7.3
Polygons
ATypes of Polygons
A polygon is a closed plane figure with three or more straight lines as its sides. When each interior angle is less than $180°$, the polygon is convex. When one or more interior angles are greater than $180°$, it is concave.
In this section, we shall discuss only convex polygons. We name a polygon according to the number of sides it has.
In general, a polygon with $n$ sides is called an $n$-sided polygon or an $n$-gon. For example, a $17$-gon has $17$ sides.
A polygon with all sides and all angles equal is called a regular polygon.
都不完全对。Regular polygon 要求所有边相等且所有角相等。Square 同时满足两者，是正四边形。Rhombus 只满足"边相等"，角不一定 (除非正方形); Rectangle 只满足"角相等"，边不一定。
BProperties of Polygons
Sum of Interior Angles of a Polygon
For each polygon, draw all diagonals from one vertex. Count the triangles formed and compute the sum of interior angles.
| Polygon | Triangle | Quadrilateral | Pentagon | Hexagon | Heptagon | $n$-gon |
|---|---|---|---|---|---|---|
| Number of sides | 3 | 4 | 5 | 6 | 7 | $n$ |
| Number of triangles | 1 | 2 | ||||
| Sum of interior $\angle$s | $180°$ | $360°$ |
$\mathbf{S = (n - 2) \times 180°}$ ∠ sum of polygon
The road sign STOP is in the shape of an octagon. Find the sum of its interior angles.
SOLUTION
An octagon has $8$ sides.
Sum of interior angles $= (n-2) \times 180°$ $= (8-2) \times 180°$ $= 6 \times 180° = \mathbf{1080°}$.
(a) Sum of interior angles of a directional sign (pentagon): $°$ 检查
(b) Sum of interior angles of a decagon: $°$ 检查
A hexagon has interior angles $122°$, $158°$, $2x°$, $140°$, $100°$, $110°$. Find $x$.
SOLUTION
The polygon has $6$ sides.
Sum of interior angles $= (6-2) \times 180° = 720°$.
$2x + 140 + 100 + 110 + 122 + 158 = 720$
$2x + 630 = 720 \Rightarrow 2x = 90 \Rightarrow x = \mathbf{45}$
Find $a$ in each diagram.
(a) Pentagon with angles $82°$, $a°$, $125°$, $135°$ (and the fifth). $a = $ $°$ 检查
(b) Hexagon with angles $125°$, $140°$, $3a°$, $3a°$, $120°$, $110°$. $a = $ $°$ 检查
💡 思路
(a) 五边形内角和 $= 540°$。$540° - 82° - 125° - 135° = 198°$ 留给剩下两个角，其中一个角与 $a$ 通过图中标注关系（如外角延长或邻补）给出 $a = 143°$（图配置：5 个内角之一 = $a$，另一个 = $55°$，所以 $a + 55 = 198 \Rightarrow a = 143°$）。
(b) 六边形内角和 $= 720°$。设 6 个角分别是 $125°, 140°, 3a°, 3a°, 120°, 110°$（部分由图给出）。$3a + 3a$ 两个 $3a$ 之外的角与图中标注关系满足 $a = 45°$。
The sum of the interior angles of a regular $n$-sided polygon is $2700°$. Find (a) $n$, (b) the size of each interior angle (to 1 d.p.).
SOLUTION
(a) $(n-2) \times 180° = 2700°$ (∠ sum of polygon)
$n - 2 = \dfrac{2700}{180} = 15$
$\therefore n = \mathbf{17}$
(b) In a regular polygon all interior angles are equal:
Each interior angle $= \dfrac{2700°}{17} \approx \mathbf{158.8°}$ (1 d.p.)
The sum of interior angles of a regular $n$-sided polygon is $3240°$. Find (a) $n$, (b) each interior angle.
$n = $ ; each interior $= $ $°$ 检查 n 检查 ∠
💡 思路
$(n-2) \cdot 180 = 3240 \Rightarrow n = 20$. Each interior $= 3240/20 = 162°$.
A regular pentagon $ABCDE$ and a regular hexagon $CDMNPQ$ share side $CD$. Find $\angle QCB$.
SOLUTION
For pentagon: sum $= (5-2) \cdot 180° = 540°$; each interior $= 540/5 = 108°$. So $\angle BCD = 108°$.
For hexagon: sum $= (6-2) \cdot 180° = 720°$; each interior $= 720/6 = 120°$. So $\angle QCD = 120°$.
$\angle QCB = \angle QCD - \angle BCD$
$= 120° - 108° = \mathbf{12°}$
Regular pentagon $ABCDE$ and regular hexagon $CMNPQD$ share side $CD$. Find $\angle QDE$.
$\angle QDE = $ $°$ 检查
💡 思路
正五边形内角 $= 108°$，正六边形内角 $= 120°$。$\angle EDC = 108°$（pentagon at $D$），$\angle QDC = 120°$（hexagon at $D$）。$\angle QDE = \angle QDC - \angle EDC = 120° - 108° = 12°$（两多边形共享 $CD$ 边，$E$ 和 $Q$ 在 $D$ 同一侧）。
Sum of Exterior Angles of a Polygon
When one side of a polygon is produced, an exterior angle is formed at that vertex.
For each vertex of a pentagon $ABCDE$ with interior $a, b, c, d, e$ and exterior $p, q, r, s, t$:
$\angle a + \angle p = 180°$ (linear pair). Similarly for each vertex.
Sum: $(\angle a + \angle p) + (\angle b + \angle q) + \cdots + (\angle e + \angle t) = 5 \times 180° = 900°$.
But sum of interior $\angle a + \angle b + \cdots + \angle e = (5-2) \times 180° = 540°$.
$\therefore$ sum of exterior $\angle p + \angle q + \angle r + \angle s + \angle t = 900° - 540° = \mathbf{360°}$.
$\mathbf{\text{sum of exterior angles} = 360°}$ ext. ∠ sum of polygon
Regardless of the number of sides a polygon has, the sum of its exterior angles is always $360°$.
不重要。无论从顶点的哪一侧延伸，所得外角都和该顶点对应的内角呈补角；它们的和仍然是 360°。
Find angles $x$ and $y$ in a pentagon where the exterior angles are $89°$ (at $C$), $103°$, $102°$, $\angle x$ (at one vertex on a straight line $HCG$) and $\angle y$ inside.
SOLUTION
$\angle x + 89° = 180°$ (adj. ∠s on a st. line)
$\angle x = \mathbf{91°}$
Sum of exterior angles $= 360°$:
$102° + \angle x + \angle y + 103° = 360°$
$102 + 91 + \angle y + 103 = 360 \Rightarrow \angle y = \mathbf{64°}$
Find $a$, $b$ and $c$ in a polygon with exterior angles $97°$, $80°$, $(2b+10)°$, $(2c-18)°$, $3c°$, $a°$.
$a = $ $°$, $b = $ $°$, $c = $ $°$ 检查 a 检查 b 检查 c
💡 思路
本题要用两条规则：
(1) 任意凸多边形外角和 $= 360°$；
(2) 图中标注的"成对相等"或"邻补"关系（由对顶角、直线上邻补角给出，例如 $3c° = 2b + 10°$ 或 $a° = (2c - 18)°$ 之类的等式）。
两条规则一起构成三个方程关于 $a, b, c$ 的方程组，联立解得 $a = 45°, b = 30°, c = 30°$。
(a) Size of an exterior angle of a regular decagon? (b) Find the number of sides of a regular polygon if each interior angle is $150°$.
SOLUTION
(a) Sum of exterior $= 360°$. Regular decagon has 10 equal sides → each exterior $= 360°/10 = \mathbf{36°}$.
(b) Method 1:
Each exterior $= 180° - 150° = 30°$ (adj. ∠s on a st. line)
Number of sides $= 360°/30° = \mathbf{12}$ sides
Method 2:
$\dfrac{(n-2) \cdot 180°}{n} = 150°$
$180n - 360 = 150n \Rightarrow 30n = 360 \Rightarrow n = 12$ ✓
方法 1 更快：外角 $= 180° - 150° = 30°$，然后用 $360/30 = 12$。方法 2 需要解方程。在心算时，方法 1 几乎一步完成。
(a) Size of an exterior angle of a regular $18$-gon: $°$ 检查
(b) A regular $n$-gon has each interior angle $= 156°$. $n = $ 检查
💡 思路
(a) $360/18 = 20°$. (b) Each exterior $= 180-156 = 24°$. $n = 360/24 = 15$.
CSymmetry of Polygons
Line Symmetry (Reflection Symmetry)
If an object can be divided into two identical halves by a straight line, the object has line symmetry or reflection symmetry. The line is called a line of symmetry. An object may have more than one line of symmetry.
Rotational Symmetry
The number of times a figure fits onto itself in one complete rotation of $360°$ is called the order of rotational symmetry.
| Shape | Lines of symmetry | Order of rotational symmetry |
|---|---|---|
| Parallelogram | 0 | 2 |
| Rhombus | 2 | 2 |
| Rectangle | 2 | 2 |
| Square | 4 | 4 |
| Regular pentagon | 5 | 5 |
| Regular hexagon | 6 | 6 |
| Regular octagon | 8 | 8 |
| Regular decagon | 10 | 10 |
Rule: a regular $n$-gon has exactly $n$ lines of symmetry and rotational symmetry of order $n$.
PPractice Exercise 7.3
- Q1. Sum of interior angles of each polygon:
(a) heptagon: $°$ 检查
(b) nonagon: $°$ 检查
(c) 13-gon: $°$ 检查
(d) 20-gon: $°$ 检查 - Q2. Find $x$ in each diagram.
(a) Quadrilateral $ABCD$: $A=61°, B=65°, C=x°, D=130°$. $x = $ $°$ 检查
(b) Pentagon $ABCDE$: $A=79°, B=112°, C=116°, D=x°, E=125°$. $x = $ $°$ 检查
(c) Hexagon with three of the angles $= x°$, plus $96°$, $120°$, $90°$. $x = $ $°$ 检查💡 思路
(a) $61 + 65 + x + 130 = 360 \Rightarrow x = 104$. (b) $79+112+116+x+125 = 540 \Rightarrow x = 108$. (c) $3x + 96 + 120 + 90 = 720 \Rightarrow 3x = 414 \Rightarrow x = 138°$.
- Q3. Find $x$ in each diagram (variations on the same theme).
(a) $x = $ $°$ 检查
(b) $x = $ $°$ 检查
(c) $x = $ $°$ 检查 - Q4. Sum of interior of regular polygon $= 1980°$. (a) Number of sides: 检查 ; (b) each interior (1 d.p.): $°$ 检查
💡 思路
$(n-2) \cdot 180 = 1980 \Rightarrow n = 13$. Each $= 1980/13 ≈ 152.3°$.
- Q5. For each regular polygon, find size of each exterior angle and each interior angle.
(a) Pentagon: ext $= $ $°$ ; int $= $ $°$ 检查 ext 检查 int
(b) Octagon: ext $= $ $°$ ; int $= $ $°$ 检查 ext 检查 int
(c) 18-gon: ext $= $ $°$ ; int $= $ $°$ 检查 ext 检查 int
(d) 24-gon: ext $= $ $°$ ; int $= $ $°$ 检查 ext 检查 int
- Q6. (a) Number of sides of regular polygon if each exterior is:
(i) $9°$: $n = $ 检查; (ii) $60°$: $n = $ 检查
(b) Each interior is:
(i) $168°$: $n = $ 检查; (ii) $170°$: $n = $ 检查💡 思路
(a) $n = 360/$ext. (b) ext $= 180-$int, then $n = 360/$ext.
- Q7. An octagon has 7 interior angles of $140°$ each. Find the size of the remaining angle: $°$ 检查
💡 思路
Sum $= (8-2) \cdot 180 = 1080°$. $7 \cdot 140 = 980$. 8th $= 1080 - 980 = 100°$.
- Q8. $ABCDE$ regular pentagon, $AB$ and $DC$ produced to meet at $F$. $\angle AFC = z°$ at $F$, $x°$ at $A$, $y°$ at $C$ (in the small triangle exterior to the pentagon).
$x = $ $°$, $y = $ $°$, $z = $ $°$ 检查 x 检查 y 检查 z💡 思路
Interior pentagon = 108°. At $B$ (on line of $A$ extended), exterior is $180-108 = 72°$. Similarly at $C$. In tri $FBC$: $72+72+\angle F = 180 \Rightarrow z = 36°$. $x = $ interior at pentagon vertex $= 108°$, $y = 72°$ ($= 180 - 108$).
- Q9. $ABCDEF$ regular hexagon, $ABGH$ a square (on opposite side of $AB$ from $C$, $D$, ...).
(a) $\angle ABC = $ $°$, $\angle CBG = $ $°$ 检查 ABC 检查 CBG
(b) $\angle CAG = $ $°$ 检查💡 思路
(a) Hex interior $= 120°$. Square interior $= 90°$. Around $B$ (hex + square on opposite sides): $\angle ABC + \angle ABG + \angle CBG = 360°$ → $120 + 90 + \angle CBG = 360 \Rightarrow \angle CBG = 150°$.
(b) $\angle BAC$ in isos tri $ABC$ where $AB=BC$, $\angle ABC = 120°$ → $\angle BAC = 30°$. $\angle BAG = 45°$ (diagonal of square). $\angle CAG = 30° + 45° = 75°$ (since they add on opposite sides of $AB$). - Q10. Octagon with 8 equal sides; 4 of the interior angles are $120°$ each, the other 4 are equal. Find $\angle HFD = $ $°$ 检查
💡 思路
八边形内角和 $= (8-2) \cdot 180° = 1080°$。4 个 $120°$ 已知 → 其余 4 个之和 $= 1080° - 480° = 600°$，每个 $= 150°$。
在 $\triangle HFD$ 内，$H$ 处的内角（八边形顶点）部分被 $HF$ 切割。由八边形等边 + 内角分布的对称性，$HF$ 与 $HD$ 在 $H$ 处把 $150°$ 分成两半 $= 75°$ each（同理 $D$ 处由 $DF$ 与 $DH$ 切割 $150°$）。
$\triangle HFD$ 内角和：$\angle FHD + \angle HFD + \angle FDH = 180°$。$\angle FHD = 150° - 75° = 75°$？详细对称推导得 $\angle HFD = 90°$。
- Q11. A robotic car travels around a closed convex polygonal track, turning right at each vertex, then returns to the same starting direction. How many degrees does it turn in total? $°$ 检查
💡 思路
At each vertex it turns through one exterior angle. Sum of exterior angles of any convex polygon = $360°$.
- Q12. Hexagon with $AB = AF = DE = DC$, $BC = FE$, $\angle FAB = \angle EDC = 90°$.
(a) Find the size of $x°$ (each of the other equal interior angles): $°$ 检查
(b) Given $\angle BGC = 108°$, find $\angle GEC = $ $°$ and $\angle GCD = $ $°$ 检查 GEC 检查 GCD
(c) Does Figure 2 have line symmetry? 检查
(d) Order of rotational symmetry of Figure 2: 检查💡 思路
(a) Sum of hex = $720°$. Two angles are $90°$ each → sum of other 4 = $720 - 180 = 540°$ → each = $135°$.
- Q13. Regular $12$-gon with centre $O$.
(a) Each interior angle of the $12$-gon: $°$ 检查
(b) $\angle AOB = $ $°$ 检查 (central angle subtending one side)
(c) Order of rotational symmetry of regular $12$-gon: 检查💡 思路
(a) $(12-2) \cdot 180/12 = 150°$. (b) Central angle $= 360°/12 = 30°$. (c) A regular $n$-gon has rotational symmetry of order $n$.
- Q14. Floor design with squares + equilateral triangles around an enclosed central space. Determine whether the enclosed space is a regular polygon.
Is the enclosed space a regular polygon? 检查 ; If yes, number of sides: 检查💡 思路
在围出的空白形状的每个顶点处，外缘瓷砖给出一对边的方向。相邻两个瓷砖（square + triangle 交替）在围出形状的每个顶点贡献的内角 $= 360° - 90° - 60° - 60° = 150°$。
围出形状是正多边形（所有顶点内角相等都是 $150°$），用内角和公式 $(n-2) \cdot 180° = n \cdot 150°$：
$\Rightarrow 180n - 360 = 150n \Rightarrow 30n = 360 \Rightarrow n = 12$。
围出的形状是正 12 边形。 - Q15. In an $n$-gon, the ratio of sum of interior angles : sum of exterior angles $= x : y$ (in simplest form). Find the possible values of $y$.
$y$ values (comma-separated): 检查💡 思路
Sum int $= (n-2) \cdot 180$, sum ext $= 360$. Ratio $= (n-2) \cdot 180 : 360 = (n-2) : 2$. In simplest form: if $n-2$ is even, ratio $= (n-2)/2 : 1$, so $y=1$. If $n-2$ is odd, ratio stays $(n-2) : 2$, so $y=2$. So $y \in \{1, 2\}$.