Appendix A — Answer Key, Chapter 1: Logic, Conditionals, and Venn Diagrams

SOL G.RLT.1 (a, b, c, d) · Covers textbook Chapter 1 and the companion workbook. Item numbers match the textbook; workbook items are the same problems, so this key serves both. Item numbers run continuously from 1 to 112 across the chapter. Reasoning answers show an acceptable response, not the only wording.

Conventions used in every answer below: $p$, $q$ stand for statements; $\sim p$ is "not $p$"; "or" is inclusive, so $p \vee q$ is true when both parts are true; a counterexample is one case with a true hypothesis and a false conclusion; in a Venn diagram a region's number counts the things in that region only; a trapezoid has exactly one pair of parallel sides.

The figures used repeatedly in the chapter, for reference:

• Figure 1 is four cards sorting sentences into statements and non-statements
• Figure 2 is the five symbolic forms with symbols, readings, and geometric examples
• Figure 3 is the truth tables for negation, conjunction, and disjunction
• Figure 4 is the hypothesis/conclusion split of "If two angles are vertical angles, then they are congruent"
• Figure 5 is the conditional truth table with the T→F row highlighted
• Figure 6 is the four related conditionals for "If it is a square, then it is a rhombus"
• Figure 7 is a square beside a non-square rectangle, the counterexample that kills the converse
• Figure 8 is a biconditional above the two conditionals it contains
• Figure 9 is union and intersection shaded
• Figure 10 is a subset drawn nested, and a negation shaded including the outside region
• Figure 11 is the quadrilateral family as nested sets
• Figure 12 is the art/band survey with a count in every region

Lesson 1.1 — Statements, Negation, and Compound Statements

Guided practice

1. a) Statement, true. b) Not a statement — a command. c) Statement, false ($12 = 3 \times 4$). d) Not a statement — a question.
2. Negation $\sim p$; conjunction $p \wedge q$; disjunction $p \vee q$; conditional $p \rightarrow q$; biconditional $p \leftrightarrow q$.
3. $\sim p$: $\angle A$ is not acute.
4. The lines are parallel and the lines are coplanar.
5. A conjunction is true only in the first row (T, T). A disjunction is false only in the last row (F, F).
6. $\sim p$ is false; $p \wedge q$ is false; $p \vee q$ is true.

Independent practice

7. a) $\triangle DEF$ is not a right triangle. b) $m\angle B \ne 55°$. c) The diagonals are congruent.
8. a) The figure is a square and it is a rectangle. b) The figure is a square or it is a rectangle. c) The figure is not a square. d) The figure is not a square and it is a rectangle.
9. a) $p \wedge \sim q$ b) $\sim p \wedge \sim q$ (accept $\sim(p \vee q)$, which is equivalent).

12. The second part — "the time is within business hours" — was false. The badge part was true and the gate still did not open, and an "and" fails when either part fails, so the false part must be the other one.
13. "$7$ is an even number" has a truth value; it is simply false. "Draw a circle" is a command, so it is neither true nor false and logic assigns it no truth value. Being false and having no truth value are different things.
14. The negation of "all rectangles are squares" is "not all rectangles are squares," which is the same as "at least one rectangle is not a square." "No rectangles are squares" is a much stronger claim — and a false one, since squares are rectangles. Negating "all" gives "at least one is not," never "none."
15. $p \wedge q$ false; $p \vee q$ false; $\sim p \wedge \sim q$ true.
16. Any disjunction with one true part, for example "A triangle has three sides or a triangle has four angles." The first part is true, so the whole statement is true.
17. Yes. The requirement is a disjunction, and the varsity-letter part is true, so "at least one" is satisfied even though the GPA part is false.
18. $\sim p$ always has the opposite truth value of $p$, and $\sim(\sim p)$ flips it back. Two flips return every row to where it started, so the columns for $p$ and $\sim(\sim p)$ are identical.

Exit ticket 1.1

19. No. It is a question, so it cannot be true or false.
20. $m\angle 1 \le 90°$. (Equivalently, "$m\angle 1$ is not greater than $90°$." Note that $90°$ itself must be included.)
21. $p \wedge q$ is false; $p \vee q$ is true.
22. An "and" is false as soon as either part is false; an "or" is true as soon as either part is true.

Lesson 1.2 — Conditional Statements

Guided practice

23. Hypothesis: two angles are vertical angles. Conclusion: they are congruent.
24. Hypothesis: a polygon has five sides. Conclusion: it is a pentagon.
25. If a figure is a square, then it is a rhombus.
26. If a number is divisible by $8$, then it is divisible by $4$. (The word "whenever" marks the hypothesis, which is why the sentence has to be reversed to reach if-then form.)
27. Row 2: $p$ true, $q$ false. That is the only false row.
28. A conditional promises a result when the condition is met. If the hypothesis is false the condition was never met, so nothing the conditional promised failed to happen. Nothing broke the promise, so the statement counts as true.

Independent practice

29. a) H: two angles are complementary. C: their measures add to $90°$. b) H: $\overline{AB} \cong \overline{CD}$. C: $AB = CD$. c) H: it is a triangle. C: a figure is a polygon. (The "if" clause comes second in this sentence; the hypothesis is still whatever follows "if.")
30. a) If a figure is a rectangle, then it is a parallelogram. b) If two lines are perpendicular, then they form right angles. c) If you have two points, then they determine exactly one line.
31. a) True. b) False — a $3 \times 7$ rectangle is a parallelogram and is not a rhombus. c) False — the two base angles of an isosceles triangle are congruent and are not vertical angles.

33. True.
34. False. That is row 2, the only false row.
35. Yes, the promise was broken. Hypothesis true (spent over $50), conclusion false (no free shipping) — row 2.
36. No. The hypothesis is false ($30 < 50$), so the promise was never tested. Rows 3 and 4 are both true, so the store's statement stands.
37. A conditional with a false hypothesis is true, no matter what the conclusion says. No triangle has four sides, so the hypothesis is never satisfied and the statement is never broken. The student is confusing "the hypothesis cannot happen" with "the statement is false."
38. A conditional claims something about every case where the hypothesis holds. Twenty confirming cases leave infinitely many unchecked. But a single case with a true hypothesis and a false conclusion is exactly row 2, which makes the statement false — so one counterexample settles it and no number of examples can.
39. Sample true: "If two angles are vertical, then they are congruent." Sample false: "If two angles are congruent, then they are vertical," with counterexample the two base angles of an isosceles triangle, which are congruent and not vertical.
40. $p$: the figure is a square. $q$: the figure has four right angles. Symbolic form: $p \rightarrow q$.

Exit ticket 1.2

41. H: a triangle is equiangular. C: each angle measures $60°$.
42. True hypothesis with a false conclusion — $p$ true, $q$ false.
43. Any non-square rectangle, for example a $4 \times 9$ rectangle. It has four right angles (hypothesis true) and is not a square (conclusion false).
44. A conditional is a promise: it is broken only when the condition is met and the result fails, and an untested promise counts as kept.

Lesson 1.3 — Converse, Inverse, and Contrapositive

Guided practice

45. Converse $q \rightarrow p$; inverse $\sim p \rightarrow \sim q$; contrapositive $\sim q \rightarrow \sim p$.
46. True: the conditional and the contrapositive. False: the converse and the inverse.
47. The dashed arrows join conditional-to-contrapositive and converse-to-inverse. The label says those pairs always agree — they are logically equivalent, so they share a truth value in every case.
48. If a polygon has five sides, then it is a pentagon.
49. If a polygon is not a pentagon, then it does not have five sides.
50. Contrapositive: If a polygon does not have five sides, then it is not a pentagon. It is true, because a conditional and its contrapositive are logically equivalent and the original is true. That equivalence is what lets you skip a separate check.

Independent practice

51. Converse: If two lines intersect, then they are perpendicular. Inverse: If two lines are not perpendicular, then they do not intersect. Contrapositive: If two lines do not intersect, then they are not perpendicular.
52. Conditional true; contrapositive true. Converse false — two lines meeting at $30°$ intersect and are not perpendicular. Inverse false, same counterexample read the other way.
53. Conditional: If a number is divisible by $6$, then it is divisible by $3$ — true. Converse: If a number is divisible by $3$, then it is divisible by $6$ — false, counterexample $9$. Inverse: If a number is not divisible by $6$, then it is not divisible by $3$ — false, counterexample $9$. Contrapositive: If a number is not divisible by $3$, then it is not divisible by $6$ — true.
54. The rectangle disproves the converse, "If it has four right angles, then it is a square." One figure is enough because a conditional claims something about every case satisfying the hypothesis; one case where the hypothesis holds and the conclusion fails is precisely the false row of the truth table.
55. The operator used the inverse ("if you are not under $48$ inches, then you may not ride") or, read another way, treated the sign as a biconditional. It is not equivalent to the sign. The sign only restricts short riders; it says nothing about tall ones.
56. The door is locked, so the conclusion is false; by the contrapositive ("if the door is not unlocked, then the alarm did not sound"), the alarm did not sound.
57. The student wrote the inverse, not the contrapositive — they negated both parts but did not swap them. The contrapositive is "If the game is not cancelled, then it did not rain."
58. A conditional restricts the hypothesis; its converse restricts the conclusion, which is a different set of cases. Sample: "If a figure is a square, then it has four congruent sides" is true, but "If a figure has four congruent sides, then it is a square" is false — a non-square rhombus is the counterexample.
59. Converse also true, sample: "If a triangle is equilateral, then it is equiangular." Converse false, sample: "If a figure is a square, then it is a rectangle."
60. Conditional: If a figure is a square, then it is a parallelogram — true. Converse: If a figure is a parallelogram, then it is a square — false, counterexample a $3 \times 7$ rectangle.
61. Start with $p \rightarrow q$. Its converse is $q \rightarrow p$; the contrapositive of that converse is $\sim p \rightarrow \sim q$, which is the inverse of the original. A statement and its contrapositive always agree, so the converse and the inverse always agree.
62. Its contrapositive is also false (they are equivalent). Its converse could be either — the truth value of a converse is independent of the original.

Exit ticket 1.3

63. Converse: If a figure is a parallelogram, then it is a rhombus. False — a $3 \times 7$ rectangle is a counterexample.
64. The contrapositive.
65. Inverse false (it agrees with the converse); contrapositive true (it agrees with the conditional).
66. It must be a single case in which the hypothesis is true and the conclusion is false.

Lesson 1.4 — Biconditionals and Definitions

Guided practice

67. If an angle is a right angle, then it measures $90°$. If an angle measures $90°$, then it is a right angle.
68. A biconditional claims the two statements always have the same truth value. If either half fails, there is a case where one holds and the other does not, and the biconditional is false in that case.
69. Forward: If a polygon is a triangle, then it has exactly three sides. Backward: If a polygon has exactly three sides, then it is a triangle.
70. True. Both halves were checked separately and both hold, which is why "exactly three sides" serves as the definition of a triangle.
71. False. The backward half fails: a rhombus need not be a square. A rhombus with $60°$ and $120°$ angles is the counterexample.
72. A line is the perpendicular bisector of a segment if and only if it is perpendicular to the segment and passes through its midpoint.

Independent practice

73. a) Forward: if an angle is acute, then $0° < m < 90°$. Backward: if $0° < m < 90°$, then the angle is acute. True. b) Forward: if a quadrilateral is a parallelogram, then both pairs of opposite sides are parallel. Backward: if both pairs are parallel, then it is a parallelogram. True. c) Forward: if a triangle is right, then it has a $90°$ angle. Backward: if a triangle has a $90°$ angle, then it is right. True.
74. Both directions hold: every even number is divisible by $2$, and every number divisible by $2$ is even. Because the two conditions pick out exactly the same numbers, one is the meaning of the other — that is what a definition does. A theorem only guarantees one direction.
75. a) True biconditional — a figure is a rectangle if and only if it is a parallelogram with a right angle. b) Not a biconditional. The backward half fails: a parallelogram need not be a rhombus. c) True biconditional — an angle is straight if and only if it measures $180°$.
76. Sample: A segment is a diameter of a circle if and only if it is a chord that passes through the center. (Accept any correct definition-style statement.)
77. No. The right-hand side is a conjunction, and the "two meetings" part is false, so the whole condition is false and the biconditional makes his standing false too.
78. The door is not secured. A biconditional works in both directions, so "light not green" gives "door not secured" — which is exactly what a one-directional conditional would not let you conclude.
79. Counterexample: a rhombus with $60°$ and $120°$ angles has four congruent sides and is not a square. Repair: "A figure is a square if and only if it has four congruent sides and four right angles."
80. "Vertical angles are congruent" has a false converse — congruent angles need not be vertical — so the two conditions do not pick out the same pairs of angles. "Right angles measure $90°$" has a true converse, so the two conditions describe exactly the same angles and can be joined.
81. Sample: "If a figure is a square, then it is a rectangle" is true and its converse is false. Because the converse fails, the statement cannot be written with "if and only if"; doing so would assert something false.
82. A definition asserts both directions at once, so reaching either side of it in a proof licenses the other. A theorem asserts only $p \rightarrow q$; arriving at $q$ tells you nothing about $p$, and using it backwards is the converse error.

Exit ticket 1.4

83. If two segments are congruent, then they have equal lengths. If two segments have equal lengths, then they are congruent.
84. False. Forward fails: a parallelogram need not be a rectangle (a $60°$–$120°$ parallelogram is a counterexample). Backward holds, but one failing half is enough.
85. A definition is a biconditional — both directions are true, so it may be used either way. A theorem guarantees only the forward direction; its converse is a separate claim that must be proved on its own.
86. Sample: "An angle is a right angle if and only if it measures $90°$," or "Two lines are perpendicular if and only if they intersect to form a right angle."

Lesson 1.5 — Venn Diagrams and Set Relationships

Guided practice

87. $A \cup B$: both circles including the overlap. $A \cap B$: only the overlap.
88. Union matches or; intersection matches and.
89. If $x$ is in $A$, then $x$ is in $B$. (Equivalently, every element of $A$ is an element of $B$.)
90. Everything inside circle $B$ but outside $A$, and the part of the rectangle outside both circles.
91. Rhombi (also Parallelograms, Rectangles, or Quadrilaterals — any set whose region contains the Squares region).
92. Art only: $18 - 7 = 11$. Neither: $30 - (11 + 7 + 7) = 5$.

Independent practice

93. $B \subseteq A$ — every right triangle is a triangle. The diagram is a small circle labelled "right triangles" drawn entirely inside a larger circle labelled "triangles."
94. a) If a figure is a rectangle, then it is a parallelogram. b) If a figure is a square, then it is a rectangle.
95. a) True — the Rhombi region lies inside the Parallelograms oval. b) False — the Parallelograms oval extends well outside Rhombi. c) True — Squares is drawn as the overlap of Rectangles and Rhombi. d) False — the Trapezoids oval does not touch the Parallelograms oval, because a trapezoid has exactly one pair of parallel sides.
96. Both: $8$. Dog only: $22 - 8 = 14$. Cat only: $17 - 8 = 9$. Neither: $40 - (14 + 8 + 9) = 9$.
97. At least one pet: $14 + 8 + 9 = 31$. Neither: $9$.
98. $\sim(\text{cat owners})$ is everyone outside the cat circle: dog only plus neither, $14 + 9 = 23$. (Check: $40 - 17 = 23$.)
99. Fill the outside first: $6$ took neither, so $50 - 6 = 44$ took at least one. Then $31 + 24 - 44 = 11$ took both. (Regions: geometry only $20$, both $11$, art only $13$, neither $6$; total $50$.)
100. Bell only $12 - 5 = 7$; both $5$; basket only $9 - 5 = 4$; neither $3$. Total $7 + 5 + 4 + 3 = 19$ bikes.
101. The student put each circle's total in its region instead of the region-only count. The overlap is already holding $7$ of the $18$ art students and $7$ of the $14$ band students. Correct diagram: art only $11$, both $7$, band only $7$, neither $5$.
102. Adding $|A|$ and $|B|$ counts every element of the overlap twice — once as a member of $A$ and once as a member of $B$. Subtracting $|A \cap B|$ removes one of those two counts, leaving each element counted exactly once.
103. Diagram: a circle labelled "squares" drawn entirely inside a circle labelled "rectangles," with points of the rectangle circle outside the square circle. Conditional: if a figure is a square, then it is a rectangle (true). Converse: if a figure is a rectangle, then it is a square (false), counterexample a $2 \times 5$ rectangle.
104. Both say that membership in $A$ forces membership in $B$, with no exceptions. The nested picture shows there is no part of $A$ outside $B$; the conditional says there is no $x$ making the hypothesis true and the conclusion false. Those are the same condition.

Exit ticket 1.5

105. Shade only the lens where the two circles overlap. It matches the word and.
106. $20 - 12 = 8$.
107. If a figure is a rhombus, then it is a parallelogram. The converse is false ($3 \times 7$ rectangle).
108. Piano only $10$, both $4$, guitar only $7$; $10 + 4 + 7 = 21$, so $25 - 21 = 4$ play neither.

Chapter 1 Review

Part A — Symbolic form

109. a) The figure is a rectangle and it is a rhombus. b) The figure is not a rectangle, or it is a rhombus. c) If the figure is a rectangle, then it is a rhombus. d) The figure is a rectangle if and only if it is a rhombus. The figure described by (a) is a square.
110. Negation: $m\angle A < 90°$.

Part B — Conditionals and their relatives

111. a) H: a triangle is equilateral. C: it is isosceles. b) Converse: If a triangle is isosceles, then it is equilateral. Inverse: If a triangle is not equilateral, then it is not isosceles. Contrapositive: If a triangle is not isosceles, then it is not equilateral. c) Conditional true (three congruent sides certainly includes two). Contrapositive true. Converse false — a triangle with sides $5$, $5$, $8$ is isosceles and not equilateral. Inverse false, same counterexample. d) A conditional and its contrapositive always agree, and a converse and an inverse always agree. So checking the conditional settles the contrapositive, and checking the converse settles the inverse — two independent checks cover all four.

Part C — Biconditionals, Venn diagrams, and context

112. a) Both: $9$. Chorus only: $34 - 9 = 25$. Drama only: $28 - 9 = 19$. Neither: $60 - (25 + 9 + 19) = 7$. b) Chorus or drama: $25 + 9 + 19 = 53$. Neither: $7$. c) No. A biconditional would require the chorus students and the drama students to be exactly the same $60$-student subset. The diagram shows $25$ students in chorus but not drama and $19$ in drama but not chorus, so each half of the biconditional fails. d) Neither set is a subset of the other, and the diagram says why: a subset would need one circle's non-overlap region to be empty, and here both non-overlap regions are non-empty ($25$ and $19$). A true subset statement about the survey is "chorus students who are also in drama $\subseteq$ chorus students," or "students in both $\subseteq$ students in chorus or drama."

Workbook-only items

Page 2, fill in the blanks. A statement is either true or false, and not both. A sentence that is false is still a statement.

Page 3, symbol frame. negation $\sim p$ / "not $p$"; conjunction $p \wedge q$ / "$p$ and $q$"; disjunction $p \vee q$ / "$p$ or $q$"; conditional $p \rightarrow q$ / "if $p$, then $q$"; biconditional $p \leftrightarrow q$ / "$p$ if and only if $q$."

Page 5, fill in the blanks. An "and" is true in exactly one row — the one where both parts are true. An "or" is false in exactly one row — the one where both parts are false.

Page 9, label the parts. The part after "if" is the hypothesis; the part after "then" is the conclusion.

Page 11, fill in the blanks. A conditional is false only when the hypothesis is true and the conclusion is false. When the hypothesis is false the conditional is true, because the promise was never tested.

Page 15, related-statement frame. Converse $q \rightarrow p$, built by swapping. Inverse $\sim p \rightarrow \sim q$, built by negating both. Contrapositive $\sim q \rightarrow \sim p$, built by swapping and negating.

Page 20, fill in the blanks. A biconditional is true exactly when the conditional and its converse are both true. Every definition in geometry is a biconditional.

Page 24, fill in the blanks. $A \cup B$ is the union — English word or. $A \cap B$ is the intersection — English word and.

Page 27, fill in the blanks. Step 1: put the both (overlap) count in the middle. Step 2: subtract to get each "only" region. Step 3: whatever is left goes outside the circles.

Pages 7 and 28, blank frames. Any assigned truth table or Venn diagram. Expected conventions: truth-table rows in the order T-T, T-F, F-T, F-F; Venn regions filled middle-first, with each number counting that region only and the outside region never left blank when a total is given.

Pages 25, 26, and 28, drawing frames. Diagrams should show a rectangle for the universe, circles labelled with set names, and nesting drawn only where one set really is contained in the other. A subset is drawn fully inside; overlapping sets that are not subsets must leave both non-overlap regions visibly non-empty.