## Sunday, January 28, 2018

### August 2017 Common Core Geometry Regents, Part 2

While I wait for the January 2018 Geometry Regents exams to become available, and for the go-ahead to publish the questions and answers, allow me to revisit the August 2017 exams, which I haven't gotten around to dealing with until now.

### August 2017, Geometry, Part II

Each correct answer is worth up to 2 credits. Work must be shown.

25. Sue believes that the two cylinders shown in the diagram below have equal volumes.

Is Sue correct? Explain why.

Answer: Sue is correct because the two cylinders have the same base area and the same height.
You didn't need to calculate the Volume to answer the question, but if you did, don't make any rounding mistakes!
V = (pi) (r)2 (h). In both cases, r = 5 and h = 11.5

26. In the diagram of rhombus PQRS below, the diagonals and intersect at point T, PR = 16, and QS = 30. Determine and state the perimeter of PQRS.

Answer: PR is perpendicular to QS, so the 4 triangles are congruent right triangles. The sides of the rhombus are the hypotenuses of the right triangles. PT = 16 / 2 = 8, QT = 30 / 2 = 15.
82 + 152 = PQ2
64 + 225 = PQ2
289 = PQ2
17 = PQ
Perimeter = 17 * 4 = 68

27. Quadrilateral MATH and its image M"A"T"H" are graphed on the set of axes below.

Describe a sequence of transformations that maps quadrilateral MATH onto quadrilateral M"A"T"H".

Answer: There a number of possible solutions.
One of them is a Rotation of 180 degrees about the axis followed by a Translation of +1, -1 (T+1,1).

Something like a reflection across y = -x or a dilation of scale factor -1 could also work in place of the rotation.

28. Using a compass and straightedge, construct a regular hexagon inscribed in circle O.

Answer: I can describe what you need to do, but showing the steps of a construction are always a problem for my blog.
Put the compass on a point on the circle. Draw an arc that goes through the center point, O, and draw an arc through the circle.
Go to the point on the circle you just found. Without changing the compass, make another arc on the circle. Repeat this until you go completely around the circle.
Your last arc should be through the point you started from.
Use the straightedge to connect the points on the circle. You have inscribed a hexagon in the circle.

The image below was taken from the state's answer key:

29. The coordinates of the endpoints of AB are A(2,3) and B(5,–1). Determine the length of A'B', the image of AB, after a dilation of 1/2 centered at the origin.

Answer: You do not need to graph the line segment or its image. The dilation with have a length of 1/2 the original.
Use the distance formula d = sqrt ( (5-2)2 + (-1-3)2 )
d = sqrt(9 + 16) = sqrt(25) = 5
length of AB is 5
Therefore, the length of A'B' is 2.5.

30. In the diagram below of triangle ABC and triangle XYZ, a sequence of rigid motions maps ∠A onto ∠X, ∠C onto ∠Z, and AC onto XZ.

Determine and state whether BC = YZ. Explain why.

Answer: Because ∠Amaps to ∠X and ∠C to ∠Z and AC to XZ, triangles ABC and XYZ are congruent by ASA. This means that BC is congruent to YZ because the corresponding parts of congruent triangles are congruent.
Additionally, rigid motions preserve side length and angle measure.

31. Determine and state the coordinates of the center and the length of the radius of a circle whose equation is x2 + y2 - 6x = 56 - 8y.

Answer: Standard form for a circle is (x -h)2 + (y - k)2 = r2. This equation needs to be rewritten and factored, by completing the square.

x2 + y2 - 6x = 56 - 8y.
x2 - 6x + y2 + 8y = 56
Half of -6 is -3, (-3)2 = 9, add 9 to both sides
Half of 8 is 4, (4)2 = 16, add 16 to both sides
x2 - 6x + 9 + y2 + 8y + 16 = 56 + 9 + 16
x2 - 6x + 9 + y2 + 8y + 16 = 56 + 9 + 16
(x - 3)2 + (y + 4)2 = 81, r2 = 81, so r = 9.
The center of the circle is (3, -4) and the radius is 9.

End of Part II

How did you do?