectors to prove that $ABCD$ is a rhombus.
(4)
(Total for Question 4 is 4 marks)
5
Higher only: The diagonals of parallelogram $ABCD$ meet at $M=(6,5)$ . The vertices include $A=(2,-1)$ , $B=(7,3)$ and $C=(k,11)$ . Work out $k$ and the coordinates of $D$ . Use vectors in your working.
(4)
(Total for Question 5 is 4 marks)

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

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

Tier 1 · Easy

Mark scheme for G1 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	Vertex $Q$	1	The side $PR$ joins vertices $P$ and $R$ . The remaining vertex is $Q$ , so $Q$ is opposite $PR$ .
2	$\angle QRS$ (or $\angle SRQ$ )	1	The vertex letter goes in the middle, so the angle is $\angle QRS$ or, in the reverse order, $\angle SRQ$ .

Tier 2 · Standard

Mark scheme for G1 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	$EF$ is perpendicular to $CD$ .	1	Because $AB\parallel CD$ , a line perpendicular to $AB$ is also perpendicular to $CD$ . Therefore $EF\perp CD$ .
2	$AB$ is an edge; $A$ is a vertex.	2	The line segment where two faces meet is an edge. A point where edges meet is a vertex, so $AB$ is an edge and $A$ is a vertex.
3	$\angle ABD$ (or $\angle DBA$ ); $BD\perp BC$	2	The common endpoint $B$ is the vertex, so it is the middle letter in $\angle ABD$ or $\angle DBA$ . The given right-angle relationship is written $BD\perp BC$ .

Tier 3 · Hard

Mark scheme for G1 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	A correctly labelled diagram with $AB\parallel CD$ , $A,E,C$ collinear, $F,E,G$ collinear, and right-angle marks at $F$ and $G$ .	3	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.
2	A correctly labelled right-angled triangle $PQR$ with $PQ=8$ cm, $PR=6$ cm, $PR\perp PQ$ , $PU=UR$ , $V$ on $PQ$ , and $UV\parallel RQ$ marked.	3	Draw $PQ=8$ cm, construct the $6$ cm perpendicular $PR$ at $P$ , and join $R$ to $Q$ . Bisect $PR$ to place $U$ . Draw the unique parallel to $RQ$ through $U$ ; it meets $PQ$ at $V$ . Add a right-angle square, equal-length ticks on $PU$ and $UR$ , and matching parallel arrows on $UV$ and $RQ$ .
3	Edge $SP$ ; vertex $S$ ; edge $RS$	3	Faces $SPQ$ and $SPR$ share vertices $S$ and $P$ , so their common edge is $SP$ . The apex $S$ is not on the base plane $PQR$ . In triangle $QRS$ , the side not touching $Q$ is $RS$ .
4	A; $PQ\parallel RS$ , $QR\perp PQ$ , and $P,T,R$ are collinear	4	In option A, $PQ$ and $RS$ are both horizontal, while $QR$ is vertical, so the parallel and perpendicular conditions hold. The line from $P(0,0)$ to $R(6,4)$ has midpoint $(3,2)$ , so $T$ lies on diagonal $PR$ . Option B puts $T$ off $PR$ , and option C makes $RS$ non-horizontal. These coordinate checks leave exactly option A.
5	A correct labelled diagram; $E(4,4)$ ; $G(4,6)$ ; $PQ\parallel SR$ ; $EG\perp SR$	5	Plotting and joining the four given points fixes the quadrilateral, with horizontal sides $PQ$ and $SR$ . A point on $PR$ has coordinates $(6t,6t)$ , while a point on $QS$ has coordinates $(8-6s,6s)$ . Equality of the coordinates gives $t=s=\frac23$ , so the diagonals meet uniquely at $E=(4,4)$ . Since $SR$ is horizontal, the unique perpendicular through $E$ is $x=4$ , which meets $SR$ at $G=(4,6)$ . Hence $PQ\parallel SR$ and $EG\perp SR$ . Every plotted point, drawn segment and requested relationship is included in the answer.

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

Tier 1 · Easy

Mark scheme for G2 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	Draw equal-radius arcs from $A$ and $B$ that meet above and below $AB$ , then join the two arc intersections.	2	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 $AB$ .
2	A circle with centre $O$ and radius $4$ cm	1	Every point at a fixed distance from $O$ lies on a circle centred at $O$ . The fixed distance is the radius, so the radius is $4$ cm.

Tier 2 · Standard

Mark scheme for G2 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	The part of the internal angle bisector from $O$ up to and including the point $6$ m from $O$ .	3	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.
2	Draw an arc centred at $P$ cutting $l$ at two points, draw equal arcs from those two points to meet, then join their intersection to $P$ .	2	With centre $P$ , draw an arc cutting $l$ at $A$ and $B$ . With the same suitable radius, draw arcs from $A$ and $B$ to meet at $Q$ on the other side of $l$ . Join $P$ to $Q$ ; $PQ\perp l$ .
3	Mark two points on $l$ equally distant from $A$ , draw equal arcs from these points to meet, then join their intersection to $A$ .	2	Draw an arc centred at $A$ to cut $l$ at $B$ and $C$ . Set a compass radius greater than $AB$ , then draw equal arcs from $B$ and $C$ to meet at $D$ . Join $A$ to $D$ ; because $AB=AC$ and $DB=DC$ , $AD$ is the perpendicular bisector of $BC$ , so $AD\perp l$ .

Tier 3 · Hard

Mark scheme for G2 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	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.	4	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.
2	The intersection of the half-plane on the $A$ side of the perpendicular bisector of $AB$ and the open strip between the two lines parallel to $l$ at distance $3$ cm from $l$ ; the three boundary lines are not included.	4	Construct the perpendicular bisector of $AB$ ; points on the $A$ side are closer to $A$ than to $B$ . Construct two lines parallel to $l$ , each $3$ cm from $l$ ; points less than $3$ cm from $l$ lie between them. Take the overlap of these regions and exclude the boundaries because both inequalities are strict.
3	An accurately constructed rhombus $ABCD$	4	Draw $AC=8$ cm. Use equal-radius arcs centred at $A$ and $C$ to construct the perpendicular bisector, which crosses $AC$ at its midpoint. Mark $B$ and $D$ at the stated distances on opposite sides and join the vertices. The diagonals bisect each other at right angles, so all four sides are equal and $ABCD$ is a rhombus.
4	Construct the two lines parallel to $l$ at perpendicular distance $3$ m and the two lines parallel to $m$ at perpendicular distance $4$ m; the four intersections are the locus, so there are $4$ positions	4	Points exactly $3$ m from $l$ lie on two parallels, one on each side of $l$ . Points exactly $4$ m from $m$ lie on another two parallels. Each line in the first pair crosses each line in the second pair once, giving four and only four positions. Taking $l$ and $m$ as the coordinate axes verifies the positions as $(\pm4,\pm3)$ , so the configuration is unique.
5	The unique point $P$ where the internal angle bisector of $\angle BAC$ meets the perpendicular bisector of $DE$ inside triangle $ABC$	5	Construct the perpendicular $AC$ at $A$ , then bisect $\angle BAC$ to obtain the locus equidistant from segments $AB$ and $AC$ inside the triangle. Construct the perpendicular bisector of $DE$ , the locus where $PD=PE$ . Their single intersection inside the triangle is $P$ . With $A=(0,0)$ , $B=(16,0)$ , $C=(0,16)$ , $D=(8,0)$ and $E=(0,6)$ , the two loci are $y=x$ and $y-3=\frac43(x-4)$ , giving $P=(7,7)$ . Its perpendicular feet $(7,0)$ and $(0,7)$ lie on the stated segments, so both segment distances are $7$ cm; $PD=PE=\sqrt{50}$ cm; and $7+7<16$ , so $P$ is inside triangle $ABC$ . The two non-parallel locus lines have exactly one intersection.

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

Tier 1 · Easy

Mark scheme for G3 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	$117^\circ$	1	Angles on a straight line total $180^\circ$ , so $x=180^\circ-63^\circ=117^\circ$ .
2	$38^\circ$	1	Vertically opposite angles are equal, so the required angle is $38^\circ$ .

Tier 2 · Standard

Mark scheme for G3 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	$x=15$ $67^\circ$	3	Alternate angles between parallel lines are equal, so $4x+7=7x-38$ . Hence $45=3x$ and $x=15$ . Substitution gives $4(15)+7=67^\circ$ .
2	$77^\circ$	2	An exterior angle equals the sum of the two opposite interior angles. Therefore the missing angle is $126^\circ-49^\circ=77^\circ$ .
3	$x=60$ ; the smallest angle is $95^\circ$	3	Angles at a point total $360^\circ$ , so $2x+10+x+35+135=360$ . This gives $3x=180$ and $x=60$ . The three angles are $130^\circ$ , $95^\circ$ and $135^\circ$ , so the smallest is $95^\circ$ .

Tier 3 · Hard

Mark scheme for G3 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	$120^\circ$	4	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 $1980^\circ-1860^\circ=120^\circ$ .
2	$\angle ABC=68^\circ$ (alternate angles, $DE\parallel BC$ ) $\angle ACB=47^\circ$ (alternate angles, $DE\parallel BC$ ) $\angle BAC=65^\circ$ (angles in a triangle total $180^\circ$ )	4	Since $DE\parallel BC$ , $\angle ABC=68^\circ$ and $\angle ACB=47^\circ$ by alternate angles. Angles in a triangle total $180^\circ$ , so $\angle BAC=180^\circ-68^\circ-47^\circ=65^\circ$ .
3	$x=56$ ; the largest interior angle is $126^\circ$	4	Exterior angles total $360^\circ$ , so $54+68+70+x+2x=360$ . Hence $3x=168$ and $x=56$ . The largest interior angle is paired with the smallest exterior angle, $54^\circ$ , so it is $180^\circ-54^\circ=126^\circ$ .
4	$x=22$ ; $\angle COD=163^\circ$	4	Angles $AOB$ and $BOC$ make the straight angle $AOC$ , so $(3x+7)+(5x-3)=180$ . This gives $8x+4=180$ and $x=22$ . Then $\angle BOC=107^\circ$ . The clockwise reflex angle from $OB$ to $OD$ passes through $OC$ , so $\angle COD=270-107=163^\circ$ . The stated ray order fixes this as the unique smaller angle.
5	$x=16$ ; $\angle DAB=73^\circ$ , $\angle ABC=122^\circ$ , $\angle BCD=58^\circ$ , $\angle CDA=107^\circ$	4	Co-interior angles on transversal $AD$ total $180^\circ$ , so $(4x+9)+(7x-5)=180$ . Hence $11x+4=180$ and $x=16$ . Therefore $\angle DAB=73^\circ$ and $\angle CDA=107^\circ$ . The given difference makes $\angle ABC=122^\circ$ . Co-interior angles on $BC$ then give $\angle BCD=180-122=58^\circ$ . For example, $A=(0,0)$ , $B=(10,0)$ , $D=(4\cot73^\circ,4)$ and $C=(10+4\cot58^\circ,4)$ give a convex trapezium with exactly these angles; the linear equation and co-interior pairs fix the four angle values uniquely.

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

Tier 1 · Easy

Mark scheme for G4 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	Trapezium	1	A quadrilateral with exactly one pair of parallel sides is a trapezium.
2	They are equal in length (accept that they bisect each other).	1	A rectangle has diagonals of equal length. As a parallelogram, it also has diagonals that bisect each other, so either property is valid.

Tier 2 · Standard

Mark scheme for G4 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	$112^\circ$ , $68^\circ$ , $112^\circ$	2	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 $180^\circ-68^\circ=112^\circ$ .
2	$36$ cm	2	Opposite sides of a parallelogram are equal, so the perimeter is $2\times7+2\times11=14+22=36$ cm.
3	$ABCD$ is a square; the right angle makes the parallelogram a rectangle, and the equal adjacent sides make it a rhombus.	3	A parallelogram with one right angle has four right angles, so it is a rectangle. A parallelogram with equal adjacent sides has all four sides equal, so it is a rhombus. A quadrilateral that is both a rectangle and a rhombus is a square.

Tier 3 · Hard

Mark scheme for G4 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	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.	3	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.
2	$\angle B=80^\circ$ and $\angle D=80^\circ$	3	The opposite angles between the unequal sides of a kite are equal, so let $B=D=x$ . The angles in a quadrilateral total $360^\circ$ , so $74+126+2x=360$ . Hence $2x=160$ and $x=80^\circ$ .
3	$\angle DAC=\angle ACB=\angle CAB$ , so $AB=BC$ and $ABCD$ is a rhombus.	4	Since $AD\parallel BC$ , $\angle DAC=\angle ACB$ by alternate angles. Since $AC$ bisects $\angle DAB$ , $\angle DAC=\angle CAB$ . Therefore $\angle CAB=\angle ACB$ , so $AB=BC$ because equal angles in a triangle face equal sides. A parallelogram with equal adjacent sides has four equal sides, so $ABCD$ is a rhombus.
4	$\angle ABE=59^\circ$ ; $\angle DAB=\angle BCD=62^\circ$ and $\angle ABC=\angle CDA=118^\circ$	4	The diagonals of a rhombus are perpendicular, so $\angle AEB=90^\circ$ . Angles in triangle $ABE$ total $180^\circ$ , giving $\angle ABE=180^\circ-90^\circ-31^\circ=59^\circ$ . Diagonal $AC$ bisects $\angle DAB$ , so $\angle DAB=2\times31^\circ=62^\circ$ . Opposite angles of a rhombus are equal, giving $\angle BCD=62^\circ$ , and adjacent angles in a parallelogram are supplementary, giving $\angle ABC=\angle CDA=118^\circ$ . Coordinates $A=(0,0)$ , $B=(1,0)$ , $D=(\cos62^\circ,\sin62^\circ)$ and $C=(1+\cos62^\circ,\sin62^\circ)$ verify existence, and the angle facts fix every requested value uniquely.
5	Kite; perimeter $=10+2\sqrt{58}$ cm; it is not a rhombus because its adjacent side pairs have different lengths	4	The four small triangles at $E$ are right-angled. Pythagoras gives $AB=AD=\sqrt{4^2+3^2}=5$ cm and $BC=CD=\sqrt{7^2+3^2}=\sqrt{58}$ cm. Two distinct pairs of equal adjacent sides make $ABCD$ a kite. Its perimeter is $2(5)+2\sqrt{58}=10+2\sqrt{58}$ cm. Since $5\ne\sqrt{58}$ , not all four sides are equal, so it is not a rhombus. Coordinates $A=(0,4)$ , $E=(0,0)$ , $C=(0,-7)$ , $B=(-3,0)$ and $D=(3,0)$ verify the unique labelled configuration.

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

Tier 1 · Easy

Mark scheme for G5 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	SSS	1	All three side lengths of one triangle match all three side lengths of the other, so the criterion is SSS.
2	SAS	1	The equal angle is included between the two equal sides in each triangle. The order in which the facts are stated does not matter, so the correct criterion is SAS.

Tier 2 · Standard

Mark scheme for G5 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	They are congruent by RHS.	2	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.
2	$\triangle JKL\cong\triangle MNP$ by ASA.	2	Two corresponding angles are equal, and the equal side $JK=MN$ lies between those angles. Therefore the triangles are congruent by ASA.
3	$\triangle ABC\cong\triangle PQR$ ; $\angle PRQ$	2	The side matches give $A\leftrightarrow P$ , $B\leftrightarrow Q$ and $C\leftrightarrow R$ , so $\triangle ABC\cong\triangle PQR$ by SSS. Therefore $\angle ACB$ , whose vertex is $C$ , corresponds to $\angle PRQ$ , whose vertex is $R$ .

Tier 3 · Hard

Mark scheme for G5 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	$BC=CD$	3	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.
2	$AD\perp BC$	4	In triangles $ABD$ and $ACD$ , $AB=AC$ , $BD=DC$ because $D$ is the midpoint, and $AD$ is common. The triangles are congruent by SSS, so $\angle ADB=\angle ADC$ . These adjacent angles form a straight line and total $180^\circ$ , so each is $90^\circ$ and $AD\perp BC$ .
3	$\triangle ACB\cong\triangle ADB$ by RHS, so $BC=BD$ .	3	Triangles $ACB$ and $ADB$ are right-angled. They have the common hypotenuse $AB$ and the equal shorter sides $AC=AD$ . Therefore the triangles are congruent by RHS, so the corresponding sides $BC$ and $BD$ are equal.
4	$\triangle ACB\cong\triangle ADB$ by ASA; $AC=AD$ and $BC=BD$ , so $A$ and $B$ lie on the perpendicular bisector of $CD$ and hence $AB$ is that perpendicular bisector	5	We have $\angle CAB=\angle DAB=45^\circ$ , $\angle CBA=\angle DBA=45^\circ$ , and common included side $AB$ , so $\triangle ACB\cong\triangle ADB$ by ASA. Corresponding sides give $AC=AD$ and $BC=BD$ . Therefore both $A$ and $B$ are equidistant from $C$ and $D$ , so both lie on the perpendicular bisector of $CD$ ; the unique line through them is $AB$ . The configuration exists uniquely: with $A=(0,0)$ and $B=(8,0)$ , the fixed $45^\circ$ rays meet once above at $C=(4,4)$ and once below at $D=(4,-4)$ .
5	$\triangle ABC\cong\triangle CDA$ by SSS; $AB\parallel CD$ and $BC\parallel AD$ , so $ABCD$ is a parallelogram	4	Triangles $ABC$ and $CDA$ have $AB=CD$ , $BC=AD$ and common side $AC$ , so they are congruent by SSS. Hence $\angle BAC=\angle DCA$ , making $AB\parallel CD$ by alternate angles. Also $\angle BCA=\angle CAD$ , making $BC\parallel AD$ . A quadrilateral with both pairs of opposite sides parallel is a parallelogram. Coordinates $A=(0,0)$ , $B=(3,1)$ , $C=(5,4)$ and $D=(2,3)$ verify a convex labelled configuration with $B$ and $D$ on opposite sides of $AC$ .

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

Tier 1 · Easy

Mark scheme for G6 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	$86^\circ$	2	Equal sides face equal angles, so angle $C$ is also $47^\circ$ . The angles in a triangle sum to $180^\circ$ , giving $A=180^\circ-47^\circ-47^\circ=86^\circ$ .
2	$8$ cm	2	Let the missing side be $x$ . By Pythagoras, $x^2+6^2=10^2$ , so $x^2=100-36=64$ and $x=8$ cm.

Tier 2 · Standard

Mark scheme for G6 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	$15$ cm	3	By Pythagoras' theorem, $c^2=9^2+12^2=81+144=225$ . Therefore $c=\sqrt{225}=15$ cm.
2	$35$ cm	2	The scale factor from the smaller triangle to the larger is $20\div8=2.5$ . The corresponding length is $14\times2.5=35$ cm.
3	$9^2+40^2=41^2$ , so the triangle is right-angled; area $=180$ cm $^2$	3	$9^2+40^2=81+1600=1681$ and $41^2=1681$ , so the triangle is right-angled by the converse of Pythagoras' theorem. The perpendicular sides are $9$ cm and $40$ cm, giving area $\frac12\times9\times40=180$ cm $^2$ .

Tier 3 · Hard

Mark scheme for G6 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	$BC=AD$	4	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.
2	$52$ cm	4	The half-diagonals are $5$ cm and $12$ cm. They form a right-angled triangle whose hypotenuse is one side of the rhombus, so the side is $\sqrt{5^2+12^2}=\sqrt{169}=13$ cm. The perimeter is $4\times13=52$ cm.
3	$DE=EC$ , so $E$ is the midpoint of $DC$	4	Compare right-angled triangles $ADE$ and $BCE$ . Their hypotenuses are equal because $AE=BE$ , and $AD=BC$ because opposite sides of a rectangle are equal. The triangles are congruent by RHS, so corresponding sides $DE$ and $EC$ are equal. Since $E$ lies on $DC$ , it is the midpoint of $DC$ .
4	Height $=12$ cm; $AC=20$ cm	5	The equal non-parallel sides make an isosceles trapezium, so the total overhang $21-11=10$ cm splits equally into $5$ cm at each end. A right triangle at an end has hypotenuse $13$ and base $5$ , giving height $\sqrt{13^2-5^2}=12$ cm. From $A$ to $C$ , the horizontal distance is $5+11=16$ cm, so $AC=\sqrt{16^2+12^2}=20$ cm. Coordinates $A=(0,0)$ , $B=(21,0)$ , $D=(5,12)$ , $C=(16,12)$ verify the unique labelled configuration.
5	$AD=12$ cm; $AB=15$ cm; $AC=20$ cm	5	The altitude creates three similar right-angled triangles. Similarity gives $AD^2=BD\times DC=9\times16=144$ , so $AD=12$ cm. Also $BC=9+16=25$ cm. From the corresponding sides, $AB^2=BD\times BC=9\times25=225$ and $AC^2=DC\times BC=16\times25=400$ . Therefore $AB=15$ cm and $AC=20$ cm. Coordinates $D=(0,0)$ , $B=(-9,0)$ , $C=(16,0)$ and $A=(0,12)$ verify all lengths and the right angle at $A$ , fixing the labelled configuration up to reflection.

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)

Tier 1 · Easy

Mark scheme for G7 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	$(2,2)$	2	Add the vector components to the coordinates: $(-3+5,4-2)=(2,2)$ .
2	$(4,3)$	1	Reflection in the $y$ -axis changes the sign of the $x$ -coordinate and keeps the $y$ -coordinate, so $(-4,3)$ maps to $(4,3)$ .

Tier 2 · Standard

Mark scheme for G7 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	$P'(1,4)$	2	A $90^\circ$ anticlockwise rotation about the origin maps $(x,y)$ to $(-y,x)$ . Therefore $(4,-1)$ maps to $(1,4)$ .
2	$(15,-6)$	2	For an enlargement about the origin, multiply both coordinates by the scale factor: $(5\times3,-2\times3)=(15,-6)$ .
3	$A'(2,-3)$ $B'(5,1)$ $C'(-2,4)$	2	Reflection in $y=x$ swaps each point's coordinates. Therefore $(-3,2)\mapsto(2,-3)$ , $(1,5)\mapsto(5,1)$ and $(4,-2)\mapsto(-2,4)$ .

Tier 3 · Hard

Mark scheme for G7 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	$A'(0,-3)$ $B'(3,-4)$ $C'(4,-1)$	4	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 $(4,-1)$ .
2	Rotation $90^\circ$ anticlockwise about $(0,2)$ (accept rotation $270^\circ$ clockwise about $(0,2)$ )	3	Relative to the centre $(0,2)$ , each vertex is rotated a quarter turn anticlockwise: the rule is $(x,y)\mapsto(2-y,\,x+2)$ , so $(3,1)\mapsto(1,5)$ , $(6,1)\mapsto(1,8)$ and $(3,4)\mapsto(-2,5)$ . Both triangles have the same orientation, so the transformation is a rotation; the centre $(0,2)$ is the fixed point of the vertex correspondences.
3	Enlargement with centre $(8,5)$ and scale factor $\frac12$	4	The object length $AB$ is $6$ and the image length $A'B'$ is $3$ , so the scale factor is $\frac12$ . The centre lies on both $AA'$ and $BB'$ . Line $BB'$ is $x=8$ ; extending $AA'$ to meet it gives $(8,5)$ . Equivalently, for scale factor $\frac12$ , the centre is $2A'-A=(8,5)$ .
4	Translation by $\begin{pmatrix}6\\-5\end{pmatrix}$ ; $B'=(7,-1)$ ; $C'=(3,-7)$ ; $D=(-6,11)$	4	The movement from $A$ to $A'$ is $(2-(-4),-2-3)=(6,-5)$ , so the translation vector is $\begin{pmatrix}6\\-5\end{pmatrix}$ . Adding it gives $B'=(1+6,4-5)=(7,-1)$ and $C'=(-3+6,-2-5)=(3,-7)$ . Subtracting the vector from $D'$ gives $D=(0-6,6-(-5))=(-6,11)$ .
5	Centre $(1,3)$ ; $Q'=(7,-1)$	4	For centre $C$ , the rule is $P'=C-2(P-C)=3C-2P$ . Hence $3C=P'+2P=(-5,7)+(8,2)=(3,9)$ , so $C=(1,3)$ . From $C$ to $Q$ the vector is $(-3,2)$ . Multiplying it by $-2$ gives $(6,-4)$ , and adding this to $C$ gives $Q'=(7,-1)$ .

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

Tier 1 · Easy

Mark scheme for G8 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	Any two of: side lengths, angle sizes, area, parallelism, or orientation remain unchanged.	2	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.
2	A vertical translation of $6$ cm upwards	2	Two reflections in parallel lines give a translation perpendicular to the lines. Its distance is twice the separation, so it is $2\times3=6$ cm from $m$ towards $n$ , which is upwards.

Tier 2 · Standard

Mark scheme for G8 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	A translation by $\begin{pmatrix}-2\\5\end{pmatrix}$ .	2	Add the translation vectors component by component: $\begin{pmatrix}3\\-2\end{pmatrix}+\begin{pmatrix}-5\\7\end{pmatrix}=\begin{pmatrix}-2\\5\end{pmatrix}$ .
2	$(-5,2)$ ; orientation is reversed	3	Reflection in the $y$ -axis sends $(2,-5)$ to $(-2,-5)$ . A $90^\circ$ clockwise rotation sends $(x,y)$ to $(y,-x)$ , giving $(-5,2)$ . One reflection reverses orientation and the rotation preserves it, so the combination reverses orientation.
3	It is congruent to the original; orientation is reversed; area $=34$ cm $^2$	3	Rotations and reflections preserve all lengths and area, so the final image is congruent and still has area $34$ cm $^2$ . Rotations preserve orientation. Each reflection reverses it, and three is an odd number of reversals, so the final orientation is reversed.

Tier 3 · Hard

Mark scheme for G8 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	A rotation of $90^\circ$ anticlockwise about the origin; orientation is preserved.	4	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.
2	A rotation of $76^\circ$ clockwise about $O$ ; any two of lengths, angle sizes, area, parallelism and orientation are invariant.	4	Two reflections in intersecting lines give a rotation about their intersection. The rotation angle is twice the directed angle from the first line to the second, so it is $2\times38^\circ=76^\circ$ clockwise about $O$ . A rotation preserves lengths, angles, area, parallelism and orientation.
3	A rotation of $180^\circ$ about $(2,2)$ ; orientation is preserved	4	The first rotation maps $(x,y)$ to $(y,-x)$ . Rotating this image $90^\circ$ clockwise about $(4,0)$ gives $(4-x,4-y)$ . This is the rule for a $180^\circ$ rotation about $(2,2)$ . A rotation preserves orientation.
4	Route 1: $A_1=(8,-2)$ and $B_1=(4,1)$ ; route 2: $A_2=(2,-2)$ and $B_2=(-2,1)$ ; the different endpoint pairs show that these transformations do not commute	5	Reflection in $x=2$ maps $(x,y)$ to $(4-x,y)$ . For route 1 this sends $A$ and $B$ to $(5,2)$ and $(1,5)$ , then the translation gives $A_1=(8,-2)$ and $B_1=(4,1)$ . For route 2 the translation first gives $(2,-2)$ and $(6,1)$ ; reflecting these gives $A_2=(2,-2)$ and $B_2=(-2,1)$ . Since corresponding final endpoints are different, reversing the order changes the result.
5	Reflection in the line $y=-2$ ; orientation is reversed; every point on $y=-2$ is fixed	4	The half-turn maps $(x,y)$ to $(6-x,-4-y)$ . Reflection in $x=3$ then maps this to $(x,-4-y)$ . This leaves the $x$ -coordinate unchanged and places the final $y$ -coordinate equally far across $y=-2$ , so it is reflection in $y=-2$ . A reflection reverses orientation, and precisely the points on its mirror line are fixed.

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

Tier 1 · Easy

Mark scheme for G9 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	$13$ cm	1	The diameter is twice the radius, so $d=2\times6.5=13$ cm.
2	Arc	1	A part of the circumference between two points is called an arc.

Tier 2 · Standard

Mark scheme for G9 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	The segment is a chord; the line is a tangent.	2	A segment whose endpoints lie on the circumference is a chord. A line meeting a circle at exactly one point is a tangent.
2	A correctly labelled circle with chord $AB$ drawn and the smaller region between chord $AB$ and the minor arc $AB$ shaded.	2	Draw the circle and label its centre $O$ . Put $A$ and $B$ on the circumference and join them with a straight chord. Shade the smaller region bounded by chord $AB$ and the minor arc $AB$ .
3	Major arc $PQ$ ; chord $PQ$ ; major segment	3	The longer part of the circumference is the major arc $PQ$ . A straight segment joining two points on the circumference is a chord. The larger region bounded by that chord and the major arc is the major segment.

Tier 3 · Hard

Mark scheme for G9 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	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 $AB$ .	3	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.
2	The major arc $AB$ , with length $36$ cm $\dfrac{7}{25}$	3	The other arc is the longer route around the circumference, so it is the major arc. Its length is $50-14=36$ cm. The shorter arc is $\dfrac{14}{50}$ of the circumference; dividing the numerator and denominator by $2$ gives $\dfrac{7}{25}$ .
3	$AB$ is both a diameter and a chord; $CD$ is a chord but not a diameter because it does not pass through $O$ .	3	Both endpoints of $AB$ are on the circumference, so it is a chord, and it passes through the centre, so it is also a diameter. Segment $CD$ also joins two points on the circumference, making it a chord, but it misses the centre and therefore is not a diameter.
4	Minor arc $=17$ cm; major arc $=14\pi-17$ cm; minor segment and major segment	5	The perimeter of a sector is two radii plus its arc, so the minor arc has length $31-2\times7=17$ cm. The full circumference is $2\pi\times7=14\pi$ cm, hence the major arc has length $14\pi-17$ cm. A region bounded by a chord and an arc is a segment: the smaller one is the minor segment and the larger one is the major segment. Taking $O=(0,0)$ , $A=(7,0)$ and $B=(7\cos(17/7),7\sin(17/7))$ verifies the minor arc because $17/7<\pi$ ; its length fixes the central angle uniquely up to rotation or reflection.
5	Minor arc $=24$ cm; major arc $=60$ cm; the smaller region is the minor segment; the larger region is the major sector	4	There are $2+5=7$ ratio parts, so each part is $84\div7=12$ cm. The arcs are therefore $2\times12=24$ cm and $5\times12=60$ cm. The smaller region bounded by chord $AB$ and its arc is the minor segment. The larger region bounded by radii $OA$ , $OB$ and the major arc is the major sector.

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

Tier 1 · Easy

Mark scheme for G10 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	$90^\circ$	1	The angle in a semicircle is $90^\circ$ , so $\angle ACB=90^\circ$ .
2	$90^\circ$	1	A radius is perpendicular to a tangent at the point of contact. Therefore $OT\perp TL$ and $\angle OTL=90^\circ$ .

Tier 2 · Standard

Mark scheme for G10 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	$62^\circ$	2	Both angles stand on the minor arc $AB$ . The angle at the centre is twice the angle at the circumference, so $\angle ACB=124^\circ\div2=62^\circ$ .
2	$x=22$ ; the angles are $78^\circ$ and $102^\circ$	3	Opposite angles in a cyclic quadrilateral total $180^\circ$ , so $3x+12+5x-8=180$ . Hence $8x+4=180$ , giving $x=22$ . The angles are $3(22)+12=78^\circ$ and $5(22)-8=102^\circ$ .
3	$\angle ACB=43^\circ$ (alternate segment theorem); $\angle BAC=65^\circ$ (angles in a triangle total $180^\circ$ )	3	By the alternate segment theorem, the angle between tangent $AT$ and chord $AB$ equals the angle in the opposite segment, so $\angle ACB=43^\circ$ . Then $\angle BAC=180^\circ-43^\circ-72^\circ=65^\circ$ .

Tier 3 · Hard

Mark scheme for G10 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	$XY=XZ$ $\angle YWZ=112^\circ$	5	Radii meet tangents at right angles, so triangles $WYX$ and $WZX$ are right-angled. They have equal hypotenuse $WX$ and equal radii $WY=WZ$ , so they are congruent by RHS. Corresponding parts give $XY=XZ$ and $\angle YXW=\angle WXZ=34^\circ$ . In triangle $WYX$ , $\angle YWX=180^\circ-90^\circ-34^\circ=56^\circ$ . Congruence gives $\angle XWZ=56^\circ$ , so $\angle YWZ=56^\circ+56^\circ=112^\circ$ .
2	$OM\perp AB$	4	Draw the perpendicular from $O$ to $AB$ , meeting it at $N$ . Right-angled triangles $OAN$ and $OBN$ have equal hypotenuses $OA=OB$ because they are radii, and the common shorter side $ON$ . RHS congruence gives $AN=NB$ , so $N$ is the midpoint of $AB$ . The midpoint is unique, hence $N=M$ and the perpendicular $ON$ is $OM$ . Therefore $OM\perp AB$ .
3	$\angle AOB=106^\circ$ ; $\angle ACB=53^\circ$	4	Radii are perpendicular to tangents, so $\angle OAP=\angle OBP=90^\circ$ . The angles in quadrilateral $OAPB$ total $360^\circ$ , giving $\angle AOB=360^\circ-90^\circ-90^\circ-74^\circ=106^\circ$ . Since $C$ is on the major arc, $\angle ACB$ and $\angle AOB$ stand on the same minor arc $AB$ . The angle at the circumference is half the angle at the centre, so $\angle ACB=53^\circ$ .
4	$\angle CPB=34^\circ$	4	The angle in a semicircle gives $\angle ACB=90^\circ$ , so $\angle ABC=180^\circ-90^\circ-28^\circ=62^\circ$ . Since $BP$ is the extension of $BA$ , $\angle CBP=180^\circ-62^\circ=118^\circ$ . By the alternate segment theorem, the angle between tangent $CP$ and chord $CB$ is $\angle BCP=\angle CAB=28^\circ$ . Hence $\angle CPB=180^\circ-118^\circ-28^\circ=34^\circ$ . On the unit circle, $A=(-1,0)$ , $B=(1,0)$ , $C=(\cos56^\circ,\sin56^\circ)$ and $P=(\sec56^\circ,0)$ verify existence with $P$ beyond $B$ ; the stated arc side and extension fix the requested angle uniquely.
5	$EC=ED=8$ cm, so $E$ is the midpoint of $CD$ ; $CD=16$ cm; $AC=AD$ , so triangle $ACD$ is isosceles	5	A radius is perpendicular to a tangent, so $AB$ is perpendicular to the tangent at $A$ . The parallel through $E$ is therefore perpendicular to $AB$ , and $OE$ is the perpendicular from the centre to chord $CD$ ; it bisects the chord, so $EC=ED$ . In right triangle $OEC$ , $EC=\sqrt{10^2-6^2}=8$ cm, hence $CD=16$ cm. Since $AB$ is the perpendicular bisector of $CD$ and $A$ lies on $AB$ , $AC=AD$ . Coordinates $O=(0,0)$ , $A=(-10,0)$ , $E=(6,0)$ , $C=(6,8)$ and $D=(6,-8)$ verify the unique labelled configuration.

G11 · Solve geometrical problems on coordinate axes

Tier 1 · Easy

Mark scheme for G11 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	$(1,5)$	2	Average the coordinates: $x=(-5+7)/2=1$ and $y=(2+8)/2=5$ . The midpoint is $(1,5)$ .
2	$9$ units	1	The points have the same $y$ -coordinate, so the segment is horizontal. Its length is $3-(-6)=9$ .

Tier 2 · Standard

Mark scheme for G11 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	$10$	3	The coordinate changes are $5-(-1)=6$ and $-4-4=-8$ . Therefore $AB=\sqrt{6^2+(-8)^2}=\sqrt{36+64}=\sqrt{100}=10$ .
2	$B=(7,-2)$	2	Let $B=(x,y)$ . For the $x$ -coordinate, $(-3+x)/2=2$ , so $x=7$ . For the $y$ -coordinate, $(4+y)/2=1$ , so $y=-2$ . Therefore $B=(7,-2)$ .
3	$42$ square units	3	$AB$ is horizontal with length $8-(-4)=12$ . The perpendicular height from $C$ to line $AB$ is $5-(-2)=7$ . Therefore the area is $\frac12\times12\times7=42$ square units.

Tier 3 · Hard

Mark scheme for G11 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	Right-angled at $B$ $AB=BC=2\sqrt{10}$ Area $=20$ square units	5	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.
2	$ABCD$ is a parallelogram; area $=30$ square units	4	$AB$ and $DC$ are both horizontal and each has length $6$ . From $A$ to $D$ and from $B$ to $C$ , the coordinate changes are both $2$ right and $5$ up, so those sides are equal and parallel. Hence $ABCD$ is a parallelogram. Taking $AB=6$ as the base, the perpendicular height is $7-2=5$ , so the area is $6\times5=30$ square units.
3	$P=(0,5)$	4	Let $P=(0,y)$ . Equating squared distances gives $3^2+(y-1)^2=5^2+(y-5)^2$ . Expanding and cancelling $y^2$ gives $10-2y=50-10y$ , so $8y=40$ and $y=5$ . Therefore $P=(0,5)$ ; indeed, both distances are $5$ .
4	Centre $(2,4)$ ; radius $3\sqrt5$ units; $C$ lies on the circle	4	The centre is the midpoint of $AB$ , namely $(({-4+8})/2,(1+7)/2)=(2,4)$ . Its squared distance to $A$ is $(-6)^2+(-3)^2=45$ , so the radius is $\sqrt{45}=3\sqrt5$ . From the centre to $C$ the change is $(3,-6)$ , whose squared length is $3^2+(-6)^2=45$ . Therefore $C$ is exactly one radius from the centre and lies on the circle.
5	$D=(1,-6)$ ; $ABCD$ is a rhombus but not a square	5	The diagonals of a parallelogram share a midpoint. Midpoint $AC$ is $(1,0)$ , so this must also be the midpoint of $BD$ ; with $B=(1,6)$ , this gives $D=(1,-6)$ . The four squared side lengths are $AB^2=BC^2=CD^2=DA^2=5^2+6^2=61$ , so all four sides are equal and the shape is a rhombus. Its diagonals have lengths $AC=10$ and $BD=12$ , so they are unequal; a square has equal diagonals, hence this is not a square. The midpoint condition makes $D$ unique.

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

Tier 1 · Easy

Mark scheme for G12 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	$5$ faces $6$ vertices	2	A triangular prism has two triangular faces and three rectangular faces, making $5$ faces. Its two triangular ends have $3+3=6$ vertices.
2	$8$ vertices (or $12$ edges)	2	A cube and a cuboid each have $8$ vertices and $12$ edges. Giving either number with the matching feature answers the question.

Tier 2 · Standard

Mark scheme for G12 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	Cone	2	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.
2	$5$ faces $8$ edges $5$ vertices	3	There is one square base and four triangular faces, giving $5$ faces. There are $4$ base edges and $4$ sloping edges, giving $8$ edges. The $4$ base vertices and the apex give $5$ vertices.
3	Cylinder; $2$ edges; $1$ curved surface	3	The two circular plane faces and one curved surface identify a cylinder. Each circle where a plane face meets the curved surface is an edge, giving $2$ edges, and the side is one continuous curved surface.

Tier 3 · Hard

Mark scheme for G12 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	$11$ faces $27$ edges $18$ vertices	3	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.
2	$10$ sides; a decagonal prism; $30$ edges and $20$ vertices	4	Subtract the two end faces from the total: $12-2=10$ , so each congruent end is a decagon and the solid is a decagonal prism. Its edge count is $3\times10=30$ , and its vertex count is $2\times10=20$ .
3	$17$ faces; $35$ edges; $22$ vertices	4	The prism has $7+2=9$ faces, $3\times7=21$ edges and $2\times7=14$ vertices. The pyramid has $7+1=8$ faces, $7+7=14$ edges and $7+1=8$ vertices. The totals are $9+8=17$ faces, $21+14=35$ edges and $14+8=22$ vertices.
4	$n=6$ ; hexagonal-based pyramid; $7$ faces, $12$ edges and $7$ vertices	4	An $n$ -sided pyramid has $n+1$ faces and $2n$ edges. The condition gives $2n=(n+1)+5$ , so $n=6$ . It is therefore a hexagonal-based pyramid. It has $6+1=7$ faces, $2\times6=12$ edges and $6+1=7$ vertices.
5	$7$ exterior faces; $12$ distinct edges; $7$ distinct vertices	5	The triangular prism has $5$ faces, $9$ edges and $6$ vertices. The triangular-based pyramid has $4$ faces, $6$ edges and $4$ vertices. Joining the triangular faces removes two exterior faces, so there are $5+4-2=7$ exterior faces. The three edges and three vertices around the join are shared, not doubled, giving $9+6-3=12$ edges and $6+4-3=7$ vertices. The check $7-12+7=2$ confirms the counts.

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

Tier 1 · Easy

Mark scheme for G13 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	A rectangle $8$ cm by $5$ cm	2	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.
2	A circle of diameter $6$ cm	2	Viewed from above, a vertical cylinder appears as its circular base. The diameter is twice the radius, so it is $2\times3=6$ cm.

Tier 2 · Standard

Mark scheme for G13 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	Front elevation, west to east: $(3,1)$ Side elevation, north to south: $(2,3)$	3	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 $\max(3,0)=3$ .
2	Plan: a rectangle $7$ cm by $4$ cm; elevation: a right-angled triangle with horizontal base $4$ cm and vertical left side $3$ cm	3	The plan shows the $7$ cm length and $4$ cm base, giving a $7$ cm by $4$ cm rectangle; the sloping face meets the top edge directly above the left-hand side of the base, so no extra line appears inside the rectangle. Looking at the end with the vertical side on the left shows the end face in true size, so the elevation is a right-angled triangle with a $4$ cm horizontal base and a $3$ cm vertical side on the left.
3	Plan: a circle of diameter $8$ cm; front elevation: an isosceles triangle with base $8$ cm and height $9$ cm	3	From above, the cone shows its circular base, whose diameter is $2\times4=8$ cm. From the front, the circular base projects to an $8$ cm line and the apex is $9$ cm above its midpoint, giving an isosceles triangle of base $8$ cm and height $9$ cm.

Tier 3 · Hard

Mark scheme for G13 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	$7$ cubes; for example, from north to south the plan rows can be $(1,2)$ and $(3,1)$ , with entries west to east.	4	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.
2	Plan: an $8$ cm by $5$ cm rectangle with a line across the full depth $3$ cm from the left end; front elevation: a step shape, $3$ cm wide and $6$ cm high on the left then $5$ cm wide and $2$ cm high; right side elevation: a $5$ cm by $6$ cm rectangle with a horizontal line $2$ cm above its base.	4	From above, both cuboids have the same depth, so the footprint is $8$ cm by $5$ cm and the top cuboid ends $3$ cm from the left. From the front, the total height under the top cuboid is $2+4=6$ cm for the first $3$ cm, while the remaining width is $8-3=5$ cm at height $2$ cm. From the right, the maximum height along the length is $6$ cm across the full $5$ cm depth; the top of the base gives a horizontal line at height $2$ cm.
3	Plan: a rectangle $10$ cm by $6$ cm; east elevation: a circle of diameter $6$ cm; south elevation: a rectangle $10$ cm by $6$ cm	4	The cylinder extends $10$ cm east-west and has diameter $2\times3=6$ cm. From above, these are the plan's two dimensions, so the plan is a $10$ cm by $6$ cm rectangle. Looking from the east is along the axis, so the elevation is the circular end of diameter $6$ cm. Looking from the south shows the length and diameter, giving a $10$ cm by $6$ cm rectangle.
4	Plan: a $10$ cm by $6$ cm rectangle with the $10$ cm ridge line midway between its long sides; east elevation: the house-shaped pentagon, $6$ cm wide with $3$ cm walls and total height $5$ cm; south elevation: a $10$ cm by $5$ cm rectangle with a horizontal eave line $3$ cm above the base	5	From above, the prism occupies a $10$ cm by $6$ cm rectangle and the roof ridge runs west-east along its centre. Looking from the east shows an end face: width $6$ cm, vertical sides to height $3$ cm and a centred apex at height $5$ cm. Looking from the south shows the $10$ cm length and maximum $5$ cm height; the wall-to-roof join gives a horizontal eave line at height $3$ cm. End-face coordinates $(0,0),(6,0),(6,3),(3,5),(0,3)$ extruded $10$ cm verify one configuration.
5	Plan: a $12$ cm by $8$ cm rectangle containing a centred $6$ cm square, with the square's four corners joined to its centre; front elevation: a $12$ cm by $4$ cm rectangle topped centrally by an isosceles triangle of base $6$ cm and height $5$ cm; side elevation: an $8$ cm by $4$ cm rectangle topped centrally by the same $6$ cm by $5$ cm triangle; total height $=9$ cm	5	From above, the cuboid gives a $12$ cm by $8$ cm rectangle. The centred pyramid base gives a $6$ cm square, and its four sloping edges project from the base corners to the central apex. From the front, the cuboid is a $12$ cm by $4$ cm rectangle and the pyramid is a centred isosceles triangle of base $6$ cm and height $5$ cm. From the side, the corresponding rectangle is $8$ cm by $4$ cm and the pyramid again projects as a $6$ cm by $5$ cm triangle. The total height is $4+5=9$ cm. Coordinates with cuboid ranges $-6\leq x\leq6$ , $-4\leq y\leq4$ , $0\leq z\leq4$ , pyramid-base corners $(\pm3,\pm3,4)$ and apex $(0,0,9)$ verify one unique centred labelled configuration.

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

Tier 1 · Easy

Mark scheme for G14 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	$3750\text{ g}$	1	There are $1000\text{ g}$ in $1\text{ kg}$ , so $3.75\times1000=3750$ . Therefore the mass is $3750\text{ g}$ .
2	$198$ minutes	1	$3$ hours is $3\times60=180$ minutes. Therefore the total time is $180+18=198$ minutes.

Tier 2 · Standard

Mark scheme for G14 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	$156$ litres	3	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.
2	$120\text{ cm}$	2	$7.5\text{ m}=750\text{ cm}$ . The pieces use $18\times35=630\text{ cm}$ , so $750-630=120\text{ cm}$ of ribbon is left.
3	$385$ litres	3	$0.42\text{ m}^3=420$ litres. The buckets add $8\times12.5=100$ litres. Therefore the volume is $420-135+100=385$ litres.

Tier 3 · Hard

Mark scheme for G14 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	£ $265.50$	5	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 £ $265.50$ .
2	£ $15.05$	3	$2.6\text{ kg}=2600\text{ g}$ . Since $2600\div375=6.93\ldots$ , the bakery must buy $7$ whole bags. The total cost is $7\times2.15=15.05$ , so it costs £ $15.05$ .
3	$3600\text{ mm}$	3	The side length is $\sqrt{0.81}=0.9\text{ m}$ . The perimeter is $4\times0.9=3.6\text{ m}$ . Since $1\text{ m}=1000\text{ mm}$ , the perimeter is $3600\text{ mm}$ .
4	$12\,060\text{ cm}^2$	3	$1.8\text{ m}=180\text{ cm}$ , so the sheet has area $180\times75=13\,500\text{ cm}^2$ . The opening measures $32\text{ cm}$ by $45\text{ cm}$ , so its area is $32\times45=1440\text{ cm}^2$ . The remaining area is $13\,500-1440=12\,060\text{ cm}^2$ .
5	$08{:}23$ on the next day	3	On the departure clock, $22{:}48+7$ hours $35$ minutes is $06{:}23$ on the next day. The destination clock is $2$ hours ahead, so the local arrival time is $08{:}23$ on the next day.

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

Tier 1 · Easy

Mark scheme for G15 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	$068^\circ$	1	A bearing is the clockwise angle from north written with three figures. Therefore $68^\circ$ is written as $068^\circ$ .
2	$135^\circ$	1	East is bearing $090^\circ$ and south is bearing $180^\circ$ . Halfway between them is $135^\circ$ .

Tier 2 · Standard

Mark scheme for G15 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	$2\text{ km}$ $037^\circ$	4	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 $037^\circ$ .
2	$4\text{ km}$	2	The map length is $4.6+3.4=8\text{ cm}$ . The actual length is $8\times50\,000=400\,000\text{ cm}=4\text{ km}$ .
3	$2.88\text{ cm}$ by $1.8\text{ cm}$	3	$72\text{ m}=7200\text{ cm}$ and $45\text{ m}=4500\text{ cm}$ . Divide each real length by $2500$ : $7200\div2500=2.88$ and $4500\div2500=1.8$ . The plan dimensions are $2.88\text{ cm}$ by $1.8\text{ cm}$ .

Tier 3 · Hard

Mark scheme for G15 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	$1:37\,500$ $4.2\text{ km}$	4	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 $11.2\times37\,500=420\,000\text{ cm}=4.2\text{ km}$ .
2	$60^\circ$	3	The bearing of the harbour from the lighthouse is the reverse bearing, $046^\circ+180^\circ=226^\circ$ . The smaller angle between bearings $226^\circ$ and $166^\circ$ is $226^\circ-166^\circ=60^\circ$ .
3	$164^\circ$ $344^\circ$ $243^\circ$	4	The bearing from $Q$ to $R$ is $038^\circ+126^\circ=164^\circ$ . The reverse bearing from $R$ to $Q$ is $164^\circ+180^\circ=344^\circ$ . Turning $101^\circ$ anticlockwise from this return direction gives $344^\circ-101^\circ=243^\circ$ .
4	$4.32\text{ km}$	4	Before the enlargement, the path was $13.5\div1.25=10.8\text{ cm}$ long on the map. Its real length is $10.8\times40\,000=432\,000\text{ cm}$ . Since $100\,000\text{ cm}=1\text{ km}$ , the real length is $4.32\text{ km}$ .
5	$\angle PQR=21^\circ$ Bearing $=339^\circ$	4	The bearing of $Q$ from $P$ is $000^\circ$ , so $\angle QPR=122^\circ$ . The angle sum of triangle $PQR$ gives $\angle PQR=180^\circ-122^\circ-37^\circ=21^\circ$ . The bearing of $P$ from $R$ is $122^\circ+180^\circ=302^\circ$ . From $R$ , the ray to $Q$ is $37^\circ$ clockwise from the ray to $P$ , so its bearing is $302^\circ+37^\circ=339^\circ$ .

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

Tier 1 · Easy

Mark scheme for G16 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	$60\text{ cm}^2$	2	Use $A=\frac12(a+b)h$ . Then $A=\frac12(7+13)\times6=\frac12\times20\times6=60\text{ cm}^2$ .
2	$84\text{ cm}^2$	1	The area of a parallelogram is base multiplied by perpendicular height, so $12\times7=84\text{ cm}^2$ .

Tier 2 · Standard

Mark scheme for G16 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	$144\pi\text{ cm}^3$	2	The circular cross-section has area $\pi r^2=\pi(4^2)=16\pi\text{ cm}^2$ . Multiply by the height: $V=16\pi\times9=144\pi\text{ cm}^3$ .
2	$14\text{ cm}$	3	Let the other parallel side be $x\text{ cm}$ . Then $76=\frac12(5+x)\times8$ , so $76=4(5+x)$ . Hence $19=5+x$ and $x=14\text{ cm}$ .
3	$12\text{ cm}$	3	Using $V=\pi r^2h$ gives $588\pi=\pi(7^2)h=49\pi h$ . Therefore $h=588\pi\div49\pi=12\text{ cm}$ .

Tier 3 · Hard

Mark scheme for G16 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	$x=10$	4	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 $x=10$ .
2	$1560\text{ cm}^3$	4	The material has cross-sectional area $12\times8-6\times3=96-18=78\text{ cm}^2$ . The volume is cross-sectional area multiplied by length, so $78\times20=1560\text{ cm}^3$ .
3	$3200\text{ cm}^3$	4	The rectangular area is $14\times10=140\text{ cm}^2$ . The removed triangular area is $\frac12\times6\times4=12\text{ cm}^2$ , so the remaining cross-sectional area is $128\text{ cm}^2$ . The volume is $128\times25=3200\text{ cm}^3$ .
4	Both volumes are $756\text{ cm}^3$ .	4	The cross-sectional area of prism $P$ is $\frac12\times12\times9=54\text{ cm}^2$ , so its volume is $54\times14=756\text{ cm}^3$ . The cross-sectional area of prism $Q$ is $7\times8=56\text{ cm}^2$ , so its volume is $56\times13.5=756\text{ cm}^3$ . Therefore the volumes are equal.
5	No, the volume increases by $15.2\%$ .	4	Cylinder volume is proportional to $r^2h$ . The new volume factor is $1.2^2\times0.8=1.152$ . The new volume is therefore $115.2\%$ of the original, so it increases by $15.2\%$ . Sam is not correct.

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

Tier 1 · Easy

Mark scheme for G17 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	$13\pi\text{ cm}$	1	Use $C=\pi d$ . With $d=13\text{ cm}$ , the circumference is $13\pi\text{ cm}$ .
2	$36\pi\text{ cm}^2$	1	Use $A=\pi r^2$ . Therefore $A=\pi(6^2)=36\pi\text{ cm}^2$ .

Tier 2 · Standard

Mark scheme for G17 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	Area $=180+\frac{25\pi}{2}\text{ cm}^2$ Perimeter $=46+5\pi\text{ cm}$	4	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 $46+5\pi\text{ cm}$ .
2	$121-16\pi\text{ cm}^2$	2	The square has area $11^2=121\text{ cm}^2$ . The circular hole has area $\pi(4^2)=16\pi\text{ cm}^2$ . Therefore the remaining area is $121-16\pi\text{ cm}^2$ .
3	$16\pi^2\text{ cm}^2$	3	The circle has perimeter $2\pi(8)=16\pi\text{ cm}$ . Each side of the square is $16\pi\div4=4\pi\text{ cm}$ . Therefore the square has area $(4\pi)^2=16\pi^2\text{ cm}^2$ .

Tier 3 · Hard

Mark scheme for G17 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	Total surface area $=96\pi\text{ cm}^2$ Volume $=96\pi\text{ cm}^3$	5	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 $\frac13\pi r^2h=\frac13\pi(6^2)(8)=96\pi\text{ cm}^3$ .
2	$\frac{686\pi}{3}\text{ cm}^3$	4	The curved surface area of a hemisphere is $2\pi r^2$ and its circular base has area $\pi r^2$ , so $3\pi r^2=147\pi$ . Hence $r^2=49$ and $r=7\text{ cm}$ . The volume is half the volume of a sphere, so $V=\frac23\pi r^3=\frac23\pi(7^3)=\frac{686\pi}{3}\text{ cm}^3$ .
3	$32\pi\text{ m}^2$	4	For the outer circle, $2\pi R=18\pi$ , so $R=9\text{ m}$ . The outer area is $81\pi\text{ m}^2$ and the pond area is $49\pi\text{ m}^2$ . Therefore the path area is $81\pi-49\pi=32\pi\text{ m}^2$ .
4	$115\pi\text{ cm}^2$	4	Only the curved surfaces are exposed. The curved surface area of the hemisphere is $2\pi r^2=2\pi(5^2)=50\pi\text{ cm}^2$ . The curved surface area of the cone is $\pi rl=\pi(5)(13)=65\pi\text{ cm}^2$ . Therefore the exterior surface area is $50\pi+65\pi=115\pi\text{ cm}^2$ .
5	$15\text{ cm}$	4	The base area is $\frac12(8+14)\times6=66\text{ cm}^2$ . If the perpendicular height is $h\text{ cm}$ , then $330=\frac13\times66\times h=22h$ . Therefore $h=15\text{ cm}$ .

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

Tier 1 · Easy

Mark scheme for G18 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	$4\pi\text{ cm}$	2	A quarter-circle has angle $90^\circ$ . Its arc length is $\frac{90}{360}\times2\pi(8)=\frac14\times16\pi=4\pi\text{ cm}$ .
2	$22+11\pi\text{ cm}$	2	The radius is $11\text{ cm}$ , so the curved edge has length $\pi r=11\pi\text{ cm}$ . Including the diameter, the perimeter is $22+11\pi\text{ cm}$ .

Tier 2 · Standard

Mark scheme for G18 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	Angle $=140^\circ$ Area $=\frac{63\pi}{2}\text{ cm}^2$	4	Use $7\pi=\frac{\theta}{360}\times2\pi(9)$ . Cancelling $\pi$ and solving gives $\theta=140^\circ$ . The sector area is then $\frac{140}{360}\times\pi(9^2)=\frac{63\pi}{2}\text{ cm}^2$ .
2	$10\pi\text{ cm}$	3	The area is $\frac12\pi r^2$ , so $\frac12\pi r^2=50\pi$ . Hence $r^2=100$ and $r=10\text{ cm}$ . The curved edge has length $\pi r=10\pi\text{ cm}$ .
3	$\frac{15\pi}{2}\text{ cm}$ (or $7.5\pi\text{ cm}$ )	3	In $25$ minutes the hand turns through $\frac{25}{60}$ of a full circle. The tip travels $\frac{25}{60}\times2\pi(9)=\frac{15\pi}{2}\text{ cm}$ .

Tier 3 · Hard

Mark scheme for G18 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	Area $=38\pi\text{ cm}^2$ Perimeter $=10+\frac{76\pi}{5}\text{ cm}$	5	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 $10+\frac{76\pi}{5}\text{ cm}$ .
2	$30+\frac{32\pi}{3}\text{ cm}$	5	Let the radius be $r\text{ cm}$ . From the area, $\frac{128}{360}\pi r^2=80\pi$ , so $\frac{16}{45}r^2=80$ . Hence $r^2=225$ and $r=15\text{ cm}$ . The arc length is $\frac{128}{360}\times2\pi(15)=\frac{32\pi}{3}\text{ cm}$ . Including the two radii, the full perimeter is $30+\frac{32\pi}{3}\text{ cm}$ .
3	Angle of sector B $=90^\circ$ Difference $=2\pi\text{ cm}$	4	Sector A has area $\frac{160}{360}\pi(9^2)=36\pi\text{ cm}^2$ . If sector B has angle $x$ , then $\frac{x}{360}\pi(12^2)=36\pi$ , giving $x=90^\circ$ . The arc lengths are $\frac{160}{360}\times2\pi(9)=8\pi\text{ cm}$ and $\frac{90}{360}\times2\pi(12)=6\pi\text{ cm}$ , so the difference is $2\pi\text{ cm}$ .
4	Radius $=18\text{ cm}$ Angle $=60^\circ$	5	For the same sector fraction, dividing area by arc length gives $\frac{\pi r^2}{2\pi r}=\frac r2$ . Hence $54\pi\div6\pi=9=\frac r2$ , so $r=18\text{ cm}$ . Now $6\pi=\frac{\theta}{360}\times2\pi(18)$ . Therefore $6=\frac{\theta}{10}$ and $\theta=60^\circ$ .
5	$9\pi\text{ cm}^2$	4	At $4{:}20$ , the minute hand is $20\times6^\circ=120^\circ$ clockwise from $12$ . The hour hand is $4\times30^\circ+\frac{20}{60}\times30^\circ=130^\circ$ clockwise from $12$ . The smaller angle is $10^\circ$ , so the sector area is $\frac{10}{360}\times\pi(18^2)=9\pi\text{ cm}^2$ .

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

Tier 1 · Easy

Mark scheme for G19 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	They are congruent by SAS.	1	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.
2	$14\text{ cm}$	1	Multiply the smaller length by the scale factor: $3.5\times4=14\text{ cm}$ .

Tier 2 · Standard

Mark scheme for G19 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	$12\text{ cm}$ corresponds to $20\text{ cm}$ $35\text{ cm}$ corresponds to $21\text{ cm}$	3	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 $35\times\frac35=21\text{ cm}$ .
2	$15\text{ cm}$	2	The length scale factor is the perimeter ratio, $36\div24=1.5$ . The corresponding larger side is $10\times1.5=15\text{ cm}$ .
3	Smaller perimeter $=52\text{ cm}$ Larger perimeter $=78\text{ cm}$	3	The length scale factor is $12\div8=1.5$ , so the larger perimeter is $1.5$ times the smaller perimeter. If the smaller perimeter is $p$ , then $1.5p-p=26$ , so $0.5p=26$ and $p=52$ . The larger perimeter is $1.5\times52=78\text{ cm}$ .

Tier 3 · Hard

Mark scheme for G19 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	Other sides $=12\text{ cm}$ and $16.5\text{ cm}$ Perimeter $=48\text{ cm}$	4	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 $12+16.5+19.5=48\text{ cm}$ .
2	$PQ=PS$ , $RQ=RS$ and $PR$ is common, so the triangles are congruent by SSS; therefore $\angle QPR=\angle SPR$ .	3	$PQ=PS$ and $RQ=RS$ are given. Also, $PR$ is the same side in both triangles. The three corresponding sides are equal, so triangle $PQR$ is congruent to triangle $PSR$ by SSS. Corresponding angles in congruent triangles are equal, hence $\angle QPR=\angle SPR$ .
3	Surface area $P : Q=9 : 25$ Volume $P : Q=27 : 125$	4	For similar solids, volume scales with $k^3$ and surface area with $k^2$ , so volume divided by surface area scales with $k$ . Hence the length ratio is $P : Q=2.4 : 4=3 : 5$ . Squaring gives the surface-area ratio $9 : 25$ , and cubing gives the volume ratio $27 : 125$ .
4	$\angle BAC=\angle DCA$ , $AB=CD$ and $AC$ is common, so the triangles are congruent by SAS; hence $\angle BCA=\angle DAC$ , so $BC$ is parallel to $AD$ and $ABCD$ is a parallelogram.	4	Because $AB$ is parallel to $CD$ , $\angle BAC=\angle DCA$ as alternate angles. Also $AB=CD$ and $AC$ is common to both triangles. The triangles are therefore congruent by SAS. Corresponding angles give $\angle BCA=\angle DAC$ , so $BC$ is parallel to $AD$ . Both pairs of opposite sides are parallel, hence $ABCD$ is a parallelogram.
5	$125\%$	4	An increase of $237.5\%$ gives a volume scale factor of $3.375=\frac{27}{8}=\left(\frac32\right)^3$ . The length scale factor is therefore $\frac32$ . The surface-area scale factor is $\left(\frac32\right)^2=\frac94=2.25$ , so the surface area is $225\%$ of the original. This is an increase of $125\%$ .

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

Tier 1 · Easy

Mark scheme for G20 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	$10\text{ cm}$	2	By Pythagoras, $c^2=6^2+8^2=36+64=100$ . Since a length is positive, $c=\sqrt{100}=10\text{ cm}$ .
2	$8\text{ cm}$	2	By Pythagoras, the missing length is $\sqrt{17^2-15^2}=\sqrt{289-225}=\sqrt{64}=8\text{ cm}$ .

Tier 2 · Standard

Mark scheme for G20 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	Height $=6.9\text{ m}$ Angle $=73.0^\circ$	4	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.
2	$16.3^\circ$	2	$\tan x=\frac{7}{24}$ , so $x=\tan^{-1}(7/24)=16.260\ldots^\circ$ . To $1$ decimal place, $x=16.3^\circ$ .
3	$16.8\text{ m}$	3	The vertical height from the hand to the kite is $25\sin38^\circ=15.3915\ldots\text{ m}$ . Adding the hand height gives $15.3915\ldots+1.4=16.7915\ldots\text{ m}$ , which is $16.8\text{ m}$ correct to $1$ decimal place.

Tier 3 · Hard

Mark scheme for G20 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	Height $=12\text{ cm}$ $\angle ABC=67.4^\circ$	4	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.
2	$17.7\text{ cm}$	3	First find a base diagonal: $\sqrt{6^2+9^2}=\sqrt{117}\text{ cm}$ . The space diagonal is then $\sqrt{117+14^2}=\sqrt{313}=17.691\ldots\text{ cm}$ , which is $17.7\text{ cm}$ correct to $1$ decimal place.
3	Other diagonal $=24\text{ cm}$ Area $=120\text{ cm}^2$	4	Half of the known diagonal is $5\text{ cm}$ . If half of the other diagonal is $x$ , then $x^2+5^2=13^2$ , so $x=12\text{ cm}$ . The full diagonal is $24\text{ cm}$ . The four right-angled triangles have total area $4\times\frac12\times5\times12=120\text{ cm}^2$ .
4	$68\text{ cm}$	4	Let the side lengths be $a\text{ cm}$ and $b\text{ cm}$ . Pythagoras gives $a^2+b^2=26^2=676$ , and the area gives $ab=240$ . Therefore $(a+b)^2=a^2+b^2+2ab=676+480=1156$ , so $a+b=34$ . The perimeter is $2(a+b)=68\text{ cm}$ .
5	$AM=4\sqrt{13}\text{ cm}$ $\angle MAB=56.3^\circ$	4	The horizontal displacement from $A$ to $M$ is $8\text{ cm}$ and the vertical displacement is $12\text{ cm}$ . Hence $AM=\sqrt{8^2+12^2}=\sqrt{208}=4\sqrt{13}\text{ cm}$ . Also $\tan\angle MAB=12/8$ , so $\angle MAB=56.3099\ldots^\circ=56.3^\circ$ correct 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°

Tier 1 · Easy

Mark scheme for G21 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	$1$	1	$\sin30^\circ=\frac12$ and $\cos60^\circ=\frac12$ . Therefore the sum is $\frac12+\frac12=1$ .
2	$45^\circ$	1	$\sin45^\circ=\cos45^\circ=\frac{\sqrt2}{2}$ , so $x=45^\circ$ .

Tier 2 · Standard

Mark scheme for G21 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	$\frac32$	3	$\sin45^\circ=\cos45^\circ=\frac{\sqrt2}{2}$ and $\cos60^\circ=\frac12$ . Therefore $2\sin45^\circ\cos45^\circ+\cos60^\circ=2(\frac{\sqrt2}{2})(\frac{\sqrt2}{2})+\frac12=1+\frac12=\frac32$ .
2	$x=30^\circ$ $\cos x=\frac{\sqrt3}{2}$	2	$\sin30^\circ=\frac12$ , so $x=30^\circ$ . The corresponding exact cosine value is $\cos30^\circ=\frac{\sqrt3}{2}$ .
3	$x=2$	3	Use $\sin30^\circ=\frac12$ and $\cos60^\circ=\frac12$ . The equation becomes $x+\frac32=\frac72$ , so $x=2$ .

Tier 3 · Hard

Mark scheme for G21 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	$18\sqrt3\text{ cm}^2$	4	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 $\frac12\times6\times6\sqrt3=18\sqrt3\text{ cm}^2$ .
2	$8\sqrt3\text{ cm}^2$	3	Let the opposite side be $h$ . Then $\tan60^\circ=\frac{h}{4}$ , so $h=4\sqrt3\text{ cm}$ . The area is $\frac12\times4\times4\sqrt3=8\sqrt3\text{ cm}^2$ .
3	No, the left side is $\frac32$ but the right side is $1$ .	3	$2\sin30^\circ+\cos60^\circ=2(\frac12)+\frac12=\frac32$ . However, $\tan45^\circ=1$ . Since $\frac32\ne1$ , Hassan is not correct.
4	$x=1$ or $x=3$	4	Use $\tan45^\circ=1$ , $\sin30^\circ=\frac12$ and $\cos0^\circ=1$ . The equation becomes $x^2-4x+3=0$ . Factorising gives $(x-1)(x-3)=0$ , so $x=1$ or $x=3$ .
5	$12\sqrt6+12\sqrt2\text{ cm}$	4	Let the side adjacent to $60^\circ$ be $x\text{ cm}$ . The opposite side is $x\tan60^\circ=x\sqrt3$ . Hence $\frac12x(x\sqrt3)=48\sqrt3$ , so $x^2=96$ and $x=4\sqrt6$ . The opposite side is $4\sqrt6\times\sqrt3=12\sqrt2$ , and the hypotenuse is $x\div\cos60^\circ=8\sqrt6$ . The perimeter is $4\sqrt6+12\sqrt2+8\sqrt6=12\sqrt6+12\sqrt2\text{ cm}$ .

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]

Tier 1 · Easy

Mark scheme for G22 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	$14\text{ cm}$	2	By the sine rule, $\frac{b}{\sin90^\circ}=\frac{7}{\sin30^\circ}$ . Hence $b=7\times\frac{1}{1/2}=14\text{ cm}$ .
2	$103.8^\circ$	3	By the cosine rule, $15^2=5^2+13^2-2(5)(13)\cos C$ . Therefore $\cos C=\frac{5^2+13^2-15^2}{2(5)(13)}=-\frac{31}{130}$ , so $C=103.795\ldots^\circ=103.8^\circ$ correct to $1$ decimal place.

Tier 2 · Standard

Mark scheme for G22 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	$\sqrt{79}\text{ cm}$	3	Using the cosine rule, $c^2=7^2+10^2-2(7)(10)\cos60^\circ=49+100-70=79$ . Therefore $c=\sqrt{79}\text{ cm}$ .
2	$B=30^\circ$	3	By the sine rule, $\frac{\sin B}{7}=\frac{\sin45^\circ}{7\sqrt2}$ . Hence $\sin B=\frac{7(\sqrt2/2)}{7\sqrt2}=\frac12$ , so $B=30^\circ$ or $150^\circ$ . Since $45^\circ+150^\circ>180^\circ$ , only $B=30^\circ$ is possible.
3	$13.4\text{ cm}$	3	By the sine rule, $\frac{b}{\sin79^\circ}=\frac{8.4}{\sin38^\circ}$ . Hence $b=\frac{8.4\sin79^\circ}{\sin38^\circ}=13.3931\ldots\text{ cm}$ , which is $13.4\text{ cm}$ correct to $1$ decimal place.

Tier 3 · Hard

Mark scheme for G22 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	$B=52.1^\circ$ , $C=92.9^\circ$ $B=127.9^\circ$ , $C=17.1^\circ$	5	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 $C=180^\circ-35^\circ-127.9^\circ=17.1^\circ$ .
2	Third side $=15.6\text{ cm}$ Angle $=33.7^\circ$	5	By the cosine rule, the third side $c$ satisfies $c^2=8^2+10^2-2(8)(10)\cos120^\circ=244$ , so $c=\sqrt{244}=15.620\ldots\text{ cm}$ . Let $B$ be the angle opposite the $10\text{ cm}$ side. By the sine rule, $\frac{\sin B}{10}=\frac{\sin120^\circ}{\sqrt{244}}$ , so $B=\sin^{-1}\left(\frac{10\sin120^\circ}{\sqrt{244}}\right)=33.670\ldots^\circ$ . The supplementary value is impossible with the $120^\circ$ angle. Therefore the answers are $15.6\text{ cm}$ and $33.7^\circ$ .
3	$52.7^\circ$	5	In triangle $ABC$ , the cosine rule gives $AC^2=7^2+9^2-2(7)(9)\cos58^\circ=63.2301\ldots$ . In triangle $ACD$ , $\cos\angle ADC=\frac{6^2+10^2-AC^2}{2(6)(10)}=0.6064\ldots$ . Therefore $\angle ADC=52.6692\ldots^\circ=52.7^\circ$ correct to $1$ decimal place.
4	$x=5$	4	By the cosine rule, $52=(x+1)^2+(x+3)^2-2(x+1)(x+3)\cos60^\circ$ . Since $2\cos60^\circ=1$ , this simplifies to $52=x^2+4x+7$ . Hence $x^2+4x-45=0$ , so $(x+9)(x-5)=0$ . The value $x=-9$ gives negative side lengths, so $x=5$ .
5	$46.5^\circ$	4	The largest angle $C$ is opposite the $11\text{ cm}$ side. The cosine rule gives $\cos C=\frac{7^2+9^2-11^2}{2(7)(9)}=\frac1{14}$ , so $C=85.9039\ldots^\circ$ . The smallest angle $A$ is opposite the $7\text{ cm}$ side, and $\cos A=\frac{9^2+11^2-7^2}{2(9)(11)}=\frac{17}{22}$ , so $A=39.4005\ldots^\circ$ . The difference is $46.5033\ldots^\circ$ , which is $46.5^\circ$ correct to $1$ decimal place.

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

Tier 1 · Easy

Mark scheme for G23 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	$17.5\text{ cm}^2$	2	$A=\frac12ab\sin C=\frac12(10)(7)\sin30^\circ=35\times\frac12=17.5\text{ cm}^2$ .
2	Triangle Q; its area is $20\text{ cm}^2$ compared with $13.5\text{ cm}^2$ for triangle P.	2	Area of P $=\frac12(6)(9)\sin30^\circ=27\times\frac12=13.5\text{ cm}^2$ . Area of Q $=\frac12(5)(8)\sin90^\circ=20\times1=20\text{ cm}^2$ . Since $20>13.5$ , triangle Q has the larger area.

Tier 2 · Standard

Mark scheme for G23 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	$x=16$	3	Use $48=\frac12(12)(x)\sin30^\circ$ . Since $\sin30^\circ=\frac12$ , this becomes $48=3x$ . Therefore $x=16$ .
2	Greatest area $=38.5\text{ cm}^2$ Included angle $=90^\circ$	3	The area is $\frac12(7)(11)\sin C=38.5\sin C$ . The greatest possible value of $\sin C$ is $1$ , which occurs at $C=90^\circ$ . Therefore the greatest area is $38.5\text{ cm}^2$ .
3	$62.2\text{ cm}^2$	4	The area of triangle $ABC$ is $\frac12(7)(12)\sin48^\circ=31.2120\ldots\text{ cm}^2$ . The area of triangle $ACD$ is $\frac12(9)(12)\sin35^\circ=30.9731\ldots\text{ cm}^2$ . Since the triangles are on opposite sides of $AC$ , their areas add to $62.1852\ldots\text{ cm}^2$ , which is $62.2\text{ cm}^2$ correct to $1$ decimal place.

Tier 3 · Hard

Mark scheme for G23 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	$60^\circ$ or $120^\circ$	4	$30\sqrt3=\frac12(10)(12)\sin C=60\sin C$ , so $\sin C=\frac{\sqrt3}{2}$ . Between $0^\circ$ and $180^\circ$ , this occurs at $C=60^\circ$ and $C=120^\circ$ .
2	Each area is $52\sin35^\circ\text{ cm}^2$ , because $\sin145^\circ=\sin35^\circ$ .	3	The first area is $\frac12(8)(13)\sin35^\circ=52\sin35^\circ\text{ cm}^2$ . The second area is $\frac12(8)(13)\sin145^\circ=52\sin145^\circ\text{ cm}^2$ . Since $145^\circ=180^\circ-35^\circ$ and $\sin(180^\circ-35^\circ)=\sin35^\circ$ , the two areas are equal.
3	Included angle $=53.1^\circ$ Third side $=16.6\text{ cm}$	5	$60=\frac12(7.5)(20)\sin C$ , so $\sin C=0.8$ . The angle is acute, so $C=\sin^{-1}(0.8)=53.1301\ldots^\circ$ . By the cosine rule, the third side $c$ satisfies $c^2=7.5^2+20^2-2(7.5)(20)\cos C$ , giving $c=16.6207\ldots\text{ cm}$ . Therefore the answers are $53.1^\circ$ and $16.6\text{ cm}$ correct to $1$ decimal place.
4	$30(\sqrt2-1)\text{ cm}^2$	4	With included angle $30^\circ$ , the area is $\frac12(8)(15)\sin30^\circ=30\text{ cm}^2$ . With included angle $45^\circ$ , the area is $\frac12(8)(15)\sin45^\circ=30\sqrt2\text{ cm}^2$ . The exact increase is $30\sqrt2-30=30(\sqrt2-1)\text{ cm}^2$ .
5	$x=6$	4	The area is $\frac12(10)(x)\sin60^\circ=5x\times\dfrac{\sqrt3}{2}=\dfrac{5\sqrt3}{2}x$ . Setting $\dfrac{5\sqrt3}{2}x=15\sqrt3$ gives $x=15\sqrt3\times\dfrac{2}{5\sqrt3}=6$ .

G24 · Describe translations as 2D vectors

Tier 1 · Easy

Mark scheme for G24 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	$\binom{5}{-5}$	2	Calculate image minus object in each coordinate: $2-(-3)=5$ and $-1-4=-5$ . The translation vector is $\binom{5}{-5}$ .
2	$P'=(1,4)$	1	Add the vector components to the coordinates: $(4-3,-1+5)=(1,4)$ . Therefore $P'=(1,4)$ .

Tier 2 · Standard

Mark scheme for G24 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	$A'=(-3,1)$ $B'=(1,4)$ $C'=(-2,7)$	3	Add $-4$ to every $x$ -coordinate and $3$ to every $y$ -coordinate. This gives $A'=(-3,1)$ , $B'=(1,4)$ and $C'=(-2,7)$ .
2	$Q=(-4,11)$	2	Reverse the translation: subtract $6$ from the image's $x$ -coordinate and add $4$ to its $y$ -coordinate. Therefore $Q=(2-6,7+4)=(-4,11)$ .
3	$\binom{6}{-7}$	3	Using any corresponding pair gives image minus object. For example, $(1-(-5),-6-1)=(6,-7)$ . The other pairs give the same displacement, so the translation vector is $\binom{6}{-7}$ .

Tier 3 · Hard

Mark scheme for G24 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	$k=11$ $\binom{7}{-6}$	4	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 $k-6=5$ , $k=11$ .
2	$\mathbf{a}=\binom{3}{4}$ $\mathbf{b}=\binom{-4}{4}$ Single vector $\binom{-1}{8}$	3	$\mathbf{a}=\binom{5-2}{3-(-1)}=\binom{3}{4}$ and $\mathbf{b}=\binom{1-5}{7-3}=\binom{-4}{4}$ . Successive translations add: the single vector is $\mathbf{a}+\mathbf{b}=\binom{3-4}{4+4}=\binom{-1}{8}$ , and indeed $P=(2,-1)$ maps directly to $(2-1,\,-1+8)=(1,7)$ .
3	No, $A$ and $B$ move by $\binom{-4}{6}$ but $C$ moves by $\binom{-4}{5}$ , so the displacement is not the same for every point.	3	The displacements from $A$ to $A'$ and from $B$ to $B'$ are both $\binom{-4}{6}$ . The displacement from $C$ to $C'$ is $\binom{-4}{5}$ . A translation must move every point by the same vector, so this mapping cannot be a translation.
4	Translation vector $\binom{7}{0}$ $A'=(3,7)$ and $B'=(15,-1)$	4	The midpoint of $AB$ is $\left(\frac{-4+8}{2},\frac{7+(-1)}{2}\right)=(2,3)$ . Its displacement to $(9,3)$ is $\binom{7}{0}$ , which is the translation vector. Adding this vector gives $A'=(3,7)$ and $B'=(15,-1)$ .
5	$k=2$ $y=2x-3$	4	The point $(5,7)$ on the image came from $(5-3,7-k)=(2,7-k)$ on the original line. Since $y=2x+1$ there, $7-k=2(2)+1=5$ , so $k=2$ . Under the translation, $x'=x+3$ and $y'=y+2$ . Thus $y'-2=2(x'-3)+1$ , which simplifies to $y'=2x'-3$ . Therefore the image line is $y=2x-3$ .

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

Tier 1 · Easy

Mark scheme for G25 Tier 1 · Easy
Question	Answer	Mark	Mark scheme
1	$\binom{-4}{-3}$	1	Add corresponding components: $3+(-7)=-4$ and $-5+2=-3$ . Therefore the sum is $\binom{-4}{-3}$ .
2	$\binom{-6}{12}$	1	Multiply both components by $-3$ : $-3\binom{2}{-4}=\binom{-6}{12}$ .

Tier 2 · Standard

Mark scheme for G25 Tier 2 · Standard
Question	Answer	Mark	Mark scheme
1	$\overrightarrow{OM}=\binom{5}{2}$ $\overrightarrow{AM}=\binom{3}{3}$	3	The midpoint position vector is $\frac12(\mathbf{a}+\mathbf{b})=\frac12\binom{10}{4}=\binom{5}{2}$ . Then $\overrightarrow{AM}=\overrightarrow{OM}-\overrightarrow{OA}=\binom{5}{2}-\binom{2}{-1}=\binom{3}{3}$ .
2	$\overrightarrow{CA}=-\mathbf a-\mathbf b$	2	$\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC}=\mathbf a+\mathbf b$ . Reversing the direction gives $\overrightarrow{CA}=-(\mathbf a+\mathbf b)=-\mathbf a-\mathbf b$ .
3	$x=3$ $y=5$	3	Equating the top components gives $2x+4=10$ , so $x=3$ . Equating the bottom components gives $-6+y=-1$ , so $y=5$ .

Tier 3 · Hard

Mark scheme for G25 Tier 3 · Hard
Question	Answer	Mark	Mark scheme
1	Resultant $=\binom{1}{12}$ Finish $=(-2,14)$ Return vector $=\binom{-1}{-12}$	4	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, $\binom{-1}{-12}$ .
2	$\overrightarrow{AB}=3\mathbf a-2\mathbf b=\frac12\overrightarrow{BC}$ , so $A$ , $B$ and $C$ are collinear.	4	$\overrightarrow{AB}=\overrightarrow{OB}-\overrightarrow{OA}=(5\mathbf a-\mathbf b)-(2\mathbf a+\mathbf b)=3\mathbf a-2\mathbf b$ . Also, $\overrightarrow{BC}=\overrightarrow{OC}-\overrightarrow{OB}=(11\mathbf a-5\mathbf b)-(5\mathbf a-\mathbf b)=6\mathbf a-4\mathbf b$ . Hence $\overrightarrow{AB}=\frac12\overrightarrow{BC}$ . The vectors are non-zero scalar multiples in the same direction and share point $B$ , so $A$ , $B$ and $C$ are collinear.
3	$p=2$ $q=3$	4	Equating components gives $2p-q=1$ and $p+3q=11$ . From the first equation, $q=2p-1$ . Substitute into the second: $p+3(2p-1)=11$ , so $7p=14$ and $p=2$ . Therefore $q=2(2)-1=3$ .
4	$\overrightarrow{AB}=\overrightarrow{DC}=\binom{5}{2}$ and $\overrightarrow{AD}=\overrightarrow{BC}=\binom{2}{5}$ , so $ABCD$ is a parallelogram; adjacent sides both have length $\sqrt{29}$ , so it is a rhombus.	4	$\overrightarrow{AB}=\binom{5}{2}$ and $\overrightarrow{DC}=\binom{8-3}{9-7}=\binom{5}{2}$ . Also, $\overrightarrow{AD}=\binom{2}{5}$ and $\overrightarrow{BC}=\binom{2}{5}$ . Therefore both pairs of opposite sides are equal and parallel, so $ABCD$ is a parallelogram. The adjacent side lengths are $\sqrt{5^2+2^2}=\sqrt{29}$ and $\sqrt{2^2+5^2}=\sqrt{29}$ . Hence all four sides are equal and $ABCD$ is a rhombus.
5	$k=10$ $D=(5,7)$	4	The diagonals of a parallelogram bisect each other, so $M$ is the midpoint of $AC$ . Hence $\frac{2+k}{2}=6$ , giving $k=10$ ; the $y$ -coordinate check is $\frac{-1+11}{2}=5$ . Also $\overrightarrow{MD}=-\overrightarrow{MB}$ . Since $\overrightarrow{MB}=\binom{7-6}{3-5}=\binom{1}{-2}$ , $\overrightarrow{MD}=\binom{-1}{2}$ and $D=(6-1,5+2)=(5,7)$ .