SME
Grade 10 Math · Geometry
Modeling with geometry: density, design and optimization
Use this for “A solid cylinder of radius 2 cm and height 5 cm has a mass of 170 g. What is its density?”
Diagram
V = π × r² × h
V = π × 2² × 5 = 20π ≈ 62.83 cm³
Mass: 170 g
Density: 2.7056 g/cm³
This is an example. Type your own values to start a new problem.
V (cm³)
=
π
×
r
2
×
h
ρ (g/cm³)
=
m (g)
÷
V (cm³)
Formulas
V
=
π
×
r
2
×
h
62.83
=
π
×
2
2
×
5
ρ
=
m
V
2.7056
=
170
62.83
Assumptions
•
Density is mass per unit of volume: ρ = m ÷ V.
•
Model the object as a solid you know: here a cylinder, V = πr²h.
•
One material has one density at any size, so ρ names the material: aluminum is about 2.7 g/cm³.
Step-by-step
WE KNOW
r = 2 cm, h = 5 cm, m = 170 g
FIND
volume (V), density (ρ)
Step 1 · Find volume (V)
Use V = π × r² × h
The base’s area πr² times the height.
V
=
π
×
2
2
×
5
V
=
4
×
5
×
π
V
=
20
×
π
V
=
62.832
cm³
Step 2 · Find density (ρ)
Use ρ = m ÷ V
Density is the mass in each cubic centimeter.
ρ
=
170
62.832
ρ
=
2.7056
g/cm³
Check
Put the numbers back in. Both sides should match.
62.832
≈
π
×
2
2
×
5
✓
2.7056
≈
170
62.832
✓
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