Perimeter, Area, and Volume
CSEC Mathematics: Measurement
Essential Understanding: Measurement is fundamental to mathematics and everyday life. Whether you’re fencing a yard, painting a room, or filling a container, you need to understand perimeter (distance around), area (surface covered), and volume (space occupied).
Understanding the Concepts
Perimeter
The total distance around the outside of a 2D shape.
Units: Linear (cm, m, km)
Think: Walking around the edge of a field
Area
The amount of surface a 2D shape covers.
Units: Square (cm², m², km²)
Think: Painting a wall or laying carpet
Volume
The amount of space inside a 3D solid.
Units: Cubic (cm³, m³, litres)
Think: Water filling a container
Perimeter of 2D Shapes
The perimeter is found by adding all the sides of a shape.
Rectangle
Square
Triangle
Circle (Circumference)
Area of 2D Shapes
Interactive Area Calculator
Perimeter: 400 units
Area: 9,600 square units
Area Formulas Reference
| Shape | Diagram | Area Formula |
|---|---|---|
| Rectangle | \( A = l \times w \) | |
| Square | \( A = s^2 \) | |
| Triangle | \( A = \frac{1}{2} \times b \times h \) | |
| Parallelogram | \( A = b \times h \) | |
| Trapezium | \( A = \frac{1}{2}(a + b) \times h \) | |
| Circle | \( A = \pi r^2 \) | |
| Sector | \( A = \frac{\theta}{360} \times \pi r^2 \) |
Circle: Arc Length and Sector Area
Interactive Sector Visualizer
Arc Length
109.96 units
\(\frac{\theta}{360} \times 2\pi r\)
Sector Area
3,848.45 units²
\(\frac{\theta}{360} \times \pi r^2\)
Arc Length and Sector Area
\[ l = \frac{\theta}{360} \times 2\pi r \]
\[ A = \frac{\theta}{360} \times \pi r^2 \]
Where \(\theta\) is the angle at the centre in degrees.
Worked Example: Sector
Problem: A sector has radius 14 cm and angle 60°. Find: (a) the arc length (b) the area of the sector. Use \(\pi = \frac{22}{7}\).
Surface Area of 3D Solids
The surface area is the total area of all the faces (outer surfaces) of a 3D solid.
Cube
\( SA = 6s^2 \)
Cuboid
\( SA = 2(lw + lh + wh) \)
Cylinder
\( SA = 2\pi r^2 + 2\pi rh \)
Cone
\( SA = \pi r^2 + \pi rl \)
Sphere
\( SA = 4\pi r^2 \)
Square Pyramid
\( SA = b^2 + 2bl \)
Volume of 3D Solids
3D Volume Visualizer
Cube
\( V = s^3 \)
Example: If s = 5 cm, V = 125 cm³
Volume Formulas Reference
| Solid | Volume Formula | Description |
|---|---|---|
| Cube | \( V = s^3 \) | Side cubed |
| Cuboid | \( V = l \times w \times h \) | Length × Width × Height |
| Prism | \( V = A_{base} \times h \) | Base area × Height |
| Cylinder | \( V = \pi r^2 h \) | Circle area × Height |
| Cone | \( V = \frac{1}{3}\pi r^2 h \) | ⅓ × Cylinder volume |
| Pyramid | \( V = \frac{1}{3} \times A_{base} \times h \) | ⅓ × Base area × Height |
| Sphere | \( V = \frac{4}{3}\pi r^3 \) | Four-thirds π r cubed |
Volume Memory Tips
- Prisms & Cylinders: \( V = \text{Base Area} \times \text{Height} \) (Think: stacking the base shape)
- Cones & Pyramids: \( V = \frac{1}{3} \times \text{Base Area} \times \text{Height} \) (One-third of corresponding prism/cylinder)
- Sphere: \( V = \frac{4}{3}\pi r^3 \) (Remember: “four-thirds pi r cubed”)
Worked Example: Cylinder
Problem: A cylindrical tank has radius 7 m and height 10 m. Calculate: (a) the volume (b) the curved surface area (c) the total surface area. Use \(\pi = \frac{22}{7}\).
Worked Example: Sphere
Problem: A sphere has radius 21 cm. Find: (a) its surface area (b) its volume. Use \(\pi = \frac{22}{7}\).
Compound Shapes
Compound shapes are made up of two or more basic shapes. To find the area or volume:
- Divide the compound shape into basic shapes
- Calculate the area/volume of each part
- Add or subtract as needed
Worked Example: Compound Shape
Problem: Find the area of the L-shaped figure below.
Rectangle A (top): 10 cm × 6 cm
Rectangle B (bottom): 5 cm × 16 cm
Area A = 10 × 6 = 60 cm²
Area B = 5 × 16 = 80 cm²
Total Area = 60 + 80 = 140 cm²
Past Paper Style Questions
A rectangular swimming pool is 25 m long and 10 m wide. The depth is 1.5 m at the shallow end and 3 m at the deep end.
(a) Calculate the area of the trapezoidal cross-section. [2 marks]
(b) Calculate the volume of water needed to fill the pool. [2 marks]
(c) If 1 m³ = 1000 litres, how many litres of water are needed? [1 mark]
(a) Cross-section is a trapezium with parallel sides 1.5 m and 3 m, width 10 m:
\( A = \frac{1}{2}(a + b) \times h = \frac{1}{2}(1.5 + 3) \times 10 = \frac{1}{2} \times 4.5 \times 10 = 22.5 \text{ m}^2 \)
(b) Volume = Cross-section area × Length = 22.5 × 25 = 562.5 m³
(c) 562.5 × 1000 = 562,500 litres
A sector of a circle has radius 12 cm and angle 150°. Calculate:
(a) the length of the arc [2 marks]
(b) the area of the sector [2 marks]
(c) the perimeter of the sector [2 marks]
Use \(\pi = 3.14\).
(a) Arc length: \( l = \frac{150}{360} \times 2 \times 3.14 \times 12 = \frac{150}{360} \times 75.36 = 31.4 \text{ cm} \)
(b) Sector area: \( A = \frac{150}{360} \times 3.14 \times 12^2 = \frac{150}{360} \times 452.16 = 188.4 \text{ cm}^2 \)
(c) Perimeter = arc + 2 radii = 31.4 + 12 + 12 = 55.4 cm
CSEC Practice Arena
Test Your Understanding
CSEC Examination Tips
- Units matter: Area uses square units (cm², m²), Volume uses cubic units (cm³, m³)
- Show formulas: Always write the formula before substituting values
- Use given π: Use the value of π given in the question (usually 22/7 or 3.14)
- Sector questions: Remember the fraction \(\frac{\theta}{360}\) appears in both arc length and sector area formulas
- Compound shapes: Break them into basic shapes, calculate separately, then add/subtract
- 3D solids: Draw and label diagrams to visualize the problem
- Check reasonableness: Does your answer make sense? Volume should be positive and appropriately sized
Summary: Essential Formulas
2D Shapes
- Rectangle: P = 2(l+w), A = lw
- Circle: C = 2πr, A = πr²
- Triangle: A = ½bh
- Trapezium: A = ½(a+b)h
Circle Parts
- Arc: l = (θ/360) × 2πr
- Sector: A = (θ/360) × πr²
3D Solids
- Cylinder: V = πr²h
- Cone: V = ⅓πr²h
- Sphere: V = (4/3)πr³, SA = 4πr²