Lesson 9: Area, Surface Area, and Volume

Area questions sound humble — how much paint, carpet, or sod? — but they train a deep idea: complicated regions can be decomposed into a few standard shapes whose areas you know. This lesson stocks the toolbox (triangles, quadrilaterals, circles, sectors, regular polygons), then practises the decompose-and-add strategy that solves real floor plans and garden plots.

The core formulas

• Triangle: A = (1/2) · b · h, where h is the altitude to the chosen base.
• Parallelogram: A = b · h (perpendicular height, not the slanted side).
• Trapezoid: A = (1/2) · h · (b₁ + b₂) — the height times the *average* of the two parallel sides.
• Circle: A = πr². A sector (pie slice with central angle θ) takes the matching fraction: A = (θ/360°) · πr², and arc length = (θ/360°) · 2πr.
• Regular polygon: A = (1/2) · P · a, where P is the perimeter and a is the apothem (center to the midpoint of a side, perpendicular to it). This works because the polygon splits into congruent triangles from the center.

Surface area and volume build on these: a prism's volume is (base area) · height, and its surface area is the sum of all face areas — Lesson 13 develops the full 3D family. Keep units straight: area is in square units, and every formula must multiply two lengths (or a length squared).

Worked example 1: Trapezoid

A trapezoid has bases 8 cm and 14 cm and height 6 cm. A = (1/2) · 6 · (8 + 14) = 3 · 22 = 66 cm². Read the formula as "height times the average base": the average base is 11, and 6 · 11 = 66 — a useful mental cross-check.

Worked example 2: A sector of a circle

A pizza has radius 10 inches; a slice is cut with a central angle of 72°. The slice is 72/360 = 1/5 of the pizza. Area of slice = (1/5) · π · 10² = 20π ≈ 62.83 in². The crust edge (arc length) is (1/5) · 2π · 10 = 4π ≈ 12.57 in. Fractions of the full circle — that is all a sector ever is.

Worked example 3: Composite figure

A house-shaped silhouette: a rectangle 12 m wide and 7 m tall, topped by a semicircle whose diameter is the 12 m top edge. Rectangle area: 12 · 7 = 84 m². Semicircle: radius 6, area (1/2) · π · 6² = 18π ≈ 56.55 m². Total ≈ 84 + 56.55 = 140.55 m². Strategy notes: the diameter of the semicircle *is* the rectangle's width (12), so the radius is 6, not 12 — the single most common slip in composite problems. Sketch, label every length, compute pieces separately, then add.

Worked example 4: Regular hexagon two ways

A regular hexagon has side length 6. Method 1 (formula): the apothem of a regular hexagon is (s√3)/2 = 3√3 ≈ 5.196; perimeter P = 36; A = (1/2) · 36 · 5.196 ≈ 93.53. Method 2 (decompose): a regular hexagon is six equilateral triangles of side 6; each has area (√3/4) · 36 = 9√3 ≈ 15.59; total 54√3 ≈ 93.53. ✓ Two decompositions, one answer — and a preview of why hexagons appear in Lesson 18's tessellations.

Worked example 5: Area on the coordinate grid

Triangle with vertices (1, 1), (5, 1) and (4, 4). Base from (1, 1) to (5, 1) is horizontal with length 4; the height is the vertical distance from (4, 4) to the base line y = 1, which is 3. A = (1/2) · 4 · 3 = 6 square units. (The shoelace method confirms it: the coordinate sum gives ½ · |25 − 13| = 6.) Choosing a horizontal or vertical side as the base makes grid areas fast.

Geometry at work

Estimating materials is applied area: flooring is sold by the square metre with a waste allowance (typically 10%) because cuts waste material; paint coverage is printed on the tin in m² per litre. Farmers compute field areas from aerial photos by decomposition, and city planners compute roof areas from satellite images the same way to plan solar panel layouts — the "roof" is treated as rectangles and triangles, exactly the strategy of Worked Example 3.

Common mistakes

• Using the diameter as the radius. Halve it before squaring: with d = 12, r² = 36, not 144.
• Forgetting the ½ in triangle and trapezoid formulas. Both come from the fact that two copies of the shape make a parallelogram.
• Using a slanted side as the height. Heights must be perpendicular to the base.
• Unit errors. An answer of "66 cm" for an area should be "66 cm²" — and if your two multiplied lengths are in different units, convert first.

Exam Experience

1. What is the area of a circle with radius 9 cm?

• A) 18π cm²
• B) 81π cm²
• C) 27π cm²
• D) 9π cm²

Answer: B. A = π · 9² = 81π cm².

2. A trapezoid has bases 5 m and 11 m and height 4 m. Its area is:

• A) 64 m²
• B) 40 m²
• C) 32 m²
• D) 22 m²

Answer: C. A = (1/2) · 4 · (5 + 11) = 2 · 16 = 32 m².

3. A sector of a circle with radius 8 has central angle 90°. Its area is:

• A) 16π
• B) 64π
• C) 8π
• D) 32π

Answer: A. (90/360) · π · 64 = (1/4) · 64π = 16π.

Constructed response. A school courtyard is a rectangle 20 m by 14 m. A semicircular stage of diameter 14 m is attached outward along one of the 14 m sides. (a) Find the total area of courtyard plus stage. (b) The stage floor is wooden and the courtyard is paved; paving costs $25 per m² and stage flooring costs $40 per m². Find the total cost.

Scoring guide (4 points):

• 1 point: Courtyard area 20 · 14 = 280 m².
• 1 point: Stage: radius 7 (half of the 14 m diameter), area (1/2) · π · 49 = 24.5π ≈ 76.97 m²; total area ≈ 356.97 m².
• 1 point: Paving cost 280 · $25 = $7,000.
• 1 point: Stage cost ≈ 76.97 · $40 ≈ $3,078.80; total ≈ $10,078.80 (accept reasonable rounding).

Going deeper: Shaded regions — areas that subtract

Many real regions are what is *left over*: a square patio with a circular fountain removed, a metal plate with holes drilled out. The method is decompose-and-add with a minus sign: compute the total, compute the removed parts, subtract. Example: a square of side 10 has a circle of radius 5 inscribed (tangent to all four sides). The region in the corners — square minus circle — has area 100 − 25π ≈ 100 − 78.54 = 21.46 square units. (Hold this number; Lesson 16 turns it into a probability.) Second example: an annulus (ring) with outer radius 9 and inner radius 5 has area π(9²) − π(5²) = π(81 − 25) = 56π ≈ 175.93. The pattern π(R² − r²) is worth remembering as "pi times the difference of squares," not the square of the difference: π(9 − 5)² = 16π would be badly wrong.

Gardeners price gravel for paths this way, and machinists compute washer blanks the same way. The habit to build: shade the target region in the diagram first — most subtraction errors are really "subtracted the wrong piece" errors, and shading prevents them.

Composite perimeters and the L-shaped room

Area decomposes by adding and subtracting; perimeter does not follow the pieces — it must be traced around the outside. An L-shaped room is a 10 m by 7 m rectangle with a 4 m by 3 m rectangular notch cut from one corner. Its area is easy by subtraction: 70 − 12 = 58 m². Its perimeter is subtler: trace the boundary and you will find the notch contributes two *new* edges (4 m and 3 m) while removing matching lengths from the original outline, so the perimeter is unchanged from the full rectangle: 2 · (10 + 7) = 34 m. That is a general and useful fact — a rectangular notch cut into a corner (or into a side, making a U) leaves the perimeter equal to the bounding rectangle's, because every inward jog is matched by the outward jog that cancels it. Skirting-board and fencing estimates depend on this: never compute a composite perimeter by adding the perimeters of the pieces (the shared edges would be counted twice and the interior edges wrongly). Trace, don't add.

Final drill: The circular segment — sector minus triangle

The region between a chord and its arc is a segment of the circle, and its area is a subtraction problem: sector minus the triangle formed by the two radii and the chord. Take radius 10 and a central angle of 90°. The sector is a quarter circle: (1/4) · π · 100 = 25π ≈ 78.54. The triangle is right-angled with legs 10 and 10: area (1/2) · 10 · 10 = 50. So the segment's area is 25π − 50 ≈ 28.54 square units. Arched window tops, tunnel cross-sections and the liquid surface in a partly filled horizontal pipe are all segments, and every one submits to this same subtraction. The strategy generalises the lesson: whenever a region resists a direct formula, build it from a region you know (sector) minus a region you know (triangle) — decomposition works with curved boundaries too, as long as one of your standard pieces carries the curve.