June 2019 Common Core Geometry Regents Part 2 (open ended)

The following are some of the multiple questions from the recent June 2019 New York State Common Core Geometry Regents exam.

June 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. Triangle A'B'C' is the image of triangle ABC after a dilation with a scale factor of 1/2 and centered at point A. Is triangle ABC congruent to triangle A'B'C'? Explain your answer.

Answer:
Triangle ABC is not congruent to triangle A'B'C' because a dilation of scale factor 1/2 does not preserve size.

It doesn't ask, but ABC would be similar to A'B'C' because shape is preserved, but size is not.

26. Determine and state the area of triangle PQR, whose vertices have coordinates P(-2,-5), Q(3,5), and R(6,1).
[The use of the set of axes below is optional.]

Answer:
To find the area of the triangle, you need a base and an altitude, which are at right angles to each other. This is trivial if the triangle is a right triangle and two sides are perpendicular.

If you graph it, if looks like angle R could be a right angle, so check the slopes of QR and RP.
QR: (5 - 1) / (3 - 6) = 4 / -3 = -4/3
RP: (1 - -5) / 6 - -2) = 6 / 8 = 3/4
The slopes are inverse reciprocals so they are perpendicular, meaning they can be used as the base and altitude.

Next, find the length of each
QR: SQRT( (3-6)² + (5-1)²) = SQRT ( (-3)² + (4)2)) = Sqrt(9+16) = sqrt(25) = 5
QR: SQRT( (6 - -2)² + (1 - -5)²) = SQRT ( (8)² + (6)2)) = Sqrt(64+36) = sqrt(100) = 10

A = 1/2(5)(10) = 25

Another method that doesn't rely on slopes and distance formula:
Graph the triangle.
Make a rectangle with P at a vertex and Q and R as points on the sides of the rectangle.
If you count boxes, you will find that the rectangle has an area of (8)(10) = 80.
There are four triangles within the rectangle, three of which are right triangles which line up with the grid. You can find the areas of these very easily: one is 10 by 5, one is 3 by 4, and the last is 6 by 8.
Those 3 have areas of 1/2(10)(5) = 25, 1/2(3)(4) = 6, 1/2(6)(8) = 24.
80 - (25 + 6 + 24 ) = 80 - 55 = 25.
This is a valid approach to take using the provided set of axes.

27. A support wire reaches from the top of a pole to a clamp on the ground. The pole is perpendicular to the level ground and the clamp is 10 feet from the base of the pole. The support wire makes a 68° angle with the ground. Find the length of the support wire to the nearest foot.

Answer:
The angle is on the ground. The distance from the clamp to the pole is the side adjacent to the pole. The length of the wire is the hypotenuse. That means using cosine.

Cos 68° = 10 / x
x (cos 68°) = 10
x = 10 / (cos 68°) =26.694...
The wire is approximately 27 feet.

28. In the diagram below, circle O has a radius of 10.

If mAB = 72°, find the area of shaded sector AOB, in terms of π.

Answer:
Area of a circle is πr².
Area of a sector of circle is (central angle/360) πr².
A = (72/360) π (10)²
A = 20 π

Note that 72/360 is 1/5, which after a year of Geometry, you should probably recognize. One-fifth of ten squared is 1/5 of 100, which is 20.

29. On the set of axes below, triangle ABC = triangle STU.

Describe a sequence of rigid motions that maps triangle ABC onto triangle STU.

Answer:
I don't like this question because students will second-guess themselves because of its simplicity. It says "a sequence of rigid motions" when, in fact, it can be done with one.

You can map ABC onto STU with a rotation of 90 degrees counterclockwise about the origin.

You might have done is about point B, for example, and then followed that with a translation of T_2,-6.

30.
In right triangle PRT, m∠P = 90°, altitude PQ s drawn to hypotenuse RT, RT = 17, and PR = 15.

Determine and state, to the nearest tenth, the length of RQ.

Answer:
RQ is the long leg of right triangle PQR. PR is the hypotenuse of PQR, and the long leg of PRT. RT is the hypotenuse of PRT.
Because PRT is similar to PQR, you can write a proportion:
(RQ / PR) = (PR / RT)
(RQ / 15) = (15 / 17)
17 RQ = (15)(15)
17 RQ = 225
RQ = 225 / 17 = 13.23...
RQ is approximate 13.2.

31. Given circle O with radius OA, use a compass and straightedge to construct an equilateral triangle inscribed in circle O.
[Leave all construction marks.]

Answer:
They have asked a very similar question before. That time it was for an inscribed hexagon, which involves many of the same steps. There are only so many construction questions that they can ask.

Say we want to make equilateral triangle ABC within circle O. That means that angle BAC will be an inscribed angle measuring 60 degrees and intercepting an arc BC with a measure of 120 degrees. The same for the other two angles.

We want the three points to be one third of the triangle apart from each other. And this is where the radius comes in. If I had a protractor (don't do this, I'm just explaining), I could draw the diameter. I could then split each semicircle in three by creating 60 degrees angles. Three 60-degree angles add up to 180 degrees. And the radii are all the same size. You could then make a hexagon by connecting the endpoints of those six radii. It would be a regular hexagon, as all the sides are the same length, and, in fact, the same length as each radius because you would have six little equilateral triangles.

Where am I going with all this? Simple:

Use the length of the radius to create six arc around the circle, starting and ending with point A. Each arc will measure 60 degrees. Mark the second one B and the fourth one C.

With the straightedge, draw AB, BC and CA. You're done.

Yes, I could have just told you to do that, but where would the fun in that be? Wouldn't you want to know why you could do it this way?

(The above image was done on a computer so it's approximate.)

End of Part II

How did you do?

Questions, comments and corrections welcome.