Geometry Problems of the Day (Geometry Regents, June 2013)

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, June 2013

Part I: Each correct answer will receive 2 credits.


1. 1 In trapezoid RSTV below with bases RS and VT, diagonals RT and SV intersect at Q.

If trapezoid RSTV is not isosceles, which triangle is equal in area to triangle RSV?

1) RQV
2) RST
3) RVT
4) SVT

Answer: 2) RST

To have the same area, the two triangles could be congruent, or they could just have the same size base and the same height.

Triangle RSV shares a base with triangle RST. Both triangles have a height equal to the distance between the parallel lines. Since they have the same base and height, their areas are equal.

2. In the diagram below, triangle XYV ≅ triangle TSV.

Which statement can not be proven?

1) ∠XVY ≅ ∠TVS
2) ∠VYX ≅ ∠VUT
3) XY ≅ TS
4) YV ≅ SV

Answer: 2) ∠VYX ≅ ∠VUT

A pentagon has 540 interior degrees, which is 180 * 3. A pentagon (like every other polygon) has a total of 360 exterior degrees. The difference between the two is 540 - 360 = 180.

Choice (1) shows vertical angles, which are always congruent.

Choice (2) shows two non-corresponding angles that are not related. This is the correct response.

Choice (3) can be shown by proving the two triangles congruent using ASA.

Choice (4) shows two corresponding sides that are congruent because of CPCTC.

3. In a park, two straight paths intersect. The city wants to install lampposts that are both equidistant from each path and also 15 feet from the intersection of the paths. How many lampposts are needed?

1) 1
2) 2
3) 3
4) 4

Answer: 4) 4

The locus of points that are equidistant from two intersecting paths would be a pair of perpendicular lines that goes through the intersection point of the paths. Fifteen feet from this intesection would be a circle.

The circle would intersect the perpendicular lines 4 times. Draw a sketch and you'll see.

4. What are the coordinates of A', the image of point A(-3,4), after a rotation of 180° about the origin?

1) (4, 3)
2) (-4, -3)
3) (3, 4)
4) (3, -4)

Answer:4) (3, -4)

The preimage is in Quadrant II. If you rotate 180 degrees, you will end up in Quadrant IV. The signs in Quadrant IV are (+, -).

Choice (4) is the only possible choice.

When you rotate 180 degrees, the coordinates (x, y) become (-x, -y). If they are already negative, they will flip and become positive.

5. Based on the construction below, which conclusion is not always true?

1) AB ⊥ CD
2) AB = CD
3) AE = DE
4) CE = DE

Answer: 2) AB = CD

The illustration shows a perpendicular bisector being constructed, which means that (1) and (3) will always be true.

Choice (4) is a byprouduct of the construction. Draw AC and BC and you will have two congruent triangles, using the Hypotenuse Leg Theorem.

Choice (2) is not true. The size of CD depends upon the size of the arc used in the construction, which is NOT based on the length of AB, except that it has to be more than 1/2 of AB.

When you rotate 180 degrees, the coordinates (x, y) become (-x, -y). If they are already negative, they will flip and become positive.

More to come. Comments and questions welcome.

More Regents problems.

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