OxfordAQA International GCSE Mathematics: Angle facts, polygons and quadrilaterals (9260)

This guide follows the OxfordAQA International GCSE Mathematics (9260) specification, Version 5.1 (for exams May/June 2018 onwards). It covers references G1 to G5 in section 3.3.1, Properties and constructions. All of this content is Core, and the specification says Core content can also be assessed on the Extension tier papers, so it applies to both tiers. There is no Extension only content in this unit.

When you have worked through it, test yourself with the revision notes and the practice questions. The course page is OxfordAQA IGCSE Mathematics, the printable list is the checklist, and you can find weak spots with the free diagnostics.

What this unit covers

Reference	What you must be able to do
G1	Use conventional terms and notation; label and refer to the sides and angles of triangles
G2	Use angles at a point, on a straight line and vertically opposite angles
G3	Use angle properties of parallel lines, triangles and quadrilaterals; name triangle types
G4	Calculate and use interior and exterior angle sums of polygons, including regular polygons
G5	Recall properties and definitions of the special quadrilaterals

Symmetry, congruence and circle theorems are separate units. This guide does not teach them.

G1: Terms and notation

A point has position only. A line is straight and has no thickness; a line segment is the part between two points. An edge is a side of a polygon, or where two faces of a solid meet. Parallel lines never meet and are marked with matching arrowheads. Perpendicular lines meet at a right angle, marked with a small square. A plane is a flat surface. A polygon is a closed shape with straight edges. Its corners are vertices. A regular polygon has all sides equal and all angles equal.

Notation to use:

Written	Meaning
AB	the line from A to B, or its length
AB ∥ CD	AB is parallel to CD
AB ⊥ CD	AB is perpendicular to CD
∠ABC or angle ABC	the angle at B, between BA and BC
triangle ABC	the triangle with vertices A, B and C

The middle letter of an angle name is the vertex. In triangle ABC, the side opposite A is labelled a, the side opposite B is b and the side opposite C is c. Equal sides carry matching tick marks.

G2: Angles at a point, on a line and vertically opposite

• Angles at a point add up to 360°.
• Angles on a straight line add up to 180°.
• Where two straight lines cross, the angles opposite each other are equal. They are vertically opposite.
• Perpendicular lines make four right angles, each 90°.

Example 1. Four angles meet at a point: 3a, (a + 35), (2a − 15) and 100°. Find a and the largest angle.

3a + (a + 35) + (2a − 15) + 100 = 360    (angles at a point)
6a + 120 = 360
6a = 240, so a = 40
Angles: 120°, 75°, 65°, 100°

The largest angle is 120°. Check: 120 + 75 + 65 + 100 = 360.

Example 2. Two straight lines cross. One angle is (5y − 20)° and the angle vertically opposite is (2y + 55)°. Find the other two angles.

5y − 20 = 2y + 55    (vertically opposite)
3y = 75, so y = 25
5y − 20 = 105°
Other angles: 180 − 105 = 75°    (angles on a straight line)

The four angles are 105°, 75°, 105° and 75°.

G3: Parallel lines, triangles and quadrilaterals

Parallel lines

A line that crosses a pair of parallel lines is called a transversal. It creates three pairs of related angles. The specification expects you to use the names alternate, corresponding and interior, and not colloquial names such as “Z angles”.

Type	Relationship	Look for
Corresponding	equal	the same position at each crossing, like an F shape
Alternate	equal	opposite sides of the transversal, between the parallel lines
Interior	add up to 180°	the same side of the transversal, between the parallel lines

Interior angles are also called co-interior angles. Always name the property you use.

Example 3. Two corresponding angles on parallel lines are (7a − 12)° and (4a + 30)°. Find a and the angle.

7a − 12 = 4a + 30    (corresponding angles are equal)
3a = 42, so a = 14
Angle = 7 × 14 − 12 = 86°

Example 4. Two interior angles on parallel lines are (3b + 20)° and (5b − 40)°. Find both.

(3b + 20) + (5b − 40) = 180    (interior angles)
8b − 20 = 180
8b = 200, so b = 25
Angles: 3 × 25 + 20 = 95° and 5 × 25 − 40 = 85°

For forming and solving equations like these, see the algebra practice.

Triangles

• Angles in a triangle add up to 180°.
• An exterior angle of a triangle equals the sum of the two interior opposite angles. This follows because the exterior angle and the interior angle beside it lie on a straight line.

Triangle	Sides	Angles
Equilateral	all 3 equal	all 60°
Isosceles	2 equal	the 2 angles opposite the equal sides are equal
Scalene	none equal	none equal
Right-angled		one angle is 90°
Acute-angled		all angles below 90°
Obtuse-angled		one angle above 90°

A triangle can be two types at once, such as isosceles and right-angled.

Example 5. Triangle ABC has AB = AC and angle BAC = 48°. Side BC is extended to D. Find angle ACD.

Angle ABC = angle ACB = (180 − 48) ÷ 2 = 66°    (isosceles triangle)
Angle ACD = 180 − 66 = 114°                      (angles on a straight line)

Check with the exterior angle fact: 48° + 66° = 114°.

Quadrilaterals

Angles in a quadrilateral add up to 360°. Drawing one diagonal makes two triangles, each with an angle sum of 180°, which gives 2 × 180° = 360° in total.

Example 6. Angles in a quadrilateral are 2x°, (x + 40)°, (3x − 25)° and 105°. Find x and the smallest angle.

2x + (x + 40) + (3x − 25) + 105 = 360
6x + 120 = 360
6x = 240, so x = 40
Angles: 80°, 80°, 95°, 105°

The smallest angle is 80°. Check: 80 + 80 + 95 + 105 = 360.

G4: Interior and exterior angles of polygons

A polygon with n sides can be split into (n − 2) triangles from one vertex.

Result	Formula
Sum of interior angles	(n − 2) × 180°
Sum of exterior angles	360°
Exterior angle of a regular polygon	360° ÷ n
Interior angle of a regular polygon	180° − exterior angle
Angle at the centre of a regular polygon	360° ÷ n

An interior angle and the exterior angle at the same vertex lie on a straight line, so they add up to 180°. The angle at the centre is the angle between two lines from the centre to neighbouring vertices.

Example 7. Find the sum of the interior angles of an 11-sided polygon.

(11 − 2) × 180 = 9 × 180 = 1620°

Example 8. Six angles of a heptagon are 118°, 134°, 126°, 141°, 152° and 129°. Find the seventh.

Sum for 7 sides = 5 × 180 = 900°
Six angles total 800°
Seventh angle = 900 − 800 = 100°

Example 9. A regular polygon has an exterior angle of 18°. Find the number of sides and the interior angle.

n = 360 ÷ 18 = 20 sides
Interior angle = 180 − 18 = 162°

Example 10. The interior angle of a regular polygon is 150°. How many sides has it?

Exterior angle = 180 − 150 = 30°
n = 360 ÷ 30 = 12 sides

Working through the exterior angle is much quicker than using (n − 2) × 180 ÷ n = 150 as an equation.

Example 11. Find the angle at the centre of a regular 9-sided polygon, and each interior angle.

Angle at centre = 360 ÷ 9 = 40°
Exterior angle = 40°
Interior angle = 180 − 40 = 140°

G5: Special quadrilaterals

Shape	Sides	Angles	Diagonals
Square	4 equal; opposite sides parallel	all 90°	equal; bisect each other at 90°; bisect the angles
Rectangle	opposite sides equal and parallel	all 90°	equal; bisect each other
Parallelogram	opposite sides equal and parallel	opposite angles equal	bisect each other
Rhombus	4 equal; opposite sides parallel	opposite angles equal	bisect each other at 90°; bisect the angles
Kite	2 pairs of adjacent equal sides	one pair of opposite angles equal	cross at 90°; one is bisected
Trapezium	one pair of parallel sides	the two angles at the ends of each non-parallel side add up to 180°	no special property

An isosceles trapezium is a trapezium with its non-parallel sides equal. Its base angles are equal in pairs.

Example 12. ABCD is a rhombus. Angle DAB = 68°. The diagonals meet at M. Find angles ABC and ABD.

Angle ABC = 180 − 68 = 112°     (interior angles, AD ∥ BC)
Diagonal BD bisects angle ABC
Angle ABD = 112 ÷ 2 = 56°

Check using triangle ABD: AB = AD, so angle ABD = (180 − 68) ÷ 2 = 56°.

Example 13. Parallelogram PQRS has angle P = (3z + 20)° and angle Q = (2z + 10)°. Find z and angle R.

Angle P + angle Q = 180    (adjacent angles between parallel sides)
5z + 30 = 180, so z = 30
Angle P = 110°, angle Q = 70°
Angle R = angle P = 110°    (opposite angles equal)

Common errors

• Giving a general reason such as “angles in a triangle” when the question needs the exact fact, such as “base angles of an isosceles triangle”.
• Treating alternate angles as adding up to 180°. They are equal. Interior angles add up to 180°.
• Using (n − 2) × 180 for an exterior angle. Exterior angles of any polygon total 360°.
• Dividing 360 by the interior angle to find n. Divide by the exterior angle.
• Calling a shape a rhombus because the sides are equal when a diagram only shows two equal sides.
• Assuming angles are as drawn. Diagrams are not always to scale, so use only the given facts.

Next steps

Use the revision notes for quick recall, then the practice questions. Try a free diagnostic afterwards to see what to revisit.

Official syllabus

OxfordAQA International GCSE Mathematics (9260), Version 5.1, for exams May/June 2018 onwards, published by OxfordAQA. Section 3.3.1, references G1 to G5.