13.3 Regular Polygons, Quadrilaterals, Parallelograms, Trapezoids, Rhombuses, Rectangles & Squares

13.3 Regular Polygons, Quadrilaterals, Parallelograms, Trapezoids, Rhombuses, Rectangles & Squares

---

1. Convex & Regular Polygons (Çokgenler ve Düzgün Çokgenler)

Universal Laws for Any Convex $n$-gon ($n \ge 3$)

• Sum of Interior Angles (İç Açılar Toplamı): Drawing all $n - 3$ diagonals from a single vertex partitions any $n$-gon into $n - 2$ non-overlapping triangles, giving:

$$\text{Sum of Interior Angles} = (n - 2) \cdot 180^\circ$$

• Sum of Exterior Angles (Dış Açılar Toplamı): Regardless of how large $n$ is, the exterior angles (one at each vertex) always sum to:

$$\text{Sum of Exterior Angles} = 360^\circ$$

• Total Number of Diagonals (Köşegen Sayısı): Each of the $n$ vertices connects to $n - 3$ non-adjacent vertices, and dividing by $2$ avoids double-counting:

$$\text{Total Diagonals} = \frac{n(n - 3)}{2}$$

Regular Polygons (Düzgün Çokgenler)

A polygon is regular (düzgün) if all $n$ sides are equal in length and all $n$ interior angles are equal. Because the exterior angle sum is always $360^\circ$, always compute the exterior angle first:

$$\alpha_{\text{ext}} = \frac{360^\circ}{n}, \qquad \alpha_{\text{int}} = 180^\circ - \frac{360^\circ}{n} = \frac{(n - 2) \cdot 180^\circ}{n}$$

Two Useful Regular Polygons: Regular Pentagon & Regular Hexagon

1. Regular Pentagon (Düzgün Beşgen, $n = 5$):
   • Angles: $\alpha_{\text{ext}} = \frac{360^\circ}{5} = 72^\circ$ and $\alpha_{\text{int}} = 108^\circ$.
   • Diagonals & Isosceles Triangles: All $\frac{5(2)}{2} = 5$ diagonals have the exact same length. Drawing any single diagonal (such as $AC$) cuts off an isosceles triangle $\triangle ABC$ with angles $36^\circ\text{-}36^\circ\text{-}108^\circ$. Drawing two diagonals from the same vertex ($AC$ and $AD$) trisects the $108^\circ$ vertex angle into $36^\circ\text{-}36^\circ\text{-}36^\circ$.
   • Axis of Symmetry: An altitude dropped from any vertex to the opposite side bisects the $108^\circ$ vertex into $54^\circ\text{-}54^\circ$ and bisects the opposite side at $90^\circ$.
2. Regular Hexagon (Düzgün Altıgen, $n = 6$):
   • Angles: $\alpha_{\text{ext}} = \frac{360^\circ}{6} = 60^\circ$ and $\alpha_{\text{int}} = 120^\circ$.
   • Short Diagonal (Kısa Köşegen): Connecting $A$ to $C$ forms a $30^\circ\text{-}30^\circ\text{-}120^\circ$ triangle $\triangle ABC$, so the short diagonal has length $a\sqrt{3}$ and meets the adjacent side $CD$ at $90^\circ$!
   • Long / Main Diagonal (Uzun Köşegen): Connecting opposite vertices ($AD, BE, CF$) passes through the center $O$, is parallel to two sides, and has length $2a$. The three main diagonals divide the hexagon into 6 congruent equilateral triangles of side $a$:

$$\text{Area}(\text{Regular Hexagon}) = 6 \cdot \frac{a^2\sqrt{3}}{4} = \frac{3a^2\sqrt{3}}{2}$$

---

2. General Quadrilaterals, Parallelograms, Rhombuses & Kites

General Quadrilaterals (Dörtgenler)

• Area via Diagonals: If diagonals $e = \lvert AC \rvert$ and $f = \lvert BD \rvert$ intersect at angle $\alpha$, then:

$$\text{Area}(ABCD) = \frac{1}{2} e \cdot f \cdot \sin\alpha$$

• Orthogonal Diagonals Law (Köşegenleri Dik Kesişen Dörtgen): If $AC \perp BD$ (where $\alpha = 90^\circ$), the sums of the squares of opposite sides are equal:

$$a^2 + c^2 = b^2 + d^2 \qquad \text{and} \qquad \text{Area}(ABCD) = \frac{e \cdot f}{2}$$

(Note: This identity also holds if $\triangle BCD$ is folded inward across diagonal $BD$ to form a concave quadrilateral with $AC \perp BD$.)

Parallelogram (Paralelkenar)

In parallelogram $ABCD$ ($AB \parallel DC$ and $AD \parallel BC$):

• Opposite sides and opposite angles are equal; consecutive angles sum to $180^\circ$; diagonals bisect each other (so $\lvert AO \rvert = \lvert OC \rvert$ and $\lvert BO \rvert = \lvert OD \rvert$).
• Adjacent Angle Bisectors: The bisectors of two consecutive angles (e.g., $\angle A$ and $\angle D$) always intersect at $90^\circ$, and their intersection point lies directly on the midline (orta taban) parallel to the bases.
• Half-Area Triangle Rule: For any point $E$ on side $DC$, the triangle $\triangle ABE$ shares base $AB$ and height $h$ with the parallelogram, so:

$$\text{Area}(\triangle ABE) = \frac{1}{2}\text{Area}(ABCD)$$

• Midpoint Trisection of the Diagonal: If segments from vertex $A$ connect to the midpoints $K$ of $DC$ and $L$ of $BC$, those two segments trisect diagonal $BD$ into three equal parts ($m, m, m$)!

Rhombus (Eşkenar Dörtgen) & Kite (Deltoid)

• Rhombus (Eşkenar Dörtgen): A parallelogram with all four sides equal ($a$). Its diagonals bisect each other at $90^\circ$ ($e \perp f$) and bisect the vertex angles. Its area is $\text{Area} = \frac{e \cdot f}{2} = a \cdot h$.
• Kite (Deltoid): Formed by two isosceles triangles sharing a common base (where $\lvert AB \rvert = \lvert AD \rvert$ and $\lvert CB \rvert = \lvert CD \rvert$). The symmetry diagonal $AC$ bisects angles $\angle A$ and $\angle C$, is perpendicular to the cross diagonal $BD$ ($e \perp f$), and bisects $BD$.

---

3. Trapezoids (Yamuk), Rectangles (Dikdörtgen) & Squares (Kare)

Trapezoid (Yamuk, $AB \parallel DC$ with lower base $a >$ upper base $c$)

• Midsegment (Orta Taban $EF$) & Diagonal-Midpoint Segment ($KL$):

$$\lvert EF \rvert = \frac{a + c}{2}, \qquad \lvert KL \rvert = \frac{a - c}{2}, \qquad \text{Area}(ABCD) = \frac{a + c}{2} \cdot h$$

• Four-Triangle Diagonal Area Law: The diagonals $AC$ and $BD$ divide any trapezoid into four triangles: $S_{\text{bottom}}$ (on base $a$), $S_{\text{top}}$ (on base $c$), and two lateral wings $S_{\text{left}}$ and $S_{\text{right}}$. Always:

$$S_{\text{left}} = S_{\text{right}} = S, \qquad S^2 = S_{\text{top}} \cdot S_{\text{bottom}}, \qquad \frac{S_{\text{top}}}{S_{\text{bottom}}} = \left(\frac{c}{a}\right)^2$$

• Isosceles Trapezoid (İkizkenar Yamuk, $\lvert AD \rvert = \lvert BC \rvert$): Dropping altitudes $DH \perp AB$ and $CK \perp AB$ from the upper vertices creates two congruent outer right triangles of base $\frac{a - c}{2}$ and leaves a remaining segment $\lvert AK \rvert = \lvert HB \rvert = \frac{a+c}{2}$, while $\lvert AH \rvert=\lvert KB \rvert=\frac{a-c}{2}$. Furthermore, if an isosceles trapezoid has perpendicular diagonals ($AC \perp BD$), then $h = \frac{a + c}{2}$ and $\text{Area} = h^2$.

Rectangle (Dikdörtgen) & Square (Kare)

• British Flag Theorem (İç/Dış Nokta Özelliği in a Rectangle or Square): For any point $P$ inside, on, or outside rectangle $ABCD$, the sum of squared distances to opposite corners is equal:

$$\lvert PA \rvert^2 + \lvert PC \rvert^2 = \lvert PB \rvert^2 + \lvert PD \rvert^2$$

• Square (Kare): All 4 sides equal $a$, all 4 corner angles are $90^\circ$, and the diagonals are equal ($e = f = a\sqrt{2}$), perpendicular ($AC \perp BD$), and bisect every corner into $45^\circ\text{-}45^\circ$. Area $= a^2 = \frac{e^2}{2}$.
  • Exam Tip for Squares: If a point $P$ lies on the diagonal of a square (which is a line of symmetry), either connect $P$ to the other two symmetric vertices (forming congruent triangles by SAS) or drop perpendiculars to the sides to form a $45^\circ\text{-}45^\circ\text{-}90^\circ$ right triangle!

---

Worked Examples

Worked Example 1: Regular Pentagon Symmetry and Shared Side with a Square

Problem: Let $ABCDE$ be a regular pentagon (düzgün beşgen) and let $ABGF$ be a square constructed inside the pentagon sharing side $AB$. Find the measure of $\angle EAF$ and $\angle EFA$.

Solution:

1. Find $\angle EAF$: Each interior angle of regular pentagon $ABCDE$ is $m(\angle EAB) = 108^\circ$, and each interior angle of square $ABGF$ is $m(\angle FAB) = 90^\circ$. Subtracting the two gives:

$$m(\angle EAF) = 108^\circ - 90^\circ = 18^\circ$$

1. Spot the hidden isosceles triangle $\triangle AEF$: Because $ABCDE$ is a regular pentagon and $ABGF$ is a square sharing side $AB$, all side lengths of both polygons are equal! In particular:

$$\lvert AE \rvert = \lvert AB \rvert = \lvert AF \rvert \implies \lvert AE \rvert = \lvert AF \rvert$$

1. Compute the base angles of $\triangle AEF$: Since $\triangle AEF$ is isosceles with apex angle $m(\angle EAF) = 18^\circ$, its base angles are:

$$m(\angle AEF) = m(\angle AFE) = \frac{180^\circ - 18^\circ}{2} = 81^\circ$$

Worked Example 2: Isosceles Trapezoid Projection + Euclid's Law

Problem: In an isosceles trapezoid $ABCD$ with $AB \parallel DC$ and $\lvert AD \rvert = \lvert BC \rvert$, the lower base is $\lvert AB \rvert = 13\text{ cm}$, the upper base is $\lvert DC \rvert = 5\text{ cm}$, and diagonal $AC$ is perpendicular to leg $BC$ ($AC \perp BC$). Find the height $h$ and the area of trapezoid $ABCD$.

Solution:

1. Drop altitude $CH \perp AB$ from vertex $C$: Because $ABCD$ is an isosceles trapezoid with bases $a = 13$ and $c = 5$, the outer segment next to $B$ is:

$$\lvert HB \rvert = \frac{a - c}{2} = \frac{13 - 5}{2} = 4\text{ cm}$$

The remaining segment to $A$ is $\lvert AH \rvert = \frac{a + c}{2} = \frac{13 + 5}{2} = 9\text{ cm}$. 2. Apply Euclid's Altitude Law in right $\triangle ABC$: Since $m(\angle ACB) = 90^\circ$ and $CH \perp AB$, we have an exact Euclid configuration:

$$h^2 = \lvert AH \rvert \cdot \lvert HB \rvert = 9 \cdot 4 = 36 \implies h = 6\text{ cm}$$

1. Compute the Trapezoid Area:

$$\text{Area}(ABCD) = \frac{a + c}{2} \cdot h = 9 \times 6 = 54\text{ cm}^2$$

Review the Method

In any rectangle (and square), the British Flag Theorem guarantees |PA|² + |PC|² = |PB|² + |PD|² for every interior or exterior point P.