IB Mathematics AI – Topic 3
Geometry & Trigonometry: 3D Shapes
Overview: Understanding 3D shapes involves calculating surface area (2D measurement in units²) and volume (3D measurement in units³). All formulas are provided in the IB formula booklet.
Key Strategy: Identify the shape → Find formula in booklet → Substitute values → Calculate with GDC
Prisms
Rectangular Prisms & Cuboids
Definition: A prism has two parallel, congruent bases connected by rectangular faces. The cross-section is constant along the length.
Rectangular Prism (Cuboid):
Dimensions: length (l), width (w), height (h)
Volume:
\[ V = l \times w \times h \]
Surface Area:
\[ SA = 2(lw + lh + wh) \]
Sum of areas of all 6 rectangular faces
Special Case - Cube:
When l = w = h = s (side length):
\[ V = s^3 \]
\[ SA = 6s^2 \]
Cylindrical Prism (Cylinder):
Dimensions: radius (r), height (h)
Volume:
\[ V = \pi r^2 h \]
Base area × height
Surface Area:
\[ SA = 2\pi r^2 + 2\pi rh \]
Two circular bases + curved surface area
Or: \(SA = 2\pi r(r + h)\)
⚠️ Common Pitfalls & Tips:
- Don't confuse diameter with radius (r = d/2)
- Surface area includes ALL faces (top, bottom, and sides)
- Units: Volume is always cubed (cm³, m³)
- Units: Surface area is always squared (cm², m²)
Pyramids
Right Pyramids
Definition: A pyramid has a polygon base and triangular faces that meet at a single point (apex). A right pyramid has the apex directly above the center of the base.
General Pyramid Formula:
\[ V = \frac{1}{3} \times \text{Base Area} \times h \]
where h is the perpendicular height from base to apex
Common Types:
1. Square-based Pyramid:
Base is a square with side length s
\[ V = \frac{1}{3}s^2h \]
2. Rectangular-based Pyramid:
Base has dimensions l × w
\[ V = \frac{1}{3}lwh \]
Surface Area:
Sum of base area and all triangular faces
For each triangular face: \(A = \frac{1}{2} \times \text{base} \times \text{slant height}\)
⚠️ Common Pitfalls & Tips:
- Key factor: 1/3 – pyramids are 1/3 the volume of corresponding prism
- Use perpendicular height (h), not slant height, for volume
- Slant height is used for surface area of faces
Cones
Right Circular Cone
Definition: A cone is like a pyramid with a circular base. The apex is directly above the center of the circular base.
Key Dimensions:
- r: radius of circular base
- h: perpendicular height
- l: slant height
Relationship (Pythagoras):
\[ l^2 = r^2 + h^2 \]
Volume:
\[ V = \frac{1}{3}\pi r^2 h \]
Curved Surface Area:
\[ A_{\text{curved}} = \pi r l \]
where l is the slant height
Total Surface Area:
\[ SA = \pi r^2 + \pi r l = \pi r(r + l) \]
Circular base + curved surface
⚠️ Common Pitfalls & Tips:
- Volume uses perpendicular height (h), not slant height (l)
- Curved surface area uses slant height (l)
- If given diameter, convert to radius first
- Use Pythagoras if you need to find missing dimension
📝 Worked Example 1: Cone Calculations
Question: A cone has radius 6 cm and slant height 10 cm.
(a) Find the perpendicular height.
(b) Calculate the volume.
(c) Calculate the total surface area.
Solution:
Given: r = 6 cm, l = 10 cm
(a) Find perpendicular height h:
Using Pythagoras: \(l^2 = r^2 + h^2\)
\[ h^2 = l^2 - r^2 = 10^2 - 6^2 = 100 - 36 = 64 \]
\[ h = \sqrt{64} = 8 \text{ cm} \]
(b) Calculate volume:
Using \(V = \frac{1}{3}\pi r^2 h\)
\[ V = \frac{1}{3}\pi (6)^2(8) = \frac{1}{3}\pi (36)(8) = 96\pi \]
\[ V \approx 301.6 \text{ cm}^3 \]
(c) Calculate total surface area:
Using \(SA = \pi r(r + l)\)
\[ SA = \pi (6)(6 + 10) = \pi (6)(16) = 96\pi \]
\[ SA \approx 301.6 \text{ cm}^2 \]
Answers: (a) 8 cm, (b) 302 cm³, (c) 302 cm²
Spheres and Hemispheres
Complete and Half Spheres
Sphere:
A perfectly round 3D shape where all points on the surface are equidistant from the center
Volume of Sphere:
\[ V = \frac{4}{3}\pi r^3 \]
Surface Area of Sphere:
\[ SA = 4\pi r^2 \]
Hemisphere (Half Sphere):
A hemisphere is exactly half of a sphere
Volume of Hemisphere:
\[ V = \frac{2}{3}\pi r^3 \]
(Half of sphere volume)
Surface Area of Hemisphere:
\[ SA = 3\pi r^2 \]
Curved surface (\(2\pi r^2\)) + circular base (\(\pi r^2\))
⚠️ Common Pitfalls & Tips:
- Sphere formulas have fractions: 4/3 for volume, not 1/3
- Hemisphere volume is 2/3πr³, not 1/3πr³
- Hemisphere SA includes the flat circular base
- Always cube the radius for volume calculations
📝 Worked Example 2: Composite Shape
Question: A toy consists of a hemisphere placed on top of a cylinder. Both have radius 5 cm. The cylinder has height 12 cm. Calculate:
(a) The total volume
(b) The total surface area
Solution:
Given: r = 5 cm, cylinder height h = 12 cm
(a) Total Volume:
Volume of cylinder:
\[ V_{\text{cyl}} = \pi r^2 h = \pi (5)^2(12) = 300\pi \text{ cm}^3 \]
Volume of hemisphere:
\[ V_{\text{hem}} = \frac{2}{3}\pi r^3 = \frac{2}{3}\pi (5)^3 = \frac{2}{3}\pi (125) = \frac{250}{3}\pi \text{ cm}^3 \]
Total volume:
\[ V_{\text{total}} = 300\pi + \frac{250}{3}\pi = \frac{900\pi + 250\pi}{3} = \frac{1150}{3}\pi \]
\[ V \approx 1204 \text{ cm}^3 \]
(b) Total Surface Area:
Curved surface of hemisphere:
\[ A_{\text{hem}} = 2\pi r^2 = 2\pi (5)^2 = 50\pi \text{ cm}^2 \]
Curved surface of cylinder:
\[ A_{\text{cyl curved}} = 2\pi rh = 2\pi (5)(12) = 120\pi \text{ cm}^2 \]
Base of cylinder:
\[ A_{\text{base}} = \pi r^2 = \pi (5)^2 = 25\pi \text{ cm}^2 \]
(Note: Top of cylinder is covered by hemisphere, so not included)
Total surface area:
\[ SA = 50\pi + 120\pi + 25\pi = 195\pi \approx 613 \text{ cm}^2 \]
Answers: (a) 1200 cm³, (b) 613 cm²
Complete Formula Summary
| Shape | Volume | Surface Area |
|---|---|---|
| Cuboid | \(V = lwh\) | \(SA = 2(lw + lh + wh)\) |
| Cube | \(V = s^3\) | \(SA = 6s^2\) |
| Cylinder | \(V = \pi r^2 h\) | \(SA = 2\pi r^2 + 2\pi rh\) |
| Pyramid | \(V = \frac{1}{3}Ah\) | Base + triangular faces |
| Cone | \(V = \frac{1}{3}\pi r^2 h\) | \(SA = \pi r^2 + \pi rl\) |
| Sphere | \(V = \frac{4}{3}\pi r^3\) | \(SA = 4\pi r^2\) |
| Hemisphere | \(V = \frac{2}{3}\pi r^3\) | \(SA = 3\pi r^2\) |
📊 Quick Reference - Key Points
Units
- Volume: Always cubed (cm³, m³)
- Surface Area: Always squared (cm², m²)
- Match units in question
Key Fractions
- Pyramid/Cone: 1/3
- Sphere: 4/3
- Hemisphere: 2/3
Common Errors
- Diameter vs radius
- Height vs slant height
- Forgetting π in formulas
Composite Shapes
- Break into simple shapes
- Calculate each separately
- Add/subtract as needed
✍️ IB Exam Strategy
- Identify the shape - read question carefully, sketch if needed
- Find formula in booklet - all formulas provided in IB formula booklet
- Check what's given: radius or diameter? height or slant height?
- Convert if needed: d → r (divide by 2), or use Pythagoras for missing dimension
- Substitute carefully into formula with correct values
- Use GDC for calculations - faster and more accurate
- Check units: Volume (cubed), Surface Area (squared)
- Round appropriately: 3 s.f. unless specified otherwise
🚫 Top Mistakes to Avoid
- Using diameter instead of radius (remember r = d/2)
- Using slant height instead of perpendicular height for volume
- Forgetting the 1/3 factor for pyramids and cones
- Using wrong fraction for sphere (4/3, not 1/3)
- Mixing up units (cm² for surface area, cm³ for volume)
- Not including all surfaces for surface area
- For composite shapes: counting shared surfaces twice
- Forgetting to square or cube the radius in formulas
- Not using π button on calculator (use π, not 3.14)
- Rounding too early - keep full accuracy until final answer