## Tuesday, July 17, 2018

### June 2018 Common Core Geometry Regents, Part II

The following are some of the multiple questions from the recent June 2018 New York State Common Core Geometry Regents exam.
The answers to Part I can be found here

### June 2018 Geometry, Part II

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

25. Triangle A'B'C' is the image of triangle ABC after a translation of 2 units to the right and 3 units up. Is triangle ABC congruent to triangle A'B'C'? Explain why

Answer:
Yes, because a translation is a rigid motion, which preserves shape and size.
The corresponing angles are congruent, and the corresponding sides are congruent, so the triangles are congruent.

26. Triangle ABC and point D(1,2) are graphed on the set of axes below.

See image below.

Graph and label triangle A'B'C', the image of triangle ABC, after a dilation of scale factor 2 centered at point D.

Answer:
See image below. No explanation is needed. Just the graph, with labels.
Each image is twice as far away from D as the original point was.
To get from D to A, go down 1 unit and 3 left. To get from A to A', do this again. A' is (-5, 0)
To get from D to B, go up 3 units and 1 left. To get from B to B', do this again. B' is (-1, 8)
To get from D to C, go down 3 units and 3 right. To get from C to C', do this again. C' is (7, -4)

27. Quadrilaterals BIKE and GOLF are graphed on the set of axes below.

Describe a sequence of transformations that maps quadrilateral BIKE onto quadrilateral GOLF.

Answer:
Notice that letters of BIKE go counterclockwise and GOLF go clockwise. A simple translation will not be good enough. A translation of T5,8 will map BIKE over GOLF; however, B doesn't map to G, and K doesn't map to L.
You need to include a reflection in your transformations. The directions ask for a sequence.
There are many possibilities, but the simplest (to me) would be a reflection over the y-axis followed by a translation of 5 units up.

28. In the diagram below, secants RST and RQP, drawn from point R, intersect circle O at S, T, Q, and P.

If RS = 6, ST = 4, and RP = 15, what is the length of RQ?

Answer:
(RS)(RT) = (RQ)(RP)
(6)(6 + 4) = (RQ)(15)
60 = 15(RQ)
RQ = 4.

29. Using a compass and straightedge, construct the median to side AC in ABC below.
[Leave all construction marks.]

Answer:
See image.
To construct a median, you need to find the midpoint of AC. You can find the midpoint by constructing a perpendicular bisector of AC. Label the midpoint D. Then use the straightedge to draw BD.

30.
Skye says that the two triangles below are congruent. Margaret says that the two triangles are similar.
Are Skye and Margaret both correct? Explain why.

Answer:
Skye and Margaret are both correct. Using the Pythagorean Theorem, 52 + 122 = 169, which is 132, so the length of the hypotenuse of the first triangle is 13. By the Hypotenuse-Leg Theorem, the two triangles are congruent.
If the triangles are congruent, then their corresponding angles are congruent, and that makes them similar triangles as well.

31. Randy’s basketball is in the shape of a sphere with a maximum circumference of 29.5 inches. Determine and state the volume of the basketball, to the nearest cubic inch.

Answer:
The circumference is 2*pi*r = 29.5
So the radius, r = 29.5 / (2 * pi) = 4.695...
Volume of a sphere = (4/3) * pi * r3 = (4/3) * pi * (4.695)3 = 433.506...
To the nearest cubic inch, the Volume is 434 cubic inches.

End of Part II

How did you do?

Questions, comments and corrections welcome.