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 2013
Part I: Each correct answer will receive 2 credits.
11. The bases of a right triangular prism are triangle ABC and triangle DEF. Angles A and D are right angles, AB = 6, AC = 8, and AD = 12. What is the length of edge BE?
1) 10
2) 12
3) 14
4) 16
Answer: 2) 12
They word it so you think that this is a right triangle problem, but it is not.
BE is the height of the prism. AD is also the height of the prism, and AD = 12. Therefore BE =12.
12. IWhat is the equation of circle O shown in the diagram below?
1) (x + 4)2 + (y - 1)2 = 3
2) (x - 4)2 + (y + 1)2 = 3
3) (x + 4)2 + (y - 1)2 = 9
4) (x - 4)2 + (y + 1)2 = 9
Answer: 3) (x + 4)2 + (y - 1)2 = 9
Once again, the formula for 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. Note that there are MINUS SIGNS in the formula. (I'm at the point where I'm literally cutting and pasting this sentence because of the HTML involved in typing it.)
The radius is 3, and 32 = 9. Eliminate Choices (1) and (2).
The center of the circle is at (-4, 1), so you want (x + 4) and (y - 1).
13. 3 The diagram below shows the construction of line m, parallel to
line ℓ, through point P.
Which theorem was used to justify this construction?
1) If two lines are cut by a transversal and the alternate interior
angles are congruent, the lines are parallel.
2) If two lines are cut by a transversal and the interior angles on
the same side are supplementary, the lines are parallel.
3) If two lines are perpendicular to the same line, they are parallel
4) If two lines are cut by a transversal and the corresponding angles
are congruent, they are parallel.
Answer: 4) If two lines are cut by a transversal and the corresponding angles are congruent, they are parallel.
The image shows a line parallel to the given line being constructed so that it passes through point P. The image uses corresponding angles.
Choice (1) mentions alternate interior angles, but that isn't shown in the image.
Choice (2) mentions interior angles, but the top angle isn't interior, and the angles aren't supplementary. Construction requires using congruency.
Choice (3) mentions perpendicular lines. Nothing states that the lines are perpendicular.
14. The lateral area of a right circular cone is equal to 120π cm2. If the
base of the cone has a diameter of 24 cm, what is the length of the slant height, in centimeters?
1) 2.5
2) 5
3) 10
4) 15.7
Answer: 3) 10
The slant height is the length from the outer edge of the base of the cone to the top. If you take a cross section of the cone, then the height, the slant height and the radius (not the diameter) form a right triangle.
The Lateral Area of a cone is A = π rl, where l is the slant height. (This formula can be found in the back of the booklet.)
120 π = π (12) l
l = 120 π / π (12) = 10
This is the expected answer from teh calculations. However, I have a problem with this question. Since the height, the slant height and the radius form a right triangle, then the slant height shouldn't be less than the radius. That would mean that such a cone as described in the question couldn't exist.
Logically, when looking at this question for the first time, I assumed the answer would be 15.7 because it was the only choice that was larger than the radius.
15. A student wrote the following equations:
2y - 3x = 6
The lines represented by these equations are
1) paralle
2) the same line
3) perpendicular
4) intersecting, but not perpendicular
Answer: 4) intersecting, but not perpendicular
Rewrite both statements in slope-intercept form.
3y + 6 = 2x
3y = 2x - 6
y = 2/3 x - 2
2y - 3x = 6
2y = 3x + 6
y = 3/2 x + 3
They don't have the same slope, so eliminate Choices (1) and (2).
They aren't perpendicular because the two slopes do not have a product of -1. (They aren't inverse reciprocals.)
Since they are not parallel, the lines will intersect, but not in a perpendicular way.
More to come. Comments and questions welcome.
More Regents problems.
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