January 2015 Geometry Regents Parts II, II, IV (Open-Ended)

Once again, here are the multiple-choice problems with explanations for today's New York State Geometry Regents exam. Since I typed this up as quickly as I could, there are no images, graphs or diagrams included. I'll edit them as I am able to.

A common theme: there were several questions involving the equation of a circle, as there always are, as well as several questions which not only involved the Pythagorean Theorem, but very specifically a 3-4-5 triangle or a multiple of it.

Link to Part I.

I hope everyone did well. Here is are the (most of the) open-ended questions. Again, sorry, no images.

Part II

 

29. The Proof. Image to be uploaded later.

Very simply, you are given two pairs of congruent sides. The included angles, BCA and DCE are congruent because they are vertical angles. Therefore ABC is congruent to DEC by SAS.

 

30. Using a compass and straightedge, construct the perpendicular bisector of side AR in triangle ART show below. [Leave all construction marks.] (image omitted)

I can’t really show this, but keep in mind that point T is irrelevant. It is NOT on the bisector. Ignore the triangle and focus on segment AR. Create the perpendicular bisector as you would with any other segment.

 

31. Determine and state the measure, in degrees, of an interior angle of a regular decagon.

 

First, even if you have this memorized, you still need to show your work.

 

Second, there are two methods, both of which are acceptable.

 

Interior:  (10 – 2) * 180 = 1440. 1440 / 10 = 144 degrees for each interior angle.

Or exterior: 360/10 = 36 degrees per exterior angle. Interior angle = 180 – 36 = 144 degrees.

 

32. Write an equation of a line that is parallel to the line whose equation I s 3y = x + 6 and that passes through the point (-3, 4).

 

Note that it says “an equation” and doesn’t say that it has to be in any specific form.

Easiest method, plug in (-3, 4) into the equation and find the new constant. (I won’t call it “b”.)

3(4) = -3 + c

12 = -3 + c

15 = c

3y = x + 15 is an acceptable answer

 

In slope-intercept form, it is y = 1/3x + 5.

 

33. In the diagram below, secants PQR and PST are drawn to a circle from point P. (image omitted)

If PR = 24, PQ = 6, and PS = 8, determine and state the length of PT.

 

Not a trick question. No extra adding or subtracting.

(PQ)(PR) = (PS)(PT)

(6)(24) = (8)(x)

144 = 8x

18 = x

PT is 18.

 

34. The slope of QR is (x -1)/4 and the slope of ST is 8/3. If QR is perpendicular to ST, determine and state the value of x.

 

If QR is perpendicular to ST, then the slope of QR is -3/8. So

(x-1)/4 = -3/8

8(x-1) = -12

8x – 8 = -12

8x = -4

X = -1/2

 

Part III

 

35. Quadrilateral HYPE have vertices H(2,3), Y(1, 7), P(-2, 7) and E(-2, 4). State and label the coordinates of the vertices of H’’Y’’P’’E’’ after the composition of transformations r _x-axis o T _{5, -3}.

 

Remember: the translation comes before the reflection. If you do it the other way around, you will lose half-credit (2 points). You MUST label the coordinates of the final points. Keep this in mind if you are using the graph!

 

H(2, 3) > H’(7, 0) > H’’(7, 0)

Y(1, 7) > Y’(6, 4) > Y’’(6, -4)

P(-2, 7) > P’(3, 4) > P’’(3, -4)

E(-2, 4) > E’(3, 1) > E’’(3, -1)

 

 

36. On the set of axes below, graph two horizontal lines whose y-intercepts are (0, -2) and (0, 6), respectively.

Graph the locus of points equidistant from these horizontal lines.

Graph the locus of points 3 units from the y-axis.

State the coordinates of the points that satisfy both loci.

 

The first locus is the horizontal line y = 2.

The second locus is comprised of the vertical lines x = -3 and x = 3.

The coordinates of the points that satisfy both conditions are (-3, 2) and (3, 2).

 

 

37. In the diagram below, a right circular cone with a radius of 3 inches has a slant height of 5 inches and a right cylinder with a radius of 4 inches has a height of 6 inches.

(image omitted)

Determine and state the number of full cones of water needed to completely fill the cylinder with water.

 

You need to know the Volume of each solid. You can leave them in terms of pi.

First, you need to know the height of the cone – you were given the slant height.

Notice that one more time the problem involves a 3-4-5 right triangle. Satisfy yourself however you have to that the height of the cone is 4 inches.  (State that it’s a right triangle with dimensions 3-4-5, using Pythagorean Theorem.)

 

The cone has V = 1/3 (pi) r² h = 1/3 (pi) (9) (4) = 12 pi

The cylinder has V = (pi) r² h = (pi) (16)(6) = 96 pi.

Divide 96pi/12pi = 8. (Pi will cancel out.)

8 cones will fill up the cylinder.

Part IV

 

38. In the diagram below, right triangle RSU is inscribed in circle O, and UT is the altitude drawn to hypotenuse RS. The length of RT is 16 more than the length of TS and TU = 15.

Find the length of TS.

Find, in simplest radical form, the length of RU.

(Image omitted)

 

You need to use the Right Triangle Altitude Theorem, and Pythagorean Theorem. And then simplify the radicals.

 

First part: x / 15 = 15 / x + 16

Cross-multiply: x² + 16x = 225

So x² + 16x – 225 = 0

(x + 25)(x – 9) = 0

X = -25 or x = 9.

Discard the negative answer and TS = 9.

 

This makes TR = 9 + 16 = 25.

RU = the square root of (25² + 15²) = the square root of (625 + 225) = the square root of (850).

The largest perfect square that is a factor of 850 is 25, which goes into 850 34 times.

RU = (25)^.5(34)^.5 = 5(34)^.5. (That is, 5 times the square root of 34.)

 

Yes, that was a crazy problem, but gettable if you worked your way through it.

I double-checked that answer twice.

 

For instance, you can find the length of US, which is the square root of 306. And the square root of (306 + 850) = square root of 1156 = 34, which is the length of the diameter of circle O. We know this because ST is 9 and TR is 25, and 9 + 25 = 34.