Semester A Review Answers. 1. point, line, and plane. 2. one. 3. three. 4. one or No, since AB BC AC 11. AC a. EG FH.

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1 1. point, line, and plane 2. one 3. three 4. one or 8 6. b 23, c No, since C C x C C x x x a. EG FH b. EG m 2 55 o 17. x m o 19. a. x 3.75 b. m C 27.5 o 20. x 20, y 80 MCPS

2 21. F x, y x, y 22. F xy, xy, 23. F xy, x 5, y 24. F xy, x, y 25. F x, y x, y C 27. a. b. y O x c. 2, 4, 3, 4 d. P x, y, P x 5, y 28. a. dog Warmblooded b. If the animal is warm-blooded, then the animal is a dog. c. If the animal is not a dog, then the animal is not warm-blooded. d. If the animal is not warm-blooded, then the animal is not a dog. e. the contrapositive MCPS

3 29. a. If a person is a teenager, then the person is between the ages of 13 and 19, (including 13 and 19). b. If a person is not between the ages of 13 and 19, (including 13 and 19), then the person is not a teenager. c. If a person is not a teenager, then the person is not between the ages of 13 and 19, (including 13 and 19). d. The converse, inverse, and contrapositive are all true. 30. a. 20 yrs old Can vote b. If you can vote, then you are inside the can vote rectangle, however, you may not be twenty years old, so you may or may not be inside the circle, so the statement is false. 31. Chris will go to the game and take Jane. 32. Triangle C is equiangular. 33. Sally did not study for the test. 34. C 35. inductive 36. deductive 37. deductive 38. inductive 39. Statement #3, statement #1, statement #2 40. Statement #4, statement #2, statement #3, statement #1 41. a. four b. eight c. infinite number MCPS

4 Property Parallelogram Rectangle Square Rhombus Trapezoid Opposite sides congruent X X X X Exactly one pair of sides parallel X Opposite angles congruent X X X X Diagonal forms 2 congruent triangles X X X X Diagonals bisect each other X X X X Diagonals congruent X X Diagonals perpendicular X X Diagonal bisects opposite angles X X ll angles are right angles X X ll sides are congruent X X 45. x x 20, y a. lines n and p; corresponding angles are congruent b. lines l and m; alternate interior angles are congruent c. lines l and m; same side interior angles are supplementary. 48. x 30, y D 50. grid in grid in grid in grid in The third side must be longer than 3 and shorter than 15. MCPS

5 55. y x x 5.5, y x x CD is a parallelogram. The slopes of and CD are 2, therefore those sides are parallel. The slopes of C and D are both 3, so those sides are parallel. Since both pairs of opposite sides are parallel, the figure is a parallelogram. Since is not perpendicular to C (slope of = 2, slope of C = 3), the figure cannot be a square or rectangle, and since the diagonals are not perpendicular (slope of C 7, slope of 1 D ), the figure is cannot a rhombus. Therefore the figure is a parallelogram The triangle is a right triangle with right angle at point. The slope of the slope of , a. Yes, SS b. No, is not a triangle congruence postulate c. Yes, S d. Yes, SSS e. No, SS is not a congruence postulate f. Yes, S (or S) 7 C, and 2 MCPS

6 64. Statements Reasons 1. D is the perpendicular bisector of C 1. Given 2. E EC 2. Definition of perpendicular bisector 3. E and EC are right angles 3. Definition of perpendicular bisector 4. E EC 4. ll right angles are congruent 5. E E 5. Reflexive property of congruence 6. E EC 6. SS 7. C C 7. CPCTC lternative Proof: Statements Reasons 1. D is the perpendicular bisector of C 1. Given 2. C 2. ny point on the perpendicular bisector of a segment is equidistant from the endpoints of the segment 3. E EC 3. Definition of perpendicular bisector 4. E E 4. Reflexive property of congruence 5. E EC 5. SSS 6. C C 6. CPCTC 65. Statements Reasons 1. D EG 1. Given 2. CF CFG 2. If two parallel lines are cut by a transversal, alternate interior angles are congruent 3. C FG 3. Given 4. CF CF 4. Reflexive property of congruence 5. CF CGF 5. SS 6. CF CGF 6. CPCTC 66. a. trapezoid b. rhombus c. rectangle d. none of these figures MCPS

7 MCPS

8 67 a. C D E Justification: Congruent circles were constructed with centers at points and. Since radii of congruent circles are congruent, C C D D; therefore CD is a rhombus. In the rhombus, the diagonals are perpendicular, therefore CD. Since CD is a parallelogram the diagonals bisect each other. Therefore E E, so CD is the perpendicular bisector of. 67 b. C D Justification: Csince they are the radii of the same circle. D DCsince they are constructed using the same compass setting. D D by the reflexive property of congruence. Therefore D CD by SSS. D CD by CPCTC, and by the definition of angle bisector D bisects C. MCPS

9 67 c. D C E F Justification: CE, D CF since they were drawn by the same compass setting. D EF since they were drawn with the same compass setting. Therefore D ECF by SSS. Therefore D ECF by CPCTC. Finally by the converse of the corresponding angles postulate, CF // D. 67 d. D C Justification. I drew segments between and and and C. I constructed the perpendicular bisector of. Every point on that line is equidistant from points and. I constructed the perpendicular bisector of C. Every point on that line is equidistant from points and C. Therefore, point D, the intersection of those two perpendicular bisectors, is equidistant from points,, and C. 68. D 69. MCPS

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