Geometry / Mr. Hansen 9/30/2003

Name: _________________________

Answer Key (Without Work) for Chapter 2 Practice Test

	Remember, you would be required to show work. This key shows only the answers. Please do not read this answer key until you have tried all the problems, under time pressure.
1-10.	ANA SAA SSA S
11.	35°
12.	Wordless answer would consist of two diagrams as in #2b on p.86, followed by the implication arrow (Þ), followed by two diagrams in which ÐYXZ and ÐCBD are both marked with a single arc. Sentence form of answer could be something like this: “If two congruent angles each have a smaller angle subtracted from them, and if those two smaller angles are congruent, then the remaining angles (i.e., the results of the subtractions) are congruent.” Or, you could write something like this: “Let ÐWXZ @ ÐABD. Assume that there exist two congruent angles, ÐWXY and ÐABC, such that point Y is in the interior of the region bounded by rays XW and XZ, and such that point C is in the interior of the region bounded by rays BA and BD. Then ÐYXZ @ ÐCBD.”
13.	For diagram, see #26 on p.108.
	Proof is as follows:
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	1. quad. EBDC, str. ÐABC	|    1. Given
	2. ext. pt. A forms ray CA, collinear w/ ray BC	|    2. Diagram based on givens
	3. ray BC bis. ÐDBE	|    3. Given
	4. ÐDBC @ ÐCBE	|    4. Def. bis.
	5. ÐABD supp. ÐDBC, ÐABE supp. ÐCBE	|    5. Def. supp. (or by diagram)
	6. ÐABD @ ÐABE	|    6. Supps. of @ Ðs are @
	Q.E.D.
14.	To be an equivalence relation, friendship would have to satisfy all three of the “R, S, T” properties: reflexive, symmetric, and transitive. Although “R” is arguably satisfied (one tries to be one’s own friend, after all), and although “S” is certainly plausible (if Fred is George’s friend, surely George is Fred’s friend), “T” causes a problem. If Al is friends with Brian and Brian is friends with Chuck, we cannot necessarily conclude that Al is friends with Chuck. Because transitivity fails, friendship is not an equivalence relation.
15.	x = 90, y = –20 \x – y = 90 – (–20) = 110
16.(a)	Let W = the event that a male drummer is well-paid Let R = the event that a male drummer has good rhythm
(b)	If a male drummer is well-paid, then he has good rhythm.
(c)	W Þ R
(d)	~W Þ ~R Alternate answer: If a male drummer is not well-paid, then he does not have good rhythm. This is probably not a true statement. Most of us know of drummers who are not well-paid but who do have good rhythm. Some of those drummers are amateurs or perform as buskers.
(e)	Zeke is not well-paid. Reason (version 1): Zeke is a clearly male, since the pronoun “his” was used. Therefore Zeke is a male drummer. The given statement is logically equivalent to its contrapositive, ~R Þ ~W. Since Zeke’s rhythm is poor, ~R is true for him, which implies ~W for poor Zeke. Q.E.D. Reason (version 2—proof by contradiction): Assume, bwoc, that Zeke is really well-paid. Then by the given implication W Þ R, we conclude that Zeke must have good rhythm. However, we know he does not, which is a contradiction (®¬). Therefore, the assumption that Zeke is well-paid must be false. Q.E.D.