Geometry - Fall Semester Extra Credit for Final Exam
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1 Geometry - Fall Semester Extra Credit for Final Exam Students need to understand and remember that the final exam counts as much as a full 6 weeks grade. The semester average is used to decide credit and grade point average is determined by averaging all three 6 weeks grades plus the grade on the final exam and dividing by four. Completing this packet over Geometry topics covered in the first semester is worth 10 additional points added to the final exam grade. The completed packet should be returned on Monday, January 7, 2013 to your Geometry teacher. You will receive an additional 3 points if your parent signs this packet cover indicating they understand this extra credit opportunity is available. THIS IS NOT REQUIRED FOR NYONE TO DO OVER THE HOLIDYS. IT IS EXTR CREDIT. Student Name Parent signature Date IMPORTNT NOTICE: Packets with only circled answers and no work shown will not receive extra credit points.
2 Name Date Period State the one reason justification for each of the following. Be specific. 1. If M is the midpoint of B, then M MB. 2. In BC, m + m B + m C = is a right, therefore, r s. 1 r s 1 4. If PQ TR, then PQ = TR 5. If B and B D, then D 6. l bisects CD, therefore, Q is the midpoint of CD. C l Q D 7. p q, therefore, 1 2 p 1 2 q therefore, p q (use picture above) 9. R and P are supplementary therefore, m R + m P = Y is a right, therefore XYZ is a right. X Y Y 11. X Z, thereforeyx YZ. Z X Z and 1 & 2 are supplementary s therefore 1 & 2 are right s. 13. In LMN and SQR, L Q & M S, therefore N R.
3 14. In isosceles BC with vertex, D bisects BC Therefore D BC and D passes thru the midpoint of BC O 1 C B OC bisects OB. Therefore If a = b then b = a & 2 are complementary; 3 and 4 are complementary; 2 4. Therefore M and N are complementary. Therefore m M + m N = m P = m Q. Therefore P Q and 2 are right angles. Therefore M O 1 2 m 1 + m 2 = m MNO N 23. If m = 50 and m + m B = 100, then 50 + m B = 100 ( n 2) is the formula for n _ 25. m 1 = m + m B _ 1 B 26. BC is isosceles. Therefore B BC _ 27. B BC. Therefore C B _ C & 2 are supp. _
4 29. m C + m D = 180. Therefore C supp to D 30. If a = b, then a + c = b + c x 1 + x M = 2, y 1 + y 2 2 is the formula for _ 32. X is a right angle. Therefore m X = BCD is a rectangle. Therefore C is a right angle. 34. X Y Z Therefore X and Z are complementary angles is the formula to find of a regular polygon. n 36. RS + ST = RT R S T If and B are supp to C, then B d = ( x 2 x 1) + ( y 2 y 1) is the formula to find the between two points m 1 + m 2 = 180 State a one reason justification for the following. 41. a ǁ b; b ǁ c, therefore a ǁ c 42. x z and y z, therefore x ǁ y 43. The slopes of ǁ lines are. 44. The slopes of lines are. 45. (n 2)180 is the formula to find the of the of a polygon. State the one reason justification of the following. Sketch a picture when necessary. 46. In BC, if m > m B, then BC > C
5 47. If l is bisector of B, then P = PB l P B Fill in the blanks. 48. The point of concurrency for the altitudes in a triangle is called the. a) cute b) right c) obtuse 49. The point of concurrency for the medians in a triangle is called the. a) cute b) right c) obtuse 50. The point of concurrency for the angle bisectors in a triangle is called the. This point is equidistant from the. a) cute b) right c) obtuse 51. The point of concurrency for the bisectors in a triangle is called the. This point is equidistant from the. a) cute b) right c) obtuse Complete the following proofs as 2 column proofs on your own paper. 56. Given: 1 2; C D Prove: KC KD 57. Given: 3 4; CB DB Prove: CB DB 58. Given: B CD ; 3 4 Prove: CKB DKB 59. Given: 1 R Prove: 2 S 1 K B C D Use this picture for T 60. Given: TX TY ; 1 R Prove: 2 R R X 1 2 Y S Use this picture for 59-60
6 61. Given: j ǁ k; l ǁ m Prove: Given: j ǁ k; l ǁ m Prove: Given: j ǁ k; 1 11 Prove: l ǁ m j k l m Use this picture for Given: l ǁ m; 1 4 Prove: l m Study Flashcards, Notes, Quizzes, Vocabulary, and Tests from this semester. You may review tests and quizzes in the classroom only.
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