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, January 2014
Part I: Each correct answer will receive 2 credits.
Which conclusion can be proven?
11. In the diagram below of quadrilateral ABCD, E and F are points on AB and CD, respectively, BE ≅ DF, and AE ≅ CF.
1) ED ≅ FBF
2) AB ≅ CD
3) ∠A ≅ ∠C
4) ∠AED ≅ ∠CFB
Answer: 2) AB ≅ CD
The Addition Postulate.
If BE = DF and AE = CF that AE + EB = CF + FD, or AB = CD.
Choice (1) would be true in a parallelogram, but there is nothing to suggest that BEDF is a parallelogram. In fact, it doesn't even look like one.
Choice (3) could only be proven if ABCD were a parallelogram or triangles ADE and CBF could be shown to be congruent. Neither of those is true, or can be shown to be true.
Choice (4) would require the triangles to be congruent, which they cannot be shown to be.
12. In the diagram below, four pairs of triangles are shown.
Congruent corresponding parts are labeled in each pair
1) A
2) B
3) C
4) D
Answer: 1) A
Each set of triangles shows a postulate or theorem of congruency, except for one.
Choice (1) Figure A shows SSA, two pairs of congruents sides and an angle that is NOT between the two sides. This is NOT a test for congruent triangles. The only exception is the Hypotenuse-Leg Theorem for right triangles (see below), but this is not labeled as a right triangle. Choice (1) is the correct answer.
Choice (2) Figure B shows the ASA Postulate for congruent triangles.
Choice (3) Figure C shows the AAS Theorem for congruent triangles.
Choice (4) Figure D shows the HL Theorem for congruent right triangles.
Each of those lines will cross the circle twice. So there would be four points that fit the criteria, one in each quadrant. You can make a little sketch to see it.
13. In ABC shown below, L is the midpoint of BC, M is the midpoint of AB, and N is the midpoint of AC.
If MN = 8, ML = 5, and NL = 6, the perimeter of trapezoid BMNC is
1) 35
2) 31
3) 28
4) 26
Answer: 1) 35
MN is congruent to BL and LC. ML is congruent to AN and NC. NL is congruent to BM and MA. Label all of these with the correct lengths.
Now add up the sides that make up the perimeter of BMNC: 6 + 8 + 5 + 8 + 8 = 35.
14. In the diagram below, and ABC are shown with m∠A = 60
and m∠ABT = 125.
What is m∠ACR?
1) 125
2) 115
3) 65
4) 55
Answer: 2) 115
A triangle has 180 degrees. Supplementary angles add up to 180 degrees. That's all you need to know, but the Remote Angle Theorem will save you a step.
∠ABC is supplementary to ∠ABT, which is 125 degrees. So m∠ABC = 180 - 125 = 55.
∠ACR is an exterior angle to the triangle and equal to the sum of the two remote angles, which are 60 and 55, which is 115 degrees.
You could have done these steps in the reverse order and found ∠ACB using the Remote Angle Theorem subtracting 125 - 60 = 65, and then found the supplementary angle, 180 - 65 = 115. Works both ways.
Likewise, if you forgot about the remote angle theorem, you could have just found all of the angles along the way.
15. Which equation represents circle O shown in the graph below?
1) x2 + (y - 2)2 = 10
2) x2 + (y + 2)2 = 10
3) x2 + (y - 2)2 = 25
4) x2 + (y + 2)2 = 25
Answer: 4) x2 + (y + 2)2 = 25
They do like asking questions about the equation of a circle. A lot of them.
The equation of a circle is (x - h)2 + (y - k)2 = r2, where (h, k) is the center of the circle, and r is the radius.
The radius is 5 (measure the vertical or horizontal distance from point O to the circle). If you square 5, you get 25, not 10. Eliminate Choices (1) and (2).
The center of the circle is (0, -2), so h = 0, and k = -2. So the formula would be
or more simply
More to come. Comments and questions welcome.
More Regents problems.
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