Level 4 · Ages 13–15
Read it again any time, including while you work through the exercises.
After lengths (1 dimension) and areas (2 dimensions), here come volumes: the room a solid takes up in space, in 3 dimensions. The base unit is the cubic metre (m³), and for small objects the cubic centimetre (cm³): a little cube with 1 cm edges.
The cuboid (a shoebox) is the simplest: we stack layers of little cubes. Volume of a cuboid = length × width × height. The cube, with all its edges equal, gives V = edge × edge × edge.
The cylinder (a tin can) follows the same stacking logic: take the area of the base disc, π × r², and multiply it by the height. V = π × r² × h.
The cone (an ice cream cone) holds a surprise: it fits exactly 3 times inside the cylinder with the same base and height. Hence V = (π × r² × h) ÷ 3. And the sphere of radius r has volume V = (4 ÷ 3) × π × r³. Notice the r cubed: the radius acts in all 3 dimensions!
One last useful connection: volumes also speak the language of liquids. 1 L = 1000 cm³ = 1 dm³, and 1 m³ = 1000 L. Handy for knowing how much water fits in an aquarium!
Compute the volume of a cuboid 5 cm long, 3 cm wide and 4 cm high.
couche : 5 × 3 = 15
One layer of little cubes at the bottom of the box: 5 × 3 = 15 cubes of 1 cm³.
4 couches : 15 × 4 = 60
The height is 4 cm: I stack 4 identical layers, so 15 × 4 = 60 cubes.
V = L × l × h = 60
That is exactly the formula: length × width × height.
Result: The volume is 60 cm³.
Compute the volume of a cylinder with radius 5 cm and height 10 cm (π ≈ 3.14).
base : 3,14 × 5² = 78,5
I compute the area of the base disc: 3.14 × 25 = 78.5 cm².
V = 78,5 × 10 = 785
I stack this base up a height of 10 cm: 78.5 × 10 = 785.
Result: The volume is about 785 cm³.