Two-column Proofs Logic Worksheet Answer Key – Geometry

Problem 1 Hypothesis and conclusion

The hypothesis follows “if” and the conclusion follows “then.”

Hypothesis: a whole number ends in 0. Conclusion: it is divisible by 10.

Problem 2 Writing a converse

Swap the parts: “If a quadrilateral has four right angles, then it is a square.” A rectangle that is not a square has four right angles, so the converse fails.

Converse: if a quadrilateral has four right angles, then it is a square. It is false.

Problem 3 Inverse and contrapositive

Inverse: negate both parts. Contrapositive: swap and negate.

The inverse is false because a sparrow is not a robin but is a bird. The contrapositive is true because anything that is not a bird cannot be a robin.

Inverse: if an animal is not a robin, then it is not a bird (false). Contrapositive: if an animal is not a bird, then it is not a robin (true).

Problem 4 Find a counterexample

The number 2 is prime (its only factors are 1 and 2), but 2 is even. The hypothesis is true and the conclusion is false.

Counterexample: 2

Problem 5 Spot the pattern

1. The differences are 5, 7, 9, which increase by 2 each time.
2. The next differences should be 11 and 13, so the next terms are 24 + 11 = 35 and 35 + 13 = 48. (Each term is a perfect square minus 1: 7² − 1 = 48.)

Next two terms: 35 and 48

Problem 6 Inductive or deductive

1. She generalizes from examples: inductive.
2. A general fact is applied to one case: deductive.
3. Generalizing from three triangles: inductive.
4. A proven fact forces the value 180 − 70 − 60 = 50: deductive.

(a) inductive, (b) deductive, (c) inductive, (d) deductive

Problem 7 Name the property

1. Symmetric Property.
2. Reflexive Property.
3. Transitive Property (or Substitution).
4. Distributive Property.
5. Division Property of Equality.

(a) symmetric, (b) reflexive, (c) transitive, (d) distributive, (e) division

Problem 8 Reasons for solving steps

Check: 4(5) + 5 = 25 and 2(8) + 9 = 25.

x = 8

Problem 9 Biconditional statements

1. Converse: if x = 7, then 3x + 1 = 22. Both are true, so: 3x + 1 = 22 if and only if x = 7.
2. Converse: if a whole number is even, then it is divisible by 4. Counterexample: 6. The converse is false, so no biconditional exists.

(a) biconditional is true; (b) no biconditional, since 6 is even but not divisible by 4

Problem 10 Given and Prove

Given: ∠1 and ∠3 are supplementary, and ∠2 and ∠3 are supplementary. Prove: m∠1 = m∠2.

Check: m∠1 = 180° − 64° = 116° and m∠2 = 180° − 64° = 116°, so they are equal.

Given: ∠1, ∠3 supplementary; ∠2, ∠3 supplementary. Prove: m∠1 = m∠2 (both 116° when ∠3 = 64°).

Problem 11 Complete a segment proof

Check: AM = 5(6) − 3 = 27 and MB = 3(6) + 9 = 27.

x = 6

Problem 12 Complements of one angle

m∠AOB = m∠COD

Problem 13 Use the vertical angles theorem

By the Vertical Angles Theorem, 8x − 17 = 5x + 19, so 3x = 36 and x = 12.

Each vertical angle is 8(12) − 17 = 79°. A neighbor forms a linear pair with 79°, so it measures 180° − 79° = 101°.

x = 12; 79° and 79°; neighbors 101°

Problem 14 Paragraph proof of perpendicular lines

Because ∠POR and ∠ROQ sit on a straight line, they are supplementary, so m∠POR + m∠ROQ = 180°. Since the angles are congruent, their measures are equal, so substitution gives 2m∠POR = 180°. Dividing by 2 gives m∠POR = 90°. The lines meet at a right angle, so by definition they are perpendicular.

The lines are perpendicular because both angles measure 90°.

Problem 15 Camp rules and logic

1. The hypothesis is true, so the range is closed.
2. This is the contrapositive: not closed means not raining. It is not raining.
3. No. The range may close for wind or a staff meeting, so the conclusion cannot be drawn.
4. Converse: if the archery range is closed, then it is raining. The rule does not guarantee it.

(a) closed, (b) not raining, (c) no conclusion, (d) converse not guaranteed

Problem 16 Disproving three statements

1. Angles of 100° and 80° drawn in different places are supplementary but not adjacent, so they are not a linear pair.
2. x = −6, since (−6)² = 36 but −6 ≠ 6.
3. 20 is divisible by 5 and ends in 0.

(a) 100° and 80° apart, (b) −6, (c) 20

Problem 17 Odd numbers and a false conjecture

Sums: 1, 1 + 3 = 4, 1 + 3 + 5 = 9, 16, 25. Conjecture: the sum of the first n odd numbers is n². For n = 12 the prediction is 144.

For the second conjecture, n = 41 gives 41² − 41 + 41 = 41² = 1,681 = 41 · 41, which is not prime. So the conjecture is false even though it works for many smaller values.

Sum = n² (144 for n = 12); counterexample n = 41 gives 1,681 = 41 · 41

Problem 18 Linear pair and bisector steps

Check: 116 + 64 = 180.

x = 15; 116° and 64°; m∠BOD = 32°

Problem 19 Equal segments with a shared piece

AC = BD

Problem 20 Bisectors of a linear pair

Verification: if m∠AOB = 76°, then m∠BOC = 104°, so m∠MOB = 38° and m∠BON = 52°, and 38° + 52° = 90°.

∠MON = 90°