Geometry Problems of the Day (Geometry Regents, August 2010)

Now that I'm caught up with the current New York State Regents exams, I'm revisiting some older ones.

More Regents problems.

Geometry Regents, August 2010

Part III: Each correct answer will receive 4 credits. Partial credit is available.


35. In the diagram below of quadrilateral ABCD with diagonal BD, m∠A = 93, m∠ADB = 43, m∠C = 3x + 5, m∠BDC = x + 19, and m∠DBC = 2x + 6. Determine if AB is parallel to DC. Explain your reasoning.

Answer:

If m∠ABD = x + 19 then the two alternate interior angles will be equal and the lines must be parallel. So naturally, you have a lot of work to do to find the size of both of those angles.

To find m∠ABD, notice that you have two of the three angles in the triangle, which has a sum of 180 degrees. In triangle BCD, you have to solve 2x + 6 + 3x + 5 + x + 19 = 180 for x and then evaluate x + 19.

m∠ABD + 43 + 93 = 180
m∠ABD + 136 = 180
m∠ABD = 44

2x + 6 + 3x + 5 + x + 19 = 180
6x + 30 = 180
6x = 150
x = 25

x + 19 = 25 + 19 = 44

Because the alternate interior angles are congruent, line AB and CD must be parallel.

36. The coordinates of the vertices of △ABC are A(1,3), B(−2,2), and C(0,−2). On the grid below, graph and label △A″B″C″, the result of the composite transformation D₂ ∘ T_3,−2. State the coordinates of A″, B″, and C″.

Answer:

The most important thing to remember here is that D₂ ∘ T_3,−2 means "the dilation of the translation". That means that you must do the translation FIRST and then do the dilation.

You must graph this for full credit. However, algebraically, these are the points:

A(1,3) -> A'(1+3,3-2) = A'(4,1) -> A"(4*2,1*2) = A"(8,2)

B(-2,2) -> B'(-2+3,2-2) = B'(1,0) -> B"(1*2,0*2) = B"(2,0)

C(0,-2) -> C'(0+3,-2-2) = C'(3,-4) -> C"(3*2,-4*2) = C"(6,-8)

You graph should look like the following. The points MUST be labels. The coordinates can be written on the graph or on the side of the graph. For instance, if three lines above are written on your paper, than just A", B" and C" are sufficient on the graph itself.

You may get credit if you get A'B'C' correct but do not finish correctly. However, A'B'C' is not required for full credit. It's just a good idea to include it.

37. In the diagram below, △RST is a 3-4-5 right triangle. The altitude, h, to the hypotenuse has been drawn. Determine the length of h.

Answer:

Label the point where the altitude intersects RS as point U. There are three right triangles: RST, RTU, and STU. These three triangles are similar and their corresponding sides are proportional.

We can find the length of b by comparing short/hypotenuse in RTU and RST:

b/3 = 3/5

5b = 9

b = 9/5

This means that a = 5 - b = 5 - 9/5 = 25/5 - 9/5 = 16/5.

The Right Triangle Altitude Theorem tells us that h² = (b)(a).

h² = (9/5)(16/5) = 144/25

h = 12/5

You could also have had the decimal equivalents a = 3.2, b = 1.8, and h = 2.4.

End of Part III.

More to come. Comments and questions welcome.

More Regents problems.

I also write Fiction! You can now order Devilish And Divine, edited by John L. French and Danielle Ackley-McPhail, which contains (among many, many others) three stories by me, Christopher J. Burke about those above us and from down below. Order the softcover or ebook at Amazon. Also, check out In A Flash 2020, by Christopher J. Burke for 20 great flash fiction stories, perfectly sized for your train rides. Available in softcover or ebook at Amazon. If you enjoy it, please consider leaving a rating or review on Amazon or on Good Reads.