Friday, June 05, 2015

New York Regents Geometry (Common Core) June 2015, Part 2

My apologies for the delay. I was literally away from the keyboard all day yesterday, between meetings, professional development and the #NYCMathTweetup last night.

Here are the questions and answers to Part 2 of Tuesday's exam. There was quite a bit of discussion about the test at last night's meet up.
Link to Part 1.

Part 2

There were seven open-ended questions in Part II, each worth 2 points. You had to show your work to get full credit.

25. Use a compass and straightedge to construct an inscribed square in circle T shown bleow. [Leave all construction marks.]

Using the straightedge, draw a diameter. Now draw a perpendicular bisector using the two points on the circle at either end of the diameter. This will create a second diameter (a secant, actually) perpendicular to the first one. Finally, using the straightedge, draw the four sides of the square by connecting the consecutive points on the circle. (I really wish I had an image here ... visually, it's a little more obvious.)

26. The diagram below shows parallelogram LMNO with diagonal LN, m<M = 118 degrees, and m<LNO = 22.
Explain why m<NLO = 40 degrees.

You have a couple of choices. First, because the figure is a parallelogram, angle O is congruent to angle M, so <O = 118o. Triangle LON has 180 degrees, so 180 = 118 + 22 + x. Solve for x, NLO = 40 degrees.

Second, using basiocally the same math as above, the consecutive angles of a parallelogram as supplementary, so 180 = 118 + MNL + 22. Angle MNL is 40 degrees. Angle NLO and MNL are alternate interior angles and are, therefore, congruent.

27.The coordinats of the endpoints of AB are A(-6, -5) and B(4, 0). Point P is on AB. Determine and state the coordinates of point P, such that AP:PB is 2:3. [The use of the set of axes below is optional.]

The ratio 2:3 means that you can divide AB into five congruent segments and that point P would be at the end of the second segment. P is located at a point that is two-fifths of the distance from A to B. You will notice that the difference in the x-coordiantes is 4 - (-6) = 10 and the difference of the y-coordinates is 0 - (-5) = 5. Two-fifths of 10 is 4; two-fifths of 5 is 2. Point P is at (-6 + 4, -5 + 2), which is P(-2, -3).

You can also graph the points and see that the slope of the line is 1/2. If you draw the triangles formed by going up 1, right 2 five times, the second triangle will bring you to point P.

28. The diagram below shows a ramp connecting the ground to a loading platform 4.5 feet above the ground. The ramp measures 11.75 feet from the ground to the top of the loading platform.

Determine and state, to the nearest degree, the angle of elevation formed by the ramp and the ground.

Trigonometry. You have the opposite side and the hypotenuse, so you need to use the sine ratio.
sin (x) = 4.5 / 11.75. Take the inverse and you get
x = sin-1 (4.5 / 11.75).
Put that in your calculator and you get 22.52 degrees, which rounds to 23 degrees.

29. In the diagram below of circle O, the area of the shaded sector AOC is 12(pi) in2 and the length of OA is 6 inches. Determine and state m<AOC.

The measure of the central angle can be found using a proportion: central angle / 360 = sector area / circle area.
The area of the entire circle is A = (pi)r2 = (pi)(6)2 = 36(pi).

Cross multiply: x / 360 = 12 / 36 and you get x = 120 degrees.
Likewise 12(pi)/36(pi) = 1/3. One-third of 360 degrees is 120.

30. After a reflection over a line triangle A'B'C' is the image of triangle ABC. Explain why triangle ABC is congruent to triangle A'B'C'.

In a reflection. the image retains the size and shape of the original. This means that AB = A'B', BC = B'C' and CA = C'A', so by SSS the two triangles are congruent.

I'm guessing that Definition of a Reflection is NOT a sufficient answer. State something that you know.

31. A flagpole casts a shadow 16.60 meters long. Tim stands at a distance of 12.45 meters from the base of the flagpole, such that the end of Tim's shadow meets the end of the flagpole's shadow. If Tim is 1.65 meters tall, determine and state the height of the flagpole to the nearest tenth of a meter.

Sketch a diagram. You will have two similar triangles, such that one is contained within the other, with two parallel, vertical lines. You can set up a proportion height : shadow = height : shadow. Notice that if Tim is 12.45 meters from the pole, that means that his shadow is 16.6 - 12.45 = 4.15 meters long.
So x / 16.60 = 1.65 /4.15. Cross-multiply, x = 6.6 meters.

Another good reason to sketch a diagram is so that you won't make the mistake of saying that Tim's shadow is 12.45 meters long because it's written in the problem and you're too busy punching keys to visual it.

That's it for Part II. Still working on the rest.


Max Nelkens said...

Your proportion is wrong!! It should be 4.15/1.65= 16.6/x. I derived at x to be 6.6 meters. It makes more sense numerically for a flagpole to be 6.6 meters tall compared to a 1.65 meter person!

Rose C said...

I know you were in a rush but the length of the Tim's shadow is NOT 12.45. You have to subtract to get the length of his shadow and will get 4.15 m. Then the height of the pole is 6.6m which is much more reasonable.

Rose C said...

Btw, do you also have parts II and IV ? And when do you expect to have the diagrams up on the blog?

Also, in no. 30' can't the student say that a reflection is a rigid motion (a la one of the MAIN CONCEPTS that was emphasized at nauseum in the common core geometry course) to explain the congruency of the triangles? Just asking.

(x, why?) said...

This is why I generally let people proofread these things, but even last night at a gathering of math teachers, I had people asking me when the rest of the answers would be up.

This is what I get for typing while covering someone else's class, sigh.

I'll fix my mistakes during my next prep period.

(x, why?) said...

I honestly have no idea what they are going for in question number 30. Basically, it is the definition of a reflection, but just saying that is not really an explanation, so give some kind of justification. But, again, I don't know what they're looking for. As long as it's something.

Bill in Boston said...

Part 1.
"bleow" [typo]
#5 "groun" [typo]

Part 2, #25. A. "Using the straightedge, draw a diameter."
Finding the first diameter requires a few more moves than that?
Unless you're given the Center O of circle T, which isn't mentioned in text; was it in diagram?
Draw any chord with straightedge, erect its perpendicular bisector (two arcs) gives your first diameter.

27. coordinats [typo]

(snippy historical comments on Trig via email.)

carlito said...

I don't understand the commentary. You correctly stated that Tim's shadow is 4.15m long and the proportion is correct, x = 6.6m.As for problem 30, can't you state that in a reflection, distance, angle, and orientation is preserved, basically an isometry?

(x, why?) said...

I edited the explanation after the comment was made. Usually, I make a note of that. This time I didn't.

Re #30: most likely that will be fine. I have no idea want kind of explanation they are looking for. As long as all the bases are covered. My mind went to "proof" as soon as I saw "explain" because I've been working with proofs with my older Geometry students who are taking the other exam next week.

(x, why?) said...

Bill, center O was given in circle O. Even so, this was not one of the usual ways that they frame construction problems. Generally, it's lines and triangles, even when most of the triangle is irrelevant and just in the way to be confusing.

Anonymous said...

anyone know when individual test results become available?

Anonymous said...

Are you planning on posting the August 2015 Common Core Geometry Regents answers? I know it's super early right now but I mean like tomorrow or maybe over the weekend? If you don't have time, it's totally cool I was just wondering. I'm just really nervous to see what things I got right and wrong and stuff. Anyway, thanks a bunch :))

(x, why?) said...

Can't post them because I don't have the exams. I didn't work the Regents this summer and I don't know anyone nearby who did. So I won't have the exams until they are published.

Anonymous said...

Wait if you don't have them until then, how do we get answers? Do you know any other teachers who immediately share answers? Thanks

(x, why?) said...

No, I don't. That's why my blog got so many hits in June.

You can check for your teachers on Twitter, or just ask in general. Some teachers check that sort of thing.

Anonymous said...

So do you happen to know when the exam will be published?

(x, why?) said...

Nope. I've been checking daily. (Although I really didn't expect them to be posted on the weekend, I checked anyway.)

Patrick Young said...
This comment has been removed by the author.