IB MYP Mathematics: Geometry: angles, circles, area, volume and metric measure (MYP Mathematics)

This guide teaches the geometry strand of the spatial reasoning branch of the IB Mathematics guide (for use from September 2020/January 2021), first assessment 2022. It is organised by framework column (MYP 1-3, then the standard and extended columns for MYP 4-5) and adds the enrichment list. MYP has no SL/HL split, so extended and enrichment items are labelled where they appear. Coordinates, transformations, similarity, congruence and trigonometry have their own units.

Use it with the revision notes and the practice set. The course hub is IB MYP Mathematics, there is a printable checklist, and you can test yourself with the free diagnostics. For how tasks are judged against criteria A to D, see criteria in practice.

What this unit covers

Framework column	What you must be able to do
MYP 1-3	Classify shapes and angles; use angle properties; angles in intersecting and parallel lines; perimeter (circumference) and area of plane figures and compound shapes; volume, surface area and nets of cubes, cuboids, cylinders and prisms
MYP 4-5 standard	Metric conversions; circle geometry and theorems (angles, radius, diameter, tangent); arc, chord, sector and segment; volume, surface area and nets of pyramids, cones and compound solids
MYP 4-5 extended	Capacity
Enrichment	Informal fractals; shapes drawn inside or around other shapes; arc and sector work in radians

The topics and skills that run through this unit are construction and manual skills, visualising and representing 3D shapes, and moving between dimensions. Criteria A (knowing and understanding) and D (applying mathematics in real-life contexts) are the usual home for these skills.

Classifying shapes and angle properties

Angles are acute (under 90°), right (90°), obtuse (between 90° and 180°), straight (180°) or reflex (over 180°). Triangles are classified by sides (equilateral, isosceles, scalene) and by angles. Quadrilaterals include the square, rectangle, parallelogram, rhombus, kite and trapezoid, each with its own symmetry and diagonal properties.

Key facts: a straight line holds 180°, a full turn holds 360°, a triangle’s three angles total 180° and a quadrilateral’s four total 360°. For an n-sided polygon the interior angles add to (n − 2) × 180°. For a regular polygon each exterior angle is 360°/n, and interior plus exterior is 180°.

Worked example 1. Each interior angle of a regular polygon is 156°. Find the number of sides.

Exterior angle = 180° − 156° = 24°. Number of sides = 360 ÷ 24 = 15.

Intersecting and parallel lines

Where two lines cross, vertically opposite angles are equal. A transversal across parallel lines gives corresponding angles equal (F shape), alternate angles equal (Z shape) and co-interior angles adding to 180° (C or U shape).

Worked example 2. Two parallel lines are cut by a transversal. Co-interior angles measure (3x + 10)° and (2x + 30)°. Find both angles.

Co-interior angles add to 180°, so 5x + 40 = 180, giving 5x = 140 and x = 28. The angles are 3(28) + 10 = 94° and 2(28) + 30 = 86°. Check: 94 + 86 = 180.

Perimeter and area of plane figures

Figure	Area	Perimeter or circumference
Rectangle	length × width	2(l + w)
Parallelogram	base × perpendicular height	sum of sides
Triangle	½ × base × height	sum of sides
Trapezoid	½(a + b)h, a and b parallel	sum of sides
Rhombus or kite	½ × d₁ × d₂ (diagonals)	sum of sides
Circle	πr²	2πr

Worked example 3. A rhombus has diagonals 10 cm and 24 cm. Find its area and perimeter.

Area = ½ × 10 × 24 = 120 cm². The diagonals bisect each other at right angles, so each side is the hypotenuse of a triangle with legs 5 and 12: side = √(25 + 144) = 13 cm. Perimeter = 4 × 13 = 52 cm.

Worked example 4 (compound shape). A noticeboard is a rectangle 90 cm wide and 60 cm high with a semicircular top of diameter 90 cm. Find its face area and the length of trim around the edge.

Radius = 45 cm. Area = 90 × 60 + ½ × π × 45² = 5400 + 3180.9 = 8581 cm² (4 s.f.). Trim: base 90, two sides of 60 and a semicircular arc of π × 45 = 141.4. Total = 210 + 141.4 = 351 cm (3 s.f.).

Method for compound shapes: split into shapes you know, or subtract a removed part from a larger shape, then add or subtract areas. For a perimeter, trace only the outer boundary and leave out any internal lines.

Metric conversions and capacity

Length steps are ×10 (cm to mm), ×100 (m to cm) and ×1000 (km to m). Area factors are the length factor squared: 1 m² = 10 000 cm². Volume factors are the length factor cubed: 1 m³ = 1 000 000 cm³.

Capacity (extended). 1 cm³ holds 1 mL, 1000 cm³ holds 1 L, and 1 m³ holds 1000 L.

Worked example 5. (a) Convert 2.5 m² to cm². (b) A tank measures 80 cm by 50 cm by 40 cm. Find its capacity in litres, and the time to fill it at 12 L per minute.

(a) 2.5 × 10 000 = 25 000 cm². (b) Volume = 80 × 50 × 40 = 160 000 cm³ = 160 L. Time = 160 ÷ 12 = 13.3 minutes (3 s.f.).

Volume, surface area and nets

Solid	Volume	Surface area
Prism or cylinder	cross-section area × length	sum of all faces; cylinder 2πr² + 2πrh
Pyramid	⅓ × base area × height	base + triangular faces
Cone	⅓πr²h	πr² + πrl

Here l is the slant height, found by Pythagoras: l² = r² + h² for a cone, or the right-angled triangle formed by the vertical height and half the base side for a square pyramid.

Worked example 6 (prism). A triangular prism has a right-angled triangle cross-section with sides 6, 8 and 10 cm, and length 15 cm. Triangle area = ½ × 6 × 8 = 24. Volume = 24 × 15 = 360 cm³. Surface area = 2 × 24 + (6 + 8 + 10) × 15 = 48 + 360 = 408 cm².

Worked example 7 (cone). A cone has radius 5 cm and vertical height 12 cm. Slant height = √(25 + 144) = 13 cm. Volume = ⅓ × π × 25 × 12 = 100π = 314 cm³. Total surface area = π × 25 + π × 5 × 13 = 90π = 283 cm².

Worked example 8 (pyramid). A square-based pyramid has base side 8 cm and height 9 cm. Volume = ⅓ × 64 × 9 = 192 cm³. Each triangular face has height √(9² + 4²) = √97 = 9.849 cm, so surface area = 64 + 4 × ½ × 8 × 9.849 = 222 cm² (3 s.f.).

Worked example 9 (compound solid). A silo is a cylinder of radius 4 m and height 9 m, topped by a cone of height 3 m with the same radius. Cylinder = π × 16 × 9 = 144π. Cone = ⅓ × π × 16 × 3 = 16π. Total = 160π = 503 m³ (3 s.f.).

Nets and visualising. A net is a solid unfolded flat. A cylinder’s net has two circles and one rectangle. The rectangle’s length equals the circumference 2πr and its width equals the height. For r = 2.5 cm and h = 11 cm the rectangle is 15.7 cm by 11 cm, with area 173 cm². A triangular prism has 2 triangles and 3 rectangles; a square pyramid has 1 square and 4 triangles. To move between dimensions, picture a 3D solid as a stack of identical 2D slices (a prism) or as shrinking slices (a pyramid or cone).

Constructions

Using a compass and straight edge, you should be able to: bisect an angle; construct a perpendicular bisector of a segment (every point on it is equidistant from both ends); construct a 60° angle; and build a triangle from three given sides by drawing arcs from each end of the base. Keep construction arcs visible, since they show your method.

Circle geometry and theorems

Name the parts: centre, radius, diameter, chord, arc, tangent, sector and segment.

Theorem	Statement
Angle at centre	Standing on one arc, the central angle is double any circumference angle
Same segment	Circumference angles on a common arc are equal
Semicircle	The angle in a semicircle is 90°
Cyclic quadrilateral	Opposite angles add to 180°
Tangent and radius	A tangent meets the radius at 90°
Two tangents	Tangents from one external point are equal in length
Alternate segment	Tangent-chord angle = inscribed angle on the far side of that chord
Chord	A line from the centre perpendicular to a chord bisects it

Worked example 10. A chord of length 24 cm lies in a circle of radius 13 cm. Find its distance from the centre.

The perpendicular from the centre meets the chord at its midpoint, 12 cm from each end. Distance² = 13² − 12² = 25, so distance = 5 cm.

Worked example 11. From a point P, a tangent touches a circle at T. The radius is 12 cm and the centre O is 37 cm from P. Find PT.

OT is perpendicular to PT, so PT² = 37² − 12² = 1225 and PT = 35 cm.

Worked example 12. The angle at the centre standing on arc AB is 124°. Find the angle at the circumference on the same arc, and the opposite angle of a cyclic quadrilateral whose other angle is 105°. Angle at circumference = 124 ÷ 2 = 62°. Opposite angle = 180 − 105 = 75°.

Arc, chord, sector and segment

For a sector with angle θ (in degrees) and radius r: arc length = (θ/360) × 2πr; sector area = (θ/360) × πr²; sector perimeter = 2r + arc. A segment is the region between a chord and its arc. Segment area = sector area − triangle area, when the angle is under 180°.

Worked example 13. A sector has radius 12 cm and angle 75°. Arc = (75/360) × 2π × 12 = 5π = 15.7 cm. Area = (75/360) × π × 144 = 30π = 94.2 cm².

Worked example 14. A circle has radius 10 cm. A chord subtends 90° at the centre. Sector area = ¼ × π × 100 = 78.54. The triangle is right-angled with legs 10, so its area is ½ × 10 × 10 = 50. Segment area = 28.5 cm² (3 s.f.). Chord length = √(10² + 10²) = 14.1 cm.

Enrichment

Arc length and sector using radians. An angle of 2π radians is a full turn, so 180° = π radians. For θ in radians, arc = rθ and sector area = ½r²θ. With r = 5 and θ = 0.8, arc = 4 and area = 10.

Inscribing and circumscribing. A regular hexagon inscribed in a circle of radius 6 has side 6, since it splits into six equilateral triangles. Its area is 3√3/2 × 36 = 93.5 and the circle’s is 113.1, so the hexagon fills 82.7%. A circle inscribed in a right-angled triangle with legs a, b and hypotenuse c has radius (a + b − c)/2; for 5, 12, 13 that is 2. A circle circumscribed about a right-angled triangle passes through all three corners; its centre is the midpoint of the hypotenuse, so for 5, 12, 13 its radius is 6.5.

Fractals (informal). In the Koch snowflake, start with an equilateral triangle of side 4 (perimeter 12). Each stage replaces every side by four sides each a third as long, so the perimeter is multiplied by 4/3: 12, 16, 21.3, 28.4. The perimeter grows without limit while the shape stays inside a bounded region.

Common errors

• Using the slanted side instead of the perpendicular height in area formulas.
• Converting areas or volumes with the length factor: 1 m² is 10 000 cm², not 100.
• Forgetting the ⅓ in pyramid and cone volumes.
• Using the vertical height in the curved surface area of a cone instead of the slant height.
• Including internal lines in a compound perimeter.
• Giving an angle answer with no reason. Name the theorem you used; this supports criterion C (communicating).
• Rounding mid-way. Keep π or the full unrounded value until the last line.

Official syllabus

IB Mathematics guide (for use from September 2020/January 2021), MYP framework, spatial reasoning: Geometry. International Baccalaureate Organization, first assessment 2022.