Thursday, January 30, 2014

January 2014 Geometry Regents Discussion Thread, Part 3: Multiple Choice

Previous threads: Part 1, Part 2.

I'm not going to type up all the questions, but I'll identify the topic addressed in the problems.

1. You are given a midpoint and an endpoint of a line segment and have to find the other endpoint. You don't really need to use the midpoint formula and solve algebraically. Just notice that from A to M the x-co-ordinate drops 2. Drop it two more and B has an x-co-ordinate of 2. There's only one choice that works, so it's the answer, and you don't have to find y, unless you want to check.

2. Which construction is used to make a 45o angle? Only two choices have 45o angles in them (or something that looks like one). Choice (3) shows a straight angle bisected and then the right angle bisected.

3. First mention of the equation of a circle: (x - h)2 + (y - k)2 = r2
If there's a "+" in the equation before h or k, it means that the co-ordinate is negative. Take the square root of the number after the equal sign.

4. If two planes are perpendicular to the same line then they are parallel to each other.

5. A triangle is transformed and the image is congruent to the original. This is true from reflections, translation and rotations but not dilations, which will expand or shrink the size of the triangle.

6. For two edges of a cube not to be coplanar, they have to point in different directions; i.e., skew lines. Remember: two edges don't have to be on the same surface of the prism for them to be coplanar.

7. The points that are 3 units away from the origin and equidistant from the x-axis and y-axis would be on the lines y = x and y = -x where they intersect the circle centered on the origin with a radius of three. Therefore, there are four points which fit both conditions.

8. The medians of a triangle meet at the centroid. The longer portion, from the vertex to the centroid, is twice the length of the other portion, from the centroid to the opposite side of the triangle. That makes the shorter piece one-third the length of the median, and the longer piece two-thirds. If the median is 12, the segments are 8 and 4.

9. Solve y = x and y = x2 - 2.
You could do substitution and factoring. Or you could just try (1, 1) and (2, 2). (1, 1) doesn't work. (2, 2) does. That leaves choice (2) (2,2) and (-1, -1). A quick double-check on (-1, -1) shows that it's the answer.

10. Parallel lines have the same slope, so first you need to know the slope of 5x + y = 6. Subtract 5x from both sides to get slope-intercept form and we see that the slope is -5. That eliminates two of the choices. Because it have to go through (5, 3), plug is 5 for x in each and see which equation gives you 3 for y.

11. It's a demonstration of the Addition Property of Equality. If you add two pairs of congruent segments, the larger segments will be congruent as well.

12. You can't use SSA to prove congruency, except in the special case of Hypotenuse-Leg.

13. The midsegments are half the length of the side they don't touch. The midsegments bisect the sides of the triangle. So there are a whole bunch of congruent segments in that diagram. The perimeter of the trapezoid is 6 + 8 + 5 + 8 + 8 = 35.

14. The exterior angle of a triangle is equal to the sum of the two remote angles. ACR = 125 - 60 = 65.

15. Again! The equation of a circle. The center is (0, -2), so h = 0, and gets omitted, and k = -2, which becomes +2 in the formula. Radius = 5, so 52 = 25.

16. The largest angle of a triangle is across from the longest side. In ABC, if angle B is the largest, AC is the longest side. Choice 1.

17. The more sides that a regular polygon has, the smaller the exterior angles will be because the sum of the exterior angles is always 360.
The fewest sides a polygon can have is 3, a triangle. Divide 360 by 3 and you get 120, which is the largest for regular polygons.
Note: 360 is an impossible answer. You should have crossed it out immediately.

18. They are looking for the lateral area of the cylinder. If it were a soup can, you'd want the area of the label, but not the top and bottom of the can.
The formula is 2*pi*r*h, which is 2*(3.141592...)*9*42, which is 2375.
I don't know why they included 2374 as an answer. If you used 3.1416 for pi, you get 2375.05. If you only used 3.14, you get a smaller answer, which isn't a choice.

19. The shape is a quadrilateral, which has 360 degrees. The angles formed by tangents and radii are right angles (90 degrees each) and you are given 38. Add 90 + 90 + 38 to get 218. Subtract 360-218 to get 142.

20. It a square has a diagonal of 3*(2)^.5, then each side of the square is 3. The perimeter is 12.

21. If you Rotate 90 degrees about the origin and they don't give you a direction then the direction is counter-clockwise!. You move from Quadrant I to Quadrant II, where x is negative and y is positive. The correct answer is (-1, 7).

22. Angles 1 and 2 are supplementary. (They are NOT congruent.) Solve 7x - 36 + 5x + 12 = 180.

12x - 24 = 180, so 12x = 204 and x = 17.

At least this one changes it up a little by making you find the radius, which is 2, so r2 = 4.

24. When a diameter bisects a chord, it is a perpendicular bisector. That means a right angle. That means you can draw an extra radius and make a right triangle and then use the Pythagorean Theorem to find whatever side isn't given.
FE = 17 - 2 = 15. Draw FC, FC = 17. To be "Pythy", CE = 8.

25. Rhombuses (rhombii?) do not have congruent diagonals, unless they are squares. Their diagonals bisect each other.

The center of the circle is (-6, 7), which is in Quadrant II. The radius is 8. That means the circle reaches across the y-axis into Quadrant I and down far enough to cross the x-axis into Quadrant III. Process of elimination, it doesn't go into IV. If you don't believe me, figure that IV is across the origin, and that the origin is (36 + 49)^(.5) = 9.22 away from the center. That's more than 8.

27. No invariant points. That means ALL the points will vary, which means they'll all move.
In other words, no image will have the same co-ordinates (be in the same spot) as the origin.
The three linear reflections all leave a point in the same spot. The point reflection (about the origin) does not.

28. There are four lines which are tangent to both circles at the same time. One touching the "tops", one on the "bottoms" and two more forming an "X" in the middle.

So how did you do?
And what did I get wrong.
Ha! Trick question! I didn't get anything wrong! I just might have made a few inadvertent typos. Yeah! That's the ticket!

January 2014 Algebra Regents Discussion Pre-Thread

This is the "pre-thread" discussion of the January 2014 New York State Regents exam.

What does that mean?

It means that I spent the day off-site grading Geometry papers, so I didn't see a copy of the exam. I have no idea what was on it (yet!). HOWEVER, if you took the exam and you would like to discuss any particular question, if you can give me the information that was on the test, I'll try to help the best that I can.

Ready? Go.

Wednesday, January 29, 2014

January 2014 Geometry Regents Discussion Thread, Part 2: The Open-Ended Questions

Link back to Part 1: Intro

It was easier to start a second thread than to edit the previous one. Actually, it wasn't -- I just decided to do this.

Unless you scored better than 75% of the multiple-choice questions correct, you need some points from the open-ended sections -- Parts II, III and IV -- to put you over the top. (If you did manage 22 multiple-choice questions correct, hey, congrats! Good for you!) I'm starting here because I'll be grading oodles of these tomorrow morning.

29. The diameter of a sphere is 5 inches. Determine and state the surface area of the sphere to the nearest hundredth of a square inch.

This is extremely easy, considering that they give you the formula on the reference table! Remember to use 2.5 as the radius, not 5. Use your calculator for pi. DO NOT use 3.14. It won't give you enough accuracy.

SA = (4)(3.141592)(2.5)^2 = 78.5398... = 78.54 sq in.

30. Using a compass and straightedge, construct the perpendicular bisector of AB.

Hint: Look at question 2. Any of those look familiar? Maybe see a good starting point?
Sorry, I'm not making the diagrams.

31. The endpoints of AB are A(3, -4) and B(7, 2). Determine and state the length of AB in simplest radical form.

Digression: (Or should I say "Tangent"?)
Everyone in my class got a question just like this one correct in practice! And everyone showed All The Steps!
And I'm sure that you didn't all copy from each other, with the first person copying the answer from the Internet!

(I'll use the abbreviation "sqrt" for the square root function.) The square root of 4-squared plus 6-squared = sqrt(16+36) = sqrt(52) = sqrt(4*13) = 2*sqrt(13)

32. A right prism has a square base with an area of 12 square meters. the volume of the prism is 84 cubic meters. Determine and state the height of the prism, in meters.

Did you look up the formula in the back of the book? Why? Okay, if you did, did you find it? If you said, "Yes", you probably got this WRONG. It's a basic formula, so it won't be there.

The volume of a prism (84 m3) is the Area of the Base (12m2) times the height. Divide 84/12 and get 7 meters.

34. Without repeating the equations, you can see by looking at them (no manipulation required) that one has a slope of positive one-half and the other has a slope of negative one-half. They are NOT inverse reciprocals -- i.e., the slopes don't have a product of -1 -- and they (obviously) are not the same, so, therefore, they are neither parallel nor perpendicular.

"YES!", you have to give an explanation. Your answer must be backed up with a reason or definition and that reason has to have some kind of work or evidence that your reason is true. Even for something this obvious.

I will be shocked -- SHOCKED -- to find that "neither" with nothing supporting it would be worth a point. Don't count on it.

34. Another locus question. (There was one in the multiple-choice questions.) Two meters from the row of corn is two vertical lines, one on either side of line c. Five meters from the scarecrow is a circle with a radius of 5, centered on point T. Important: Six point T is 6 meters from line c, then T is only 4 meters from the locus line on the left side of c. The circle will intersect that line in two places, but it will NOT cross c. (And the other vertical line is right out.) There should be two X's on the paper.

Part III: Four credit questions

35. I can't copy the picture. (It isn't online yet.) So let me describe it.
Triangle ABC is isosceles with AB = BC. CA is extended through A to point D, and DB is drawn.
The measure of: <D = x, <DAB = 5x - 30, <DBA = 3x - 60, AB = 6y - 8, BC = 4y - 2.
Only algebraic answers will receive full credit.

Find m<D: Find x where x + 5x - 30 + 3x - 60 = 180
9x - 90 = 180,
9x = 30
x = 30, m<D = 30 degrees.

Find m<BAC, which is supplementary to <DAB, which is 5(30) - 30 = 120 degrees.
So BAC = 180 - 120 = 60 degrees.

Find the length of BC: Solve 6y - 8 = 4y - 2
2y = 6
y = 3, so BC = 4(3) - 2 = 12 - 2 = 10

Find the length of DC: SAY WHAT? How are we supposed to figure that out? Is A the midpoint of DC? Do we know AC? Can we get AC?
No to all of that.
Use what you already know -- and if you made a mistake, GUESS.
I say that only because this problem makes sense only if BAD is a right triangle. Two of the angles are 30 and 60, so angle DBC is 90 degrees.
You can't use Pythagorean Theorem because we don't know the length of BD and can't figure it out. BUT... we do know that in a 30-60-90 right triangle, the hypotenuse is twice the size of the shorter leg, and BC is the shorter leg. And BC, we just found, has a length of 10. Therefore, DC is 20.

EDIT: On further review, the smaller triangle is equilateral and the triangle on the left is isosceles, so you can show that those four smaller line segments all have a length of 10, therefore DC is 20. I would still require some work to get this, but I've been informed that no work is necessary because you "can assume" that they did the work because how else would they know. Okay, it wasn't quite stated in that fashion, but it might as well have been.

36. When doing a Composition of Transformations (two transformations) you have to go RIGHT TO LEFT. If you do it the other way, that is a Conceptual Error and you will lose 2 points right off the top. You have to find the Translation OF THE Reflection, so the reflection gets done first.

Flip the triangle over the y-axis: the x-co-ordinates change signs, but the y-co-ordinates will remain the same.
Next Translate x co-ordinates +4 (to the right) and y-co-ordinates -5 (down) to get your final answer.

37. Right Triangle Altitude Theorem. AD/DC=DC/DB, and DC is 6. Additionally, AD:AB = 1:5.

Let AD be x. That makes AB = 5x, so DB = 4x.
So x/6 = 6/4x.
Cross-multiply: 4x2 = 36, x2 = 9, so x = 3.
AD = 3, DB = 4x = 4(3) = 12.
(Algebraic solution.)

37. A very strange proof. They gave you the Statements. You needed to supply (some of) the Reasons. You have to prove a Tangent-Secant rule: (RS)2 = RA * RT.

Basically, St is perpendicular to RS because tangent lines are perpendicular to the radius (and therefore the diameter) and the point of tangency. Angle RAS is a right angle because it is an inscribed angle for a semicircle, so it is half of 180 degrees, which is 90 degrees. RST is congruent to RAS because all right angles are congruent. Triangles RAS and RST are similar because of the AA Theorem. Because they are similar, their corresponding sides are proportional. Since they are proportional, the product of the means is equal to the product of the extremes, which gives us the equation which we were looking for.

Questions? Comments? Corrections? (Surely, I'll get those! And I'll call you "Surely" if I feel like it!) Debate?

Multiple-choice as I get to them in the next thread.

January 2014 Geometry Regents Discussion Thread, Part 1

Note: At the moment, this is most a placeholder for a later discussion.

The New York State Geometry Regents exam has ended in most of the state, excepting unusual situations, such as late rooms, time extensions and conflict rooms. Because of these special circumstances, I'll refrain from talking specifics for a couple more hours. At some point later this afternoon, I'll edit this thread and start posting information about the exam itself.

Don't let that get in the way of leaving any comments or impressions you have about the exam.

Overall, it was a "doable" test. (I hate that word. But "test" isn't a much better word.) There were three questions about the equation of a circle, like there always seems to be.

The final proof was a little ridiculous in the way it was presented. The statements were given, you just had to come up with the reasons. I have to admit, that I had to go back to the book to find the "proper" wording for the very last statement because I kept looking at it from an Algebra lens not a Geometry lens. I knew what it was supposed to say, but I just wasn't 100% sure on how to say it. But, that's me. Hopefully, that wasn't you. The one good thing about the fill in the blank approach is that students who would otherwise max out at 1 or 2 points had a shot at half-credit or more.

And, yes, I still look through the prism of getting more students to 65 than I do getting them "the best score possible" because the latter students don't need my help as much and the outcome generally isn't as critical. The "best bang for the buck" is getting as many kids to pass as possible.

So before I sign off: does anyone have any comments or questions to focus on? Should I start with open-ended or multiple-choice? Your call.

Tuesday, January 28, 2014

Internet's Back ... Now, about that Algebra II / Trigonometry Regents ...

My Internet connection appears to be working ... for now. I'm having phone line issues at home. We'll figure it out.

I was busy proctoring the early exam at work and then filling out miscellaneous paperwork, so I didn't get a chance to wish Good Luck (or "Good Skill" because you don't need "luck") to all the students taking today's Algebra II/Trigonometry Regents exam in New York.

I haven't seen the exam ... I wasn't around the side of the school where it was being administered where I could sneak it and peruse a copy of it. So I'll have to wait for a copy to be uploaded or hope that a copy is floating around the office tomorrow.

Odds are that I won't review the entire thing. Like last year, I'll focus on the questions which could be worked out by my Algebra I class (or any Alegbra review class, which has finished a complete year of Algebra.) I'll play it be ear when I see the exam.

On the other hand, I might skip it for the time being and just focus on the Geometry and Algebra Regents being given over the next two days.

Sunday, January 26, 2014

Offline... Using Borrowed Internet

I've spent (and will spend) a few days with my iPad hanging out at friends' houses and fast-food/coffee places with free internet. It gets me my mail and news, but isn't an optimum connection for making updates. Even this one is getting too long to handle. Hopefully, the issues will be resolved soon or better alternatives may present themselves.

Tuesday, January 21, 2014

Finding Numbers

(Click on the cartoon to see the full image.)

(C)Copyright 2014, C. Burke.

Subtract the constant and divide! You'll find your x, and you'll know why!

Hey, it's catchy!

Tuesday, January 14, 2014

Another Obituary

We received some shocking news late last week: my niece passed away. She was 31. No one could have prepared us for any sort of tragedy like this, and the family is in shock and disbelief. You can never tell how much time you have, so don't waste it. I wish I had spent more time (and had more time to spend) being the Favorite Uncle. Now, I have memories, and wish that there had been more of them.

Rest in Peace, Lisa Tarver. You'll remain forever in our hearts. You will be missed every day.

Thursday, January 09, 2014

Night Out

(Click on the cartoon to see the full image.)

(C)Copyright 2014, C. Burke.

No, seriously. I really need to know.

These things keep me awake at night.

Friday, January 03, 2014

White Out

(Click on the cartoon to see the full image.)

(C)Copyright 2014, C. Burke.

It's just about that white outside.

Excuse the commentary, but this was a no-brainer of a decision if only for the predicted temperatures in the city, not the snowfall amount. Why this decision couldn't have been made the evening before instead of scheduling the decision for a time after many have already started their day -- especially on a day where one has to allow extra time for traveling -- is just silly.

In fairness, the decision was made earlier than planned and before I dressed for work, but after my alarm went off so that I could check.

My admiration for those of you living in the affected areas who made it in to work, or attempted to do so. Be careful getting home. Be aware of the conditions around you, and watch out for the crazies, particularly the ones who think they're acting rationally.

Wednesday, January 01, 2014

Happy New Year 2014!

(Click on the cartoon to see the full image.)

(C)Copyright 2013, C. Burke.

Okay, so a lot less elaborate than last year, but also more applicable to what I'm teaching.

And I do have to adjust my watch on the first day of any month after a month with fewer than 31 days. Unless I wear the digital watch, of course.