7.1 Surface Area & Volume of 3D Prisms & Cylinders

7.1 Surface Area & Volume of 3D Prisms & Cylinders

Quick Summary: Surface area and volume of prisms and cylinders are two of the six fundamental knowledge and skills in the MPT's Measurement dimension, and every measurement question is tagged Application or Thinking — the formula sheet supplies the formula, so the marks are in choosing and substituting correctly. This section covers the geometric properties, surface areas, and volumes of right prisms (rectangular and triangular) and right circular cylinders. You will learn to construct and interpret 2D net diagrams, apply base-times-height volume formulas, and solve practical multi-step word problems involving packaging optimization, painting coverage, and liquid container capacities.

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3D Geometric Solids & Right Prisms

A three-dimensional (3D) solid occupies space and possesses three dimensions: length, width, and height (or depth). In the Ontario curriculum, primary focus is placed on right prisms and right cylinders.

What is a Right Prism?

A right prism is a 3D solid bounded by two congruent, parallel polygonal bases connected by rectangular lateral faces that meet the bases at right angles ($90^\circ$).

• Bases: Two identical (congruent) polygons that lie in parallel planes. The shape of the base defines the name of the prism (e.g., rectangular prism, triangular prism, pentagonal prism).
• Lateral Faces: The flat side faces connecting the two bases. In a right prism, all lateral faces are rectangles.
• Edges & Vertices: Straight line segments where two faces meet are edges; corner points where three or more edges intersect are vertices.

graph TD
    subgraph PrismTypes["Types of Right 3D Solids"]
        RP["Rectangular Prism<br/>(Base: Rectangle)"]
        TP["Triangular Prism<br/>(Base: Triangle)"]
        CYL["Right Circular Cylinder<br/>(Base: Circle)"]
    end

Key Differences Between Prisms and Pyramids

Candidates must distinguish right prisms from pyramids and cones:

• Prism: Has two parallel congruent bases connected by rectangular faces. Uniform cross-section throughout its height.
• Pyramid: Has one polygonal base tapering up to a single point called an apex. Lateral faces are triangles.

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Visualizing 3D Figures with 2D Nets

A 2D net is a flat, two-dimensional pattern that can be folded along its edges to form a closed three-dimensional figure without overlaps or gaps.

Why Nets Matter

1. Surface Area Calculation: Finding the total surface area of a complex solid is equivalent to summing the areas of all individual 2D polygons in its net.
2. Spatial Reasoning: Sketching the net is the most reliable way to avoid the two classic errors on surface-area items — omitting a face, or counting one twice. The blueprint lists surface area of prisms and cylinders and volume of prisms and cylinders; matching a solid to a printed net is a classroom skill rather than a listed MPT skill, but the net is still the best scratch-paper tool for getting the listed skills right.

Nets of Primary Solids:

• Rectangular Prism Net: Consists of $6$ rectangles arranged in $3$ pairs of identical opposite rectangles (front/back, top/bottom, left/right).
• Triangular Prism Net: Consists of $2$ congruent triangles (the bases) and $3$ rectangles (the lateral faces).
• Cylinder Net: Consists of $2$ congruent circles (top and bottom bases) and $1$ large rectangle (the unrolled lateral side surface).

Important Concept: The length of the unrolled rectangular lateral face of a cylinder is exactly equal to the perimeter (circumference) of the circular base: $L_{\text{rectangle}} = C_{\text{circle}} = 2\pi r$.

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Surface Area Formulas & Calculations

Total Surface Area ($SA$) is the total two-dimensional area of all outer faces of a 3D solid, expressed in square units ($\text{cm}^2$, $\text{m}^2$).

1. Rectangular Prism Surface Area

A right rectangular prism has dimensions length ($l$), width ($w$), and height ($h$). $SA_{\text{rect}} = 2lw + 2lh + 2wh = 2(lw + lh + wh)$ Derivation: The top and bottom faces each have area $lw$; front and back faces each have area $lh$; left and right faces each have area $wh$.

2. Right Triangular Prism Surface Area

A right triangular prism has two triangular bases of base length $b$, height $h_b$, and side lengths $s_1, s_2, s_3$, connected by a prism length/height $h$. $SA_{\text{tri}} = 2 \cdot A_{\text{base}} + A_{\text{lateral}}$ $SA_{\text{tri}} = 2 \left( \frac{1}{2} b h_b \right) + (s_1 + s_2 + s_3) h = b h_b + (P_{\text{base}} \cdot h)$ Where $P_{\text{base}} = s_1 + s_2 + s_3$ is the perimeter of the triangular base.

3. Right Circular Cylinder Surface Area

A cylinder of radius $r$ and height $h$ has two circular bases and a lateral surface wrapper.

• Base Area (each): $A_{\text{base}} = \pi r^2$
• Lateral Area: $A_{\text{lateral}} = 2\pi r h$
• Total Surface Area (Closed Cylinder): $SA_{\text{cyl}} = 2\pi r^2 + 2\pi r h = 2\pi r (r + h)$

Open Cylinder Variation: If a question describes an open-topped cylindrical container (like a bucket or cup), exclude one base: $SA_{\text{open}} = \pi r^2 + 2\pi r h$. If it is an open pipe, exclude both bases: $SA_{\text{pipe}} = 2\pi r h$.

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Volume Formulas & Uniform Cross-Sections

Volume ($V$) is the amount of three-dimensional space enclosed inside a 3D solid, expressed in cubic units ($\text{cm}^3$, $\text{m}^3$).

The Universal Prism & Cylinder Volume Formula

For any solid with a uniform cross-section along its height (all right prisms and right cylinders), volume is calculated as: $V = A_{\text{base}} \times h$

1. Rectangular Prism Volume

$V_{\text{rect}} = (l \times w) \times h = l w h$

2. Triangular Prism Volume

$V_{\text{tri}} = \left( \frac{1}{2} b h_b \right) \times h$ Note: Do not confuse the height of the triangular base ($h_b$) with the length/height of the prism solid ($h$).

3. Right Circular Cylinder Volume

$V_{\text{cyl}} = (\pi r^2) \times h = \pi r^2 h$

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Real-World Applications & Contextual Problems

MPT questions link surface area and volume to practical scenarios:

1. Packaging & Material Optimization

Determining the minimum square centimetres of cardboard required to manufacture a shipping box or product container requires calculating total surface area plus any specified percentage allowance for overlapping flaps (e.g., $+10%$).

2. Surface Coating & Painting Costs

Painting walls, water tanks, or floors involves surface area. Paint coverage is given in square metres per Litre ($\text{m}^2/\text{L}$). $\text{Litres Required} = \frac{\text{Total Surface Area to Paint}}{\text{Coverage Rate per Litre}}$

3. Liquid Capacity, Filling, and Draining

Converting cubic geometric volume into liquid capacity relies on standard metric conversion bridges: $1 \text{ cm}^3 = 1 \text{ mL} \quad | \quad 1,000 \text{ cm}^3 = 1 \text{ Litre (L)} \quad | \quad 1 \text{ m}^3 = 1,000 \text{ Litres (L)}$ $\text{Time to Fill/Drain} = \frac{\text{Total Capacity (L)}}{\text{Flow Rate (L/min)}}$

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Detailed Step-by-Step Worked Examples

Example 1: Triangular Prism Net & Surface Area

Problem: A wedge-shaped doorstop is designed as a right triangular prism. The triangular face is a right-angled triangle with leg lengths $b = 9\text{ cm}$ and $h_b = 12\text{ cm}$, and hypotenuse $s_3 = 15\text{ cm}$. The width (height of the prism) is $h = 20\text{ cm}$.

1. Calculate the area of the two triangular bases.
2. Calculate the lateral area of the three rectangular faces.
3. Determine the total surface area of the doorstop.

Solution:

1. Step 1: Calculate Triangular Base Area $A_{\text{base}} = \frac{1}{2} b h_b = \frac{1}{2} (9\text{ cm}) (12\text{ cm}) = 54\text{ cm}^2$ Since there are two bases: $2 \times A_{\text{base}} = 2 \times 54\text{ cm}^2 = 108\text{ cm}^2$

2. Step 2: Calculate Lateral Area The perimeter of the triangular base is: $P_{\text{base}} = 9 + 12 + 15 = 36\text{ cm}$ The lateral surface area is: $A_{\text{lateral}} = P_{\text{base}} \times h = 36\text{ cm} \times 20\text{ cm} = 720\text{ cm}^2$

3. Step 3: Sum for Total Surface Area $SA = 2 A_{\text{base}} + A_{\text{lateral}} = 108\text{ cm}^2 + 720\text{ cm}^2 = 828\text{ cm}^2$

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Example 2: Cylindrical Storage Tank Capacity & Painting

Problem: A closed cylindrical fuel storage tank has a diameter of $4\text{ metres}$ (radius $r = 2\text{ m}$) and a height of $10\text{ metres}$.

1. Calculate the exact volume in cubic metres and storage capacity in Litres.
2. Calculate the total exterior surface area (including both circular ends) to be painted (use $\pi \approx 3.14$).
3. If one can of rust-inhibiting paint covers $25\text{ m}^2$, how many full cans must be purchased?

Solution:

1. Step 1: Calculate Volume and Capacity $V = \pi r^2 h = \pi (2\text{ m})^2 (10\text{ m}) = 40\pi\text{ m}^3$ Using $\pi \approx 3.14$: $V = 40 \times 3.14 = 125.6\text{ m}^3$ Since $1\text{ m}^3 = 1,000\text{ Litres}$: $\text{Capacity} = 125.6 \times 1,000 = 125,600\text{ Litres}$

2. Step 2: Calculate Exterior Surface Area $SA = 2\pi r^2 + 2\pi r h = 2(3.14)(2)^2 + 2(3.14)(2)(10)$ $SA = 2(3.14)(4) + 2(3.14)(20) = 25.12 + 125.6 = 150.72\text{ m}^2$

3. Step 3: Calculate Paint Cans Required $\text{Cans needed} = \frac{150.72\text{ m}^2}{25\text{ m}^2/\text{can}} = 6.0288\text{ cans}$ Since paint is purchased in whole cans, round up to $7$ cans.