Now that I'm caught up with the current New York State Regents exams, I'm revisiting some older ones.
More Regents problems.
Geometry Regents, August 2012
Part I: Each correct answer will receive 2 credits.
6.In the diagram of tirangle JEA below, m∠JEA = 90 and m∠EAJ = 48.
Line segment MS connects points M and S on the triangle, such that m∠EMS = 59.
What is m∠JSM?
1) 163
2) 121
3) 42
4) 17
Answer: 4) 17
There are two good approaches to this problem. Either find the size of angle J, which is complementary to A, which is 48 degrees. Then find angle JMS, which is supplementary to 59 degrees. You can then find JSM since a triangle has 180 degrees. (Slightly shorter: 59 = Angle J + Angle JSM.)
Alteratively, you can find angle MSA knowing that a quadrilateral has 360 degrees and then find JSM, which is supplemenatry to MSA.
90 + 59 + 48 = 197; 360 - 197 = 163; 180 - 163 = 17.
90 - 48 = 42; 180 - 59 = 129; 180 - 129 - 42 = 17
If you tried to "eyeball" it, you can see that it has to be small, smaller than 48 degrees, and 17 makes the most sense.
7. In triangle AED with shown in the diagram below, EB and EC are drawn.
If AB ≅ CD, which statement could always be proven?
1) AC ≅ DB
2) AE ≅ ED
3) AB ≅ BC
4) EC ≅ EA
Answer: 1) AC ≅ DB
The Addition Property of Congruency. If two line segments are congruent and you add a congruent segment to the first two congruent segements, the longer segments will also be congruent.
We don't know anything about AE and ED, about AB and BC, or about EC and EA. They may appear to be congruent in this example but there is nothing to prove that they are in fact congruent. (And appearances can be deceiving.)
8. Given that ABCD is a parallelogram, a student wrote the proof below to show that a pair of its opposite angles are congruent.
What is the reason justifying that ∠B ≅ ∠D?
1) Opposite angles in a quadrilateral are congruent.
2) Parallel lines have congruent corresponding angles.
3) Corresponding parts of congruent triangles are congruent.
4) Alternate interior angles in congruent triangles are congruent.
Answer: 3) Corresponding parts of congruent triangles are congruent.
A very common final reason in proofs involving parts of triangles being congruent.
Choice (1) is incorrect. The reason only applies to parallelograms, not to quadrilaterals. Additionally, this is the statement that is being proven. It would be circular logic to state it as a Reason.
Choice (2) is gibberish. There's no mention of a transversal to create the corresponding angles, and the angles listed are NOT corresponding angles.
Choice (4) is also gibberish. There are not "alternate interior angles in congruent triangles". That is a meaningless phrase, even if it sounds legitimate.
9. The equation of a circle with its center at (-3,5) and a radius of 4 is
1) (x + 3)2 + (y - 5)2 = 4
2) (x - 3)2 + (y + 5)2 = 4
3) (x + 3)2 + (y - 5)2 = 16
4) (x - 3)2 + (y + 5)2 = 16
Answer: 3) (x + 3)2 + (y - 5)2 = 16
The equation of a circle is given by the formula (x - h)2 + (y - k)2 = r2, where (h,k) in the center of the circle and r is the radius. Note that there are MINUS signs in the formula, so the signs will be flipped.
The radius is 4, and 4 squared is 16. Eliminate Choices (1) and (2).
Since the signs are flipped the correct answer is Choice (3).
10.In the diagram below of triangle DAE and triangle BCE, AB and CD intersect at E, such that AE ≅ CE and ∠BCE ≅ ∠DAE
Triangle DAE can be proved congruent to triangle BCE by
1) ASA
2) SAS
3) SSS
4) HL
Answer: 1) ASA
You are given a pair of congruent angles and a pair of congruent sides. There is also a pair of vertical angles and vertical angles are always congruent. Each congruent side is included within the two angles, so it is ASA.
It cannot be SAS or SSS because we only have one pair of sides.
It cannot be HL because there is nothing to suggest that those are right triangles, and again, we only know one side, and HL requires two.
More to come. Comments and questions welcome.
More Regents problems.
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