Angles, parallel lines, triangles, polygons & symmetry – IGCSE Maths exam questions with mark scheme

1Grade 1–3FoundationCalculator4 marks

(a) Write down the type of angle for each of these sizes.

(i) $128^\circ$ (ii) $213^\circ$ (iii) $64^\circ$ [2]

(b) A quadrilateral has exactly one pair of parallel sides.
Write down the mathematical name of this quadrilateral. [1]

(c) A different quadrilateral has four sides of equal length but none of its angles is a right angle.
Write down the mathematical name of this quadrilateral. [1]

(a) at least two of (i) obtuse, (ii) reflex, (iii) acute correctB1
(a) all three correct: (i) obtuse (ii) reflex (iii) acuteB1
(b) trapezium (accept trapezoid; ignore minor misspelling if the word is recognisable)B1
(c) rhombus (do not accept diamond or square)B1
Answer: (a) (i) obtuse (ii) reflex (iii) acute; (b) trapezium; (c) rhombus
Examiner tip: An angle between $180^\circ$ and $360^\circ$ is reflex, not obtuse, and 'diamond' is not a mathematical name, so it scores nothing in (c).

2Grade 1–3FoundationCalculator4 marks

The diagram shows two parallel straight lines crossed by a third straight line at the points $P$ and $Q$.

(a) (i) Write down the value of $x$.
(ii) Give a reason for your answer. [2]

(b) (i) Work out the value of $y$.
(ii) Give a reason for your answer. [2]

(a)(i) $x = 62$B1
(a)(ii) alternate angles are equal (accept 'alternate'; do not accept 'Z angles' or 'corresponding')B1
(b)(i) $y = 118$ (ft $180 - x$ from (a))B1
(b)(ii) co-interior (allied) angles add up to $180^\circ$, or angles on a straight line add up to $180^\circ$ (using $x$); accept 'interior angles' if 'add to 180' is statedB1
Answer: (a) $x = 62$ (alternate angles are equal); (b) $y = 118$ (co-interior angles add up to $180^\circ$)
Examiner tip: The reason must name the angle fact in words: 'Z angles' or just 'parallel lines' does not earn the reason mark.

3Grade 4–5Calculator4 marks

Each interior angle of a regular polygon is $156^\circ$.

(a) Work out the number of sides of this polygon. [2]

Kemal says, 'I have drawn a regular polygon in which each interior angle is $145^\circ$.'

(b) Explain why Kemal cannot be correct. [2]

(a) $180 - 156 = 24$ (exterior angle) or $360 \div (180-156)$, or $\frac{(n-2)\times180}{n} = 156$ set upM1
(a) $15$A1
(b) exterior angle $180 - 145 = 35$ and $360 \div 35$ attempted (or $\frac{(n-2)\times 180}{n} = 145$ solved)M1
(b) $360 \div 35 = 10.28...$ (accept $10.3$ or $\frac{72}{7}$), which is not a whole number, so the number of sides cannot be found / polygon cannot existC1
Answer: (a) 15 sides; (b) exterior angle would be $35^\circ$ and $360 \div 35 = 10.28...$ is not a whole number
Examiner tip: Work with the exterior angle: number of sides $= 360 \div$ exterior angle, and the explanation must say the result is not a whole number.

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4Grade 1–3FoundationCalculator3 marks

Complete the table.

Shape	Number of lines of symmetry	Order of rotational symmetry
kite	1	1
parallelogram (not a rectangle or rhombus)
rhombus (not a square)
regular octagon

[3]

parallelogram: 0 lines of symmetry and order 2B1
rhombus: 2 lines of symmetry and order 2B1
regular octagon: 8 lines of symmetry and order 8B1
Answer: parallelogram: 0, 2; rhombus: 2, 2; regular octagon: 8, 8
Examiner tip: A parallelogram has no lines of symmetry (its diagonals are not mirror lines), and the smallest order of rotational symmetry is 1, never 0.

5Grade 4–5Calculator4 marks

$ABC$ is a triangle in which $AB = AC$.

$BCD$ is a straight line.

Angle $ACD = 116^\circ$

Work out the size of angle $BAC$.
Give a reason for each stage of your working. [4]

angle $ACB = 180 - 116 = 64$ (may be marked on diagram)M1
angle $ABC = 64$ and $180 - 2\times64$ attempted, or $116 - 64$ (exterior angle)M1
$52$A1
reasons appropriate to the method used, e.g. angles on a straight line add up to $180^\circ$ AND base angles of an isosceles triangle are equal AND angles in a triangle add up to $180^\circ$ (or exterior angle of a triangle equals the sum of the two interior opposite angles)C1
Answer: angle $BAC = 52^\circ$
Examiner tip: Every step needs its own reason in full: 'isosceles' alone is not enough, say 'base angles of an isosceles triangle are equal'.

6Grade 4–5Calculator5 marks

$ABCD$ is a quadrilateral.

The sizes of its angles, in degrees, are shown on the diagram.

(a) Work out the value of $x$. [3]

(b) Show that $AB$ is parallel to $DC$.
Give a reason for your answer. [2]

(a) $4x + 5 + 3x + 35 + 2x + 20 + 5x - 50 = 360$ (all four angles summed and equated to $360$)M1
(a) $14x + 10 = 360$ or $14x = 350$M1
(a) $x = 25$A1
(b) angle $A = 105$ and angle $D = 75$ (or angle $B = 110$ and angle $C = 70$), ft their $x$M1
(b) $105 + 75 = 180$ (or $110 + 70 = 180$) with reason: co-interior (allied) angles add up to $180^\circ$, so $AB$ is parallel to $DC$; dep on correct $x$C1
Answer: (a) $x = 25$; (b) $A = 105^\circ$, $D = 75^\circ$, $105 + 75 = 180$, co-interior angles add up to $180^\circ$ so $AB \parallel DC$
Examiner tip: In (b) you must calculate a pair of angles between the two lines and state the co-interior angle fact: angles that just look right score nothing.

7Grade 6–7HigherCalculator5 marks

$ABCDEF$ is a regular hexagon.

$ABGHI$ is a regular pentagon.

$F$ is joined to $I$ by a straight line.

Work out the size of angle $AFI$.
Give a reason for each stage of your working. [5]

interior angle of hexagon $= 120$ (e.g. $180 - 360\div6$ or $720 \div 6$)M1
interior angle of pentagon $= 108$ (e.g. $180 - 360\div5$ or $540\div5$)M1
angle $FAI = 360 - 120 - 108\ (= 132)$M1
$24$A1
reasons: angles at a point add up to $360^\circ$ AND triangle $AFI$ is isosceles because $AF = AB = AI$ (sides of regular polygons with the common side $AB$) so base angles are equal AND angles in a triangle add up to $180^\circ$ (accept a correct method for the interior angles of a regular polygon as the reason for 120 and 108)C1
Answer: angle $AFI = 24^\circ$
Examiner tip: The key step is spotting that $AF = AI$ because both equal $AB$; without stating why triangle $AFI$ is isosceles the reason mark is lost.

8Grade 8–9HigherCalculator5 marks

$P$ is a regular polygon with $n$ sides.

$Q$ is a regular polygon with $(n + 3)$ sides.

Each interior angle of $Q$ is $6^\circ$ larger than each interior angle of $P$.

(a) Show that $n^2 + 3n - 180 = 0$ [3]

(b) Work out the number of sides of polygon $Q$. [2]

(a) exterior angles $\frac{360}{n}$ and $\frac{360}{n+3}$, or interior angles $180 - \frac{360}{n}$ and $180 - \frac{360}{n+3}$, or $\frac{180(n-2)}{n}$ and $\frac{180(n+1)}{n+3}$M1
(a) correct equation, e.g. $\frac{360}{n} - \frac{360}{n+3} = 6$ or $\frac{180(n+1)}{n+3} - \frac{180(n-2)}{n} = 6$M1
(a) completes with no errors, e.g. $360(n+3) - 360n = 6n(n+3) \Rightarrow 1080 = 6n^2 + 18n \Rightarrow n^2 + 3n - 180 = 0$A1
(b) solves $n^2 + 3n - 180 = 0$, e.g. $(n + 15)(n - 12) = 0$ or $n = 12$ (ignore $-15$)M1
(b) $15$ (sides of $Q$); $12$ alone scores M1 A0A1
Answer: (a) shown; (b) $Q$ has 15 sides
Examiner tip: Use exterior angles: they make the algebra much easier, and in (b) remember to add 3 to $n$ because the question asks for $Q$.

9Grade 6–7HigherCalculator5 marks

$AB$ is parallel to $CD$.

$E$ is a point between the two parallel lines.

Angle $ABE = (x + 18)^\circ$, angle $CDE = (2x - 6)^\circ$ and angle $BED = 105^\circ$

Work out the value of $x$.
Give a reason for each stage of your working. [5]

a line through $E$ parallel to $AB$ drawn or implied, splitting angle $BED$ into $(x+18)$ and $(2x-6)$; or $BE$ extended to meet $CD$ with the angle there marked $(x+18)$M1
$x + 18 + 2x - 6 = 105$ (oe)M1
$3x + 12 = 105$ or $3x = 93$M1
$x = 31$A1
reasons appropriate to the method: alternate angles are equal (both pairs, with the parallel line through $E$); or alternate angles are equal AND the exterior angle of a triangle equals the sum of the interior opposite anglesC1
Answer: $x = 31$
Examiner tip: Draw a line through $E$ parallel to $AB$: then angle $BED$ splits into two alternate angles, and 'alternate angles are equal' is the reason you must write.

10Grade 4–5FoundationCalculator3 marks

Here are four triangles, $\mathbf{A}$, $\mathbf{B}$, $\mathbf{C}$ and $\mathbf{D}$.

(a) Two of the triangles are congruent.
Write down the letters of these two triangles. [1]

(b) Lena says, 'Triangles $\mathbf{A}$ and $\mathbf{D}$ are congruent because they both have a $50^\circ$ angle and a $6$ cm side.'
Explain why Lena is wrong. [1]

(c) Omar says, 'If the diagonals of a quadrilateral cross at right angles, the quadrilateral must be a rhombus.'
Is Omar correct? Give a reason for your answer. [1]

(a) $\mathbf{A}$ and $\mathbf{C}$ (either order)B1
(b) the other side next to the $50^\circ$ angle is $9$ cm in $\mathbf{A}$ but $8$ cm in $\mathbf{D}$, so they are not the same size (accept 'one side is 9 cm and the other is 8 cm'; do not accept 'they look different')B1
(c) No, with a correct counter-example, e.g. a kite also has diagonals that cross at right angles (a square is not a counter-example because a square is a rhombus)C1
Answer: (a) A and C; (b) A has sides 9 cm and 6 cm with the 50° angle between them but D has 8 cm and 6 cm, so they are not the same size; (c) No, e.g. a kite also has perpendicular diagonals
Examiner tip: Congruent means exactly the same shape and size, so every matching side and angle must be equal, even if one triangle is turned round.

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