4.2 Area, Rectangular Volume & Elapsed Time

Area, Rectangular Volume & Elapsed Time

While perimeter measures the linear boundary surrounding a figure, area and volume measure spatial content: area quantifies the two-dimensional surface within a boundary, and volume quantifies the three-dimensional space enclosed within a solid. Alongside spatial measurement, the measurement of time and elapsed duration governs schedules, transportation logistics, and everyday planning.

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The Concept of Area vs. Perimeter

Understanding the fundamental distinction between perimeter and area is essential:

• Perimeter: A 1-dimensional boundary measurement. Expressed in linear units ($\text{cm}, \text{m}, \text{km}$).
• Area: A 2-dimensional surface measurement. Expressed in square units ($\text{cm}^2, \text{m}^2, \text{km}^2$), defined as the number of unit squares of a given size required to tile a surface completely without gaps or overlaps.

The Independence of Area and Perimeter

A common misconception is that a larger perimeter automatically implies a larger area. Rectangles with identical perimeters can enclose drastically different areas:

Consider four different rectangles, each with a fixed perimeter of $24\text{ m}$ ($2l + 2w = 24$, so $l + w = 12\text{ m}$):

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Standard Area Formulas

1. Square and Rectangle

• Square: $A = s \times s = s^2$
• Rectangle: $A = l \times w$ (Length $\times$ Width)

2. Triangles: The Half-Rectangle Rule

Every triangle can be inscribed inside a bounding parallelogram or rectangle that shares the identical base ($b$) and perpendicular height ($h$). Because the diagonal divides a parallelogram into two congruent triangles, the area of any triangle is exactly half the area of the corresponding rectangle:

$A = \frac{1}{2} \times b \times h \quad \text{or} \quad A = \frac{b \times h}{2}$

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Composite Figures: Additive and Subtractive Decomposition

Real-world architectural and land survey layouts are rarely simple rectangles. To calculate the area of an irregular rectilinear shape, apply one of two decomposition strategies:

Method 1: Additive Decomposition (Splitting)

Divide the irregular figure into non-overlapping standard rectangles ($R_1, R_2, \dots$), calculate the individual area of each sub-region, and add them together: $A_{\text{total}} = A(R_1) + A(R_2)$

Method 2: Subtractive Decomposition (Cutout)

Enclose the entire figure within a large overall bounding box rectangle, calculate its total area, and subtract the area of the empty missing "cutout" corner: $A_{\text{total}} = A_{\text{enclosing rectangle}} - A_{\text{cutout}}$

Worked Example: Tiling a School Library Floor in Portmore

Problem: A school library reading area in Portmore has an L-shaped floor layout. The overall length is $12\text{ m}$, the overall width is $9\text{ m}$, one inner alcove has a length of $7\text{ m}$, and its corresponding width is $4\text{ m}$. Ceramic tiles measuring $50\text{ cm} \times 50\text{ cm}$ are used to cover the entire floor. How many ceramic tiles are required?

          12 m
     +------------+
     |            | 
 9 m |     R1     | 5 m
     |            |
     +------+-----+ 
       5 m  | R2  | 4 m
            +-----+
              7 m

Step-by-Step Solution:

1. Calculate the Floor Area Using Additive Decomposition:

   • Split into Rectangle 1 ($R_1$) and Rectangle 2 ($R_2$):
   • $R_1$: Length $= 12\text{ m}$, Width $= 5\text{ m}$ (since $9\text{ m} - 4\text{ m} = 5\text{ m}$). $A(R_1) = 12\text{ m} \times 5\text{ m} = 60\text{ m}^2$
   • $R_2$: Length $= 7\text{ m}$, Width $= 4\text{ m}$. $A(R_2) = 7\text{ m} \times 4\text{ m} = 28\text{ m}^2$
   • Total Floor Area: $A_{\text{total}} = 60\text{ m}^2 + 28\text{ m}^2 = 88\text{ m}^2$

2. Verify with Subtractive Decomposition:

   • Enclosing rectangle: $12\text{ m} \times 9\text{ m} = 108\text{ m}^2$.
   • Cutout corner: Length $= (12 - 7) = 5\text{ m}$, Width $= 4\text{ m}$. $A_{\text{cutout}} = 5\text{ m} \times 4\text{ m} = 20\text{ m}^2$
   • Total Area: $108\text{ m}^2 - 20\text{ m}^2 = 88\text{ m}^2$. (Both methods match!)

3. Calculate the Area of One Ceramic Tile:

   • Convert tile dimensions to metres: $50\text{ cm} = 0.5\text{ m}$.
   • Area of one tile: $0.5\text{ m} \times 0.5\text{ m} = 0.25\text{ m}^2$.

4. Calculate the Number of Tiles Required: $\text{Number of tiles} = \frac{\text{Total Floor Area}}{\text{Area of One Tile}} = \frac{88\text{ m}^2}{0.25\text{ m}^2} = 88 \times 4 = 352\text{ tiles}$

5. Final Answer: The floor area is $88\text{ m}^2$, requiring 352 tiles.

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Volume of Rectangular Prisms and Liquid Capacity

Volume measures the three-dimensional space occupied by a solid object, expressed in cubic units ($\text{cm}^3, \text{m}^3$). A rectangular prism (or cuboid) is a 3D solid bounded by six rectangular faces.

Volume Formulas for Prisms

$\text{Volume} = \text{Length} \times \text{Width} \times \text{Height} = l \times w \times h$ $\text{Volume} = \text{Base Area} \times \text{Height} = A_{\text{base}} \times h$

The Fundamental Equivalence: Volume and Liquid Capacity

In the metric system, solid cubic volume and liquid capacity are directly linked:

• $1\text{ cm}^3 = 1\text{ mL}$ (one cubic centimetre holds exactly one millilitre of liquid).
• $1,000\text{ cm}^3 = 1,000\text{ mL} = 1\text{ L}$.
• $1\text{ m}^3 = 1,000\text{ L}$ (one cubic metre holds one thousand litres, or one kilolitre).

Worked Example: Concrete Cistern Water Storage in Mandeville

Problem: Due to seasonal rainfall variations, a community center in Mandeville, Manchester installs a concrete rainwater harvesting cistern. The interior dimensions of the cistern are:

• $\text{Length} = 3.5\text{ m}$
• $\text{Width} = 2.0\text{ m}$
• $\text{Depth (Height)} = 1.6\text{ m}$

Calculate:

1. The interior volume of the cistern in cubic metres ($\text{m}^3$).
2. The maximum liquid capacity of the cistern in litres ($\text{L}$).
3. If the cistern is currently filled to $\frac{3}{4}$ of its maximum capacity, how many litres of water are in storage?

Step-by-Step Solution:

1. Calculate the Volume: $V = l \times w \times h = 3.5\text{ m} \times 2.0\text{ m} \times 1.6\text{ m} = 7.0 \times 1.6 = 11.2\text{ m}^3$
2. Convert Cubic Metres to Litres:
   • Since $1\text{ m}^3 = 1,000\text{ L}$: $11.2\text{ m}^3 \times 1,000 = 11,200\text{ L}$
3. Calculate Water at $\frac{3}{4}$ Capacity: $11,200\text{ L} \times \frac{3}{4} = 2,800 \times 3 = 8,400\text{ L}$
4. Summary Answer: The volume is $11.2\text{ m}^3$, the full capacity is $11,200\text{ L}$, and the volume at $\frac{3}{4}$ capacity is $8,400\text{ L}$.

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Units of Time and Clock Conventions

Time is measured in hierarchical units based on rotational and orbital astronomy:

• $60\text{ seconds (s)} = 1\text{ minute (min)}$
• $60\text{ minutes} = 1\text{ hour (h)}$
• $24\text{ hours} = 1\text{ day}$
• $7\text{ days} = 1\text{ week}$
• $12\text{ months} = 1\text{ year} = 365\text{ days}$ (or $366\text{ days}$ in a leap year; $52\text{ weeks} + 1\text{ day}$)

12-Hour vs. 24-Hour Time Systems

• 12-Hour Clock: Splits the 24-hour day into two 12-hour halves designated by AM (ante meridiem = before midday/noon) and PM (post meridiem = after midday/noon).
• 24-Hour Clock (Transportation/Standard Time): Expresses time as a continuous four-digit number from $0000$ (midnight) to $2359$ (one minute before the next midnight).

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Elapsed Time Timelines and Transit Timetable Analysis

Calculating elapsed time (the duration between a start time and an end time) across hour or noon/midnight boundaries requires systematic decomposition rather than standard vertical subtraction, because time operates in base 60 (60 minutes per hour) rather than base 10.

The Number Line Jump Method

To calculate elapsed time from $8:45\text{ AM}$ to $2:20\text{ PM}$:

1. Jump to the next whole hour: From $8:45\text{ AM}$ to $9:00\text{ AM} = 15\text{ minutes}$.
2. Jump across whole hours to noon: From $9:00\text{ AM}$ to $12:00\text{ PM} = 3\text{ hours}$.
3. Jump from noon to the target hour: From $12:00\text{ PM}$ to $2:00\text{ PM} = 2\text{ hours}$.
4. Jump the remaining minutes: From $2:00\text{ PM}$ to $2:20\text{ PM} = 20\text{ minutes}$.
5. Sum the segments: $(3\text{ h} + 2\text{ h}) + (15\text{ min} + 20\text{ min}) = 5\text{ hours } 35\text{ minutes}$

Worked Example: Inter-City Express Bus Schedule (Kingston to Montego Bay)

Problem: Consider the following sample inter-city coach timetable connecting Kingston to Montego Bay:

Answer the following:

1. How long does the express coach travel between Kingston and Ocho Rios?
2. What is the total duration of the scheduled passenger layover in Ocho Rios?
3. What is the total elapsed journey time from initial departure in Kingston to final arrival in Montego Bay?

Step-by-Step Solution:

1. Leg 1 Duration (Kingston to Ocho Rios: $08:15$ to $09:55$):
   • From $08:15$ to $09:15 = 1\text{ hour}$.
   • From $09:15$ to $09:55 = 40\text{ minutes}$.
   • Duration = $1\text{ hour } 40\text{ minutes}$.
2. Layover Duration in Ocho Rios ($09:55$ to $10:20$):
   • From $09:55$ to $10:00 = 5\text{ minutes}$.
   • From $10:00$ to $10:20 = 20\text{ minutes}$.
   • Total Layover = $5 + 20 =$ $25\text{ minutes}$.
3. Total Journey Elapsed Time ($08:15$ to $12:20$):
   • Jump $08:15$ to $12:15 = 4\text{ hours}$.
   • Jump $12:15$ to $12:20 = 5\text{ minutes}$.
   • Total Elapsed Time = $4\text{ hours } 5\text{ minutes}$.

Choosing Area Units, Counting Cube Layers and Tiling a Plane

Which Unit of Area?

Volume by Layers of Unit Cubes

A rectangular prism built from 1 cm cubes can be counted layer by layer. If the bottom layer has $4$ rows of $3$ cubes, it holds $12$ cubes. With $5$ layers, the prism holds $12 \times 5 = 60$ cubes, so its volume is $60\text{ cm}^3$. This is why the formula $V = l \times w \times h$ works: $l \times w$ counts the cubes in one layer and $h$ counts the layers. To find how many more cubes would complete a partly built cuboid, work out the full volume and subtract the cubes already in place.

Area of the Surfaces of a Cuboid

A cuboid has six rectangular faces in three matching pairs, so its total surface area is

$\text{Surface area} = 2(lw + lh + wh)$

For a box $5\text{ cm}$ long, $3\text{ cm}$ wide and $2\text{ cm}$ high: $2(15 + 10 + 6) = 2 \times 31 = 62\text{ cm}^2$ of card is needed.

Shapes That Cover a Plane (Tessellation)

A shape tessellates when copies of it cover a flat surface with no gaps and no overlaps, like floor tiles. The angles that meet at each corner point must add up to exactly $360^\circ$.

Rectangles also tessellate, which is why most floor and wall tiles are rectangles or squares.