Cambridge IGCSE Mathematics 0580 – Geometry Notes (2025‑2027)
1. Terminology Box (C4.1)
| Term |
Definition / Example |
| Point |
A location in space with no size (e.g., A, B). |
| Line |
Infinite straight path extending in both directions; denoted by two points (AB) or a single letter with a bar (‾AB‾). |
| Ray |
Starts at a point and extends infinitely in one direction (e.g., AB ray starts at A). |
| Plane |
A flat, two‑dimensional surface that extends infinitely in all directions (e.g., the surface of a sheet of paper). |
| Angle |
Space between two rays sharing a common endpoint; measured in degrees (°). |
| Polygon |
Closed figure formed by straight line segments (e.g., triangle, quadrilateral). |
| Solid (3‑D shape) |
Cube, cuboid, prism, cylinder, pyramid, cone, sphere, hemisphere, frustum. |
| Bearing (three‑figure) |
Direction measured clockwise from north; always written with three digits (e.g., 045° = 45° east of north). |
| Scale factor |
Ratio of a length in a drawing to the corresponding real length (e.g., scale 1 cm : 5 m → factor = 1/500). |
| Similarity |
Two figures have the same shape but possibly different sizes; corresponding angles equal, corresponding sides proportional. |
| Line‑symmetry |
A figure can be reflected in a line and coincide with itself (e.g., a rectangle has two lines of symmetry). |
| Rotational symmetry |
A figure can be rotated about its centre and match its original position. Order n means it matches after a rotation of 360°/n (e.g., a regular hexagon has order 6). |
| Perpendicular bisector (of a segment) |
A line that is at right‑angles to a segment and passes through its midpoint. |
| Angle bisector (of an angle) |
A ray that divides an angle into two equal smaller angles. |
2. Angles (C4.6)
3. Parallel Lines and Transversals (C4.2)
When a transversal cuts two parallel lines, four angle relationships arise.
| Relationship |
Equality / Sum |
| Corresponding angles |
∠₁ = ∠₂ |
| Alternate interior angles |
∠₃ = ∠₄ |
| Alternate exterior angles |
∠₅ = ∠₆ |
| Consecutive interior (same‑side interior) angles |
∠₇ + ∠₈ = 180° |
These properties are used to prove lines are parallel or to find unknown angles.
4. Polygons (C4.4)
5. Circles (C4.7)
- Basic definitions
- Radius \(r\) – distance from centre to any point on the circle.
- Diameter \(d\) – twice the radius, \(d=2r\).
- Circumference \(C\) – \(C=2\pi r=\pi d\).
- Area \(A\) – \(A=\pi r^{2}\).
- Chord – segment joining two points on the circle.
- Arc – part of the circumference between two points.
- Sector – region bounded by two radii and the intercepted arc.
- Angle theorems
- Central angle = measure of the intercepted arc (in degrees).
- Inscribed angle = ½ × central angle.
\[
\angle_{\text{inscribed}}=\tfrac12\;\angle_{\text{central}}
\]
- Tangent‑chord theorem: the angle between a tangent and a chord equals ½ × the intercepted arc.
\[
\angle_{\text{tangent‑chord}}=\tfrac12\;\text{arc opposite the chord}
\]
- Why the theorems work (proof outline)
- Draw the radii to the ends of the intercepted arc; they form an isosceles triangle.
- Show that the exterior angle at the centre equals the sum of the two base angles.
- Since the base angles are equal, each equals half the central angle – giving the inscribed‑angle result.
- For the tangent‑chord case, extend the radius to the point of tangency; the radius is perpendicular to the tangent, creating a right‑angled triangle that leads to the same ½‑arc relationship.
- Arc length and sector area
- Arc length \(l = \dfrac{\theta}{360^\circ}\times 2\pi r\).
- Sector area \(A_{\text{sector}} = \dfrac{\theta}{360^\circ}\times \pi r^{2}\).
Example: In a circle of radius 5 cm, find the length of a 60° arc and the area of the corresponding sector.
Solution:
- Arc length \(l = \frac{60}{360}\times2\pi(5)=\frac{1}{6}\times10\pi\approx5.24\text{ cm}\).
- Sector area \(A_{\text{sector}} = \frac{60}{360}\times\pi(5)^{2}= \frac{1}{6}\times25\pi\approx13.1\text{ cm}^{2}\).
6. Constructions (C4.2)
All constructions use only a straight‑edge (unmarked ruler) and a compass.
Standard construction checklist
- Perpendicular bisector of a segment AB (required)
- With the compass at A, draw arcs above and below AB.
- Without changing the radius, repeat from B; the arcs intersect at C and D.
- Draw CD – this is the perpendicular bisector.
- Angle bisector of ∠ABC (required)
- Place the compass at B and draw an arc cutting both sides of the angle.
- From the two intersection points, draw arcs of equal radius that intersect at P.
- Draw BP – it bisects ∠ABC.
- Triangle from three given sides (SSS) (required)
- Draw the longest side as base AB.
- With centre A and radius equal to the second side, draw an arc.
- With centre B and radius equal to the third side, draw a second arc intersecting the first at C.
- Join AC and BC – triangle ABC is constructed.
- Regular polygon (n‑gon) inscribed in a circle (required)
- Draw a circle with centre O.
- Set the compass to the radius.
- Starting from a point on the circle, step around the circumference marking n points (where n = number of sides).
- Join successive points – the result is a regular n‑gon.
Extension constructions (useful enrichment)
- Parallel line through a point P not on line l (using corresponding angles).
- Perpendicular line through a point on a given line.
- Nets of a cuboid or prism (see Section 9 for a worked example).
7. Scale Drawings & Bearings (C4.3)
Scale drawings
- Scale is written as “1 cm : k units” or “1 unit = k units”.
- To convert a real length \(L\) to a drawing length \(d\):
\[
d = \frac{L}{k}
\] (same units as the drawing).
- Area scales with the square of the linear scale:
\[
\text{Area}_{\text{drawing}} = \frac{\text{Area}_{\text{real}}}{k^{2}}
\]
- Example: A garden is 120 m long. On a 1 cm : 5 m scale, the length on paper is \(d = \frac{120}{5}=24\) cm.
Bearing basics
8. Similarity (C4.4)
- Definition: Two figures are similar if all corresponding angles are equal and the ratios of all corresponding sides are the same (scale factor \(k\)).
- Criteria for triangles
- AA (two angles equal) → triangles are similar.
- SAS (two sides in proportion and the included angle equal).
- SSS (all three sides in proportion).
- Using the scale factor
If \(k = \dfrac{\text{corresponding side in larger figure}}{\text{corresponding side in smaller figure}}\), then every length in the larger figure is \(k\) times the matching length in the smaller.
- Worked example
In similar triangles ΔABC and ΔA'B'C', AB = 6 cm, AC = 9 cm, and A'B' = 10 cm. Find A'C'.
Solution:
- Scale factor \(k = \dfrac{A'B'}{AB} = \dfrac{10}{6}= \dfrac{5}{3}\).
- Therefore \(A'C' = k \times AC = \dfrac{5}{3}\times9 = 15\) cm.
9. Symmetry (C4.5)
10. Solids (C4.1 – vocabulary & calculations)
- Cube – six equal square faces.
- Cuboid (rectangular prism) – opposite faces equal rectangles.
- Prism – two parallel, congruent bases joined by rectangular faces.
- Cylinder – circular bases, curved surface.
- Pyramid – polygonal base and triangular faces meeting at a vertex.
- Cone – circular base and a single curved surface meeting at a vertex.
- Sphere – set of points equidistant from a centre.
- Hemisphere – half of a sphere.
- Frustum – portion of a cone or pyramid cut by a plane parallel to the base.
Surface‑area and volume formulas
| Solid |
Surface Area (SA) |
Volume (V) |
| Cube (side \(a\)) |
\(6a^{2}\) |
\(a^{3}\) |
| Cuboid (length \(l\), width \(w\), height \(h\)) |
\(2(lw+lh+wh)\) |
\(lwh\) |
| Cylinder (radius \(r\), height \(h\)) |
\(2\pi r(h+r)\) |
\(\pi r^{2}h\) |
| Sphere (radius \(r\)) |
\(4\pi r^{2}\) |
\(\dfrac{4}{3}\pi r^{3}\) |
| Cone (radius \(r\), height \(h\)) |
\(\pi r(r+\sqrt{r^{2}+h^{2}})\) |
\(\dfrac{1}{3}\pi r^{2}h\) |
| Pyramid (base area \(B\), height \(h\)) |
\(B + \dfrac{1}{2}Pl\) (where \(P\) is perimeter of base, \(l\) slant height) |
\(\dfrac{1}{3}Bh\) |
Net‑to‑solid example (cuboid)
- Draw a rectangle for the base (e.g., 6 cm × 4 cm).
- Attach four rectangles of size 6 cm × 3 cm (height = 3 cm) to each side of the base – these become the lateral faces.
- Attach a second 6 cm × 4 cm rectangle opposite the base – this is the top.
- Fold along the edges; the net folds into a 6 cm × 4 cm × 3 cm cuboid.
- Check using the formulas:
SA = 2(6·4 + 6·3 + 4·3) = 2(24+18+12)=108 cm²,
V = 6·4·3 = 72 cm³.
11. Summary Checklist
- Know all geometric terminology, including the newly added terms “plane”, “perpendicular bisector”, and the three‑figure bearing rule.
- Recall the four angle relationships for a transversal of parallel lines.
- Calculate interior and exterior angle sums for any polygon; apply the exterior‑angle theorem.
- Use circle formulas: circumference, area, arc length, sector area, and the three main theorems (central, inscribed, tangent‑chord) with a brief proof idea.
- Perform the required constructions (perpendicular bisector, angle bisector, SSS triangle, regular polygons) and recognise extension constructions.
- Convert between real dimensions and scaled drawings; read and write three‑figure bearings accurately.
- Apply similarity criteria (AA, SAS, SSS) and use the scale factor to find missing lengths.
- Identify line‑symmetry and rotational symmetry; visualise symmetry lines for common shapes.
- Recall surface‑area and volume formulas for all listed solids; be able to sketch a net and relate it to the solid.