Proofs of Parallel and Perpendicular Lines of Parallel and Perpendicular Lines ... and Perpendicular Lines Answer Section ... REF: 3-1 Properties of Parallel Lines OBJ: 3-1.2 Properties of ...

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<ul><li><p>Name: ________________________ Class: ___________________ Date: __________ ID: A</p><p>1</p><p>Proofs of Parallel and Perpendicular Lines</p><p>Short Answer</p><p> 1. Give the missing reasons in this proof of the Alternate Interior Angles Theorem.</p><p>Given: l n</p><p>Prove: 4 6</p><p>Statments Reasons</p><p>1. l n</p><p>2. 2 6</p><p>3. 4 2</p><p>4. 6 4</p><p>1. Given</p><p>a. ?</p><p>b. ?</p><p>c. ?</p></li><li><p>Name: ________________________ ID: A</p><p>2</p><p> 2. State the missing reasons in this proof.</p><p>Given: 1 5</p><p>Prove: p r</p><p>Statements Reasons</p><p>1. 1 5</p><p>2. 4 1</p><p>3. 4 5</p><p>4. p r</p><p>Given</p><p>a.____</p><p>b.____</p><p>c.____</p><p> 3. The 8 rowers in the racing boat stroke so that the angles formed by their oars with the side of the boat all stay </p><p>equal. Explain why their oars on either side of the boat remain parallel.</p><p> 4. Suppose you have four identical pieces of wood like those shown below. If mb = 40 can you construct a </p><p>frame with opposite sides parallel? Explain.</p><p> 5. Find the measure of each interior and exterior angle. The diagram is not to scale.</p></li><li><p>Name: ________________________ ID: A</p><p>3</p><p> 6. The fireworks technician has two rocket launchers, each with a base and stand in the form of an L. A </p><p>diagonal trough on which the technician places a rocket joins the ends of each L. One launcher has a 4-inch </p><p>base and 10-inch stand. The other has a 6-inch base and a 15-inch stand. Explain why two rockets launched </p><p>from the two devices could follow parallel paths.</p><p>Essay</p><p> 7. Write a paragraph proof of this theorem: In a plane, if two lines are perpendicular to the same line, then they </p><p>are parallel to each other.</p><p>Given: r s, t s </p><p>Prove: r t</p><p> 8. Write a two-column proof.</p><p>Given: 2 and 5 are supplementary.</p><p>Prove: l m</p></li><li><p>Name: ________________________ ID: A</p><p>4</p><p> 9. Find the values of the variables. Show your work and explain your steps. The diagram is not to scale.</p><p>Other</p><p> 10. Given m1 = m2, what can you conclude about the lines l, m, and n? Explain.</p><p> 11. Justify the statement algebraically.</p><p>In a triangle, if the sum of the measures of two angles is equal to the measure of the third angle, then the </p><p>triangle is a right triangle.</p><p> 12. Line p contains points A(1, 4) and B(3, 5). Line q is parallel to line p. Line r is perpendicular to line q. </p><p>What is the slope of line r? Explain.</p></li><li><p> ID: A</p><p>1</p><p>Proofs of Parallel and Perpendicular Lines</p><p>Answer Section</p><p>SHORT ANSWER</p><p> 1. ANS: </p><p>a. Corresponding angles.</p><p>b. Vertical angles.</p><p>c. Transitive Property.</p><p>PTS: 1 DIF: L2 REF: 3-1 Properties of Parallel Lines</p><p>OBJ: 3-1.2 Properties of Parallel Lines NAT: NAEP 2005 M1f | ADP K.2.1</p><p>STA: NJ 4.1.12 B.1 | NJ 4.2.12 A.3a | NJ 4.2.12 A.4a TOP: 3-1 Example 3</p><p>KEY: alternate interior angles | Alternate Interior Angles Theorem | proof | reasoning | two-column proof | </p><p>multi-part question</p><p> 2. ANS: </p><p>a. Vertical angles.</p><p>b. Transitive Property.</p><p>c. Alternate Interior Angles Converse.</p><p>PTS: 1 DIF: L2 REF: 3-2 Proving Lines Parallel</p><p>OBJ: 3-2.1 Using a Transversal NAT: NAEP 2005 M1e | NAEP 2005 M1f | ADP K.2.3</p><p>STA: NJ 4.1.12 B.1 | NJ 4.2.12 A.1 | NJ 4.2.12 A.3a | NJ 4.2.12 A.4a </p><p>TOP: 3-2 Example 1 </p><p>KEY: two-column proof | proof | reasoning | corresponding angles | multi-part question</p><p> 3. ANS: </p><p>The rowers keep corresponding angles congruent.</p><p>PTS: 1 DIF: L3 REF: 3-2 Proving Lines Parallel</p><p>OBJ: 3-2.1 Using a Transversal NAT: NAEP 2005 M1e | NAEP 2005 M1f | ADP K.2.3</p><p>STA: NJ 4.1.12 B.1 | NJ 4.2.12 A.1 | NJ 4.2.12 A.3a | NJ 4.2.12 A.4a </p><p>TOP: 3-2 Example 1 KEY: transversal | word problem | reasoning | parallel lines</p><p> 4. ANS: </p><p>No. Explanations may vary. Sample:</p><p>Placing three pieces together forms same-side interior angles with measures of 80. Since 80+ 80 180, they </p><p>are not supplementary, so the sides are not parallel.</p><p>PTS: 1 DIF: L3 REF: 3-3 Parallel and Perpendicular Lines</p><p>OBJ: 3-3.1 Relating Parallel and Perpendicular Lines </p><p>NAT: NAEP 2005 M1e | NAEP 2005 M1f | ADP K.2.1 </p><p>STA: NJ 4.2.12 A.1 | NJ 4.2.12 A.3a | NJ 4.2.12 A.4a TOP: 3-3 Example 1</p><p>KEY: word problem | problem solving | parallel lines</p></li><li><p> ID: A</p><p>2</p><p> 5. ANS: m1 = m2 = m3 = 90,m4 = 122,m5 = m6 = 58,m8 = 32,m7 = m9 = 148</p><p>PTS: 1 DIF: L3 REF: 3-4 Parallel Lines and the Triangle Angle-Sum Theorem</p><p>OBJ: 3-4.2 Using Exterior Angles of Triangles </p><p>NAT: NAEP 2005 G3b | NAEP 2005 G3f | ADP J.5.1 | ADP K.1.2 </p><p>STA: NJ 4.1.12 B.1 | NJ 4.2.12 A.1 | NJ 4.2.12 A.3a | NJ 4.2.12 A.4b </p><p>KEY: Triangle Angle-Sum Theorem | exterior angle</p><p> 6. ANS: </p><p>Pointed in the same direction, the two launchers have equal slopes, so the rockets would be set up to follow </p><p>parallel paths.</p><p>PTS: 1 DIF: L3 REF: 3-7 Slopes of Parallel and Perpendicular Lines</p><p>OBJ: 3-7.1 Slope and Parallel Lines </p><p>NAT: NAEP 2005 A1h | NAEP 2005 A2a | ADP J.4.1 | ADP J.4.2 | ADP K.10.2</p><p>STA: NJ 4.1.12 B.1 | NJ 4.2.12 A.1 | NJ 4.2.12 A.3a | NJ 4.2.12 C.1b | NJ 4.2.12 C.1d | NJ 4.2.12 C.1e | NJ </p><p>4.3.12 C.1a | NJ 4.3.12 C.2 </p><p>KEY: slope | slopes of parallel lines | word problem | problem solving | writing in math</p><p>ESSAY</p><p> 7. ANS: </p><p>[4] By the definition of , r s implies m2 = 90, and t s implies m6 = 90. Line s </p><p>is a transversal. 2 and 6 are corresponding angles. By the Converse of the </p><p>Corresponding Angles Postulate, r || t.</p><p>[3] correct idea, some details inaccurate</p><p>[2] correct idea, not well organized</p><p>[1] correct idea, one or more significant steps omitted</p><p>PTS: 1 DIF: L4 REF: 3-3 Parallel and Perpendicular Lines</p><p>OBJ: 3-3.1 Relating Parallel and Perpendicular Lines </p><p>NAT: NAEP 2005 M1e | NAEP 2005 M1f | ADP K.2.1 </p><p>STA: NJ 4.2.12 A.1 | NJ 4.2.12 A.3a | NJ 4.2.12 A.4a TOP: 3-3 Example 2</p><p>KEY: paragraph proof | proof | reasoning | extended response | rubric-based question | perpendicular lines</p></li><li><p> ID: A</p><p>3</p><p> 8. ANS: </p><p>[4] Statements Reasons</p><p>1. 2 and 5 are supplementary</p><p>2. 3 2</p><p>3. 3 and 5 are supplementary</p><p>4. l m</p><p>1. Given</p><p>2. Vertical angles</p><p>3. Substitution</p><p>4. Converse of Same-Side</p><p>Interior Angles Theorem</p><p>[3] correct idea, some details inaccurate</p><p>[2] correct idea, some statements missing</p><p>[1] correct idea, several steps omitted</p><p>PTS: 1 DIF: L4 REF: 3-2 Proving Lines Parallel</p><p>OBJ: 3-2.1 Using a Transversal NAT: NAEP 2005 M1e | NAEP 2005 M1f | ADP K.2.3</p><p>STA: NJ 4.1.12 B.1 | NJ 4.2.12 A.1 | NJ 4.2.12 A.3a | NJ 4.2.12 A.4a </p><p>KEY: two-column proof | proof | extended response | rubric-based question | parallel lines | supplementary </p><p>angles</p><p> 9. ANS: </p><p>[4] w + 31 + 90 = 180, so w = 59. Since vertical angles are congruent, y = 59. Since </p><p>supplementary angles have measures with sum 180, x = v = 121. z + 68 + y = z </p><p>+ 68 + 59 = 180, so z = 53.</p><p>[3] small error leading to one incorrect answer</p><p>[2] three correct answers, work shown</p><p>[1] two correct answers, work shown</p><p>PTS: 1 DIF: L3 REF: 3-4 Parallel Lines and the Triangle Angle-Sum Theorem</p><p>OBJ: 3-4.2 Using Exterior Angles of Triangles </p><p>NAT: NAEP 2005 G3b | NAEP 2005 G3f | ADP J.5.1 | ADP K.1.2 </p><p>STA: NJ 4.1.12 B.1 | NJ 4.2.12 A.1 | NJ 4.2.12 A.3a | NJ 4.2.12 A.4b </p><p>KEY: Triangle Angle-Sum Theorem | vertical angles | supplementary angles | extended response | </p><p>rubric-based question </p><p>OTHER</p><p> 10. ANS: </p><p>l and m are both perpendicular to n. Explanation: Since l and m are parallel, 1 and 2 are supplementary </p><p>by the Same-Side Interior Angles Theorem. It is given that m1 = m2, so </p><p>180 = m1 + m2 = m1 + m1 = 2m1, and m1 = 90 = m2. Since 1 and 2 are right angles, l is </p><p>perpendicular to n and m is perpendicular to n.</p><p>PTS: 1 DIF: L3 REF: 3-1 Properties of Parallel Lines</p><p>OBJ: 3-1.2 Properties of Parallel Lines NAT: NAEP 2005 M1f | ADP K.2.1</p><p>STA: NJ 4.1.12 B.1 | NJ 4.2.12 A.3a | NJ 4.2.12 A.4a </p><p>KEY: perpendicular lines | reasoning | writing in math</p></li><li><p> ID: A</p><p>4</p><p> 11. ANS: </p><p>m1 + m2 + m3 = 180. Given m1 + m2 = m3, by substitution, m3 + m3 = 180. 2m3 = 180, and </p><p>m3 = 90. Thus, 3 is a right angle, and the triangle is a right triangle.</p><p>PTS: 1 DIF: L4 REF: 3-4 Parallel Lines and the Triangle Angle-Sum Theorem</p><p>OBJ: 3-4.1 Finding Angle Measures in Triangles </p><p>NAT: NAEP 2005 G3b | NAEP 2005 G3f | ADP J.5.1 | ADP K.1.2 </p><p>STA: NJ 4.1.12 B.1 | NJ 4.2.12 A.1 | NJ 4.2.12 A.3a | NJ 4.2.12 A.4b </p><p>KEY: Triangle Angle-Sum Theorem | reasoning | writing in math </p><p> 12. ANS: </p><p>4</p><p>9; Line r is perpendicular to line p because a line perpendicular to one of two parallel lines is also </p><p>perpendicular to the other. Thus, the slope of line r is the opposite reciprocal of the slope of line p.</p><p>PTS: 1 DIF: L3 REF: 3-7 Slopes of Parallel and Perpendicular Lines</p><p>OBJ: 3-7.2 Slope and Perpendicular Lines </p><p>NAT: NAEP 2005 A1h | NAEP 2005 A2a | ADP J.4.1 | ADP J.4.2 | ADP K.10.2</p><p>STA: NJ 4.1.12 B.1 | NJ 4.2.12 A.1 | NJ 4.2.12 A.3a | NJ 4.2.12 C.1b | NJ 4.2.12 C.1d | NJ 4.2.12 C.1e | NJ </p><p>4.3.12 C.1a | NJ 4.3.12 C.2 </p><p>KEY: perpendicular lines | parallel lines | slopes of parallel lines | slopes of perpendicular lines | reasoning | </p><p>writing in math</p></li></ul>

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