Lesson. Warm Up deductive 2. D. 3. I will go to the store; Law of Detachment. Lesson Practice 31
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1 Warm Up 1. deductive 2. D b. a and b intersect 1 and 2 are supplementary 2 and 3 are supplementary 3. I will go to the store; Law of Detachment Lesson Practice a and 2 are and 3 are. 3. m 1 + m m 1 = m m 3 + m and 2 are. 5. Substitution 6. Linear Pair Definition of supplementary m 1 = m Linear Pair m 1 + m 2 = 180º m 2 + m 3 = 180º c. m 1 + m 2 = m 2 + m and 4 are. 2. m 1 + m and 3 are. 2 and 4 are. 4. m 1 = m 3, m 2 = m 4 5. m 2 + m and 3 are. Definition of supplementary Transitive Subtraction Vertical Angles Substitution 6. Supplemental Publishers Inc. All rights reserved. LSN 1 Saxon Geometry
2 d. It is given that s and t are perpendicular lines. By the definition of perpendicular lines, 1 and 2 are right. Angles 1 and 2 are by 5-3: All right are. Angles 1 and 2 are adjacent by the definition of adjacent. Therefore, 1 and 2 are adjacent. Supplemental Publishers Inc. All rights reserved. LSN 2 Saxon Geometry
3 Practice AE CE, DE BE 2. AE = CE, DE = BE 3. m AED = m CEB 4. AED CEB segments 3. Vertical Angles 4. SAS 5. AD CB 5. CPCTC cm By the Vertical Angles, 1 2. Then by the Same- Side Interior Angles, 2 and 3 are supplementary. By the Alternate Interior Angles, 3 4. By substitution 2 and 4 are supplementary. By the definition of a linear pair, 4 and 5 are supplementary. Using the Congruent Supplements, 2 5. Finally, 1 5 by the Transitive Congruence no in. Supplemental Publishers Inc. All rights reserved. LSN 3 Saxon Geometry
4 m 1 = m 3 m 2 = m BD bisects ABC 1 2 m 1 = m 2 Substitution bisector 12. A X, B _ Y, C Z, AB XY, BC YZ, CA ZX _ ; Transitive Congruence A = 25, 1. l = 5 2. A = lw 2. Area of a rectangle = 5w 3. Substitution 4. _ 25 = _ 5w 4. Division w = 5 5. Simplify 6. ABCD is a square 6. a square 14. If two numbers are opposites, then they have midpoint 0 on a number line; Both the statement and the converse are true. 15. C 16. a: x = -3; b: y = There is a good chance of wind and rain later in the day. Supplemental Publishers Inc. All rights reserved. LSN 4 Saxon Geometry
5 18. Gayle is correct because according to the ASA or the AAS Postulate, corresponding sides must be in order for the tri to be. 19. Sample: The equation x + 1 = x + 2 has no solutions AD BC, 1. AD BC 2. AD = BC 2. segments 3. m ADE = m CBE 4. m AED = m CEB 3. Alternate Interior Angles 4. Vertical Angles 5. ADE 5. AAS CBE 6. AE = CE 6. CPCTC Lesson _ 21. Since AD bisects CB, CD BD _ by definition of _ bisector. _ Because AD CB, ADC and ADB are right by definition of perpendicular. Then ADC ADB, because all right are. From the drawing, AD _ AD _ by the Reflexive Property. Finally, ACD ABD by SAS Postulate. Supplemental Publishers Inc. All rights reserved. LSN 5 Saxon Geometry
6 22. ((x - 3) - 1) 2 + ((4x + 1) - 1) 2 = 5 Distance Formula (x - 4 ) 2 + (4x ) 2 = 5 Simplify. x 2-8x x 2 = 5 Expand exponents. 17 x 2-8x + 16 = 5 Simplify. ( 17 x 2-8x + 16 ) 2 = 5 2 Square both sides. 17 x 2-8x + 16 = 25 Simplify. 17 x 2-8x = Subtraction 17 x 2-8x - 9 = 0 Simplify. (17x + 9)(x - 1) = 0 Factor. x - 1 = 0 x > 0 x = 1 Addition x = 1 Simplify. The coordinates are (-2, 5). 23. y - 2 = 4(x + 3) or y = 4x + 14 Supplemental Publishers Inc. All rights reserved. LSN 6 Saxon Geometry
7 24. LP MN LO MO LP = MN LO = MO 26. a. 0.9 m 2 b m 27. a. p or q congruency congruency b. p and q 25. yes; LP - LO = MN - MO OP = ON OP ON 1. B and E are right, AB = DE, AC = DF Subtraction Property of Simplify congruency A B 2 + B C 2 = AC 2 2. Pythagorean 3. D E 2 + E F 2 = DF 2 3. Pythagorean 4. A B 2 + B C 2 = D E 2 + E F 2 4. Substitution 5. B C 2 = E F 2 5. Subtraction 6. BC = EF 6. Square root 7. m ABC = m DEF 7. All right are 8. ABC DEF 8. SAS c. p and q 28. interior: TSZ, WVU; exterior: AZY, SZB, and WVC 29. No, the circumference is cm. 30. x = 5 Supplemental Publishers Inc. All rights reserved. LSN 7 Saxon Geometry
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