*The following are the questions and answers (and commentary) for part of the New York State Algebra Regents exam. If you have any questions or comments (or corrections), please add them in the*

**Comments**section.My apologies for typos, particularly if they are in the questions, because then the answers are subject to change.

The answers to Part II can be found here.

The answers to Part III can be found here.

### June 2017, Geometry (Common Core), Part IV

**35.***Quadrilateral PQRS has vertices P(-2, 3), Q(3, 8), R(4, 1) and S(-1, -4).
Prove that PQRS is a rhombus.
[The use of the set of axes is optional.]*

This is NOT a two-column proof. You could write one, but it will not be worth any points if you don’t use the coordinates and the formulas to show the work. You need to back up everything you say with numbers. (And those numbers have to be correct.)

There are a number of ways to show that a quadrilateral is a rhombus. The easiest is that the four sides have equal lengths. You must have a concluding statement that says this and show the work. (Radical 50 or 5 radical 2)

You can show that the diagonals are perpendicular bisectors of each other. So you need to find the slopes of the lines and the midpoint of each line. The slopes must be inverse reciprocals and therefore are perpendicular (that last part is important!) and the midpoints must be the same point.

If you only prove that the slopes are perpendicular, you MUST also show that the figure is a parallelogram. Otherwise, your proof is incomplete. A kite, for example, has perpendicular diagonals.

To show that it is NOT a square, you can show that two sides don’t have right angles: find the slopes and show that they are not perpendicular. You can show that the diagonals are not congruent: find the lengths of the two diagonals. You can even use Pythagorean Theorem to show that two sides and one diagonal do not form a right triangle.

Note that you did not have to simplify your radicals when you use the distance formula, but you did have to have accurate numbers.

Seem the image below.

Points: Basically, you got 4 points for proving and stating the figure was a rhombus and 2 points for proving that it was not a square, for a total of 6 points.

**36.*** Freda, who is training to use a radar system, detects an airplane flying at a constant speed and heading in a straight line to pass directly over her location. She sees the airplane at an angle of elevation of 15 ^{o} and notes that it is maintaining a constant altitude of 6250 feet. One minute later, she sees the airplane at an angle of elevation of 52^{o}. how far has the airplane traveled, to the nearest foot?
*

*Determine and state the speed of the airplane, to the nearest mile per hour.*

You have to find the horizontal distance to the plane from the first sighting, and then the horizontal distance of the second sighting. And then subtract the two values to find how far the plane traveled.

Finally, you have to convert the number of feet it moved per second instead miles per hour.

This could also be done with the **Law of Sines** but I'll save that for another post, where I can go more in-depth into the problem.

Sketch a figure, or two, to show the plane's location. In each right triangle, you have the height of 6250, which is opposite the angle (either 15 or 52). And you need the ground distance, which will be adjacent to the angle. You don't need the hypotenuse in either triangle. That means using tangent.

So *Tan 15 ^{o} = 6250 / x* and

*Tan 52*. And the distance traveled will be

^{o}= 6250 / y*x - y*.

x = 6350 / tan 15 and y = 6250 / tan 52

x = 23325/32 and y = 4883.03

x - y = 18442 feet

No matter what number you got in the top half of the problem, you can get point for converting it into milers per hour correctly.

To change feet per second into miles per hour, multiply by 60 and divide by 5280.

18442.28... x 60 / 5280 = 1106536.94... / 5280 = 297.57... = 210 mph.

You might have run into problems if your calculator was in *Radians* mode. Hopefully, the negative numbers alerted you to a problem.

If you used sine or cosine (other than using the Law of Sines), you would have lost two points from the top portion of the problem.

End of **Part IV**

How did you do?

Comments, questions, corrections and concerns are all welcome.

Typos happen.

## 4 comments:

Wow

Was that a good wow or a bad wow?

Well then fudge my life. I was using ratios and got 240 or maybe 210 I'm not sure

Well, 210 is the correct answer, so maybe you did it right.

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