7.2 Perimeter, Circumference & Area of Plane and Composite Figures
Key Takeaways
- Every parallelogram area formula uses the perpendicular height, never the slanted side length.
- The trapezoid area formula A = ½(b₁ + b₂)h averages the two parallel bases and multiplies by the perpendicular distance between them.
- A rhombus or kite can be found with A = ½d₁d₂, half the product of the diagonals.
- Circumference is C = 2πr = πd and area is A = πr², so a problem giving diameter requires halving before squaring.
- Composite figures are handled by decomposition into known shapes or by subtraction of a removed region, and shared interior edges never count toward perimeter.
7.2 Perimeter, Circumference & Area of Plane and Composite Figures
Because test 025 supplies no reference sheet, these formulas must be memorized outright. Skill 2 names triangles, rectangles, trapezoids, parallelograms, and rhombi specifically; skill 4 adds perimeter and circumference.
The core formulas
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| FIGURE AREA PERIMETER |
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| Rectangle A = lw P = 2l + 2w |
| Square A = s^2 P = 4s |
| Triangle A = (1/2)bh P = a + b + c |
| Parallelogram A = bh P = 2a + 2b |
| Trapezoid A = (1/2)(b1 + b2)h P = sum of all four sides |
| Rhombus A = bh or (1/2)d1*d2 P = 4s |
| Kite A = (1/2)d1*d2 P = 2a + 2b |
| Circle A = pi*r^2 C = 2*pi*r = pi*d |
| Regular n-gon A = (1/2)*a*P (apothem) P = n*s |
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The height trap
[!WARNING] In every area formula, h is the perpendicular height, measured at a right angle to the base — never the length of a slanted side.
A parallelogram with base 10 cm and a slanted side of 8 cm does not have area 80 cm². If its perpendicular height is 6 cm, the area is 10 · 6 = 60 cm². The slant side is used for perimeter; the perpendicular height is used for area. Items supply both numbers precisely to see which one you pick.
The same applies to triangles. In an obtuse triangle the height may fall outside the triangle, dropping to an extension of the base. The formula A = ½bh still holds; the height is simply measured to the extended base line.
For a trapezoid, h is the perpendicular distance between the two parallel sides, and the legs are irrelevant to area.
Why the formulas are related
Understanding the derivations makes them harder to forget and directly supports Competency 5 items about building conceptual understanding.
- Parallelogram → rectangle. Cut a right triangle from one end and slide it to the other; the result is a rectangle with the same base and height. Hence A = bh.
- Triangle → parallelogram. Two congruent triangles joined along a side form a parallelogram, so a triangle is half of one: A = ½bh.
- Trapezoid → parallelogram. Two congruent trapezoids rotated together form a parallelogram of base (b₁ + b₂) and height h, so one trapezoid is ½(b₁ + b₂)h. Equivalently, the area equals the average of the bases times the height.
- Circle → parallelogram. Slice a circle into thin sectors and interleave them; the shape approaches a parallelogram of height r and base πr (half the circumference), giving A = πr².
Circles
The two circle formulas are confused more than any other pair.
- Circumference C = 2πr = πd — a length, measured in linear units
- Area A = πr² — a region, measured in square units
When a problem gives the diameter, halve it before using r. A circle of diameter 14 has r = 7, so C = 14π ≈ 43.98 and A = 49π ≈ 153.94. Using d in place of r in the area formula gives 196π, four times too large — and that value always appears as a distractor.
Arc length and sector area are proportional parts of the whole:
arc length = (θ/360°) · 2πr and sector area = (θ/360°) · πr²
A 90° sector of a circle with r = 6 has arc length (1/4)(12π) = 3π and area (1/4)(36π) = 9π.
Composite figures
Two strategies, and choosing the right one saves time.
Decomposition (add). Split the figure into non-overlapping familiar shapes and sum their areas.
An L-shaped room: a 12 ft × 8 ft rectangle joined to a 5 ft × 6 ft rectangle. Area = 96 + 30 = 126 ft².
Subtraction (remove). Find the area of a bounding shape and subtract the removed region.
A 20 m × 14 m rectangular garden with a circular fountain of radius 3 m removed. Area = 280 − 9π ≈ 280 − 28.27 = 251.73 m².
A running track shape: a 40 m × 25 m rectangle with a semicircle of diameter 25 m on each end. The two semicircles form a full circle of radius 12.5 m. Area = (40)(25) + π(12.5)² = 1,000 + 490.87 = 1,490.87 m².
Perimeter of a composite figure
Perimeter follows the outer boundary only. Interior edges where two pieces meet are not part of the perimeter, even though they were used to compute area.
For the track above, the perimeter is the two straight 40 m sides plus the two semicircular arcs, which together form a full circumference:
P = 2(40) + π(25) = 80 + 78.54 = 158.54 m
The 25 m segments where each semicircle joins the rectangle are interior and are not counted. Adding them is the most common composite-perimeter error.
Working backward from area
Items often give the area and ask for a dimension.
A triangle has area 84 cm² and base 24 cm. Find its height. 84 = ½(24)h → 84 = 12h → h = 7 cm.
A circle has area 64π. Find its circumference. πr² = 64π → r² = 64 → r = 8, so C = 2π(8) = 16π.
Do not take the square root of the area to find a radius directly — divide by π first.
A parallelogram has a base of 15 inches, a slanted side of 10 inches, and a perpendicular height of 8 inches. What are its area and perimeter?
A circular pizza has a diameter of 18 inches. What is its area, to the nearest square inch?
A patio consists of a 16 ft by 9 ft rectangle with a semicircle of diameter 9 ft attached to one 9-foot end. What is the perimeter of the patio, to the nearest tenth of a foot?