6.2 Perimeter and Area of Rectangles, Triangles, Circles, and Composite 2D Shapes

School budgeting and facilities projects frequently require two-dimensional geometry: fencing a playground perimeter, marking an athletics track, carpeting an L-shaped library, or ordering paint for a gymnasium wall with windows and doors. Distinguishing between one-dimensional boundary length (perimeter) and two-dimensional surface coverage (area) is the first essential step.

Perimeter of Polygons and Deducing Missing Side Lengths

The perimeter ($P$) of any closed plane figure is the total distance around its outer boundary, measured in linear units ($\text{mm}$, $\text{cm}$, $\text{m}$, or $\text{km}$).

The Missing-Side Trap on L-Shaped and T-Shaped Floor Plans

When a diagram shows an L-shaped, U-shaped, or stepped school building, two of the interior edges are often left unlabelled. A common error is adding only the numbers printed on the diagram. Because all corners meet at right angles ($90^\circ$), you must deduce the missing sides first using the Parallel Segment Rule:

• Horizontal Balance: The sum of all horizontal segments facing one way (e.g., top edges) equals the longest overall horizontal width at the base:

$$W_{\text{total}} = w_1 + w_2 + \dots$$

• Vertical Balance: The sum of all vertical segments on one side equals the total vertical height on the opposite side:

$$H_{\text{total}} = h_1 + h_2 + \dots$$

Always write the deduced lengths directly onto the unlabelled edges and verify that the number of terms in your perimeter sum matches the total number of sides of the polygon (for example, $6$ sides for an L-shape, $8$ sides for a T-shape or U-shape).

---

Circles, Semicircles, Quarter-Circles, and Running Tracks

For a circle of radius $r$ and diameter $d = 2r$, the boundary distance is called the circumference ($C$) and the interior surface is the area ($A$):

$$C = 2\pi r = \pi d \quad \text{and} \quad A = \pi r^2$$

• Diameter vs. Radius Check: Before substituting into $A = \pi r^2$, confirm whether the question states the diameter (distance right across the circle through the centre) or the radius (distance from the centre to the rim). If $d = 14\text{ m}$, halve it immediately to get $r = 7\text{ m}$.
• Standalone Semicircles and Quarter-Circles: The curved arc of a semicircle has length $\pi r$ (half of $2\pi r$), and its area is $\frac{1}{2}\pi r^2$. However, if you are fencing a standalone semicircular garden bed, its total perimeter includes the straight diameter edge: $P_{\text{semicircle}} = \pi r + d = \pi r + 2r$.
• Stadium Running Tracks: A standard school athletics track consists of two parallel straight sides of length $L$ joined by two semicircular ends of diameter $d$ (radius $r = \frac{d}{2}$). Because the two semicircular ends join together to form one complete circle and the interior straight section forms a rectangle of $L \times d$:

$$P_{\text{track}} = 2L + \pi d \quad \text{and} \quad A_{\text{track}} = (L \times d) + \pi r^2$$

---

Area Formulas for Core 2D Polygons

---

Composite Area by Addition vs. Subtraction and Material Costing

Complex 2D surfaces are solved using one of two strategies:

1. Addition (Splitting): Slice an L-shaped room, gable wall (rectangle + triangle), or running track (rectangle + two semicircles) into non-overlapping simple shapes and add their areas.
2. Subtraction (Cut-Outs and Borders):
   • Walls with windows or doors: Calculate the gross wall area and subtract the area of every unpainted window, door, or whiteboard: $A_{\text{net}} = A_{\text{gross}} - A_{\text{cut-outs}}$.
   • External Path or Frame of Uniform Width $w$: When a path of width $w$ surrounds all four sides of a rectangular garden bed ($L \times W$), the path adds $w$ on both the left and right ends and on both the top and bottom edges:

$$L_{\text{outer}} = L + 2w, \quad W_{\text{outer}} = W + 2w, \quad A_{\text{path}} = (L + 2w)(W + 2w) - LW$$

Converting Area into Paint, Turf, or Tile Orders

When ordering paint or surface coatings:

1. Multiply the net area by the number of coats to obtain the total coated area.
2. Divide the total coated area by the coverage rate (in $\text{m}^2\text{/L}$) to find the exact number of litres needed.
3. Divide by the container size and always round up to the next whole tin so you do not run short.

---

Worked Examples in School Contexts

Worked Example 1: School Athletics Track Boundary and Turf Cost

A school sports field is shaped like a stadium running track with two straight parallel sides each $80\text{ m}$ long and two semicircular ends each with a diameter of $42\text{ m}$. Using $\pi \approx \frac{22}{7}$, find (a) the length of one complete lap around the outer boundary, and (b) the cost to re-turf the entire interior field at $14.50 per $\text{m}^2$.

1. Identify radius and diameter: $d = 42\text{ m}$, so $r = 42 \div 2 = 21\text{ m}$.
2. Calculate perimeter (one lap): The two straight sides plus one full circle of circumference $\pi d$ give:

$$P = 2(80) + \frac{22}{7} \times 42 = 160 + 22 \times 6 = 160 + 132 = 292\text{ m}$$

1. Calculate total interior area: Add the central $80\text{ m} \times 42\text{ m}$ rectangle and one full circle of area $\pi r^2$:

$$A = (80 \times 42) + \left(\frac{22}{7} \times 21 \times 21\right) = 3{,}360 + (22 \times 3 \times 21) = 3{,}360 + 1{,}386 = 4{,}746\text{ m}^2$$

1. Calculate turf cost: $4{,}746 \times 14.50 = 68{,}817.00$ ($68,817.00).

Worked Example 2: Painting a Hall Wall with Window Cut-Outs

A rectangular school hall wall is $15\text{ m}$ long and $4\text{ m}$ high. It contains two windows each measuring $2.5\text{ m}$ by $1.2\text{ m}$ and one double door measuring $2.0\text{ m}$ by $2.2\text{ m}$. The wall requires $2$ coats of washable acrylic paint with a spread rate of $12\text{ m}^2\text{/L}$, sold in $4\text{ L}$ tins at $68 per tin. What is the total cost of the paint tins required?

1. Gross wall area: $15 \times 4 = 60\text{ m}^2$.
2. Cut-out areas: Two windows $= 2 \times (2.5 \times 1.2) = 6.0\text{ m}^2$; double door $= 2.0 \times 2.2 = 4.4\text{ m}^2$. Total cut-outs $= 6.0 + 4.4 = 10.4\text{ m}^2$.
3. Net area for 1 coat and 2 coats: $A_{\text{net}} = 60 - 10.4 = 49.6\text{ m}^2$. For $2$ coats, total coverage $= 2 \times 49.6 = 99.2\text{ m}^2$.
4. Litres and tins required: $99.2 \div 12 = 8.267\text{ L}$. Dividing by $4\text{ L}$ per tin gives $8.267 \div 4 = 2.067$ tins, which rounds up to $3$ tins.
5. Total cost: $3 \times 68 = 204$ ($204).