Wednesday, June 03, 2015

New York Regents Geometry (Common Core) June 2015, Part 1

Yesterday was the first-ever Common Core Geometry Regents exam. I could not get a copy of it until this morning. What follows are the questions from Part 1. (If the images are omitted, come back soon -- they'll be added at some point.) Come back over the next few days for the rest of the exam.

Part 1 only had 24 questions, instead of 28, and they were a little more challenging that those found on the old exam. One thing that struck me was that there was only one Eqaution of a Circle question, not 3 or 4, but they made it tougher. I hope you paid attention in Common Core Algebra last year.

Part I -- Multiple Choice

1. What object is formed when right triangle RST is rotated around leg RS?

(4) A cone. Notice that it doesn't mention an image. This isn't a transformation. An object is being created, and it will be created in three dimensions. That object will be a cone as every point on the hypotenuse spins about in a circle.

2. The vertices of triangle JKL have coordinates J(5, 1), K(-2, -3) and L(-4, 1). Under which transformation is the image triangle J'K'L' not congruent to JKL?

(4) A dilation with a scale factor of 2 and centered on the origin. Dilations produce images that are different sizes from the original, and therefore are NOT congruent. (They are similar.)

3. The center of circle Q has coordinates (3, -2). If circle Q passes through R(7, 1), what is the length of its diameter?

(3) 10. QR is a radius and it has length 5. The difference of the x coordinates is 4; the difference of the y coordinates is 3. The square root of the sum of 4 squared plus 3 squared is 5. The diameter is twice the radius, which is 10.

4. In the image below, congruent figures 1, 2, and 3 are shown.

Which sequence of transformations maps figure 1 onto figure 2 and then figure 2 onto figure 3?

(4) A translation followed by a rotation. Neither move is a reflection, and the translation happens first.

5. As shown in the diagram below (omitted), the angle of elevation from a point on the groun to the top on the tree is 34 degrees.

If the point is 20 feet from the base of the tree, what is the height of the tree, to the nearest tenth of a foot?

(3) 13.5. Trigonometry. The height is opposite the angle. The ground is adjacent to the angle. Opposite over adjacent is Tangent. Tan (34) = x/20, so x = 20 * Tan (34). Put that in your calculator and you get 13.490..., which rounds to 13.5.

6. Which figure can have the same cross section as a sphere?

(2). A cone. The cross section of a sphere will always be a circle, no matter how you slice it. It is possible to slice a cone so that the cross-section is also a circle. The others are prisms and pyramids with polygonal (not circular) bases.

7. A shipping container is in the shape of a right rectangular prism with a length of 12 feet, a width of 8.5 feet, and a height of 4 feet. The container is completely filled with contents that weigh, on average, 0.25 pound per cubic foot. What is the wight, in pounds, of the contents in the container?

(3) 102. The weight of the contents is the weight of one cubic foot times the number of cubic feet. The number of cubic feet is the Volume, which is length times width times height. So Weight = 12 X 8.5 X 4 X 0.25, which is 102 pounds. If you noticed, 4 x 0.25 = 1, so you could have canceled those two numbers.

8. In the diagram of circle A shown below (omitted), chords CD and EF intersect at G, and chords CE and FD are drawn.

Which statement is NOT always true?

(1) CG is congruent to FG. There is no reason that those segments must be congruent. On the other hand, inscribed angles C and F intercept the same arc, so they are congruent. Moreover, inscribed angles E and D intercept the same arc, and angle CGE and DGF are vertical angles, so the two triangles are similar. The proportion in choice (3) is true because the triangles are similar.

9. Which equation represents a line that is perpendicular to the line represented by 2x - y = 7?

(1) y = -1/2 x + 6. The slope of the given line is 2, which you can find be re-writing the equation in slope-intercept form, or recognizing that in Standard form, Ax + By = C, the slope of the line is -A/B, which is -2/-1 = 2. The slope of the perpendicular line is the negative reciprocal, which is -1/2.

10. Which regular polygon has a minimum rotation of 45 degrees to carry the polygon onto itself?

(1) Octagon. To carry the polygon onto itself means that as you rotate the polygon, you will at some point get an image that is the same as the original. A rectangle would have to turn 180 degrees. An equilateral triangle would have to rotate 120 degrees. A square would rotate 90 degrees. Etc. Divide 360 / 45 = 8. A polygon with 8 sides will carry onto itself every 45 degrees. Note that 45 degrees is also the size of the exterior angles of an octagon, a figure which is found by dividing 360 / 8. The two concepts are related.

11. In the diagram of triangle ADC below, EB || DC, AE = 9, ED = 5, and AB = 9.2

What is the length of AC, to the nearest tenth?

(3) 14.3. Create a proportion 9/9.2 = 5/x, and cross-multiply. 9x = 46, and x = 5.111..., or 5.1. The length of AC is 9.2 + 5.1 = 14.3.

12. In scalene triangle AbC shown in the diagram below, m<C = 90.

Which equation is always true?

(4) sin A = cos B. This is a basic Trig identity. What is opposite A is adjacent to B. Choices (1) and (2) are only true if angles A and B are congruent (e.g., 45 degrees). Choice (3) is silly as you can't take the sine of the right angle.

13. Quadrilateral ABCD has diagonals AC and BD. Which information is NOT sufficient to prove ABCD is a parallelogram?

(4) AB = CD and BC || AD. This choice could also be an isosceles trapezoid. The other choices show properties of parallelograms.

14. The equation of a circle is x2 + y2 + 6y = 7. What are the coordinates of the center and the length of the radius of the circle?

(2) Center (0, -3) and radius 4. This is not the standard form for the equation of a circle. To fix it, you have to complete the square.

x2 + y2 + 6y + 9 = 7 + 9
x2 + (y + 3)2 = 16

The center is (0, -3) and the radius is 4.

15. Triangles ABC and DEF are drawn below

If AB = 9, BC = 15, DE = 6, EF = 10 and <B = <E, which statement is true?

(3) Triangle ABC ~ Triangle DEF. The two pairs of sides are proportional and the included angle is congruent, so by SAS the two triangles are similar. The other statements do NOT match corresponding parts, so they are NOT true.

16. If triangle ABC is dilated by a scale factor of 3, which statement is true of the image A'B'C'?

(2)B'C' = 3BC. The sides of the new triangle will be three times as long.

17. Steve drew line segments ABCD, EFG, BF, and CF as shown in the diagram below.

Triangle BFC is formed.
Which statement will allow Steve to prove ABCD || EFG?

(1) <CFG = <FCB. These two angles are Alternate Interior Angles. None of the other pairs are alternate interior angles (or corresponding for that matter).

18. In the diagram below, CD is the image of AB after a dilation of scale factor k with center E.

Which ratio is equal to the scale factor k of the dilation?

(1) EC/EA. A scale factor of 2 moves all the points to a distance twice as far from the center of dilation. So the ratio of the distances, in that case, would be 2. Likewise, EC/EA will give the scale factor k of the problem.

19. A gallon of paint will cover approximately 450 square feet. An artist wants to paint all the outside surfaces of a cube measuring 12 feet on each edge. What is the least number of gallons of paint he must buy to paint the cube?

(2) 2. The amount of paint needed is 6 X 12 X 12 = 864. Divide that by 450 and you get a number between 1 and 2, so you need 2 cans of paint to complete the task.

20. In circle O shown below, AC is perpendicular to CD at point C, and chords AB, BC, AE and CE are drawn.

Which statement is NOT always true?

(1) <ACB = <BCD. Choice (1) implies that chord CB bisects angle ACD (which is a right angle), but there is nothing stated or shown that proves this. As for the other choices: Angles ABC and ACD are both right angles because the first is an inscribed angle that intercepts a semicircle and the latter is a formed by a diameter and a tangent. Angles BAC and DCB intercept the same arc, and both are half the size of that arc. Angles CBA and AEC are right angles, which intercept a semicircle.

21. In the diagram below, triangle ABC ~ DEC.

If AC = 12, DC = 7, DE = 5 and the perimeter of triangle ABC is 30, what is the perimeter of triangle DEC?

(4) 17.5. In the smaller triangle, DC corresponds to AC in the larger triangle. That gives a ratio of 7/12. The smaller triangle's perimeter is unknown, but the larger triangle has a perimeter of 30, so the ratio is x/30. That gives the proportional 7/12 = x/30.
Cross-multiply and you will find that x = 17.5.

22. The line 3y = -2x + 8 is transformed by a dilation centered at the origin. Which linear equation could be its image?

(1) 2x + 3y = 5. A dilation will not change an image's orientation, which means that the slope of the line is preserved. You had to find the slope of the line, but you didn't need to convert anything into Slope-Intercept form. All four choices are in Standard form, so rewrite the given equation in Standard form by adding 2x to both sides of the equation. That gives you 2x + 3y = 8. Only one choice has 2x + 3y in it, and that is Choice (1).

23. A circle with a radius of 5 was divided into 24 congruent sectors. The sectors were then rearranged, as shown in the diagram below.

To the nearest integer, the value of x is

(2) 16. The sectors are arranged into what looks almost like a parallelogram. As you divide the circle into more and more sectors, the base of the parallelogram gets closer and closer to one-half the circumference (with the other base being the other half of the circumference). The radius is 5. Circumference is 2 X pi X r, and half of that is just pi X r, or (3.14)(5) = 15.70, or roughly 16.

24. Which statement is sufficient evidence that triangle DEF is congruent to triangle ABC?

(3) There is a sequence of rigid motions that maps AB onto DE, BC onto EF, and AC onto DF. Side-side-side. Choice (1) only gives two sides. Choice (2) proves that they are similar, but not that they are congruent. Choice (4) does not have enough information.

Part 2 when I can get to it. Hopefully, soon.


Anonymous said...

Honestly, this exam wasn't as difficult as I was going to expect. Especially since the new Common Core curriculum for geometry has new trig lessons such as Law of Sines/Cosines. Aside from the trickiness and wording for the exam, it was definitely more than doable.

carlito said...

Totally fooled me. Didn't even considered reviewing trig with the students. Is the CC geometry moving away from geometry? What made this exam seem difficult, at least to my English Language Learners, was some of the wording.

(x, why?) said...

It was definitely a gettable test, but if you came at it after studying old Geometry Regents, you would not have been prepared well.

Wording was an issue. For instance, two teachers at my school read the first question as a transformation before realizing what was being asked, so it's likely that many of the students made the same mistake.

Rose C said...

One more question --- does anyone have any information (sample questions, major changes, good texts to use, etc) on the coming 11th year math course Common Core that will be rolled out next year? If so, please let us know - via this website or anywhere else that is accessible to all teachers and students?

Anonymous said...

Does anyone know where to find the last question with the water tower? I think that was a hard question and I aready know I got it wrong, but I want to see where I messed up.

Anonymous said...

I love what you have already done for the non Common Core exam. Will you be posting the entire common core exam with explanations and figures as well?

Anonymous said...

Ps. And images for part 1 C C geometry too??

(x, why?) said...

I'll add images when the tests become available online. On the non Common Core, I scanned in the pics from my sample test because it was just too much to type up quickly.

Anonymous said...

I think it is totally pathetic that the common core Geometry regents (June 2016) was so difficult that experienced math teachers are unable to get a 100 on it if they took the exam themselves. This common core crap is killing morale among students. Students who are scoring 90 and above on Regents in every other subject are struggling to get into the 70s on these common core exams. What a crock of shit.

(x, why?) said...

I won't disagree with you because I don't personally know anyone who is getting over 90 in every other Regents, except math. I don't have the advanced class at my current location.

The other one who comes to mind is a friend's daughter for whom math is her subject of interest and she's not settling for less than 100.