Module 7
3D shapes & Pythagoras
Quick-reference revision notes for parents.
7.1 Properties of 3D shapes
| Shape | Faces | Edges | Vertices |
|---|---|---|---|
| Cube | 6 | 12 | 8 |
| Cuboid | 6 | 12 | 8 |
| Triangular prism | 5 | 9 | 6 |
| Square-based pyramid | 5 | 8 | 5 |
| Tetrahedron | 4 | 6 | 4 |
| Cylinder | 3* | 2 | 0 |
| Sphere | 1 | 0 | 0 |
*counting curved surfaces.
Euler's rule for shapes with flat faces: F + V − E = 2.
7.2 Nets
A net is a 2D layout that folds up into the 3D shape. A cube has 11 different nets. The key: every face of the 3D shape appears once in the net.
It must have 6 squares. Mentally fold it — opposite faces should not be next to each other in the net.
7.3 Surface area and volume of a cuboid
Volume:
V = length × width × height
Surface area (sum of all six rectangles):
SA = 2(lw + lh + wh)
V = 5 × 3 × 4 = 60 cm³
SA = 2(5×3 + 5×4 + 3×4) = 2(15 + 20 + 12) = 2 × 47 = 94 cm²
Volume uses cm³, area uses cm², length uses cm. Mixing them up loses easy marks.
7.4 Pythagoras' theorem
For a right-angled triangle with legs a, b and hypotenuse c (the longest side, opposite the right angle):
a² + b² = c²
Finding the hypotenuse
Add the squares of the two shorter sides, then take the square root.
c² = 3² + 4² = 9 + 16 = 25
c = √25 = 5
Finding a shorter side
Subtract the squares.
a² = 13² − 5² = 169 − 25 = 144
a = √144 = 12
7.5 Applying Pythagoras directly
Many problems hide a right-angled triangle. Look for ladders, distances, diagonals.
A 5m ladder leans against a wall, with its base 1.4m from the wall. How high up the wall does it reach?
h² = 5² − 1.4² = 25 − 1.96 = 23.04
h = √23.04 = 4.8 m
Quick reference
- Cuboid volume: l × w × h
- Cuboid surface area: 2(lw + lh + wh)
- Pythagoras: a² + b² = c² (c is the hypotenuse)
- Hypotenuse is opposite the right angle and the longest side
- Always identify which side is the hypotenuse before applying the formula