Geometry Problems of the Day (Geometry Regents, June 2012)

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 2012

Part I: Each correct answer will receive 2 credits.


21. In circle O shown below, diameter DB is perpendicular to chord AC at E.

If DB = 34, AC = 30, and DE > BE, what is the length of BE?

1) 8
2) 9
3) 16
4) 25

Answer: 2) 9

The first thing you should do is draw OA and OC. These are both radii with a length of 17, which is half of 34. Since the chord is perpendicular to the diameter, then OEA and OEC are both right triangles, each with a leg of 15 and a hypotenuse of 17.

If you didn't recognize the Pythagorean Triple, then you can work out

(OE)² + 15² = 17²
which will tell you that OE = 8 because 8-15-17 is a triple.

However, the question didn't ask for OE. It asked for BE. BE = OB - OE = 17 - 8 = 9.

22. In parallelogram ABCD shown below, diagonals AC and BD intersect at E.

Which statement must be true?

1) AC ≅ DB
2) ∠ABD ≅ ∠CBD
3) triangle AED ≅ triangle ∠CED
4) triangle DCE ≅ triangle ∠BCE

Answer:3) triangle AED ≅ triangle ∠CED

The diagonals of a parallelogram bisect each other. They are NOT congruent unless you know that the parallelogram is also a rectangle. You can prove using SSS that the opposite triangles are congruent -- that is, the ones that share an entire diagonal or the ones on the opposites sides of the vertex E.

Choice (1) MAY be true if ABCD is a rectangle, but it is NOT true otherwise. Eliminate Choice (1).

Choice (2) MAY be true if ABCD is a rhombus where the diagonals are angle bisectors, but it is NOT true otherwise. Eliminate Choice (2).

Choice (3) can be shown to be true using SSS -- the opposite sides of the parallelogram are congruents and the diagonals bisect each other. This is the answer.

Choice (4) cannot be shown to be true with the information given. It would only be true for a rhombus.

23. Which equation of a circle will have a graph that lies entirely in the first quadrant?

1) (x - 4)² + (y - 5)² = 9
2) (x + 4)² + (y + 5)² = 9
3) (x + 4)² + (y + 5)² = 25
4) (x - 5)² + (y - 4)² = 25

Answer: 1) (x - 4)² + (y - 5)² = 9

They changed up the question a little bit. It's nice to see some variety.

For the circle to be in the first quadrant, it would have to have a center (h,k) in the first quadrant, and the radius would have to be smaller than either h or k.

Choice (1) has a center at (4,5) because the signs are flipped. It has a radius of 3. This means that it has no points to the left of (1,5) or below (4,2). This is the solution.

Choice (2) has its center in Quadrant III at point (-4, -5).

Choice (3) has its center in Quadrant III at point (-4, -5).

Choice (4) has its center at (5,4) but the radius is 5. That means that (5,-1) is on the circle, and that is in Quadrant IV.

24. In the diagram below, triangle ABC ∼ triangle RST.

Which statement is not true?

1) ∠A ≅ ∠R
2) AB/RS = BC/ST
3) AB/BC = ST/RS
4) (AB + BC + AC) / (RS + ST + RT) = AB/RS

Answer: 3) AB/BC = ST/RS

The corresponding angles of similar triangles are congruent. The corresponding sides over of similar triangles are proportional to the scale factor.

Choice (1) says that two corresponding angles are congruent. This is true. Eliminate (1).

Choice (2) shows a proportion of two pairs of corresponding sides. This is true. Eliminate (2).

Choice (3) has a incorrect proportion. The second ratio should be RS/ST, not ST/RS. This is the answer.

Choice (4) shows that the ratio of the two perimeters is the same as the ratio of two corresponding sides, which is the scale factor. This is true. Eliminate (4).

If you need to, you could convert to slope-intercept form:

20x - 2y = 6
-2y = -20x + 6
y = 10x - 3

The slope of a line perpendicular to a line with a slope of 10 would be -1/10.

More to come. Comments and questions welcome.

More Regents problems.

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