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 2012
Part I: Each correct answer will receive 2 credits.
6. In triangle FGH, m∠F = 42 and an exterior angle at vertex H has a
measure of 104. What is m∠G?
1) 34
2) 62
3) 76
4) 146
Answer: 2) 62
The Remote Angle Theorem tells us that the sum of angles F and G will equal the measure of the exterior angle at H. So 104 - 42 = 62.
If you didn't know this, you could find the interior angle of H, which is supplementary to the exterior: 180 - 104 = 76. Then find G using 42 + 76 + G = 180, which gives you G = 62.
Note that 76 is one of the incorrect choices. That isn't a coincidence.
As for the other incorrect choices: 146 + 42 = 146 (Choice 4), and 180 - 146 = 34, Choice (1). There is no reason to do either of those things.
18x - 12 = 6x + 42
12x = 54
x = 54/12 = 4.5
You also could have plugged in each choice into the two expressions to find which value made them equal.
7. Which diagram represents a correct construction of equilateral triangle ABC, given side
AB?
Answer: 1) [See Image]
The make it easy in that only one image even looks like an equilateral triangle!
To construct an equilateral triangle from a side, you measure the side and create an arc of that length from each of the endpoints. Then draw the two sides from the endpoint to the intersection of the two arcs.
Choice (2) shows an isosceles triangle, but it's missing the extra mark to show that it's equilateral.
Choice (3) is a triangle, but there were no construction methods used in its making. Just two random arcs drawn.
Choice (4) appears to be a right trianle, but it is just a triangle of unknown size. It is not constructed to be a right triangle.
8. In the diagram below, triangle ABC is circumscribed about circle O and
the sides of triangle ABC are tangent to the circle at points D, E, and F.
If AB = 20, AE = 12, and CF = 15, what is the length of AC?
1) 8
2) 15
3) 23
4) 27
Answer: 4) 27
Since the sides of the triangle are tangent to the circle, then AD = AE, BE = BF, and CD = CF.
AC = AD + CD = 12 + 15 = 27. The information about AB = 20 is not important.
9. In triangle ABC and triangle DEF, AC/DF = CB/FE. Which additional information
would prove triangle ABC ~ triangle DEF?
1) AC = DF
2) CB = FE
3) ∠ACB ≅ ∠DFE
4) ∠BAC ≅ ∠EDF
Answer: 3) ∠ACB ≅ ∠DFE
Two triangles can be shown to be similar if two pairs of corresponding angles are congruent. They can also be shown to be similar if all three pairs of corresponidng sides are proportional. Finally, it can be shown using SAS if two pairs of sides are proportional and the included angle is congruent. We need to use the last case for this problem.
The included angle between AC and CB is angle ACB. The included angle between DF and FE is angle DFE. So Choice (3) is the correct answer.
Choices (1) and (2) would be useful when trying to show that the traingles were congruent, not just similar. However, by themselves neither choice is enough to do that.
Choice (4) is not the included angle, so you would have SSA, which is not a postulate or theorem.
10. The angles of triangle ABC are in the ratio of 8:3:4. What is the
measure of the smallest angle?
1) 12°
2) 24°
3) 36°
4) 72°
Answer: 3) 36°
Place an x after each number in the ratio. Then sum them to 180 and solve for x.
15x = 180
x = 12
The smallest angle is NOT 12. It's 3x, which is 3(12) = 36.
The triangle would be 96 + 36 + 48 = 180.
My assumption for Choices (2) and (4) is if you used 360 instead of 180, then x would be 24 and the smallest angle would be 72.
More to come. Comments and questions welcome.
More Regents problems.
I also write Fiction!You can now preorder Devilish And Divine, edited by John L. French and Danielle Ackley-McPhail, which contains (among many, many others) three stories by me, Christopher J. Burke about those above us and from down below. Preorder the softcover or ebook at Amazon. Also, check out In A Flash 2020, by Christopher J. Burke for 20 great flash fiction stories, perfectly sized for your train rides. Available in softcover or ebook at Amazon. If you enjoy it, please consider leaving a rating or review on Amazon or on Good Reads. |
 |
 |
No comments:
Post a Comment