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The extended-time tests are finished, so I assume that it's safe to talk about the
New York State Geometry Regents exam given this morning.
By my count, in the multiple-choice, 15 problems were definitions, theorems, postulates, formulas, etc, with no calculations involved. That is, you either knew it, or you didn't, but there wasn't anything to work out. The other 13 required some kind of calculation or working through of steps. (Your opinion of what counts as working through steps may vary.)
For the test as a whole, I notice four questions regarding the equation of a circle. If you didn't know
(x - h)2 + (y - k)2 = r2
then there were a bunch of points lost. Yes, that means if you forget that subtracting a negative number means adding a positive number, then you had problems. (Don't worry, I won't make you write the
contrapostitive of that statement.)
Additionally, there was a question involving the vertex form of a parabola, just to be different.
There were two transformation problems: one multiple-choice, one open-ended. The open-ended was also a composition. You did do it right to left, right? You did the Translation of the Dilation, and not a Translation followed by a Dilation. You'll probably get a point for doing the wrong thing, if it's done correctly.
Two locus questions.
Two constuctions: one multiple-choice, which hopefully provided some assistance to the open-ended question. For an added twist, you needed to make the equilateral triangle with sides congruent to the diagonal of a given triangle. That just tells you how wide to make the compass. I really wonder how many scorers are going to measure those triangles.
Two questions involving right triangles and one more that used the distance formula.
And there was one question using quadratic equations. (Two, if you didn't know the vertex form of a parabola and had to work it out.)
Personally, I had two problems when I took the test. (Yes, I take the tests along with the students. Sometimes I can't assess the difficulty of a question until I actually do it. It may or may not look at difficult as it is.)
My biggest problem was question 24, the similarity problem:
24. In trangles ABC and DEF, AB = 4, AC = 5, DE = 8, DF = 10 and [angle A is congruent to angle D].
Which method could be used to prove [triangle]ABC ~ [triangle]DEF?
The choices are AA, SAS, SSS, and ASA.
This is a STUPID question.
The only postulate or theorem which gets used to prove similarity is AA, because two triangles are similar if their angles are congruent. But we only know about one pair of angles, and we don't even know their size. There's no way to find information on any other.
Now, you are given information about the pairs of sides which include the congruent pair of angles. That much is true. But those sides are NOT congruent. SAS is used for congruence. If two triangles are congruent, then, of course, they are similar. But these aren't congruent. The sides are proportional, as corresponding sides of similar sides should be, but that in itself isn't proof. (For one thing, we don't know about the third pair of sides.)
So I have a problem with this question. They were looking for Number 2. The only answer which makes sense is Number 1, and that one isn't useful, either.
You may argue and disagree all you like, but SAS for similarity is NOT in the curriculum.
My second problem was question 36. They laid a trap and I fell into it. Three times. I wouldn't have gotten the right answer, despite the fact that I knew I had the wrong answer.
How did I know? Test-taking Strategies and Number Sense. Basically, the part that said "Determine the length of OA" gave me trouble because I got an irrational answer. However, the answer COULD NOT be irrational because they neither stated "round to the nearest ####" or "give the answer in simplest radical form".
What was the confusion? The setup was complicated enough, and I just through all the hoops fine, except for the last one. It gave CF = y + 10 and CD = 4y - 20. I read the latter part as FD. So on top of all the radii you had to pencil in, and using the Pythagorean Theorem, and knowing that when a radius is perpendicular to a chord it bisects the chord (into two congruent pieces), you had to know enough to double y + 10 into 2y + 20 before setting it equal to 4y - 20.
So I had y = 10, instead of y = 20, which was correct, but not the answer. You then had to substitute to get the length of DF. You know, the segment that I thought was 4y - 20, but was really half as big. So DF = 30.
Now the last answer was obvious to me (once corrected) because 16 and 30 are part of a Pythagorean Triple. That being said, just because the answer is obvious doesn't mean that you don't show the work!!! OA is 34. (I don't have to show it. Exercise left to the reader and all that.)
So how did you do?
Any questions?
