Surface area of solids
Interactive curriculum lessons, worked examples, and geometric problem-solving techniques designed to help Kenyan students master CBC, KPSEA, KCSE, and IGCSE mathematics.
First Principles
Objective: Find the total and lateral surface area of 3D prisms and cylinders by analyzing their 2D nets.
First Principle: Surface area is the total 2D area needed to cover the entire outer boundary of a three-dimensional solid. If you cut along the seams of any solid and flatten it out without overlapping, you produce a flat 2D blueprint called a net. The surface area of the solid is identical to the total area of all the polygons and circles making up its net.
1. The Geometry of Unfolding (Nets)
Consider two fundamental solid families:
- Prisms (Rectangular & Triangular): When unfolded, every prism yields two congruent base shapes (e.g., triangles or rectangles) joined by rectangular lateral faces. The total lateral area forms one giant rectangle whose length equals the perimeter of the base and whose width is the height of the prism.
- Right Cylinders: A cylinder consists of two parallel circular bases and a smooth curved lateral wall. When you slit the curved wall vertically and unfold it, it flattens into a rectangle. The width of this rectangle is the cylinder height \(h\), and its length is the exact circumference of the circular base \(C = 2\pi r\).
2. Interactive Net Explorer
Use the interactive tool below to visualize how a 3D cylinder and rectangular prism unfold into flat 2D nets.
Rectangular Prism: 6 rectangular faces (3 pairs of congruent faces).
Key Formulas
1. Cube Surface Area
\[ SA_{\text{cube}} = 6s^{2} \]Where \(s\) is the edge length. Six identical square faces of area \(s^2\).
2. Rectangular Prism (Cuboid) Surface Area
\[ SA_{\text{total}} = 2(lw + lh + wh) \] \[ SA_{\text{lateral}} = 2h(l + w) = P_{\text{base}} \times h \]Where \(l\) is length, \(w\) is width, and \(h\) is height.
3. Right Circular Cylinder
\[ SA_{\text{lateral}} = 2\pi r h \] \[ SA_{\text{total}} = 2\pi r^{2} + 2\pi r h = 2\pi r(r + h) \]Where \(r\) is the radius of the circular base and \(h\) is the height.
4. Right Triangular Prism
\[ SA_{\text{total}} = 2\left(\frac{1}{2} b h_{\text{tri}}\right) + (a + b + c)L \]Where \(a, b, c\) are the side lengths of the triangular base, \(h_{\text{tri}}\) is the perpendicular height of the triangle, and \(L\) is the prism length.
Worked Examples
Example 1 (Easy): Surface Area of a Storage Crate
Problem: A wooden packing crate in a Nairobi warehouse is a cube with an edge length of \(1.2\text{ m}\). Find the total external surface area of the crate.
- Identify the net: A cube has \(6\) identical square faces.
- Area of one face: \[ A_{\text{face}} = s^2 = (1.2\text{ m})^2 = 1.44\text{ m}^2 \]
- Total surface area: \[ SA = 6 \times 1.44 = 8.64\text{ m}^2 \]
Final Answer: \(\boxed{8.64\text{ m}^2}\)
Example 2 (Medium): Plastering an Open Water Tank
Problem: A rectangular rainwater tank in Machakos has an internal length of \(5\text{ m}\), width of \(3\text{ m}\), and depth (height) of \(2\text{ m}\). The top of the tank is open. Find the total internal surface area that requires waterproof plastering (the base plus the 4 interior vertical walls).
- Deconstruct the net: Since the top is open, the net consists of \(1\) base rectangle and \(4\) wall rectangles. \[ SA = \text{Area of Base} + \text{Area of 4 Walls} \]
- Base area: \[ A_{\text{base}} = l \times w = 5 \times 3 = 15\text{ m}^2 \]
- Lateral wall area: \[ A_{\text{lateral}} = 2(lh + wh) = 2(5 \times 2 + 3 \times 2) = 2(10 + 6) = 32\text{ m}^2 \]
- Total plaster area: \[ SA = 15 + 32 = 47\text{ m}^2 \]
Final Answer: \(\boxed{47\text{ m}^2}\)
Example 3 (Hard): Metal Sheet for a Closed Grain Cylinder
Problem: An agricultural cooperative builds a closed cylindrical grain silo with a radius of \(1.4\text{ m}\) and a height of \(6\text{ m}\). Using \(\pi = \frac{22}{7}\), calculate the total area of galvanized steel sheeting required to fabricate the closed silo.
- Identify the net components: Two circular disks (top and bottom) and one rectangular curved wall sheet. \[ SA = 2\pi r^2 + 2\pi r h = 2\pi r (r + h) \]
- Calculate base circular ends (2 circles): \[ 2 \times \pi r^2 = 2 \times \frac{22}{7} \times (1.4)^2 = 2 \times \frac{22}{7} \times 1.96 = 2 \times 6.16 = 12.32\text{ m}^2 \]
- Calculate lateral curved wall: \[ 2\pi r h = 2 \times \frac{22}{7} \times 1.4 \times 6 = 2 \times 4.4 \times 6 = 52.8\text{ m}^2 \]
- Sum both components: \[ SA_{\text{total}} = 12.32 + 52.8 = 65.12\text{ m}^2 \]
Final Answer: \(\boxed{65.12\text{ m}^2}\)
Common Mistakes
Misconception 1: Confusing Lateral Area with Total Surface Area
Misconception 2: Using the Radius or Height as the Width of the Unfolded Cylinder Wall
Misconception 3: Adding Linear Edges or Multiplying Dimensions (Volume vs Area)
Real World
1. Paint Estimation for Silos in Rift Valley
Farmers in Nakuru store grain in cylindrical silos. To protect the galvanized iron from rust, farmers compute the total surface area to buy the exact number of paint tins needed. At a paint coverage of \(12\text{ m}^2\) per litre, accurate net calculations prevent expensive material waste.
2. Corrugated Iron for Shipping Containers in Mombasa Port
A standard 20-foot ISO shipping container is a rectangular prism measuring \(6.0\text{ m} \times 2.4\text{ m} \times 2.6\text{ m}\). Logistics and refurbishment companies calculate its total external surface area \(SA = 2(6.0 \times 2.4 + 6.0 \times 2.6 + 2.4 \times 2.6) = 72.48\text{ m}^2\) to estimate sandblasting and marine-grade anti-corrosion coating costs.
3. Manufacturing Cylindrical Water Tanks (Roto Tanks)
Rotational molding manufacturers determine the exact weight of high-density polyethylene (HDPE) polymer resin needed per tank by calculating the total surface area and multiplying by the target wall thickness.
Practice