Here are the questions and answers to Parts 3 and 4 of Tuesday's exam. There was quite a bit of discussion about the test at last night's meet up.

Links to Part 1 and Part 2 .

### Part 3

There are 3 problems, each worth 4 credits.

**32. ** In the diagram below, EF intersects AB and CD at G and H, respectively, and GI is drawn such that GH = IH.

If m<EGB = 50

If m<EGB = 50

^{o}and m<DIG = 115^{o}, explain why AB || CD.You could write this as proof, but it just says "explain", so writing your answer as a paragraph is fine.

Angle DIG is 115 degrees. This means that angle HIG is 65 degrees because it is supplementary to DIG. Because GH is congruent to HI, GHI is an isosceles triangle with HIG and HGI as the base angles. Therefore, HGI has a measure of 65 degrees. The sum of the angles in triangle GHI is 180 degrees, so angle GHI is 180 - 65 - 65 = 50. Angle GHI is congruent to angle EGB, so AB || CD because the corresponding angles are congruent.

**33. ** Given: Quadrilateral ABCD is a parallelogram with diagonals AC and BD intersecting at E. Prove triangle AEB is congruent to triangle CEB.

Describe a single rigid motion that maps AED onto GEB.

Due to formatting, I can't write a proof right now. I'll update it later as an image. But, basically ...

AE = EC and BE = DE because the diagonals of parallelograms bisect each other. Angles AED and CEB are congruent because vertical angles are congruent. Triangle AED is congruent to CEB because of SAS.

A rigid motion that would map AED on CEB is a 180 degree rotation about point E.

**34. ** In the diagram below, the line of sight from the park ranger station, P, to the lifeguard chair, L, on the beach of a lake is perpendicular to the path joining the campground, C, and the first aid station, F. The campground is 0.25 mile from the lifeguard chair. The straight paths from both the campground and first aid station to the park ranger station are perpendicular.

If the path from the park ranger station to the campground is 0.55 mile, determine and state, to the nearest hundredth of a mile, the distance between the park ranger station and the lifeguard chair.

Gerald believes the distance from the first and station to the campground is at least 1.5 miles. Is Gerald correct? Justify your answer.

If the path from the park ranger station to the campground is 0.55 mile, determine and state, to the nearest hundredth of a mile, the distance between the park ranger station and the lifeguard chair.

Gerald believes the distance from the first and station to the campground is at least 1.5 miles. Is Gerald correct? Justify your answer.

The length of PL is simple **Pythagorean Theorem**: x^{2 + 0.252 = 0.552. So x2 = .3025 - .0625 = .24, and x = .489... or .49 to the nearest hundredth.
}

To answer the second part, you have your choice of ratios. Either you can compare the base and hypotenuse of PLC (.25/.55) to the base and hypotenuse of PFC (.55/x) or you can use the **Right Triangle Altitude Theorem** to find FL and then add FL + LC to get FC. (I would recommend the first, but given the way the diagram is drawn, some will be drawn to the second method even if it is longer because it's more obvious to them.)

Using the first method: .25/.55 = .55 / x means that x = .55 * .55 / .25 = 1.21 miles, for Gerald is incorrect. It is shorter than 1.5 miles.

Using the second method: .25 / .49 = .49 / y means that y = .49 * .49 / .25 = 0.96. Add 0.96 + 0.25 = 1.21 miles. Gerald is still incorrect.

### Part 4

There is 1 problem, worth 6 credits. And what a doozy of a problem. I have warned my students about area problems where you have to break the irregular figure down into parts, and that's essentially what this question was, but in three dimensions.

**35. ** The water tower in the picture below is modeled by the two-dimensional figure beside it. The water tower is composed of a hemisphere, a cylinder and a cone. Let C be the center of the hemisphere and let D be the center of the base of the cone.

If AC = 8.5 feet, BF = 25 feet and m<EFD = 47

The water tower was constructed to hold a maximum of 400,000 pounds of water. If water weighs 62.4 pounds per cubic foot, can the water tower be filled to 85% of its volume and not exceed the weight limit? Justify your answer.

If AC = 8.5 feet, BF = 25 feet and m<EFD = 47

^{o}, determine and state, to the nearest cubic foot, the volume of the water tower.The water tower was constructed to hold a maximum of 400,000 pounds of water. If water weighs 62.4 pounds per cubic foot, can the water tower be filled to 85% of its volume and not exceed the weight limit? Justify your answer.

You have the Volume formulas for the cylinder and cone in the reference table. Also, you have the formula for a sphere, which you will take half of.

If AC is 8.5, then BC and DF are also 8.5, so we have our radii for the formulas.
** Cylinder:** V = pi*r

^{2}h = (3.141592...)(8.5)

^{2}(25) = 5674.5017...

**V = (1/2)(4/3)pi*r**

*Hemisphere:*^{3}= (2/3)(3.141592...)(8.5)

^{3}= 1286.2204...

**V = (1/3) pi*r**

*Cone:*^{2}h -- except we need to find the height! Trigonometry, again.

The side *adjacent* to the 47^{o} angle is 8.5. The height, which we are looking for, is the side *opposite* the angle. That means that we need to use **Tangent** to find the height.

Tan(47) = x/8.5, so x = 8.5 * tan(47), which is 9.12. Soooooooo ....
*Cone:* V = (1/3) pi*r^{2}h = (1/3)(3.141592...)(8.5)^{2}(9.12) = 690.0194

The total volume of the water tower is 5674.5017 + 1286.2204 + 690.0194 = 7650.7415, which rounds to 7651.

And that's just part of the answer! If you got that far, you probably got a lot of points. Loss of points for computational errors. (I wonder how many I made.)

Second part: .85 * 7651 * 62.4 = 405,809 pounds. Therefore, the water tower cannot be filled to 85% of its volume and not exceed 400,000 pounds.
*Note:* If you got an incorrect answer to the first part of the question, you can still receive credit for this portion if you mistake is consistent.

**36. ** In the coordinate plane, the vertices of triangle RST are R(6, -1), S(1, -4) and T(-5, 6). Prove that RST is a right triangle.

State the coordinates of point P such that quadrilateral RSTP is a rectangle.

Prove that your quadrilateral RSTP is a rectangle.

[The use of the set of axes is optional.]

You can prove that a triangle is a right triangle by showing that two sides form a right angle. You can prove two sides form a right angle by showing that they are perpendicular, which will be true if their slopes are *negative reciprocals*, that is, they have a product of -1.

**Slope of RS:** (-4 - -1)/(1 - 6) = 3/5.
**Slope of ST:** (6 - -4)/(-5 - 1) = -5/3.
**Slope of TR:** (-1 - 6)/(6 - -5) = -7/11.

The slopes of RS and ST are negative reciprocals, so RS is perpendicular to ST. Therefore <S is a right angle and triangle RST is a right triangle.

Finding point P: They gave you one hint. The rectangle has to be RSTP, so P has to be above R and to the right, and will be opposite from S.

The properties of rectangles are that the opposite sides are the same length and have the same slope. To get from point S to point R, you have to go up 5 and to the right 3 units. Or to use a **rigid motion** expression, if you S is moved be a translation of (x + 5, y + 3), it will be mapped onto R.

The image of T(-5, 6) after a translation of T_{+5,+3} is P(0, 9).

Proving that it's a rectangle: First, show that it's a parallelogram by showing that the opposite sides are parallel.
**Slope of RP:** (9 - -1)/(0 - 6) = -5/3.
**Slope of TP:** (9 - 6)/(0 - -5) = 3/5.

RP || ST because they have the same slope.

TP || RS because they have the same slope. Therefore, RSTP is a parallelogram.

Angle S is right angle. Therefore, RSTP is a rectangle.

...

And that's that. The rest of the Regents Exams are over a week away. Answers are farther away than that. :)

## 16 comments:

What about the last question in which you were given coordinates of a triangle that you had to prove was a right triangle, and then had to state a point that would make it a rectangle and prove that it was a rectangle?

For the volume of the water tower I got 7650?

If the student doesn't round off the height (or anything) until the very last total, as we have always taught the students to do, he/she will get a total volume for the water tower to be 7650.373374 which rounds off to 7650 to the nearest cubic foot.

Then when the student gets the total lbs for 85% of the water in full tower, and multiplies by the density, he/she should get 405756 lbs which is still too much for the tower's capacity.

I'm doing these problems without seeing the diagrams, so I'm using your interpretation of the diagrams. Still haven't seen the test --- retired math teacher.

I was using wolframalpha.com while sitting at the computer because I didn't have a graphing calculator handy. I had more numbers on the screen, but only typed what I did for the sake of brevity. I'll check my numbers to see if I had a transcription error. I don't think the rounding should have resulted in a big enough difference.

The water tower question should be 7650. Don't round until the end. The height of the cone is 9.11513403521, which does round to 9.12, but is not incredibly close like 9.119897. This changes the answer by 1.

Also, in #33, can the student (just asking) use the other properties (besides the diagonals bisecting each other) such as "the opposite sides of a parallelogram are parallel and congruent") in order to prove the triangles congruent by other means (SAS or SSS or AAS, etc.?

Can they also use other rigid motions (more than one) to show the triangles will map onto each other? (Again, this topic of rigid motions was taught in the common core course ad nauseum, so that is why I'm asking.)

And there was another part IV question for 6 points --- coordinate geometry proof?

Thank you for doing the work of putting up these problems for those of us who wanted to see this NEW exam but do NOT have access to it yet...when, I wonder?

The rounding before the end (specifically the height here)DOES result in a change of answer so it should be 7650 cubic feet.

Regarding rounding:

Yes, I ran the numbers again. This is going to be a problem for students, I think. I'm guilty of a mistake that I tell my students not to make. Not so much the rounding in the middle, but rounding to the wrong number of places (in the middle).

Two errors I see in 8-10th graders: first, changing 1/3 into .3 or .33 or sqrt(40) into 6.3 in the middle of a problem when they're likely going to multiply by 3 or square the number again before they're finished. A second one is to put 3.14 into the calculator for pi

even though there is a pi key on the calculator.In the past few years on the Algebra Regents, they have given questions where using 3.14 would not provide enough significant digits.

In this case, I used "pi", not 3.14, in WolframAlpha's entry field. For two out of three parts, I rounded to 4 decimals, which was more than enough. However, for the cone, the height had to be found first. In that instance, I only rounded to two places, which is as bad as if I had used 3.14 for pi.

There was no way I would even think of putting "8.5tan(47)" in for the height in the equation, so some kind of rounding was going to occur. So I think that there will be a lot of 7651 responses -- in relation to the number of 7650 answers, that is. I sure that there will be a lot of completely wrong or blank answers as well.

I just read what you said about the rounding error. Putting 8.5 tan (47) in for the height is easy using either the "copy and paste" feature of the new graphing calculators or the old feature of "storing" using the STO key on the older models. So, really students shouldn't be rounding off until the very last step in problems. Thus, I really believe the answer of 7651 should get a deduction of some points (perhaps only one) on this regents. Also, students should be specifically warned ALL year to be sure to use The "pi" key instead of an approximation as the regents states in its regents rules for the past number of years.

Don't you think If we allow kids to use calculators, they should be taught to use them very carefully? (Again, I'm retired and just asking what others think)

Sorry for the delay.

Requiring use of the features of the calculator just adds one more burden in teaching all the functions of the calculators -- assuming that everyone has the same kind. Most students will not own these and many will only have them during class time, and not all the time.

So, yes, it would be great if they are taught to use them carefully, but this is not happening and they are suffering for it. I don't believe I've encountered any time in first year Algebra where we used the STO key or cut and paste. (I beleive the Nspire has that ability, do the TI-83s?)

This is a new exam, and there's a learning curve in both writing and grading it. Considering the tests are being marked in Albany, we'll never know their reasoning if they decide to accept 7651. By this I mean, was it originally an acceptable answer, or did enough students get that result that they decided to accept it.

A question this complex, students are likely to break it down into smaller bits that they can handle. Because of this, many of those who feel somewhat overwhelmed by this task may choose to find the height first and -- as I did -- not consider just how many decimal places they need in their answer to achieve an accurate final answer to the nearest unit.

If only 7650 is acceptable, so be it. I'll be curious as to the percent of students who finished the question correctly.

Yes, I agree that using the calculator means having to teach it and teach it well --- but then, it is such a helpful tool. I'm so old that I remember teaching the slide rule as well as the tables for logarithms and for finding the trig. functions of an angle given in degrees and minutes. So, I believe teaching the calculator is a whole lot easier and the kids love it.

As for the STO key, we used to use it way back when I was teaching and it is on the TI-83's. The copy/paste feature is on the TI-84 plus I believe and both should be taught since they are so useful.

I didn't understand when you said that "we'll never know their reasoning if they decide to accept 7651. By this I mean, was it originally an acceptable answer, or did enough students get that result that they decided to accept it?" Do they now accept answers even those that are incorrect simply because a lot of students made the same error? Wow --- times have changed. The question should have been asked in scaffold form ---i.e. first it should have been worded to find the height to the nearest hundredth and then to use that value in the next part of the problem to find the volume, etc. However, wrong answers should never be acceptable even if the large majority of students make the same error.

Yes, times have changed. If significant numbers of students get a question wrong, they will re-evaluate the question for fairness, syntax, whatever. Perhaps it was just not a good question.

Teachers should be doing this, too. Maybe you don't need to reteach the topic, you may just need to rethink how you assessed it.

I had a slide rule when I was a kid. Probably second grade. It was a gift from an older brother in college. Unfortunately, I didn't learn how to use it in any meaningful way before it was lost. I did do the logs and trig functions with tables and no calculators.

By all means, they should re-evaluate how they worded a question if it was poorly worded but I still feel that they should NOT accept incorrect answers even if a lot of students put an incorrect answer. By the way, teachers should always be looking at test results and re-evaluating their own teaching techniques to help students.

This question was not, in my opinion, poorly worded --- it is just that in an effort to make it multi-steps and ONLY 6 points, they don't allow for small deductions for small errors (in this case the rounding error). My feeling is that since these exams are not based on 100 points, this will always be a problem.

Why, oh why, do they not go back to the 100 point system and have 65 as a passing grade? --- This would eliminate those 2 point questions, which essentially come down to no real partial credit, and questions that cannot be judged correctly since there are not enough points to award for partially correct responses. Of course, I know the reason --- they want to curve the tests since otherwise too many students throughout the State would fail the regents.

Yes, times have changed --- but they should not have, in my humble opinion.

can you post the answers for the algebra 2 trig test?

Unfortunately, I don't usually post Alg 2 tests, except sometimes for questions that my Algebra 1 students can handle, and, moreover, I wasn't able to grab a copy of the exam, so I won't be able to get a copy until sometime next week.

I took the other Geometry regents( 2005 standards) yesterday June 19 2015, I was just wondering if you have the answers yet, so maybe you could post them. Thank you for your time

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