Geometry / Mr. Hansen 12/1/2003

Name: _________________________

Solution Key to Practice Test
(Chapter 5 and PBT)

1.	S (would be A if the word congruent were replaced by supplementary)
2.	A (in fact, one diagonal is the ^ bisector of the other)
3.	A
4.	S
5.	S (true for a rhombus, but not for other types of kites)
6.	S (true for a rhombus, but not for other types of parallelograms)
7.	S (could also be an “ugly nothing”)
8.	S (could just be a plain kite with 1 or 2 right angles)
9.	A
10.	N (they have to be congruent, but if they were congruent and supp., they would be right angles, and then we would have a rectangle, not a trapezoid)
11.	S (we need a second point to determine the ^ bisector)
12.	A (this is a disguised form of PBT)
13.	S (the word “coplanar” is missing)
14.	A (true by converse of PBT, since segment CM is the ^ bisector of segment AB)
15.	[already done]
16.	Three noncollinear points determine a plane.
17.	Two coplanar lines are parallel iff they never intersect.
18.	The following is a list of all possible ways that a sphere and a plane can intersect: (1) as a circle, (2) null set (i.e., no intersection at all), xor (3) in a single point.
19.	All of the following are possibilities for a quadrilateral that has two congruent consecutive angles: kite, rectangle, non-isosceles trapezoid, “ugly nothing.” [Note: There is no point in saying “parallelogram,” since a parallelogram with two congruent consecutive angles would have to be a rectangle. There is no point in saying “rhombus,” since a rhombus with two congruent consecutive angles would have to be a square, hence also a rectangle. A kite-looking kite could have two congruent consecutive angles—actually three, since at least one pair of opposite angles must be congruent—but that’s all we can say. Finally, a trapezoid with two congruent consecutive angles could not be isosceles for reasons similar to those given in #10.]
20.	quadrilateral
21.	Let 5x = larger Ð. Let 3x = smaller Ð. Since consec. Ðs of a parallelogram are supp., 5x + 3x = 180 Þ 8x = 180 Þ x = 22.5 Þ larger Ð = 5x = 5(22.5) = 112.5 Answer: 112°30¢
22.(a)	quadrilateral (could be either a parallelogram or an isosceles trapezoid)
(b)	rectangle
(c)	parallelogram
(d)	quadrilateral (could be a square xor a non-isosceles trapezoid)
23.(a)	Presumably you can do this.
(b)	msAC = Dy/Dx = (0 – (–4))/(11 – 3) = 4/8 = 1/2 msBD = Dy/Dx = (–4 – 0)/(8 – 6) = –4/2 = –2
24.	Since the diagonals of a rhombus are ^, the slopes should be negative reciprocals of each other.
25.	Given: rhombus ABCD with AC = BD Prove: ABCD is a square _______________________________________________________________________________________
	1. ABCD is a rhombus	|    1. Given
	2. ABCD is a parallelogram	|    2. Every rhombus is a parallelogram
	3. AC = BD	|    3. Given
	4. ABCD is a rectangle	|    4. Prop.: parallelogram w/ @ diags. Þ rect.
	5. ABCD is a square	|    5. Prop.: rhombus & rectangle Þ square
	(Q.E.D.)
	The last step makes use of what Harrison called the “rhomtangle”: The only way a quadrilateral can be both a rhombus and a rectangle is for the quadrilateral to be a square.