Angles: parallel lines, triangles, polygons (interior/exterior) – GCSE Maths exam questions with mark scheme

1Grade 1–3FoundationNon-calculator2 marks

The diagram shows triangle ABC.

(a) Work out the size of the angle marked $x$. [1]

(b) Give a reason for your answer. [1]

(a) $x = 65°$B1
(b) angles in a triangle add up to $180°$ (accept equivalent wording, e.g. 'angles in a triangle sum to 180°')C1
Answer: $x = 65°$; angles in a triangle add up to $180°$
Examiner tip: Write the reason in full words — 'triangle = 180' on its own often loses the reason mark.

2Grade 4–5Non-calculator3 marks

AB and CD are parallel straight lines. PQ is a straight line that crosses both of them.

Work out the size of the angle marked $x$.
Give a reason for each stage of your working. [3]

finds a correct first angle with its position clear, e.g. angle DQP (above CD, right of PQ) $= 64°$, or angle BPQ $= 180 - 64 = 116°$M1
$x = 116°$A1
fully correct reasons for their method, e.g. 'corresponding angles are equal' and 'angles on a straight line add up to $180°$' (or 'co-interior angles add up to $180°$' and 'vertically opposite angles are equal'); 'F angles'/'Z angles' are not acceptedC1
Answer: $x = 116°$
Examiner tip: Use the proper names: 'corresponding', 'alternate', 'co-interior' — 'F', 'Z' and 'C' angles score no reason marks.

3Grade 4–5Non-calculator4 marks

(a) Work out the size of each interior angle of a regular 12-sided polygon. [2]

(b) Each exterior angle of a different regular polygon is $24°$.
How many sides does this polygon have? [2]

(a) $360 \div 12$ $(=30)$ then $180 - 30$, or $(12-2)\times 180 \div 12$ (sum $= 1800$ seen)M1
(a) $150°$A1
(b) $360 \div 24$M1
(b) $15$A1
Answer: (a) $150°$ (b) $15$ sides
Examiner tip: The 360 ÷ n rule gives the EXTERIOR angle — students often stop at 30° and forget to subtract from 180°.

4Grade 6–7HigherNon-calculator4 marks

ABCD is a quadrilateral. The sizes of its angles, in degrees, are shown in the diagram.

Show that AD is parallel to BC. [4]

forms an equation using angle sum of a quadrilateral, e.g. $3x+5+2x+25+4x-12+x+42=360$ (or $10x+60=360$)M1
$x = 30$A1
substitutes to find angle A $= 95°$ and angle B $= 85°$ (or angles D $=72°$ and C $=108°$)M1
$95 + 85 = 180$ so co-interior angles add up to $180°$, hence AD is parallel to BC (conclusion with reason required)C1
Answer: $x=30$; $\angle A = 95°$, $\angle B = 85°$, $95+85=180$ so co-interior angles sum to $180°$ ⇒ AD ∥ BC
Examiner tip: A 'show that' needs a stated conclusion — finish with the angle facts AND the words 'co-interior angles add to 180°, so AD is parallel to BC'.

5Grade 4–5Non-calculator3 marks

The diagram shows a regular pentagon and a regular hexagon. They share the side AB.
E is a vertex of the pentagon and F is a vertex of the hexagon.

Work out the size of the angle marked $y$. [3]

finds one interior angle: pentagon $= 108°$ (e.g. $180 - 360\div5$ or $540 \div 5$) or hexagon $= 120°$; or one exterior angle: pentagon $= 72°$ or hexagon $= 60°$M1
$360 - (108 + 120)$ (both interior angles correct, angles at a point used) or $72 + 60$ (both exterior angles correct)M1
$y = 132°$A1
Answer: $y = 132°$
Examiner tip: Angles around point A add up to 360°, not 180° — check which angles actually meet at A.

6Grade 8–9HigherNon-calculator4 marks

Polygon Q is a regular polygon with $n$ sides.
Polygon P is a regular polygon with twice as many sides as polygon Q.

Each interior angle of P is $30°$ larger than each interior angle of Q.

Work out the value of $n$. [4]

exterior angle of Q $=\dfrac{360}{n}$ (or interior angle $180-\dfrac{360}{n}$)P1
exterior angle of P $=\dfrac{360}{2n}$ (or interior angle $180-\dfrac{180}{n}$)P1
forms a correct equation, e.g. $\dfrac{360}{n}-\dfrac{180}{n}=30$ (the exterior angle of Q is $30°$ larger than that of P)P1
$n = 6$A1
Answer: $n = 6$ (Q is a hexagon, P a regular 12-gon: 120° and 150°)
Examiner tip: If interior angles differ by 30°, the exterior angles also differ by 30° — the other way round. Check: 150 − 120 = 30.

7Grade 6–7HigherNon-calculator5 marks

PQ is parallel to RS. A lies on PQ, and B and C lie on RS.
ABC is an isosceles triangle with AB = AC.

Angle PAB $= 3x$ and angle BAC $= 2x + 12$.

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

angle ABC $= 3x$ (may be marked on the diagram)M1
angle ACB $= 3x$M1
forms an equation, e.g. $3x + 3x + 2x + 12 = 180$ (angles in a triangle), or $3x + (2x + 12) + 3x = 180$ using angle QAC $= 3x$ (angles on a straight line at A)M1
$x = 21$A1
all reasons for their method given, e.g. alternate angles are equal; base angles of an isosceles triangle are equal; angles in a triangle add up to $180°$ (or angles on a straight line add up to $180°$ if the straight-line route is used); 'Z angles' not acceptedC1
Answer: $x = 21$
Examiner tip: Mark each angle you find on the diagram — examiners can then award method marks even if your equation goes wrong.

8Grade 1–3FoundationNon-calculator4 marks

AB and CD are straight lines that cross at the point O.

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

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

(a)(i) $x = 47°$B1
(a)(ii) vertically opposite angles are equal (accept 'opposite angles are equal'; do not accept 'X angles' or 'it is opposite')C1
(b)(i) $y = 133°$B1
(b)(ii) angles on a straight line add up to $180°$ (accept 'sum to 180°'; dependent on a correct or follow-through value of $y$; do not accept 'it is a straight line' on its own)C1
Answer: (a) $x = 47°$ (vertically opposite angles are equal) (b) $y = 133°$ (angles on a straight line add up to $180°$)
Examiner tip: A reason must name the angle fact in full: 'opposite' or 'straight line' on their own score nothing.

9Grade 1–3FoundationNon-calculator5 marks

(a) Write down the mathematical name of a quadrilateral that has exactly one pair of parallel sides. [1]

(b) Write down the special name for an angle of $215°$. [1]

square rectangle rhombus kite parallelogram

Write down the quadrilateral from the list whose four sides are all equal but whose angles are not all right angles. [1]

(d) For a regular octagon, write down
(i) the number of lines of symmetry,
(ii) the order of rotational symmetry. [2]

(a) trapezium (accept trapezoid; ignore 'isosceles' or 'right-angled' if written in front)B1
(b) reflex (accept 'reflex angle')B1
(c) rhombus (do not accept square)B1
(d)(i) $8$B1
(d)(ii) $8$ (accept 'order 8')B1
Answer: (a) trapezium (b) reflex (c) rhombus (d)(i) 8 (ii) 8
Examiner tip: Spelling must be clear enough to recognise the shape, and 'order 8' means it looks the same 8 times in one full turn, not 'every 8°'.

10Grade 4–5Non-calculator3 marks

ABCD is a kite.
AB = AD and CB = CD.

Angle BAD $= 76°$ and angle BCD $= 48°$.

Work out the size of the angle marked $x$. [3]

$360 - 76 - 48$ $(= 236)$, using angles in a quadrilateral add up to $360°$M1
(dep) their $236 \div 2$, using angle ABC = angle ADC (the kite's equal angles)M1
$x = 118°$A1
Answer: $x = 118°$
Examiner tip: The two equal angles of a kite are the ones between the unequal sides, not the angles at A and C, so halve what is left after subtracting 76° and 48°.

11Grade 4–5Calculator4 marks

A hexagon has six angles. The sizes of the angles, in degrees, are

$130 \qquad 150 \qquad x \qquad x \qquad 2x \qquad 2x + 20$

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

(b) Work out the size of the largest angle of the hexagon. [1]

sum of interior angles of a hexagon $= (6 - 2) \times 180 = 720$ (may be seen as $4 \times 180$ or $720$ in an equation)M1
$130 + 150 + x + x + 2x + 2x + 20 = 720$ or $6x + 300 = 720$ (ft their angle sum if it came from a valid method)M1
$x = 70$A1
(b) $160°$ (ft $2 \times$ their $x + 20$ provided this is the largest angle)B1
Answer: (a) $x = 70$ (b) $160°$
Examiner tip: Use $(n-2)\times180$ for the angle sum; using $360°$ or $540°$ for a hexagon loses every mark.

12Grade 4–5Non-calculator4 marks

(a) Ravi says,

'A regular polygon can have interior angles of $100°$.'

Is Ravi correct?
You must show how you get your answer. [2]

(b) Regular hexagons fit together around a point with no gaps.
Show that regular pentagons cannot fit together around a point with no gaps or overlaps. [2]

(a) exterior angle $= 180 - 100 = 80$ and $360 \div 80$ attempted, or $\frac{180(n-2)}{n} = 100$ solved for $n$ (also accept interior angles of a regular quadrilateral ($90°$) and a regular pentagon ($108°$) both found)M1
(a) No, with $360 \div 80 = 4.5$ (or $n = 4.5$) and a statement that the number of sides must be a whole number (or with $100$ shown to lie between $90°$ and $108°$, so there is no whole number of sides between 4 and 5)C1
(b) interior angle of a regular pentagon $= 108°$ (from $540 \div 5$ or $180 - 360 \div 5$)M1
(b) $360 \div 108 = 3.33\ldots$, not a whole number (or $3 \times 108 = 324$ leaves a gap of $36°$, $4 \times 108 = 432 > 360$ overlaps), so they cannot fitC1
Answer: (a) No: exterior angle would be $80°$, $360 \div 80 = 4.5$ sides, not a whole number (b) $108°$; $360 \div 108$ is not a whole number
Examiner tip: 'Show' needs the number AND the conclusion: writing 4.5 without saying it is not a whole number loses the communication mark.

13Grade 6–7Non-calculator3 marks

ABC is a triangle.
PAQ is a straight line through A that is parallel to BC.

The angles of the triangle are $a$, $b$ and $c$, as shown in the diagram.

Prove that $a + b + c = 180°$.
You must give a reason for each stage of your proof. [3]

angle PAB $= b$ and reason: alternate angles are equal (or angle QAC $= c$, alternate angles)M1
both angle PAB $= b$ and angle QAC $= c$ (alternate angles)M1
angle PAB + angle BAC + angle QAC $= 180°$ because angles on a straight line add up to $180°$, so $b + a + c = 180°$: complete proof with all reasons (do not accept 'Z angles'; the proof must not use 'angles in a triangle add up to $180°$')C1
Answer: Angle PAB $= b$ and angle QAC $= c$ (alternate angles are equal); $b + a + c = 180°$ (angles on a straight line add up to $180°$), so $a + b + c = 180°$
Examiner tip: You cannot quote 'angles in a triangle add up to 180°' as a reason: that is the thing you are proving.

14Grade 6–7HigherNon-calculator5 marks

ABCD is a parallelogram.
E is the point on DC such that DE = DA.

Angle DAB $= 116°$.

Work out the size of the angle marked $x$.
You must give a reason for each stage of your working. [5]

angle ADE (angle ADC) $= 180 - 116 = 64°$M1
angle DAE $=$ angle DEA $= (180 - 64) \div 2 = 58°$M1
$x = 116 - 58$ or $x =$ angle DEA (alternate angles)M1
$x = 58°$A1
all reasons for their method, e.g. co-interior angles add up to $180°$ (or adjacent angles of a parallelogram add up to $180°$); base angles of an isosceles triangle are equal; angles in a triangle add up to $180°$; alternate angles are equal (do not accept 'F', 'Z' or 'C angles')C1
Answer: $x = 58°$
Examiner tip: Spot that triangle ADE is isosceles from DE = DA: the equal angles are the two opposite the equal sides, at A and E, not the angle at D.

15Grade 8–9HigherNon-calculator4 marks

The diagram shows part of a regular polygon.
A, B, C and D are four consecutive vertices of the polygon.

The diagonals AC and BD meet at X.
Angle AXB $= 40°$.

Work out the number of sides of the polygon. [4]

uses triangle ABC (or BCD) being isosceles to find a base angle, e.g. angle BCA $= \frac{180 - \text{interior}}{2}$ or $= \frac{e}{2}$ where $e$ is the exterior angle (may be in terms of $n$)P1
angle XBC $=$ angle XCB (by symmetry or congruent triangles ABC and DCB), so angle BXC $= 180 - 2 \times$ base angle, or angle AXB $=$ angle XBC $+$ angle XCB (exterior angle of triangle BXC)P1
exterior angle of the polygon $= 40°$ (or interior angle $= 140°$, or equation such as $2 \times \frac{180 - I}{2} = 40$ leading to $I = 140$)P1
$9$ sides (from $360 \div 40$)A1
Answer: $9$ sides (exterior angle $= 40°$)
Examiner tip: Mark the equal angles in the isosceles triangles on the diagram first: the 40° turns out to equal the exterior angle, which unlocks $360 \div 40$.

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