Jump to heading🔗 Module 6-03 Quadrilateral

Jump to heading🔗 1.Parallelogram

The lengths of the two sides of the parallelogram are $a,b$, the height with $b$ as the base is $h$, the area is $S=bh$, and the perimeter is $C=2(a+b)$.

Jump to heading🔗 2.Rectangle

The lengths of the two sides of the rectangle are $a,b$, the area is $S=ab$, the perimeter is $C=2(a+b)$, and the diagonal is $l=\sqrt{a^2+b^2}$.

Jump to heading🔗 3.Rhombus

The length of the four sides of rhombus is $a$, the height with $a$ as the base is $h$, the area is $S=ah=\frac{1}{2}{l}_1{l}_2$, where $l_1,l_2$ are the lengths of the diagonals respective, and the perimeter is $C=4a$.

$S=\frac{1}{2}\times \frac{l_1}{2}\times \frac{l_2}{2}\times 4=\frac{l_1\cdot l_2}{2}$

Jump to heading🔗 4.Square

The length of the four sides of a square is $a$, the four internal angles are ${90}^{\circ }$, the area is $S=a^2$, and the perimeter is $C=4a$.

According to the Pythagorean theorem $c=\sqrt{a^2+b^2}$, diagonal is $\sqrt{2}$.

Jump to heading🔗 5.Trapezoid

• The upper base of the trapezoid is $a$, the lower base is $b$, the height is $h$, the median is $l=\frac{1}{2}(a+b)$, and the area is $S=\frac{1}{2}(a+b)h$.
  

• Note: There are two special types of trapezoids — the isosceles trapezoid and the right trapezoid.
  

Jump to heading🔗 6.Butterfly Theorem

The butterfly theorem provides us with a way to solve the area problem of irregular quadrilaterals by constructing a model; on the one hand, the area relationship of the irregular quadrilateral can be linked to the triangles inside the quadrilateral, on the other hand, the proportional relationship of the diagonal corresponding to the area can also be obtained.

Jump to heading🔗 $\boxed{\text{1}}$Proportional Relations in Any Quadrilateral (Figure 6–23)

Jump to heading🔗 $\boxed{\text{2}}$The Butterfly Theorem of Trapezoid and Similar Proportions (Figure 6–24)

• By combining the above four, the unified proportion is obtained: $S_1:S_3:S_2:S_4=a^2:b^2:ab:ab.$
  

  Jump to heading🔗 Formula derivations

  ${\mathbf{S}}_{\mathbf{1}}\mathbf{:}{\mathbf{S}}_{\mathbf{3}}\mathbf{:}{\mathbf{S}}_{\mathbf{2}}\mathbf{:}{\mathbf{S}}_{\mathbf{4}}\mathbf{=}{\mathbf{a}}^{\mathbf{2}}\mathbf{:}{\mathbf{b}}^{\mathbf{2}}\mathbf{:}\mathbf{ab}\mathbf{:}\mathbf{ab}$

  $\begin{array}{l}\frac{S_1}{S_3}=\frac{a_2}{b_2}\\ \frac{S_1}{S_2}=\frac{a}{b}\\ \frac{S_1}{S_2}=\frac{a}{b}\times \frac{a}{a}=\frac{a^2}{ab}& S_1=a^2\\ S_1:S_3:S_2:S_4=a^2:b^2:ab:ab\end{array}$

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Jump to heading🔗 7.Focus 1

Square

• The length of the four sides is $a$, the four interior angles are ${90}^{\circ }$, the area is $S=a^2$, and the perimeter is $C=4a$.

Jump to heading🔗 Solution

Jump to heading🔗 Conclusion

• Derived Solution

  $\mathbf{(}\mathbf{B}\mathbf{)}$
  According to the Solution, get $S_{EFGH}=\frac{1}{2}$, so choose $B$.

• Formula used

  $\begin{array}{ll}{c}^2=a^2+b^2\Rightarrow c=\sqrt{a^2+b^2}& \text{Pythagorean theorem}\\ S=a^2& \text{Square area formula}\end{array}$

• Derivation: Connecting the midpoints of the sides of any quadrilateral forms a new quadrilateral whose area is half the area of the original one

  
  In any triangle, the triangle formed by joining the midpoints of two sides has an area equal to $\frac{1}{4}$ of the original triangle.

  Example: $\mathrm{△}ABD$ (The same is true for other triangles)
  $\begin{array}{l}E=AB\text{ midpoint}\\ H=AD\text{ midpoint}\\ EH=\text{ Midline}\\ \mathbf{\text{Because }}\mathbf{EF}\mathbf{\text{ is a midline and triangle }}\mathrm{△}\mathbf{AEH}\mathbf{\text{ shares vertex }}\mathbf{A}\mathbf{\text{ with }}\mathrm{△}\mathbf{ABD}\mathbf{,}\\ \mathbf{\text{its base and height are each half of those of }}\mathrm{△}\mathbf{ABD}\mathbf{.}\\ S=\frac{1}{2}ah\\ S_{\mathrm{△}ABD}=\frac{1}{2}\times (1\cdot a)\times (1\cdot h)\\ S_{\mathrm{△}AEH}=\frac{1}{2}\times (\frac{1}{2}\cdot a)\times (\frac{1}{2}\cdot h)\\ \text{Properties of Similar triangles}\\ \frac{\mathrm{△}AEH}{\mathrm{△}ABD}=\frac{\frac{1}{8}\cdot a\cdot h}{\frac{1}{2}\cdot a\cdot h}=\frac{\frac{1}{8}}{\frac{1}{2}}=\frac{1}{4}\end{array}$

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  $\begin{array}{l}\{\begin{array}{l}{S}_{\mathrm{△}ABD}=x\\ S_{\mathrm{△}BCD}=y\end{array}\\ \{\begin{array}{l}{S}_{\mathrm{△}DAC}=x\\ S_{\mathrm{△}ABC}=y\end{array}\\ \boxed{{\displaystyle \begin{array}{l}{S}_{\mathrm{△}AEH}=\frac{1}{4}{S}_{\mathrm{△}ABD}\\ S_{\mathrm{△}CFG}=\frac{1}{4}{S}_{\mathrm{△}BCD}\end{array}}}\\ \boxed{{\displaystyle \begin{array}{l}{S}_{\mathrm{△}DHG}=\frac{1}{4}{S}_{\mathrm{△}ACD}\\ S_{\mathrm{△}BEF}=\frac{1}{4}{S}_{\mathrm{△}ABC}\end{array}}}\\ (S_{\mathrm{△}AEH}+S_{\mathrm{△}CFG})+(S_{\mathrm{△}DHG}+S_{\mathrm{△}BEF})=\frac{1}{4}(\mathrm{△}ABD+\mathrm{△}BCD)+\frac{1}{4}(\mathrm{△}ACD+\mathrm{△}ABC)\\ S_{\mathrm{△}ABD}+S_{\mathrm{△}BCD}=S_{ABCD}\\ S_{\mathrm{△}ACD}+S_{\mathrm{△}ABC}=S_{ABCD}\\ \frac{1}{4}{S}_{ABCD}+\frac{1}{4}{S}_{ABCD}=\frac{2}{4}{S}_{ABCD}=\frac{1}{2}{S}_{ABCD}\end{array}$

Jump to heading🔗 Solution

Jump to heading🔗 Conclusion

• Derived Solution

  $\mathbf{(}\mathbf{E}\mathbf{)}$
  According to the Solution, get $GE=\sqrt{65}$, so choose $E$.

• Formula used

  $\begin{array}{ll}{c}^2=a^2+b^2\Rightarrow c=\sqrt{a^2+b^2}& \text{Pythagorean theorem}\\ (3,4,5)& \text{Commonly used pythagorean numbers}\end{array}$

Jump to heading🔗 Solution

Jump to heading🔗 Conclusion

• Derived Solution

  $\mathbf{(}\mathbf{B}\mathbf{)}$
  According to the Solution, get $S_{ABCD}=74$, so choose $B$.

• Formula used

  $\begin{array}{ll}{c}^2=a^2+b^2\Rightarrow c=\sqrt{a^2+b^2}& \text{Pythagorean theorem}\\ \{\begin{array}{l}\text{Find a side and an angle (Acute angle)}\\ \text{Find their two sides}\end{array}& \text{Right triangle congruence}\\ S=a^2& \text{Square area formula}\end{array}$

Jump to heading🔗 8.Focus 2

Rectangle

• The rectangle has two pairs of opposite sides that are perpendicular to each other, and its diagonals bisect each other.
• The problem of a straight line appearing in a rectangle and dividing it into several triangles.

Jump to heading🔗 Solution

Jump to heading🔗 Conclusion

• Derived Solution

  $\mathbf{(}\mathbf{B}\mathbf{)}$
  According to the Solution, get $S_{\text{Shaded}}=28$, so choose $B$.

• Diagonals of a rectangle divide it into four congruent right triangles, each with $\frac{1}{4}$ of the rectangle's area

  
  $S_1=S_2=S_3=S_4$

• Jump to heading🔗 A triangle with base and height equal to those of a rectangle has $\frac{1}{2}$ the area of the rectangle

  
  $S_{\mathrm{△}}=\frac{1}{2}{S}_{ABCD}$

Jump to heading🔗 Solution

Jump to heading🔗 Conclusion

• Derived Solution

  $\mathbf{(}\mathbf{C}\mathbf{)}$
  According to the Solution, get $C$, so choose $S_{PFCG}=8$.

• Formula used

  $\begin{array}{ll}S=\frac{1}{2}ah& \text{Triangle area formula}\end{array}$

Jump to heading🔗 Solution

Jump to heading🔗 Conclusion

• Derived Solution

  $\mathbf{(}\mathbf{E}\mathbf{)}$
  According to the Solution, get $S\ge 550$, so choose $E$.

• Formula used

  $\begin{array}{ll}{c}^2=a^2+b^2\Rightarrow c=\sqrt{a^2+b^2}& \text{Pythagorean theorem}\\ S=ab& \text{Rectangle area formula}\\ 2ab=(a+b{)}^2-(a^2+b^2)& \text{Rectangle area identity}\\ y^2=x⟹\sqrt[2]{x}& \text{Definition of square root}\end{array}$

• Additionally, Cuboid surface area identity

  $2ab+2bc+2ac=(a+b+c{)}^2-(a^2+b^2+c^2)$

• Inequality sign flips

  
  

  Operation	Inequality sign changes?
  Multiply $\vee$ divide by negative	✅ Flip the sign
  Multiply $\vee$ divide by positive	❌ Do not flip the sign
  Add $\vee$ subtract both sides	❌ Do not flip the sign

Jump to heading🔗 9.Focus 3

Rhombus

• In a rhombus, the two diagonals bisect each other at right angles, and its area equals half the product of the diagonals.

Jump to heading🔗 Solution

Jump to heading🔗 Conclusion

• Derived Solution

  $\mathbf{(}\mathbf{B}\mathbf{)}$
  According to the Solution, get $S=120$, so choose $B$.

• Formula used

  $\begin{array}{ll}S=\frac{l_1\cdot l_2}{2}& \text{Rhombus area formula}\end{array}$

Jump to heading🔗 Solution

Jump to heading🔗 Conclusion

• Derived Solution

  $\mathbf{(}\mathbf{B}\mathbf{)}$
  According to the Solution, get $S=120$, so choose $B$.

• Formula used

  $\begin{array}{ll}(5,12,13)& \text{Commonly used pythagorean numbers}\\ S=\frac{l_1\cdot l_2}{2}& \text{Rhombus area formula}\end{array}$

Jump to heading🔗 Solution

Jump to heading🔗 Conclusion

• Derived Solution

  $\mathbf{(}\mathbf{C}\mathbf{)}$
  According to the Solution, get $PM+PN=PM+P{N}^{\prime }\ge 5$, so choose $C$.

• Formula used

  $\begin{array}{ll}(3,4,5)& \text{Commonly used pythagorean numbers}\\ a+b>c& \text{Trilateral relations}\end{array}$

• Reflectional symmetry

  
  After reaching the orange line by the shortest distance, return $B$. $A\to C\to B$

Jump to heading🔗 10.Focus 4

Parallelogram

• The two pairs of opposite sides of a parallelogram are parallel and equal. The core point of a parallelogram is the diagonal. In addition, if there are no other requirements, the parallelogram can be specialized into a rectangle or square to find the answer.

Jump to heading🔗 Solution

Jump to heading🔗 Conclusion

• Derived Solution

  $\mathbf{(}\mathbf{B}\mathbf{)}$
  According to the Solution, get $S_{\mathrm{△}PAC}=3$, so choose $B$.

Jump to heading🔗 11.Focus 5

Trapezoid

• Analyze, according to the area formula and properties of trapezoid, paying attention to the two special types of trapezoids: right trapezoids and isosceles trapezoids.

Jump to heading🔗 Solution

Jump to heading🔗 Conclusion

• Derived Solution

  $\mathbf{(}\mathbf{B}\mathbf{)}$
  According to the Solution, get $S_{ABCD}=24$, so choose $B$.

• Formula used

  $\begin{array}{ll}\{\begin{array}{l}\mathrm{\angle }1=\mathrm{\angle }4\\ \mathrm{\angle }2=\mathrm{\angle }4\\ \mathrm{\angle }1+\mathrm{\angle }4={180}^{\circ }\end{array}& \text{Angle relationship}\\ S=\frac{a+b}{2}h& \text{Trapezoid area formula}\end{array}$

Jump to heading🔗 Solution

Jump to heading🔗 Conclusion

• Derived Solution

  $\mathbf{(}\mathbf{B}\mathbf{)}$
  According to the Solution, get $frac{S}_{AMNB}{S}_{MDCN}=1$, so choose $B$.

Jump to heading🔗 12.Focus 6

Other quadrilaterals

• When encountering other quadrilaterals, you can divide them into triangles to solve the problem or analyze them using the properties of special quadrilaterals.

Jump to heading🔗 Solution

Jump to heading🔗 Conclusion

• Derived Solution

  $\mathbf{(}\mathbf{B}\mathbf{)}$
  According to the Solution, get $S_{Shaded}=15+6=21$, so choose $B$.

• Formula used

  $\begin{array}{ll}{S}_1:S_3:S_2:S_4=a^2:b^2:ab:ab& \text{Butterfly theorem(Trapezoid) \& Similarity proportions}\end{array}$

Jump to heading🔗 Solution

Jump to heading🔗 Conclusion

• Derived Solution

  $\mathbf{(}\mathbf{B}\mathbf{)}$
  According to the Solution, get $S_{OFBC}=9$, so choose $B$.

• Formula used

  $\begin{array}{ll}{S}_1\times S_3=S_2\times S_4& \text{Top}\times \text{Bottom}=\text{Left}\times \text{Right}\\ S_1:S_3:S_2:S_4=a^2:b^2:ab:ab& \text{Butterfly theorem(Trapezoid) \& Similarity proportions}\\ S_{\mathrm{△}}=\frac{1}{2}{S}_{ABCD}& \text{Half area rule}\end{array}$

Jump to heading🔗 13.Focus 7

Polygon

• When encountering polygon, you can draw auxiliary lines (Usually by connecting the diagonals) to divide it into several triangles for solving.

• Jump to heading🔗 Regular Hexagon

  $\begin{array}{l}\text{Divided into }6\text{ equilateral triangles}\\ S=\frac{\sqrt{3}}{4}{a}^2& \text{Equilateral triangle area formula}\\ S=\frac{\sqrt{3}}{4}{a}^2\times 6=\frac{3\sqrt{3}}{2}{a}^2\end{array}$

• Interior Angles of Common Regular Polygons

  $\text{Each interior angle}=\frac{(n-2)\times {180}^{\circ }}{n}$

  Regular Polygon	Number of Sides(n)	Each Interior Angle(${}^{\circ }$)
  Equilateral Triangle	3	$\frac{(3-2)\times {180}^{\circ }}{3}={60}^{\circ }$
  Square	4	$\frac{(4-2)\times {180}^{\circ }}{4}={90}^{\circ }$
  Regular Pentagon	5	$\frac{(5-2)\times {180}^{\circ }}{5}={108}^{\circ }$
  Regular Hexagon	6	$\frac{(6-2)\times {180}^{\circ }}{6}={120}^{\circ }$
  Regular Octagon	8	$\frac{(8-2)\times {180}^{\circ }}{8}={135}^{\circ }$
  Regular Dodecagon	12	$\frac{(12-2)\times {180}^{\circ }}{12}={150}^{\circ }$

Jump to heading🔗 Solution

Jump to heading🔗 Conclusion

• Derived Solution

  $\mathbf{(}\mathbf{D}\mathbf{)}$
  According to the Solution, get $S=6\sqrt{3}$, so choose $D$.

• Formula used

  $\begin{array}{ll}S=\frac{3\sqrt{3}}{2}{a}^2& \text{Regular hexagon area formula}\end{array}$