IB Mathematics AI
From a sector to a cone
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Unit 7 · Geometry and trigonometry

From a sector
to a cone

Volume, surface area and container design

90 minutes · break · 45 minutes

SL and HL · 135 minutes

Today’s lesson

Before the break · 90 min

Compare cones made from sectors. Use a calculator table or graph to choose a container design.

Then compare the volumes and surfaces of three solids.

After the break · 45 min

Calculate storage capacity, paint requirements and the mass of a wax display.

Finish with an individual exit ticket.

Have your GDC, notebook and a small paper slip ready. Work in pairs, then groups of three.

All tasks are for SL and HL unless marked HL only.

0–8 min · 5 min alone, 3 min compare and correct

Start with these three questions

SL / common

  1. A sector has radius 8 cm and angle 135°. Find its arc length.
  2. Write the perimeter of the sector.
  3. A right triangle has hypotenuse l = 13 cm and one leg r = 5 cm. Find the other leg h.

HL only · Radians

  1. Convert 135° to radians.
  2. r = 8 cm and s = 12 cm. Find θ in radians.
  3. Find the area of the same sector.
Check · Correct in another colour

Answers

SL / common

6π cm

16 + 6π cm

h = 12 cm

HL only

3π4

θ = 1.5 rad

A = 48 cm²

8–14 min · Predict alone, then compare in pairs

Three sectors. Three cones.

Every sector has radius 12 cm. Rank the cones from smallest to largest volume. Roll each sector so its straight edges meet, without overlap.

90°

12 cm

180°

12 cm

355°

12 cm
Write first: write your ranking and one sentence of reasoning.

Diagrams are not drawn to a common scale. All three radii are 12 cm.

14–19 min · Discuss in pairs

What becomes what?

larc s
rhl

Match the sector radius and arc to lengths on the cone. Which length is the vertical height?

arc length = base circumference   ·   sector radius = slant height
19–34 min · Pairs

Build the model

Sector radius l = 12 cm. The straight edges meet without overlap. Find the cone radius r, vertical height h and volume V. Keep full calculator values.

Sector angle αArc lengthCone radius rHeight hVolume V
90°
180°
270°
330°
One partner explains the geometry; the other calculates and checks units. Swap roles after two rows. Both record the table.
Need a starting point?

s = (α/360) × 2πl; s = 2πr; h² + r² = l²; V = πr²h/3.

34–41 min · Compare methods and write a model

Compare your results

αr (cm)h (cm)V (cm³)
90°311.62109.5
180°610.39391.8
270°97.94673.3
330°114.80607.7

As α increases…

the base radius increases.

But also…

the vertical height decreases.

SL and HL · α is measured in degrees

Can one rule replace four calculations?

2πr = α360(2πl)
r = αl360

When l = 12, this becomes r = α ÷ 30.

SL and HL · Cone model

The volume model

Radius

r = α30

Height

h = √(144 − r²)

Volume

V = 13πr²h
Domain: what values of α make a real, non-flat cone? Explain both endpoints.
Check the domain

0° < α < 360°. At 0° there is no base; at 360° the height is zero. Neither endpoint gives a usable cone.

41–53 min · Pairs · GDC

Find the greatest capacity

Let x be the sector angle in degrees. Enter:

Y₁ = (π/3)(x/30)²√(144 − (x/30)²)
  1. Use a graph window 0 ≤ x ≤ 360, 0 ≤ y ≤ 750. Explain why only 0 < x < 360 represents a cone.
  2. Locate the maximum numerically. Check nearby angles with a table.
  3. Record the angle, radius, height and capacity. Explain why volume falls near 360°.

Use your GDC’s maximum tool or narrow the table step from 10° to 1° to 0.1°. No differentiation is needed. The next slide offers a slider check.

SL and HL · Check with the slider

Where is the greatest volume?

α (degrees)270°
r (cm)9.00
h (cm)7.94
V (cm³)673.3
Record your best angle, radius, height and volume. Then explain the shape of the graph.

Sector

12

Cone cross-section

hrbase width = 2r
Table + graph

Compare your maximum

α

≈ 293.9°

r and h

9.80 cm, 6.93 cm

Vmax

≈ 696.5 cm³

Pairs · Explain your graph

Why does the graph rise, then fall?

At first

The increasing r² has the stronger effect.

The base grows quickly while the cone is still tall.

Near 360°

The cone becomes very wide but almost flat.

The shrinking h drives the volume back towards zero.

Return to your prediction. What did you track? What did you overlook?
Check the opening ranking

90°: 109.5 cm³; 180°: 391.8 cm³; 355°: 292.3 cm³. Smallest to largest: 90°, 355°, 180°.

53–58 min · Heights and surfaces

Which height belongs where?

hlr

Volume

Use the perpendicular vertical height h.

Curved surface

Use the slant height l.

h² + r² = l²

53–58 min · How much card is used?

The sector becomes the curved surface

Asector = 12ls
s = 2πr
Acurved cone = πrl

The base is extra only when it is present and exposed.

58–69 min · Two pairs join

A cone must hold 600 mL

Keep the sector radius at 12 cm. The open cone must hold at least 600 mL. Use as little card as possible.

  1. Use the graph or table to find two sector angles with capacity about 600 cm³. Compare their curved areas.
  2. Choose a whole-degree angle that meets the capacity requirement. Check it and the next smaller whole-degree angle.
  3. Give your recommendation with capacity, card area and one limitation of the model.

V = (π/3)(α/30)²√(144 − (α/30)²)   ·   A = 0.4πα

1 mL = 1 cm³. Ignore thickness and overlap initially. Rotate calculator, diagram, recording and checking roles.

SL and HL · Compare recommendations

Which design uses less card?

The 600 cm³ intersections are approximately 240.2° and 331.2°. The smaller angle uses less card.

Choose 241°: V ≈ 602.4 cm³ and A ≈ 302.8 cm². At 240°, V ≈ 599.5 cm³, so it is too small.

Real seams and card thickness reduce capacity. A practical design needs extra allowance and a physical fill test.

Would the greatest-volume cone also use the least card? Explain using your results.

5-minute hinge ticket

Record both answers

On one small slip, write your class identifier and two letters: H1 and H2.

Work alone. Show the slip together on the signal. Hand it in.

69–74 min · Individual check · H1

The sector radius stays 12 cm while its angle increases.

Once rolled into a cone, what must happen?

A r increases and h increases
B r increases and h decreases
C r decreases and h increases
D r and h are unchanged

On a paper slip, write your usual class identifier and H1/H2 choices. Show together when asked; hand in before checking.

H2 · Record one letter

A cone has r = 5 cm and l = 13 cm.

Only its curved surface is made from card. What area is required?

A 60π cm²
B 65π cm²
C 90π cm²
D 100π cm²
Check · Then follow the instruction

Check H1 and H2

Show answers

H1: B. With fixed l, r increases and h = √(l² − r²) decreases.

H2: B. Curved area = πrl = 65π cm²; the base is not included.

Explain one correction to a partner. If both were correct, explain why option C in H2 includes too much card.

74–77 min · SL and HL · Formula reference

Formula toolkit

Cone

V = ⅓πr²h

CSA = πrl

Pyramid

V = ⅓Bh

Faces: triangles

Sphere

V = ⁴⁄₃πr³

SA = 4πr²

Hemisphere

V = ⅔πr³

curved SA = 2πr²

Before substituting: vertical or slant height? curved or total area? exposed or hidden surface?

B = base area. h = perpendicular height. l = cone slant height. Cylinder: V = πr²h; curved area = 2πrh.

77–90 min · Groups of three

Calculate, then teach your group

  1. A solid cone has r = 5 cm and l = 13 cm. Find its vertical height, volume, curved area and total area.
  2. An open hemispherical bowl has internal radius 6 cm. Find its capacity and inner curved area. Express the capacity in mL.
  3. A right square pyramid has base side 10 cm and vertical height 12 cm. Find its volume and total surface area. Distinguish the triangular face height from the edge joining the apex to a corner.
Take one problem each (5 min). Teach your method to the group (6 min), then check together (2 min). Everyone records all three methods.
Group check · Before the break

Answers

1 · Cone

h = 12 cm

V = 100π cm³

CSA = 65π cm²

TSA = 90π cm²

2 · Bowl

V = 144π cm³

CSA = 72π cm²

Capacity ≈ 452 mL

3 · Pyramid

face slant height = 13 cm

V = 400 cm³

TSA = 360 cm²

90 minutes completed

Break

Keep your group’s three worked solutions. After the break, start with the silo problem.

After break · 0–14 min · Groups of three

A silo

3 m8 m

A silo is a cylinder of radius 3 m and height 8 m, topped by a hemisphere of radius 3 m. Model the walls as having negligible thickness.

  1. Find the internal volume.
  2. Find the external area painted above ground.
  3. Paint covers 12 m² per litre per coat. For two coats, how many 5 L tins must be bought? Assume no wastage.
Sketch and shade only the painted surfaces. Explain whether the shared circle or floor belongs in your calculation.
Compare expressions

Silo solution

Volume

π(3²)(8) + 23π(3³) = 90π ≈ 282.7 m³

Painted area

2π(3)(8) + 2π(3²) = 66π ≈ 207.3 m²

Shared and hidden circles are not exposed surfaces.

Check the purchase

2 × 66π ÷ 12 ≈ 34.6 L. Buy 7 tins (35 L). Rounding down to 6 tins leaves too little paint.

After break · 14–29 min · Pairs

A solid wax display

6 cm8 cm10 cm

A cone is joined to a hemisphere of the same radius. The shared circle is internal.

Wax density: 0.90 g cm−3

  1. Find the total volume.
  2. Find the mass.
  3. There are 700 g of wax available. Is that enough for one display?
Same display · Part of the 15-minute pair task

Surface and sector

Cone: r = 6 cm, h = 8 cm, l = 10 cm. Hemisphere: r = 6 cm.

  1. Find the external surface area to be coated.
  2. A thin paper wrap covers only the cone’s curved surface. It is cut from a sector of radius 10 cm. Find the sector angle. Ignore overlap.

Surface check

Which circle is shared?

Which surfaces can the coating touch?

Sector check

What is the sector’s arc length after rolling?

Application · Verify

Answers

Volume

240π

≈ 754.0 cm³

Mass

216π

≈ 678.6 g

External area

132π

≈ 414.7 cm²

Sector angle

216°

Check your working: point to the exact place where you excluded the shared circle.

700 g is enough: about 21.4 g remains, assuming no wax is lost.

After break · 29–37 min · Compare and extend

SL · Same capacity

A cylinder has the same radius (6 cm) and volume (240π cm³) as the wax display. Find its height.

If its top and bottom are both exposed, compare its total area with 132π cm². Which needs less coating?

HL only · Radians

Convert 216° to radians.

Verify s = lθ = 2πr.

For fixed l, complete:

r = ______ ,   0 < θ < ______

θ = 6π5,   r = lθ2π,   0 < θ < 2π

Check SL

h = 240π/(36π) = 20/3 ≈ 6.67 cm. Total area = 2π(6)(20/3) + 2π(6²) = 152π cm². The original display needs 20π ≈ 62.8 cm² less coating.

After break · 37–45 min · Work alone and hand in

Exit ticket

  1. A cone has r = 6 cm, h = 8 cm and l = 10 cm. State its volume and curved surface area.
  2. A student adds the total surface areas of a cone and a hemisphere joined at their circular bases. Explain the error and state what must be subtracted.
  3. HL only: give the sector angle that forms this cone in radians.

V = 96π cm³; curved area = 60π cm²; HL only: θ = 6π/5 rad.

Subtract both copies of the shared circular face: 2πr².