January 2019 Common Core Geometry Regents, Part II

The following are some of the multiple questions from the recent January 2019 New York State Common Core Geometry Regents exam.
The answers to Part I can be found here
The answers to Parts III and IV can be found here

January 2019 Geometry, Part II

Each correct answer is worth up to 2 credits. Partial credit can be given. Work must be shown or explained.

25. Write an equation of the line that is parallel to the line whose equation is 3y + 7 = 2x and passes through the point (2,6).

Answer:
There are several approaches you can take.
You can rewrite the expression into slope-intercept form, and then solve for "b" using the point (2, 6).

3y + 7 = 2x
3y = 2x - 7
y = 2/3 x - 7/3
y = 2/3 x + b
(6) = 2/3 (2) + b
6 = 4/3 + b
b = 14/3 = 4 2/3
y = 2/3 x + 14/3

You can rewrite the expression into slope-intercept form to get the slope, and then use point-slope form.
(as above, slope is 2/3)
y - 6 = 2/3 (x - 2)

You could replace the 7 with b, and use (2, 6) to find the value that would make the line parallel:

3y + b = 2x
3(6) + b = 2(2)
18 + b = 4
b = -14
3y - 14 = 2x
If you check, you'll see that all three of these are equivalent forms.

26. Parallelogram ABCD is adjacent to rhombus DEFG, as shown below, and FC intersects AGD at H.
If m∠B = 118° and m∠AHC = 138°, determine and state m∠GFH.

Answer:
See image below.
Both figures are parallelograms. Angle B is congruent to angle CDG because opposite angles are congruent, so Angle D = 118.
Angle DGF is congruent to Angle CDG because the alternate interior angles are congruent because the lines are parallel. Angle DGF = 118.
Angle GHF is supplementary to Angle AHC. Supplementary angles have a sum of 180 degrees. 180 - 138 = 42, so Angle GHF = 42.
Triangle FGH has a total of 180 degrees, two of which are 118 and 42, so x + 118 + 42 = 180.
x + 160 = 180, x = 20. Angle GFH = 20 degrees.

27. As shown in the diagram below, secants PWR and PTS are drawn to circle O from external point P.

If m∠RPS = 35°and mRS = 121°, determine and state mWT .

Answer:
The measure of the angle is one-half of the difference of the two arcs it intercepts.
This means that angle P will be 1/2 (arc RS - arc WT)
Substitute the values and solve for x.
35 = 1/2 (121 - x)
70 = 121 - x
-51 = -x
x = 51 degrees.

28. On the set of axes below, triangle ABC is graphed with coordinates A(-2,-1), B(3,-1), and C(-2,-4). Triangle QRS, the image of triangle ABC, is graphed with coordinates Q(-5,2), R(-5,7), and S(-8,2).

Describe a sequence of transformations that would map triangle ABC onto triangle QRS.

Answer:
Two methods come to mind:
First, a counterclockwise rotation of 90 degrees about the origin will give you A'(1, -2), etc. Then, reflect over the y-axis, which gives you A"(-1, -2), etc. Finally, a translation to left 4, up 4. T_{(-4, 4)}. (There are other combinations of reflection, rotation and translation that will work, depending on the order you do them in.)
Second method: if you reflect over the line y = x, A'B'C' will have the correct orientation, and then will just need a translation. A' will be at (1, -2), and then do T_{(-4, 4)}, as in the first method.

29. Given points A, B, and C, use a compass and straightedge to construct point D so that ABCD is a parallelogram.
[Leave all construction marks.]

Answer:
Construction to be added later. Sorry.
First, the parallelogram is ABCD, so point D must be to the left of point C.
The opposite sides of a parallelogram are congruent.
Starting at point B, open the compass to make an arc through point A. Then move the compass to point C and make an arc to the left side.
Starting at point B, open the compass to make an arc through point C. Then move the compass to point A and make another arc that goes through the first arc you make.
Label the point at the intersection of those two arcs point D. Use the straightedge to make lines CD and AD.

30.
On the set of axes below, triangle DEF has vertices at the coordinates D(1, -1), E(3,4), and F(4,2), and point G has coordinates (3,1). Owen claims the median from point E must pass through point G. Is Owen correct? Explain why.

Answer:
If EG is a median, then G must be the midpoint of DF. If G is the midpoint, Owen is correct. If it is not, the Owen is not correct. You MUST state one of these.
The x- and y- coordinates of the midpoint must be halfway between those of D and F. You can label this on the graph by counting boxes, or you can use the formula.
To find the middle (the average), add up the two numbers and divide by 2.
x-coordinate: (1 + 5) / 2 = 6 / 2 = 3
x-coordinate: (1 + 4) / 2 = 5 / 2 = 2.5
y-coordinate: (-1 + 2) / 2 = 1/2
The midpoint is (3, 1/2). The midpoint is (2.5, 1/2). Point G is at (3, 1). G is not the midpoint, so the median does not go through G. Owen is not correct.

31. A walking path at a local park is modeled on the grid below where the length of each grid square is 10 feet. The town needs to submit paperwork to pave the walking path. Determine and state, to the nearest square foot, the area of the walking path.

Answer:
For the two straight pieces, count the boxes and then multiply by 10 * 10 (the scale): 18 * 10 * 10 = 1800.
For the curved pieces: notice that the two semicircles create one complete circle. To find the area of the path, find the area of the bigger circle and subtract the area of the smaller circle.
The larger circle has a radius of 30. The smaller one has a radius of 20.
A = (30)²(pi) - (20)²(pi) = 900(pi) - 400(pi) = 500(pi) = 500(3.141592) = 1570.796
1800 + 1570.796 = 3370.796 = 3371 square feet.

End of Part II

How did you do?

Questions, comments and corrections welcome.