Honors Geometry Chapter 1 Comprehensive Study Guide

Class Guidelines and Chapter 1 Assignment Overview

• General Course Requirements:

  • Keep the Ch 1 Assignment Sheet at the beginning of the Geometry section in your binder.
  • Show all work necessary to arrive at solutions; mental math without written support is unaccepted.
  • Calculators are not permitted for Chapter 1 problems.
  • Check answers to odd-numbered textbook problems in the back of the book prior to class.

• Chapter 1 Assignment Schedule:

  • Assignment 1: Review Class Expectations, obtain required materials, read Section 1.1, complete Worksheet Chapter 1 Algebra Review.
  • Assignment 2: Section 1.1 (Finish Classwork pp. 7–8 #1, 3, 5abcd, 11a; Homework p. 7 #5efg, 6ab, 7, 8, 11b, 12, 14b; Read Sections 1.7–1.8 and take notes).
  • Assignment 3: Sections 1.7 & 1.8 (pp. 42–43 #5a[i, iv], 5b, 8, 9; pp. 47–48 #1b, 5, 7, 10).
  • Assignment 4: Sections 1.7 & 1.8 (Finish Deductive Reasoning Worksheet both sides; pp. 42–43 #10; p. 48 #9; Read Section 1.2).
  • Assignment 5: Section 1.2 (pp. 14–20 #1a, 2b, 4, 5b, 8, 16, 17, 21).
  • Assignment 6: Section 1.2 (Finish Systems Practice; pp. 14–20 #9cd).
  • Assignment 7: Review (Complete Chapter 1 Quiz Review Worksheet; study past assignments).
  • Assignment 8: Quiz / Section 1.3 (Read Section 1.3; 1.3 Collinearity, Betweenness, and Assumptions Examples #1–9; pp. 21–22 #3, 6, 8, 10, 13).
  • Assignment 9: Section 1.4 (pp. 26–27 #2, 3, 5, 7, 14, 15; p. 22 #11).
  • Assignment 10: Section 1.5 (pp. 32–35 #5, 9, 10, 12, 16, 18, 20a, 22; p. 47 #3).
  • Assignment 11: Section 1.5 / 1.9 (Finish Introduction to Geometry: Additional Practice 2 #1, 2, 5, 6; p. 27 #11; pp. 33–35 #6, 7, 11, 14, 20b, 21; Read Section 1.9; p. 51 #1, 7).
  • Assignment 12: Section 1.9 (pp. 51–52 #6, 8, 10, 11, 14, 15; Chapter 1 Systems & Quadratics Worksheet evens).
  • Assignment 13: Review (Chapter 1 Systems & Quadratics Worksheet odds; pp. 54–59 #1h–l, 4, 8, 13, 15, 19, 21c).
  • Assignment 14: Review (pp. 54–59 #21b, 22, 24b, 26, 27, 29, 32, 33, 35, 36; comprehensive review).
  • Assignment 15: Chapter 1 Test.

Section 1.1: Geometric Fundamentals & Definitions

• Fundamental Geometric Elements:
  • Points: Locations in space represented by dots and named using single capital letters (e.g., $A$, $B$, $C$).
  • Lines: Continuous, infinitely long straight paths extending in two opposite directions. A line is named using any two points on it with a double-headed arrow above them (e.g., $\overleftrightarrow{AB}$).
  • Line Segments: Parts of a line bounded by two distinct end points. Named using the endpoints with a straight bar above them (e.g., $\overline{AB}$).
  • Rays: Parts of a line that begin at an initial endpoint and extend infinitely in one direction. Named starting with the endpoint followed by another point on the ray with a single-headed arrow (e.g., $\overrightarrow{AB}$).
  • Angles: Figures formed by two rays sharing a common endpoint called the vertex. Named using three points with the vertex in the middle (e.g., $\angle ABC$), the vertex letter alone, or a numbered label.

• Planes: Infinite two-dimensional flat surfaces. Named using a script capital letter (e.g., plane $\mathcal{P}$) or three non-collinear points.

• Triangles:
  • Definition: A triangle is the union ($\cup$) of three line segments joined at non-collinear endpoints.
  • Vertex Definition: The intersection ($\cap$) of any two sides of a triangle is a vertex.
  • Intersections of Triangles and Rays/Lines:
  • Single point intersection: $\triangle ABC \cap \overrightarrow{DE} = F$

- Two point intersection: $\triangle ABC \cap \overrightarrow{DE} = \{F, G\}$

- Three point intersection: $\triangle ABC \cap \overrightarrow{DE} = \{F, G, H\}$

Section 1.2: Measurement of Segments and Angles

• Degree Measure Conversions:

  • $1^\circ = 60'$ (minutes)
  • $1' = 60''$ (seconds)
  • $1^\circ = 3600''$
  • Degree Conversion Example: Convert $35^\circ 40' 40''$ to fractional degrees:     $35^\circ + \left(\frac{40}{60}\right)^\circ + \left(\frac{40}{3600}\right)^\circ = 35^\circ + 0.6667^\circ + 0.0111^\circ = 35.6778^\circ$

• Angle Measure Addition and Subtraction:

  • Example Problem: Given $m\angle ABC = 48^\circ 51' 50''$, $m\angle EBD = 67^\circ 45' 12''$, and $\angle ABE, \angle EBD, \angle DBC$ together form a straight angle ($180^\circ$). Find $m\angle DBC$.
  • Add given adjacent angles:       $48^\circ 51' 50'' + 67^\circ 45' 12'' = 115^\circ 96' 62'' = 116^\circ 37' 02''$
  • Subtract from $180^\circ$ (borrowing $179^\circ 59' 60''$):       $179^\circ 59' 60'' - 116^\circ 37' 02'' = 63^\circ 22' 58''$

• Restrictions on Variable Angle Expressions:

  • An acute angle must measure strictly between $0^\circ$ and $90^\circ$: $0^\circ < m\angle P < 90^\circ$.
  • Example: Given acute $\angle P = (4x - 20)^\circ$, set up inequality bounds:     $0 < 4x - 20 < 90$$20 < 4x < 110$$5 < x < 27.5$

Section 1.3: Collinearity, Betweenness, and Diagram Assumptions

• Fundamental Definitions:

  • Collinear Points: Points that lie on the exact same straight line.
  • Betweenness of Points: Point $B$ is between $A$ and $C$ if and only if $A, B, C$ are collinear and $AB + BC = AC$.

• Rules for Diagram Assumptions:

• Diagram Assumptions Example Analysis:

• $V$ is between $Q$ and $T$: Yes (Assumed from lines/betweenness)
• $\angle PTS$ is a straight angle: Yes (Assumed from straight line)
• $\angle QTS$ is a right angle: No (Right angles cannot be assumed)
• $\angle PQS$ is a right angle: No (Right angles cannot be assumed)
• $T$ is between $P$ and $S$: Yes (Assumed from betweenness)
• $P, R, T$ are noncollinear: Yes (Assumed from relative position)
• $\overline{QR} \cong \overline{RS}$: No (Segment congruence cannot be assumed)
• Line $QS$ is a straight line: Yes (Assumed straight line)
• $QV = VT$: No (Relative sizes/congruence cannot be assumed)

Section 1.4: Beginning Proofs and Basic Theorems

• Definitions & Postulates:

  • Theorem: A mathematical statement that can be proved using logical deduction, definitions, postulates, and prior theorems.

• Theorem 1: Right Angle Congruence Theorem

  • Statement: All right angles are congruent.

• Proof:
  • Given: $\angle A$ is a right angle; $\angle B$ is a right angle.
  • Prove: $\angle A \cong \angle B$

• Theorem 2: Straight Angle Congruence Theorem
  • Statement: All straight angles are congruent.

• Proof:
  • Given: $\angle ABC$ is a straight angle; $\angle DEF$ is a straight angle.
  • Prove: $\angle ABC \cong \angle DEF$

• Sample Proof using Diagram Assumptions:

• Given: Diagram as shown.
• Prove: $\angle EFG \cong \angle HFJ$

Section 1.5: Division of Segments and Angles (Bisection & Trisection)

• Segment and Angle Bisection:

  • Segment Bisector: A point, line segment, ray, or line that divides a segment into two congruent segments.
  • Midpoint: The specific point where a segment is bisected.
  • Impossibility of Line/Ray Bisection: Lines and rays extend infinitely in length, so they cannot be bisected or trisected.
  • Consequences of Bisection: If segment $XY$ bisects $\overline{AC}$ at point $B$, then $B$ is the midpoint of $\overline{AC}$, meaning $\overline{AB} \cong \overline{BC}$.

• Trisection:

  • Segment Trisection: Two points divide a segment into three equal segments.
  • Angle Trisection: Two rays divide an angle into three equal angles.

• Sample Division Problems:

  • Problem 1: Segment $\overline{AD}$ is divided by points $B$ and $C$ in the ratio $5:2:4$ from left to right. If $AD = 77$, find $BC$.
  • Let $\overline{AB} = 5k$, $\overline{BC} = 2k$, $\overline{CD} = 4k$
  • $5k + 2k + 4k = 11k = 77 \implies k = 7$
  • $BC = 2k = 2(7) = 14$
  • Problem 2: Given ray $\overrightarrow{EG}$ bisects $\angle FEH$. If $m\angle FEH = 23^\circ 19' 2''$, find $m\angle GEH$
  • $m\angle GEH = \frac{1}{2} m\angle FEH$
  • Convert $23^\circ 19' 2''$ to an even degree: $22^\circ 79' 2'' = 22^\circ 78' 62''$
  • Divide each unit by 2: $11^\circ 39' 31''$

Section 1.7 & 1.8: Deductive Structure and Symbolic Logic

• Conditional Forms:

  • Declarative Sentence Example: "All students in this Geometry class passed Algebra 1."
  • Conditional Form ($p \rightarrow q$): If a student is in this Geometry class, then they passed Algebra 1.
  • Converse ($q \rightarrow p$): If a student passed Algebra 1, then they are in this Geometry class.
  • Inverse ($\sim p \rightarrow \sim q$): If a student is not in this Geometry class, then they did not pass Algebra 1.
  • Contrapositive ($\sim q \rightarrow \sim p$): If a student did not pass Algebra 1, then they are not in this Geometry class.
  • Logical Equivalence: A conditional statement is always logically equivalent to its contrapositive ($p \rightarrow q \equiv \sim q \rightarrow \sim p$).

• Chain Rule (Hypothetical Syllogism):

  • If $p \rightarrow q$ and $q \rightarrow r$, then $p \rightarrow r$

• Chain Rule Logic Examples:

  • Problem 1 (Symbolic):

  • Given: $\sim g \rightarrow e$, $t \rightarrow w$, $\sim t \rightarrow \sim e$

  • Contrapositive of statement 3: $e \rightarrow t$

  • Reordered Chain: $\sim g \rightarrow e \rightarrow t \rightarrow w$

  • Final Conclusion: $\sim g \rightarrow w$

  • Problem 2 (People Logic Puzzle):

  • Given Statements:

    1. If Fran is funny ($f$), then Doris is not dizzy ($\sim d$). ($f \rightarrow \sim d$)
    2. If Alice is absolutely wonderful ($a$), then Barbara is not beautiful ($\sim b$). ($a \rightarrow \sim b$)
    3. If Doris is not dizzy ($\sim d$), then Edward is not everything they say he is ($\sim e$). ($\sim d \rightarrow \sim e$)
    4. If Calvin is not corny ($\sim c$), then Barbara is beautiful ($b$). ($\sim c \rightarrow b$)
    5. If Calvin is corny ($c$), then Edward is everything they say he is ($e$). ($c \rightarrow e$)

  • Contrapositives used:

    • $\sim c \rightarrow b \equiv \sim b \rightarrow c$
    • $\sim d \rightarrow \sim e \equiv e \rightarrow d$
    • $f \rightarrow \sim d \equiv d \rightarrow \sim f$

  • Logical Order: $a \rightarrow \sim b \rightarrow c \rightarrow e \rightarrow d \rightarrow \sim f$

  • Conclusion (Symbolic): $a \rightarrow \sim f$

  • Conclusion (Words): If Alice is absolutely wonderful, then Fran is not funny.

  • Problem 3 (Study Routine Chain):

  • Statements:

    1. If you pay attention in Geometry class ($p$), you will understand the material ($u$).
    2. If the homework is not easy for you ($\sim e$), then you do not understand the material ($\sim u$).
    3. If you complete the homework quickly ($q$), you will have more time ($t$).
    4. If you can get a job ($j$), you can make lots of money ($m$).
    5. If you have more time ($t$), you can get a job ($j$).
    6. If the homework is easy for you ($e$), you will complete the homework quickly ($q$).

  • Chain Ordering: $p \rightarrow u \rightarrow e \rightarrow q \rightarrow t \rightarrow j \rightarrow m$

  • Conclusion: $p \rightarrow m$ ("If you pay attention in Geometry class, you can make lots of money.")

  • Problem 4 (Color Logic Puzzle):

  • Given: $r \rightarrow b$, $\sim g \rightarrow y$, $\sim r \rightarrow \sim y$, $p \rightarrow \sim b$

  • Rearranged Chain: $p \rightarrow \sim b \rightarrow \sim r \rightarrow \sim y \rightarrow g$

  • Conclusion: $p \rightarrow g$ ("If p is purple, then g is green.")

Section 1.9: Geometric Probability

• Definition: Geometric probability determines the likelihood that a randomly selected point on a geometric figure falls within a specified sub-region.
• Formula:   $P(\text{Event}) = \frac{\text{Length or Area of Favorable Region}}{\text{Total Length or Area of Entire Region}}$

Clock Geometry: Angle Calculations & Rate Formulas

• Minute Hand Angular Movement Rate:
  • Moves $360^\circ$ in $1\text{ hr} = 60\text{ min}$.

• Formula: $\frac{360^\circ}{60\text{ min}} = 6^\circ \text{ per minute}$.

  • Hour Hand Angular Movement Rate:

• Moves $\left(\frac{1}{12}\right)(360^\circ) = 30^\circ$ in $1\text{ hr} = 60\text{ min}$.

• Formula: $\frac{30^\circ}{60\text{ min}} = 0.5^\circ \text{ per minute}$.

  • Algorithm for Finding Angle Between Clock Hands at Given Time:

• Calculate initial minute gap assuming the hour hand stays fixed at the base hour tick mark ($1\text{ minute gap} = 6^\circ$).

• Calculate the hour hand displacement over elapsed minutes: $\text{Minutes} \times 0.5^\circ/\text{min}$.

• Adjust total angle: Add displacement if the hour hand moves away from the minute hand; Subtract displacement if the hour hand moves toward the minute hand.

  • Worked Clock Examples:

• Example 1: 7:25

  • Fixed hour 7 to minute 25 (5 mark) = $10\text{ min gap}$.
  • Fixed angle = $(10\text{ min})(6^\circ/\text{min}) = 60^\circ$.
  • Hour hand displacement = $(25\text{ min})(0.5^\circ/\text{min}) = 12.5^\circ$
  • Direction: Hour hand moves away from minute hand (towards 8).
  • Final Angle: $60^\circ + 12.5^\circ = 72.5^\circ$

• Example 2: 5:50

  • Fixed hour 5 (25 mark) to minute 50 = $25\text{ min gap}$.
  • Fixed angle = $(25\text{ min})(6^\circ/\text{min}) = 150^\circ$.
  • Hour hand displacement = $(50\text{ min})(0.5^\circ/\text{min}) = 25^\circ$
  • Direction: Hour hand moves toward minute hand.
  • Final Angle: $150^\circ - 25^\circ = 125^\circ$

• Example 3: 5:49

- Fixed hour 5 (25 mark) to minute 49 = $24\text{ min gap}$.
- Fixed angle = $(24\text{ min})(6^\circ/\text{min}) = 144^\circ$.
- Hour hand displacement = $(49\text{ min})(0.5^\circ/\text{min}) = 24.5^\circ$
- Direction: Hour hand moves toward minute hand.
- Final Angle: $144^\circ - 24.5^\circ = 119.5^\circ$

• Example 4: 7:00

  • Angle = $7 \times 30^\circ = 210^\circ$
  • Smaller acute/obtuse interior angle: $360^\circ - 210^\circ = 150^\circ$

• Example 5: 1:20

  • Fixed hour 1 (5 mark) to minute 20 = $15\text{ min gap}$.
  • Base angle = $(15\text{ min})(6^\circ/\text{min}) = 90^\circ$
  • Hour hand displacement = $(20\text{ min})(0.5^\circ/\text{min}) = 10^\circ$
  • Moved toward minute hand: $90^\circ - 10^\circ = 80^\circ$

• Example 6: 7:43

  • Fixed hour 7 (35 mark) to minute 43 = $8\text{ min gap}$.
  • Base angle = $(8\text{ min})(6^\circ/\text{min}) = 48^\circ$
  • Hour hand displacement = $(43\text{ min})(0.5^\circ/\text{min}) = 21.5^\circ$
  • Moved toward minute hand: $48^\circ - 21.5^\circ = 26.5^\circ$

• Example 7: 2:32

  • Fixed hour 2 (10 mark) to minute 32 = $22\text{ min gap}$.
  • Base angle = $(22\text{ min})(6^\circ/\text{min}) = 132^\circ$
  • Hour hand displacement = $(32\text{ min})(0.5^\circ/\text{min}) = 16^\circ$
  • Moved toward minute hand: $132^\circ - 16^\circ = 116^\circ$

• Example 8: 8:20

  • Fixed hour 8 (40 mark) to minute 20 = $20\text{ min gap}$.
  • Base angle = $(20\text{ min})(6^\circ/\text{min}) = 120^\circ$
  • Hour hand displacement = $(20\text{ min})(0.5^\circ/\text{min}) = 10^\circ$
  • Moved away from minute hand: $120^\circ + 10^\circ = 130^\circ$

• Example 9: 4:40

  • Fixed hour 4 (20 mark) to minute 40 = $20\text{ min gap}$.
  • Base angle = $(20\text{ min})(6^\circ/\text{min}) = 120^\circ$
  • Hour hand displacement = $(40\text{ min})(0.5^\circ/\text{min}) = 20^\circ$
  • Moved toward minute hand: $120^\circ - 20^\circ = 100^\circ$

• Example 10: 1:24

  • Fixed hour 1 (5 mark) to minute 24 = $19\text{ min gap}$.
  • Base angle = $(19\text{ min})(6^\circ/\text{min}) = 114^\circ$
  • Hour hand displacement = $(24\text{ min})(0.5^\circ/\text{min}) = 12^\circ$
  • Moved toward minute hand: $114^\circ - 12^\circ = 102^\circ$

Algebraic Applications in Geometry

• Algebra Review: Polynomial Multiplication:

  1. $(7y - 1)(2y - 26) = 14y^2 - 182y - 2y + 26 = 14y^2 - 184y + 26$
  2. $(10a + 3)(2a - 5) = 20a^2 - 50a + 6a - 15 = 20a^2 - 44a - 15$
  3. $(2m + 3n)(5m + 2n) = 10m^2 + 4mn + 15mn + 6n^2 = 10m^2 + 19mn + 6n^2$
  4. $(1 - x)^2 = 1 - 2x + x^2$
  5. $(h + 11)^2 = h^2 + 22h + 121$
  6. $(3w + 2)^2 = 9w^2 + 12w + 4$
  7. $(7 - 4m)^2 = 49 - 56m + 16m^2$
  8. $(3w - 10x)^2 = 9w^2 - 60wx + 100x^2$
  9. $(4x + 7y)(4x - 7y) = 16x^2 - 49y^2$
  10. $(3m - 8p)(3m + 8p) = 9m^2 - 64p^2$

• Algebra Review: Splitting the Middle Term Factoring:

  1. $2x^2 + 5x + 2 = (2x + 1)(x + 2)$
  2. $2n^2 - 7n + 3 = (2n - 1)(n - 3)$
  3. $5y^2 - 9y - 2 = (5y + 1)(y - 2)$
  4. $4y^2 - 5y + 1 = (4y - 1)(y - 1)$
  5. $2a^2 + 11a + 5 = (2a + 1)(a + 5)$
  6. $2x^2 - 3x - 9 = (2x + 3)(x - 3)$
  7. $4x^2 - 8x - 5 = (2x + 1)(2x - 5)$
  8. $-x^2 - 3x + 10 = -(x^2 + 3x - 10) = -(x + 5)(x - 2)$
  9. $12x^2 - 8x + 1 = (6x - 1)(2x - 1)$
  10. $2x^2 - xy - 3y^2 = (2x - 3y)(x + y)$
  11. $3a^2 + 6ab + 3b^2 = 3(a^2 + 2ab + b^2) = 3(a + b)^2$
  12. $16x^4 - 24x^3 + 9x^2 = x^2(16x^2 - 24x + 9) = x^2(4x - 3)^2$
  13. $20x^3 - 28x^2 + 8x = 4x(5x^2 - 7x + 2) = 4x(5x - 2)(x - 1)$
  14. $6x^2 + 22xy - 8y^2 = 2(3x^2 + 11xy - 4y^2) = 2(3x - y)(x + 4y)$
  15. $20 - 60x + 45x^2 = 5(9x^2 - 12x + 4) = 5(3x - 2)^2$
  16. $6x^2 - 18xy + 12y^2 = 6(x^2 - 3xy + 2y^2) = 6(x - y)(x - 2y)$
  17. $2n^2 + 15n + 7 = (2n + 1)(n + 7)$
  18. $18x^3 - 24x^2 + 8x = 2x(9x^2 - 12x + 4) = 2x(3x - 2)^2$
  19. $9x^2 - 20x + 4 = (9x - 2)(x - 2)$

• Systems of Linear Equations in Geometry:

  • Problem 1:

- Given line $RT$ intersecting line $WV$ at $S$, with $\angle RST = (3x + 10)^\circ$, $\angle TSV = (2y + 30)^\circ$, and $\angle VSW = (2x + y)^\circ$.
- Set up system using vertical angles and supplementary angles:
  - Vertical angles: $3x + 10 = 2x + y \implies x - y = -10$
  - Supplementary linear pair: $(3x + 10) + (2y + 30) = 180 \implies 3x + 2y = 140$
- Solve system:
  - Multiply first equation by 2: $2x - 2y = -20$
  - Add equations: $5x = 120 \implies x = 24$
  - Substitute: $24 - y = -10 \implies y = 34$
- Angle measures: $\angle RST = 3(24) + 10 = 82^\circ$; $\angle TSV = 2(34) + 30 = 98^\circ$.

• Problem 2:

- Given intersecting lines forming angles $(3y + 6)^\circ$, $(2x + 32)^\circ$, and $(x + 2y)^\circ$
- Vertical angles: $3y + 6 = x + 2y \implies x - y = 6$
- Supplementary linear pair: $(3y + 6) + (2x + 32) = 180 \implies 2x + 3y = 142$
- Solve system:
  - $3x - 3y = 18$
  - Add equations: $5x = 160 \implies x = 32$
  - Substitute: $32 - y = 6 \implies y = 26$

• Problem 3:

- Given $\angle EBA$ is a right angle ($90^\circ$), $m\angle CBD = (x + 3y)^\circ$, $m\angle EBA = (3x + 2y + 8)^\circ$, $m\angle DBA = (x + 4y - 5)^\circ$.
- Set up equation: $3x + 2y + 8 = 90 \implies 3x + 2y = 82$

• Quadratic Applications in Geometry:

  • Problem 1: Straight Angle Calculation

  • Angle $C = (2x^2 - 5x + 30)^\circ$. Find $x$ such that $\angle C$ is a straight angle ($180^\circ$).

  • $2x^2 - 5x + 30 = 180$

  • $2x^2 - 5x - 150 = 0$

  • $(2x + 15)(x - 10) = 0 \implies x = 10$ or $x = -7.5$

  • Problem 2: Right Angle Calculation

  • Angle $B = (x^2 - 3x + 20)^\circ$. Find $x$ such that $\angle B$ is a right angle ($90^\circ$).

  • $x^2 - 3x + 20 = 90$

  • $x^2 - 3x - 70 = 0$

  • $(x - 10)(x + 7) = 0 \implies x = 10$ or $x = -7$

  • Problem 3: Angle Bisector Word Problem

  • Ray $\overrightarrow{OP}$ bisects $\angle EOM$. $m\angle EOP = (x^2 - 6x)^\circ$ and $m\angle POM = (10 - 3x)^\circ$.

  • $x^2 - 6x = 10 - 3x$

  • $x^2 - 3x - 10 = 0 \implies (x - 5)(x + 2) = 0$

  • Check $x = 5$: $m\angle EOP = 25 - 30 = -5^\circ$ (Invalid angle measure).

  • Check $x = -2$: $m\angle EOP = (-2)^2 - 6(-2) = 4 + 12 = 16^\circ$; $m\angle POM = 10 - 3(-2) = 16^\circ$ (Valid).

  • Result: $x = -2$, creating two acute angles measuring $16^\circ$ each (total angle = $32^\circ$).

  • Problem 4: Supplementary Angle Quadratic

- $\angle ADC$ is a straight angle ($180^\circ$). $\angle ADB = (x^2 + 5x + 4)^\circ$ and $\angle CDB = (x^2 - 3x - 4)^\circ$
- $(x^2 + 5x + 4) + (x^2 - 3x - 4) = 180$
- $2x^2 + 2x = 180 \implies x^2 + x - 90 = 0$
- $(x + 10)(x - 9) = 0 \implies x = 9$ or $x = -10$
- For $x = 9$: $m\angle ADB = 81 + 45 + 4 = 130^\circ$; $m\angle CDB = 81 - 27 - 4 = 50^\circ$
- For $x = -10$: $m\angle ADB = 100 - 50 + 4 = 54^\circ$; $m\angle CDB = 100 + 30 - 4 = 126^\circ$
- Both values of $x$ yield valid supplementary angle pairs.

• Problem 5: Right Angle Partition Quadratic

- Given $m\angle ACB = (x^2 - 5x + 4)^\circ$, $m\angle BCD = (2x^2 + 2x + 26)^\circ$, and $m\angle ACD = 90^\circ$
- $(x^2 - 5x + 4) + (2x^2 + 2x + 26) = 90$
- $3x^2 - 3x + 30 = 90 \implies 3x^2 - 3x - 60 = 0 \implies x^2 - x - 20 = 0$
- $(x - 5)(x + 4) = 0 \implies x = 5$ or $x = -4$
- Solution 1 ($x = 5$): $m\angle ACB = 25 - 25 + 4 = 4^\circ$; $m\angle BCD = 2(25) + 10 + 26 = 86^\circ$
- Solution 2 ($x = -4$): $m\angle ACB = 16 + 20 + 4 = 40^\circ$; $m\angle BCD = 2(16) - 8 + 26 = 50^\circ$