Geometry / Mr. Hansen
1/8/2002

Name: ________________________

2. 

S

4.

A

6.

N

8.

S

10.

N

12.

S

14.

S

16.

S

18.

N

20.

A

22.

A

24.

C

26.

153 [work required]

28.

C

30.

–3/2 [work required]

32.

270 [work required; be sure to use a pair of "Let" statements]

34.

E

36.

B

38.

20/3 [must show work, using Pythagorean Theorem]

40.

D [heptagon]

42.

54 [work required]

44.

perimeter = 10 + 7Ö2 [work and Pythagorean Theorem required]

Easiest way to prove figure is a trapezoid is to superimpose a coordinate grid with origin at the point that is 4 units below Q and 3 units to the left of P. Then compute slopes of segments PS and QR (both 1, hence ||) and the slopes of segments RS and QP (–3/4 and –4/3, respectively, hence not ||). Since a trapezoid by definition is a quadrilateral with two sides || and the other two not ||, we are finished.

The figure is, in fact, an isosceles trapezoid since the legs are both 5, but the problem did not ask for that. The diagonals (RP and QS) are both 7 by inspection.

46.

E [since exterior angle measures of any regular polygon will add up to 360]

48.

E [next year you will be able to prove that mÐPQR » 98.3, but since you can see by even a sloppy sketch that the angle is quite large and probably obtuse, E is the only choice that is even remotely reasonable]

50.

33/2 [work required]

52.

B

54.

Paragraph proof: Let y = OP, so that PA = 16 – y. Since AB = 20 [by Pythagorean Thm.], apply ABT to get y/12 = (16 – y)/20. Solve this proportion for y, and the conclusion is immediate.

56.

Paragraph proof: If we let x = mÐPBO, then mÐABO = 2x, mÐBPO = 90 – x, and mÐBAO = 90 – 2x. Mark all of these in your diagram. Since AP = 10 [from problem #54] and AM = 10 [by def. of midpt. and our knowledge that AB = 20], DAMP must be isosceles. Thus base Ðs are @. Use that fact and a little bit of algebra [left for you to do] to show that mÐAPM = 45 + x. Now we have three angles (namely Ð APM, Ð BPM, and Ð BPO) that combine to form a straight angle. A couple more lines of algebra [again, left for you to do as an exercise] lead to the conclusion.

There are other ways to do this problem involving auxiliary lines and complicated algebra. However, the way outlined above is probably the best.

I hear you asking, "Will anything this hard be on the actual midterm?" The answer is probably not, but you should be able to understand this proof and should supply the missing portions above. Any single part of this proof would be fair game on the midterm. This problem is only hard because it involves combining several non-obvious steps, each of which requires work.

58.

B

60.

A, C [both answers must be given]

62.

E [trick question]

D is the closest of the wrong answers, but the word "committed" should be changed to "convicted." We will try to avoid trick questions on the real midterm exam. In the event that a trick question arises inadvertently, we will attempt to be lenient in the grading and will take your demonstrated knowledge into account. That is another reason why you should always show your work, even on problems that seem not to require any!

64.

rhombus

66.

quadrilateral

68.

square

70.

860,000 miles [by similar Ds; diagram and work required]