August 2014 Geometry Regents, Part 2 and 3

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Continuing with the answers to yesterday's exam. If you missed Part 1, it's here.
There were six two-point questions, three four-point questions, and one six-point question. Partial credit was available for all of them, and work was required.

Part 2

29. Triangle ABC has coordinates A(-2, 1), B(3, 1) and C(0, -3). On the set of axes below [OMITTED], graph and label Triangle A'B'C', the image of triangle ABC after a dilation of 2.

You only needed the image, but it helped to graph the original first. A dilation of 2, centered on the origin (because nothing else was stated), means that you double all the x- and y-coordinates. You had to label them A', B', and C', but you didn't need to write the actual co-ordinates.
If you didn't do the graph, but wrote and labeled the coordinates of the image, you scored one point. If you Dilated by something other than 2, or did a different Transformation, you may get one point if you carried it through correctly to the end of the problem. (I saw kind a few T_-2,-2.)

30. In the diagram below of triangle ABC, DE and DF are midsegments.

If DE = 9 and BC = 17, determine and state the perimeter of quadrilateral FDEC.

If DE is a midsegment then E is the midpoint of BC. That makes EC half of 17, which is 8.5. (Don't round!) Because DF is a midsegment, it is half the length of BC, so it is also 8.5. AC is twice as big as DE, so 2 X 9 = 18. CF is half of 18, which is 9 again. Quadrilateral FDEC is a parallelogram (hopefully, you knew that!), and the perimeter is the sum of its four sides: 9 + 9 + 8.5 + 8.5 = 35.

I saw a couple of 34 and 36 from people who apparently didn't like fractions or decimals.

31. The image of triangle ABC under a translation is triangle A'B'C'. Under this translation, B(3, -2) maps onto B'(1, -1). Using this translation, the coordinates of image A' are (-2, 2). Determine and state the coordinates of point A.

Note: They gave you A', and you need to work backward. First, calculate the translation: to go from 3 to 1, you subtract 2. To go from -2 to -1, you add 1. The transformation is T_{-2, 1}. You have to do the opposite to A' to get A. (-2 + 2, 2 - 1) gets you A(0, 1). The most common incorrect answer I saw was (-4, 3).

32. As shown in the diagram below, quadrilateral DEFG is inscribed in a circle and m<D = 86. [Diagram coming soon.]
Determine and state mGFE (arc)
Determine and state m<F.

Angle D is an inscribed angle. It measure 86 degrees, so the arc GFE is double that amount or 172 degrees.

Angle F is also an inscribe angle. If arc GFE is 172, then arc EDG is 360-172 = 188 degrees. The angle is half as much, which is 94 degrees.

Note that the shape inscribed in the circle is neither a parallelogram nor a trapezoid. It's just an oddly-shaped quadrilateral. If you made a mistake on part one, but were consistent throughout the problem, you got half credit.

33. In the diagram below, QM is a median of triangle PQR and point C is the centroid of triangle PQR.

If QC = 5x and CM = x + 12, determine and state the length of QM.

The centroid is two-thirds of the way from the angle to the midpoint of the opposite side. The equation you would write is 5x = 2(x + 12). Don't forget to double the expression on the right because it's only half as long.

5x = 2x + 24
3x = 24
x = 8
So QM = 5(8) + (8) + 12 = 40 + 8 + 12 = 60

34. The sum of the interior angles of a regular polygon is 540 degrees. Determine and state the number of degrees in one interior angle of the polygon.

You should have known that the polygon is a pentagon because the sum of the interior angles of a pentagon is 540. You could have calculated any way you wanted to if you didn't know, making a chart if you needed to. (I don't think you needed to. You could have just stated it as a fact because you can't learn about interior angles and not know a pentagon has 540 degrees.)

Divide 540 by 5 and you get 108 degrees, which you also probably knew before you did the problem, but you needed to show the work. Did you have to state or show that the polygon had five sides? I don't know. I would say "No". There was nothing in the scoring chart that says so (other than a note that if you calculated that it had 5 sides, but did no further work, you got one point.)

Part 3

35. Given: MT and HA intersect at B, MA || HT, and MT bisects BA.
Prove: MA = HT.

You had two choices here, one using AAS and another using ASA. Make sure you provided the correct reasons for the one you chose.
In paragraph form:
MA is parallel HT, which is given. This means that angle M is congruent to angle T AND angle A is congruent to angle H because they are pairs of alternate interior angles. MB = TB because MT is bisected by HA. So by AAS, triangle AMB is congruent to triangle HMT. Then by the definition of congruent polygons (or CPCTC, corresponding parts of congruent triangles are congruent), MA must be congruent to HT.

If you used the fact that angle MBA is congruent to angle TBH because they are vertical angles, you could have used ASA, instead of AAS.

If you thought that MT and HA bisected each other and that MB was congruent to TB, you are incorrect. However, if you made only this mistake and then used SAS (NOT SSA) to prove the triangles congruent and finished with CPCTC, you should have gotten at least half credit for the problem. I don't think that thinking the lines bisected each other is a "conceptual" mistake or just a logical one, so I can't say if you would lose one or two points on that.

36. A right circular cone has an altitude of 10 ft and the diameter of the base is 6 ft as shown in the diagram below. [COMING SOON] Determine and state the lateral area of the cone, to the nearest tenth of a square foot.

The good news: the formula was in the back of the book. Lateral area of a cone is L = Pi * r * l, where l is the slant height. Note that r is NOT squared.
Now the bad news: if you used r = 6 or l = 10, you messed up. If you did both, you probably scored zero points. The radius is 3, which is half the diameter. The height of the cone is not the slant height, which is the length from the tip of the cone to the edge of the base. The slant height can be found by making a right triangle with the height and the radius, which is actually shown in the picture. Using Pythagorean Theorem, we find that the slant height is the square root of 109. If you thought you made a mistake because it was so terrible, you didn't. I hope you didn't erase it, thinking you got it wrong.

The equation you should have written was L = (3.141592)(3)(108)^(1/2), which gives 98.4, to the nearest tenth.
Warning: if you used 3.14, instead of the pi key, you will be off by a tenth. Will you lose a point for that? Very likely, yes. Maybe not.

37. The construction question:

To divide the line into four equal segments using only a compass and straightedge (and NOT a ruler, measuring the line), you needed to construct a perpendicular bisector. Then construct a second between the left endpoint and the midpoint and finally construct a third one between the midpoint and the right endpoint. You have four equal pieces.

To be continued . . .

EDIT: continuing

38. I have to make another copy of the sketch of the locus problem. It was a little complicated, if only because they have NEVER asked a question like that in that way. I had told me review class the day before that questions like this one always have a vertical or a horizontal line, and to make sure they make the correct one. This one had a line with a slope of -1, and you have to write the equations.

I hope to have a sketch posted soon, but in the meantime:

The equation of the circle was (x - 3)² + (x + 2)² = 25
The two points given had a midpoint of (1, -5) and the locus was a line with a slope of -1. The equation of that line was y = -x - 4.
The line intersected the circle at the leftmost and bottommost points of the circle (-2, -2) and (3, -7), which you had to state, not simply mark with an X.

UPDATE: Here's a sketch. Obviously, your sketch doesn't have to have a perfect circle drawn. If you did, you might have noticed that the circle goes through (0, -6), but it is NOT one of the solutions.

The question stated On the set of axes, graph the locus of points 5 units from the point (3, -2). Write an equation that represents this locus. On the same set of axes, graph the locus of points equidistant from the points (0, -6) and (2, -4). Write an equation that represents this locus. State the coordinates of all points that satisfy both conditions.

If you sketched either incorrectly, you needed to state whatever points satisfied both conditions in your graph. If your graph didn't overlap, you had to say that No Points satisfied the conditions, and so on.

I don't know where all the points came from, but you basically got one point for each locus, one point for each equation, and one point from the two points satisfying the conditions. Where's the sixth point? I don't know, but if you had the other five, you had all the points.