Sunday, September 26, 2021

Geometry Problems of the Day (Geometry Regents, January 2013)



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, June 2013

Part I: Each correct answer will receive 2 credits.


11. Triangle ABC is shown in the diagram below. true?


If DE joins the midpoints of ADC and AEB, which statement is not true?

1) DE = 1/2 CB
2) DE || CB
3) AD/DC = DE/CB
4) Triangle ABC ~ AAED

Answer: 3) AD/DC = DE/CB


DE is a midsegment of triangle AB. That means that DE || CB and DE = 1/2 CB. Since the lines are parallel the corresponding angles along the transverals are congruent, which means that the triangles are similar. Eliminate Choices (1) and (2) and (4) because these are ALWAYS true by definition.

Choice (3) looks correct, but it's deceiving. What is true is that AD/AC = DE/CB, which is 1/2. The ratio AD/DC = 1/1 because D is the midpoint of AC.





12. The equations x2 + y2 = 25 and y = 5 are graphed on a set of axes.
What is the solution of this system?

1) (0,0)
2) (5,0)
3) (0,5)
4) (5,5)

Answer: 3) (0,5)


The second equation literally says that y = 5, so the y-coordinate MUST BE 5. Eliminate Choices (1) and (2).

(0)2 + (5)2 = 0 + 25 = 25. Choice (3) is the solution.

(5)2 + (5)2 = 25 + 25 = 50 =/= 25. Choice (4) is incorrect.





13. Square ABCD has vertices A(-2,-3), B(4,-1), C(2,5), and D(-4,3). What is the length of a side of the square?

1) 2 SQRT(5)
2) 2 SQRT(10)
3) 4 SQRT(5)
4) 10 SQRT(2)

Answer: 2) 2 SQRT(10)


Use the Distance Formula of Pythagorean Theorem. Pick any two consecutive points. I'll pick the two with the fewest minus signs.

SQRT( (4 - 2)2 + (-1 - 5)2) = SQRT( 22 + (-6)2)
= SQRT(40) = SQRT(4) * SQRT(10) = 2 SQRT(10)





14. The diagram below shows ABD, with ray ABC, BE ⊥ AD, and ∠EBD ≅ ∠CBD.
If m∠ABE = 52, what is m∠D?



1) 26
2) 38
3) 52
4) 64

Answer: 1) 26


Since ∠EBD ≅ ∠CBD and ∠ABE + ∠EBD + ∠CBD = 180, we can solve for ∠EBD.

52 + x + x = 180
52 + 2x = 180
2x = 128
x = 64

Since EBD = 64 and BED is a right angle (90), then we can solve for D:

64 + 90 + y = 180
154 + y = 180
y = 26





15. As shown in the diagram below, FD and CB intersect at point A and ET is perpendicular to both FD and CB at A.


Which statement is not true?

1) ET is perpendicular to plane BAD.
2) ET is perpendicular to plane FAB.
3) ET is perpendicular to plane CAD.
4) ET is perpendicular to plane BAT.

Answer: 4) ET is perpendicular to plane BAT.


ET is shown as a vertical line, so it will be perpendicular to the horizontal plane. That plane can be named using any three of the following points: A, B, C, D, F.

Choice (4) using the point T. Line AT is the same as line ET, so ET is contained in that plane BAT and cannot be perpendicular to it.




More to come. Comments and questions welcome.

More Regents problems.

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