7.3 Perimeter, Area, Volume & Applied Geometry

Perimeter, Area, Surface Area, and Volume Calculations

Exam Focus: Perimeter, area and volume problems sit in the Geometry & Measurement cluster of Mathematics Problem Solving and in TASK Mathematics. At the Advanced and TASK levels Pearson provides a mathematics reference sheet with the formulas needed to solve problems, so the skill being tested is choosing and applying the right formula. Typical items are applied word problems about fencing, carpeting, tiling, framing and packaging. Items test one-dimensional boundary perimeter, two-dimensional surface area, three-dimensional solid capacity, decomposition of composite irregular figures, and the mathematical effects of proportional scaling.

Applied geometry questions on the Stanford 10 connect algebraic formulas to practical, physical situations. Success requires distinguishing between linear boundary measurements (units), flat surface coverage (square units), and spatial capacity (cubic units), while recognizing when problem diagrams require decomposing complex shapes into simpler geometric components.

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Perimeter of Polygons and Circumference of Circles

Perimeter is the total distance around the outside boundary of a two-dimensional closed figure. It is a linear, one-dimensional measurement expressed in standard linear units ($\text{in}$, $\text{ft}$, $\text{cm}$, $\text{m}$).

Polygon Perimeter Formulas

• General Polygon: Sum of all exterior side lengths: $P = s_1 + s_2 + s_3 + \dots + s_n$
• Rectangle: Two lengths plus two widths: $P = 2l + 2w = 2(l + w)$
• Square / Rhombus: Four congruent sides: $P = 4s$
• Regular $n$-gon: Number of sides multiplied by side length: $P = n \cdot s$

Circumference of a Circle

The perimeter of a circle is designated as its circumference ($C$). The mathematical constant $\pi$ (pi) represents the invariant ratio of a circle's circumference to its diameter: $\pi = \frac{C}{d} \approx 3.14159\dots$ (approximated as $3.14$ or $\frac{22}{7}$). $C = \pi d = 2\pi r$

Test Trap: Perimeter of a Semicircle: Items may ask for the total perimeter of a semicircular garden or window. The perimeter consists of the curved arc ($\frac{1}{2}\pi d$ or $\pi r$) plus the straight diameter base ($d$): $P_{\text{semicircle}} = \pi r + 2r = \frac{1}{2}\pi d + d$ Omitting the straight baseline is one of the most common student mistakes.

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Area Formulas for Two-Dimensional Polygons and Circles

Area measures the amount of two-dimensional surface covered by a closed shape, expressed in square units ($\text{in}^2$, $\text{ft}^2$, $\text{cm}^2$, $\text{m}^2$).

The Critical Role of Perpendicular Height

In parallelograms, triangles, and trapezoids, the height ($h$) must be strictly perpendicular to the chosen base ($90^\circ$ angle):

• In a right triangle, the two perpendicular legs serve directly as the base and height ($A = \frac{1}{2} a b$).
• In an obtuse triangle, the altitude from the obtuse vertex often falls outside the triangle's body along an extension of the base line.
• In parallelograms, the slant side is always longer than the perpendicular height; using the slant side produces an incorrect, inflated area.

Circle Area Calculations

When calculating circle area ($A = \pi r^2$):

• If the problem provides the diameter, first divide by 2 to determine the radius ($r = \frac{d}{2}$).
• Square the radius before multiplying by $\pi$.
• Example: For a circle with diameter $14\text{ cm}$, $r = 7\text{ cm}$. Using $\pi \approx \frac{22}{7}$: $A = \frac{22}{7} \times (7)^2 = \frac{22}{7} \times 49 = 22 \times 7 = 154\text{ cm}^2$

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Decomposing Irregular Composite Two-Dimensional Shapes

Standardized tests routinely feature irregular figures such as L-shaped floor plans, cross layouts, or shaded border regions. These shapes are solved using two primary decomposition strategies.

Decomposition Strategies:
├── Additive Decomposition (Figure Partitioning)
│   ├── Step 1: Divide irregular shape into non-overlapping rectangles/triangles.
│   ├── Step 2: Deduce missing boundary lengths using parallel sides.
│   └── Step 3: Compute individual component areas and sum them: A_total = A_1 + A_2.
└── Subtractive Decomposition (Border / Void Removal)
    ├── Step 1: Calculate the gross area of the outer bounding figure.
    ├── Step 2: Calculate the area of the unshaded inner cut-out / void.
    └── Step 3: Subtract inner area from outer area: A_shaded = A_outer - A_inner.

Worked Example: Additive Decomposition of an L-Shaped Room

An L-shaped living room has a bottom horizontal edge of 12 meters, a left vertical edge of 10 meters, a top horizontal edge of 5 meters, and a right vertical edge of 4 meters.

1. Find missing dimensions:
   • The inner horizontal edge is $12 - 5 = 7$ meters.
   • The inner vertical edge is $10 - 4 = 6$ meters.
2. Partition into two non-overlapping rectangles:
   • Vertical cut: Left rectangle is $5\text{ m} \times 10\text{ m} = 50\text{ m}^2$. Right rectangle is $7\text{ m} \times 4\text{ m} = 28\text{ m}^2$.
3. Sum the areas: $A_{\text{total}} = 50\text{ m}^2 + 28\text{ m}^2 = 78\text{ m}^2$

Worked Example: Subtractive Decomposition of a Border Walkway

A rectangular garden measures 16 feet by 10 feet. A brick border 2 feet wide surrounds the garden on all four sides. What is the area of the brick border alone?

1. Inner garden area: $A_{\text{inner}} = 16 \times 10 = 160\text{ ft}^2$.
2. Outer dimensions (adding 2 feet to BOTH sides of each dimension): $\text{Outer Length} = 16 + 2 + 2 = 20\text{ ft}, \quad \text{Outer Width} = 10 + 2 + 2 = 14\text{ ft}$
3. Outer gross area: $A_{\text{outer}} = 20 \times 14 = 280\text{ ft}^2$.
4. Border area: $A_{\text{border}} = A_{\text{outer}} - A_{\text{inner}} = 280 - 160 = 120\text{ ft}^2$

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Three-Dimensional Solids: Surface Area and Volume

Three-dimensional geometry measures both the boundary skin (surface area) and internal capacity (volume).

Surface Area of Rectangular Prisms and Cubes

Surface area ($SA$) is the sum of the areas of all external two-dimensional polygonal faces:

• Rectangular Prism: Composed of 6 faces arranged in 3 congruent opposite pairs (top/bottom, front/back, left/right): $SA = 2lw + 2lh + 2wh = 2(lw + lh + wh)$
• Cube: Composed of 6 identical square faces: $SA = 6s^2$

Volume of Rectangular Prisms and Composite Solids

Volume ($V$) measures the number of unit cubes that completely fill the interior of a three-dimensional solid:

• General Prism Formula: Base area ($B$) multiplied by height ($h$): $V = B \times h$
• Rectangular Prism: Length multiplied by width multiplied by height: $V = l \times w \times h$
• Cube: Side length cubed: $V = s^3$

Unit Cube Packing

A common problem type asks how many small cubes of side length $\frac{1}{2}$ inch can fit inside a rectangular box measuring 4 inches by 3 inches by 2 inches.

• Method: Determine how many cubes fit along each linear edge: $\text{Length: } 4 \div \frac{1}{2} = 8 \text{ cubes}, \quad \text{Width: } 3 \div \frac{1}{2} = 6 \text{ cubes}, \quad \text{Height: } 2 \div \frac{1}{2} = 4 \text{ cubes}$
• Total cubes $= 8 \times 6 \times 4 = 192\text{ cubes}$.

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Dimensional Scaling Effects on Perimeter, Area, and Volume

One of the highest-yield concepts on standardized tests is how geometric measures change when linear dimensions are scaled proportionally by a factor of $k$.

Crucial Rule: If all linear dimensions of a solid are multiplied by $k$, its surface area is multiplied by $k^2$, and its volume is multiplied by $k^3$. Doubling the dimensions of a water tank multiplies its perimeter by 2, quadruples its surface area ($2^2 = 4$), and octuples its volume capacity ($2^3 = 8$).