Chapter 11: Area of Polygons and Circles
Learning Objectives
- Apply area formulas for triangles, parallelograms, trapezoids, rhombuses, and kites
- Use the SAS triangle area formula $A = \frac{1}{2}ab\sin C$
- Find the area of regular polygons using the apothem formula $A = \frac{1}{2} \cdot a \cdot P$
- Calculate arc length and sector area using degree and radian measures
- Find the area of circular segments (sector minus triangle)
- Decompose composite figures into simpler shapes to compute total area
11.1 Areas of Triangles and Quadrilaterals
Area Formula Summary
| Shape | Formula | Key Variables |
|---|---|---|
| Triangle | $A = \frac{1}{2}bh$ | $b$ = base, $h$ = height |
| Triangle (SAS) | $A = \frac{1}{2}ab\sin C$ | $a$, $b$ = two sides, $C$ = included angle |
| Parallelogram | $A = bh$ | $b$ = base, $h$ = perpendicular height |
| Rectangle | $A = lw$ | $l$ = length, $w$ = width |
| Rhombus / Kite | $A = \frac{1}{2}d_1 d_2$ | $d_1$, $d_2$ = diagonals |
| Trapezoid | $A = \frac{1}{2}(b_1 + b_2)h$ | $b_1$, $b_2$ = parallel bases, $h$ = height |
The height of any polygon is always the perpendicular distance between a base and the opposite vertex (or parallel side). A slant side is never the height.
Example 11.1 — Triangle Area
A triangle has base $b = 12$ and height $h = 9$. Find the area.
$$A = \frac{1}{2}bh = \frac{1}{2}(12)(9) = \frac{108}{2} = \mathbf{54 \text{ sq units}}$$Example 11.2 — Trapezoid Area
A trapezoid has parallel bases $b_1 = 8$ and $b_2 = 14$, with height $h = 6$. Find the area.
$$A = \frac{1}{2}(b_1 + b_2)h = \frac{1}{2}(8 + 14)(6) = \frac{1}{2}(22)(6) = \mathbf{66 \text{ sq units}}$$Example 11.3 — Rhombus Area
A rhombus has diagonals of length $10$ and $16$. Find the area.
$$A = \frac{1}{2}d_1 d_2 = \frac{1}{2}(10)(16) = \mathbf{80 \text{ sq units}}$$A triangle has two sides $a = 7$, $b = 10$ with included angle $C = 60°$. Find the area using the SAS formula.
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Four common shapes — triangle, parallelogram, trapezoid, and rhombus — each with dashed height lines showing the perpendicular dimension used in the area formula.
Figure 11.1 — Area Formulas: Triangle, Parallelogram, Trapezoid, Rhombus
11.2 Regular Polygons and Apothem
Regular Polygon and Apothem
A regular polygon has all sides equal and all interior angles equal. The apothem $a$ is the perpendicular distance from the center of the polygon to any side.
The area of a regular $n$-gon with side length $s$ and apothem $a$ is:
$$A = \frac{1}{2} \cdot a \cdot P = \frac{1}{2} \cdot a \cdot ns$$where $P = ns$ is the perimeter. Each interior angle of a regular $n$-gon measures $\dfrac{(n-2) \cdot 180°}{n}$.
To find the apothem of a regular polygon with side $s$ and $n$ sides, use the relationship:
$$a = \frac{s}{2} \cdot \cot\!\left(\frac{180°}{n}\right) = \frac{s}{2\tan(180°/n)}$$Example 11.4 — Regular Hexagon Area
A regular hexagon has side length $6$. Find the apothem and the area.
For a regular hexagon ($n = 6$), the apothem $= \frac{s\sqrt{3}}{2} = \frac{6\sqrt{3}}{2} = 3\sqrt{3}$.
Perimeter $= 6 \times 6 = 36$.
$$A = \frac{1}{2}aP = \frac{1}{2}(3\sqrt{3})(36) = 54\sqrt{3} \approx \mathbf{93.53 \text{ sq units}}$$Example 11.5 — Regular Pentagon Area
A regular pentagon has side length $s = 8$ and apothem $a = 5.5$. Find the area.
Perimeter $= 5 \times 8 = 40$.
$$A = \frac{1}{2}(5.5)(40) = \frac{1}{2}(220) = \mathbf{110 \text{ sq units}}$$A regular octagon has side length $s = 4$ and apothem $a \approx 4.83$. Find the area.
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Hexagon shortcut: For a regular hexagon with side $s$, the apothem is always $\frac{s\sqrt{3}}{2}$ and the area is $\frac{3s^2\sqrt{3}}{2}$. Memorize this for the SAT/ACT — it appears frequently.
Regular hexagon with center marked and one dashed apothem drawn to the midpoint of a side.
Figure 11.2 — Regular Hexagon with Apothem
11.3 Circumference, Arc Length, and Sector Area
Arc Length and Sector Area Formulas
- Circumference: $C = 2\pi r = \pi d$
- Arc Length (degrees): $L = \dfrac{\theta}{360°} \cdot 2\pi r$
- Arc Length (radians): $L = r\theta$ (when $\theta$ is in radians)
- Sector Area (degrees): $A = \dfrac{\theta}{360°} \cdot \pi r^2$
- Sector Area (radians): $A = \dfrac{1}{2}r^2\theta$
- Circular Segment Area: $A_{\text{segment}} = A_{\text{sector}} - A_{\triangle}$
Example 11.6 — Arc Length and Sector Area
A circle has $r = 10$ and central angle $\theta = 120°$. Find the arc length and sector area.
$$L = \frac{120}{360} \cdot 2\pi(10) = \frac{1}{3} \cdot 20\pi = \frac{20\pi}{3} \approx \mathbf{20.94}$$ $$A_{\text{sector}} = \frac{120}{360} \cdot \pi(10)^2 = \frac{1}{3} \cdot 100\pi = \frac{100\pi}{3} \approx \mathbf{104.72}$$Example 11.7 — Circular Segment Area
Find the area of a circular segment where $r = 8$ and the central angle $= 90°$.
Sector area: $A_{\text{sector}} = \frac{90}{360} \cdot \pi(8)^2 = \frac{1}{4} \cdot 64\pi = 16\pi$
Triangle area: The triangle formed by the two radii and the chord is a right isosceles triangle with legs $8$: $A_{\triangle} = \frac{1}{2}(8)(8) = 32$
$$A_{\text{segment}} = 16\pi - 32 \approx 50.27 - 32 = \mathbf{18.27 \text{ sq units}}$$A pizza slice (sector) has $r = 9$ inches and central angle $= 40°$. Find the area of the slice and the arc length of the crust.
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Sector area $= \dfrac{40}{360} \cdot \pi(9)^2 = \dfrac{1}{9} \cdot 81\pi = 9\pi \approx \mathbf{28.27}$ sq inches.
Radians vs. Degrees: On the AP exam, know both forms. If $\theta$ is in radians: arc length $= r\theta$ and sector area $= \frac{1}{2}r^2\theta$. If $\theta$ is in degrees, use the fraction $\frac{\theta}{360}$. Converting: $\theta_{\text{rad}} = \theta_{\text{deg}} \cdot \frac{\pi}{180}$.
Circle with a 120° sector shaded — two purple radii and the arc marking the sector boundary.
Figure 11.3 — Circle with Sector from 0° to 120°
11.4 Composite Figures
A composite figure is made up of two or more basic shapes. To find its area, decompose the figure into recognizable shapes, compute each area separately, then add (or subtract if a piece is removed).
Strategy for Composite Areas
- Draw and label the figure, identifying all component shapes.
- List each component with its area formula and known dimensions.
- Compute each area separately.
- Combine: add areas for additive pieces; subtract areas for cut-outs or holes.
Example 11.8 — Running Track (Rectangle + Semicircles)
A running track consists of a rectangle $100\text{ m} \times 50\text{ m}$ with two semicircles of radius $r = 25\text{ m}$ at each end.
Rectangle area: $A_1 = 100 \times 50 = 5{,}000\text{ m}^2$
Two semicircles $=$ one full circle: $A_2 = \pi(25)^2 = 625\pi \approx 1{,}963.5\text{ m}^2$
$$A_{\text{total}} = 5{,}000 + 1{,}963.5 \approx \mathbf{6{,}963.5 \text{ m}^2}$$Example 11.9 — Annulus (Circle with Hole)
Find the shaded area of a large circle with $r = 8$ from which a small circle with $r = 3$ has been removed.
$$A = \pi(8)^2 - \pi(3)^2 = 64\pi - 9\pi = 55\pi \approx \mathbf{172.79 \text{ sq units}}$$A house shape consists of a square $8 \times 8$ with an equilateral triangle roof of side $8$ on top. Find the total area.
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Equilateral triangle (side 8): $A_2 = \frac{\sqrt{3}}{4}(8)^2 = \frac{64\sqrt{3}}{4} = 16\sqrt{3} \approx 27.71$.
Total $\approx 64 + 27.71 = \mathbf{91.71}$ sq units.
AP Tip: For composite areas on the AP exam, always draw and label the figure first. Write out each component's formula before plugging in numbers — this prevents errors and earns partial credit even if the arithmetic goes wrong.
Practice Problems
A parallelogram has base $15$ and height $8$. Find its area.
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A kite has diagonals $d_1 = 14$ and $d_2 = 9$. Find its area.
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A regular hexagon has side length $s = 10$. Find the exact area.
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A circle has $r = 12$. Find the arc length and sector area for a central angle of $150°$.
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Find the area of a circular segment with $r = 10$ and central angle $= 60°$.
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A regular octagon has side $s = 6$ and apothem $a \approx 7.24$. Find the area.
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A composite figure is a rectangle $10 \times 6$ with a semicircle of diameter $6$ on one end. Find the total area.
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A square of side $12$ has a circle of radius $4$ inscribed at its center (removed). Find the remaining area. Then find what fraction of the square the circle occupies. Express the fraction in terms of $\pi$.
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Circle area $= \pi(4)^2 = 16\pi$.
Remaining area $= 144 - 16\pi \approx 144 - 50.27 \approx \mathbf{93.73}$ sq units.
Fraction occupied by circle $= \dfrac{16\pi}{144} = \dfrac{\pi}{9} \approx 34.9\%$.
📋 Chapter Summary
Area Formulas
$A = \frac{1}{2}bh$ — base times height divided by 2. Height must be perpendicular to base.
$A = bh$ — base times perpendicular height. Note: height is not the slant side.
$A = \frac{1}{2}(b_1 + b_2)h$ — average of the two parallel bases times height.
$A = \frac{1}{2}ap$ where $a$ = apothem (center to midpoint of side) and $p$ = perimeter.
Circle and Composite
$A = \pi r^2$, $C = 2\pi r$. Memorize both — area uses $r^2$, circumference is linear in $r$.
Break into simpler shapes (rectangles, triangles, circles). Add areas for combined figures; subtract for holes or removed sections.