IB Math AI SL · Topic 3
Geometry and trigonometry
The topic that gives Applications its identity: measurement and location problems solved with triangles, circles and — uniquely to this course — Voronoi diagrams, where geometry answers real questions like where to put a hospital.
What the syllabus covers
Six assessed sections, SL 3.1 to 3.6.
How it is examined
Voronoi questions have opened recent Paper 2s — long, structured, and generous if your working is legible.
A real Paper 2 opener ran a restaurant floorplan as a Voronoi problem: read coordinates off the diagram, find midpoints, derive perpendicular bisectors, then interpolate waiting times at an unseen point.
Constructing a labelled diagram from written statements carries its own marks — it is listed in the syllabus, and missing labels are visible losses.
Composite solids combine volumes and surface areas of pyramids, cones, spheres and hemispheres in one question.
Angles between a line and a plane test 3D visualisation inside otherwise routine trigonometry.
This is the topic where the booklet is most generous: every volume, surface-area, sine-rule, cosine-rule and sector formula above is printed in it. What is examined is selection — reading 'two sides and the included angle' and reaching for the cosine rule without hesitation.
The arc and sector formulae are printed with θ in degrees, and at SL the topic never leaves degree mode — a calculation that suddenly goes wrong usually means the calculator did.
A marked question, mark by mark
A Voronoi-style bisector question, marked as the scheme would.
Practise Voronoi and trigonometry as papers ask them
Fresh generated exercises on every SL 3.1–3.6 skill — bisectors, sectors, composite solids — marked against the same scheme, with your diagram-described working read line by line.
Start free trialNew to the codes? M1, A1, R1 explained →
Perpendicular bisectors · SL 3.5 / 3.6
Seed 9F44
Two food stalls stand at P(2, 2) and Q(8, 10). Customers walk to the nearer stall.
- (a)Find the equation of the perpendicular bisector of [PQ].
- (b)Determine which stall is nearer to a customer at (9, 2).
Your work
midpoint = (5, 6), m(PQ) = 8/6 = 4/3
bisector: y − 6 = −¾(x − 5)
y = −¾x + 39/4
(9,2): dist P = 7, dist Q ≈ 8.06 → stall P
Marks awarded
- M1Mark awarded. Midpoint found and perpendicular gradient applied via m₁ × m₂ = −1.
- A1Mark awarded. y = −¾x + 39/4 correct.
- R1Mark not awarded. Distances quoted as bare GDC output — no working shown linking the customer's position to either side of the bisector.
Where marks are actually lost
Geometry punishes invisible method harder than any other topic.
Diagrams left unlabelled, or never drawn despite the wording describing a scenario.
CostDiagram construction is directly assessable; an unlabelled figure forfeits those marks outright.
Calculator left in radian mode from a previous question — every formula in this topic assumes degree measure at SL.
CostAll downstream values shift, and follow-through only rescues marks if the written method was right.
Graphing bisectors entirely on the GDC without showing the midpoint-gradient algebra.
CostWhen the plotted line is misread, there is no written method left to award.
Answering volume where surface area was asked in composite-solid questions.
CostA complete, accurate calculation of the wrong quantity scores almost nothing.
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