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Diwali maths activities for IB Maths SL and HL

A ready-to-teach Diwali lesson built around rangoli patterns and diyas: a 5-minute starter, a 35-minute main activity, an HL extension and full worked answers.

Level
IB Maths SL and HL (AA and AI)
Time
40 minutes, plus a 10-minute extension
Topics
Radians, arcs and sectors; Area of a triangle and regular polygons; Volumes, including volumes of revolution (AA); Angles of regular polygons; Complex numbers and roots (HL extension)
Equipment
The starter is non-calculator. A GDC is useful for the main activity.

Download student sheet (PDF)Answers (PDF)

Suggested timings

PartTimeWhat
Starter5 minQuick questions on the board
Main: task A10 minAn eight-point rangoli
Main: task B8 minRangoli petals
Main: task C9 minThe diya
Main: task D8 minTessellations
Extension10 minFast finishers or homework

Starter (5 minutes)

No GDC.

  1. Find the area of a sector of radius 6 with angle π/3.
  2. Find the length of an arc of radius 10 with angle 0.8 radians.
  3. Write down the image of (2, 5) after reflection in the line y = x.
  4. Find each interior angle of a regular hexagon.

Main activity (35 minutes)

Task A: An eight-point rangoli (10 min)

Eight points of a rangoli are equally spaced on a circle of radius 4 centred at the origin. The first point is (4, 0).

  1. Find the exact coordinates of the second point, anticlockwise.
  2. Find the distance between neighbouring points, to 3 significant figures.
  3. The eight points are joined to make a regular octagon. Find its exact area.
  4. Find the area inside the circle but outside the octagon, to 3 significant figures.

Task B: Rangoli petals (8 min)

A petal is made of two circular segments placed back to back along a chord. Each segment is cut from a circle of radius 6 by a chord that subtends an angle of 2π/3 at the centre.

  1. Find the exact area of one segment.
  2. Find the area of the petal, to 3 significant figures.

Task C: The diya (9 min)

A diya is modelled as a hemispherical bowl of inner radius 4 cm.

  1. Find the volume of oil the diya holds when full, to 3 significant figures.
  2. (AA) The bowl is the region swept out when the curve x = √(16 − y2), −4 ≤ y ≤ 0, is rotated through 2π about the y-axis. Oil is poured in to a depth of 2 cm. Use integration to find the volume of oil.
  3. What fraction of the full diya is this?

Task D: Tessellations (8 min)

Rangoli floors often tile regular polygons with no gaps.

  1. Two copies of a regular n-gon and one equilateral triangle meet exactly at a point. Find n.
  2. A regular polygon has interior angles of 140°. How many sides does it have? Can it tessellate on its own?

Extension (10 minutes)

HL: the rangoli in the complex plane.

  1. (HL) The eight rangoli points are z = 4 cis(kπ/4), k = 0, 1, …, 7. Show that they are the roots of z8 = 65 536. Find their sum and their product.

For teachers

Teacher notes and full worked answers
  • AI students can skip the integration in C (b) and use the spherical cap volume as a given formula instead; AA students should integrate.
  • In A (b), check students are in radian mode before they find sin(π/8).
  • A nice discussion for the extension: why is the sum of the eight points zero? (Symmetry: they balance about the origin.)

Starter

  1. 6π
    • ½ × 36 × π/3 = 6π
  2. 8
    • s = rθ = 10 × 0.8 = 8
  3. (5, 2)
    • Reflection in y = x swaps the coordinates.
  4. 120°
    • 180° − 360°/6 = 120°

Task A: An eight-point rangoli

  1. (2√2, 2√2)
    • Rotate by 2π/8 = π/4: (4 cos π/4, 4 sin π/4) = (2√2, 2√2).
  2. 3.06
    • Chord = 2 × 4 × sin(π/8) = 8 sin(π/8) = 3.061…
  3. 32√2 ≈ 45.3
    • Eight triangles, each ½ × 4 × 4 × sin(π/4) = 4√2.
    • Area = 8 × 4√2 = 32√2 = 45.25…
  4. 5.01
    • 16π − 32√2 = 50.27 − 45.25 = 5.01

Task B: Rangoli petals

  1. 12π − 9√3
    • Sector: ½ × 36 × 2π/3 = 12π. Triangle: ½ × 36 × sin(2π/3) = 9√3.
    • Segment = 12π − 9√3
  2. 44.2
    • 2(12π − 9√3) = 24π − 18√3 = 44.22…

Task C: The diya

  1. 134 cm3
    • ½ × 4⁄3π × 43 = 128π/3 = 134.04…
  2. 40π/3 ≈ 41.9 cm3
    • The oil fills −4 ≤ y ≤ −2.
    • V = ∫−4−2 π(16 − y2) dy = π[16y − y3/3]−4−2
    • = π[(−32 + 8/3) − (−64 + 64/3)] = 40π/3 = 41.9
  3. 5/16
    • (40π/3) ÷ (128π/3) = 40/128 = 5/16

Task D: Tessellations

  1. n = 12
    • 2(180 − 360/n) + 60 = 360 gives 360/n = 30, so n = 12.
  2. 9 sides; no
    • Exterior angle = 40°, so n = 360 ÷ 40 = 9.
    • 360 ÷ 140 is not a whole number, so it cannot tessellate on its own.

Extension

  1. Sum 0; product −65 536
    • (4 cis(kπ/4))8 = 48 cis(2kπ) = 65 536, by De Moivre’s theorem.
    • z8 − 65 536 = 0 has no z7 term, so the sum of the roots is 0.
    • Product of the roots = (−1)8 × (−65 536) ÷ 1 = −65 536.

The same answers are in the answers PDF. Every answer was recomputed by computer before publishing.

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