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.


11. In the diagram of triagnle ABC shown below, D is the midpoint of AB, E is the midpoint of BC, and F is the midpoint of AC.

If AB = 20, BC = 12, and AC = 16, what is the perimeter of trapezoid ABEF?

1) 24
2) 36
3) 40
4) 44

Answer: 4) 44

Add all of AB to half of AC and half of BC. Then add half of AB, which is the length of EF. So 20 + 6 + 8 + 10 = 44

Choice (1) is the size of triangle DEF which connects the midpoints.

Choices (2) and (3) are the perimeters of the other two trapezoids.

12. In the diagram below, LMO is isosceles with LO = MO.

If m∠L = 55 and m∠NOM = 28, what is m∠N?

1) 27
2) 28
3) 42
4) 70

Answer: 1) 27

Since m∠L = 55 then m∠LMO = 55. This makes m∠OMN = 125 becuase it is supplementary.

Triangle MNO has 180 degrees (like all triangles), so m∠N = 180 - 28 - 125 = 27

Choice (2) is there in case you thought MNO was isosceles. There is nothing that suggests that it is (because it is NOT).

13. If AB is contained in plane P, and AB is perpendicular to plane R, which statement is true?

1) AB is paralel to plane R.
2) Plane P is parallel to plane R.
3) AB is perpendicular to plane P.
4) Plane P is perpendicular to plane R.

Answer: 4) Plane P is perpendicular to plane R.

AB is in P. AB is ⊥ R. Therefore P ⊥ R.

It's that simple.

If it were more complicated than that, you would need a lot more information to make a decision. But it isn't and you don't.

14. In the diagram below of triangle ABC, AE ≅ BE, AF ≅ CF, and CD ≅ BD.

Point P must be the

1) centroid
2) circumcenter
3) incenter
4) ortocenter

Answer: 1) centroid

Points D, E, and F are all midpoints. That makes AB, BC, and CA medians. Medians meet at the centroid.

The circumcenter is the concurrence point for three perpendicular bisectors of a triangle. AD, BF, and CE are NOT shown to be perpendicular to the lines they bisect.

The incenter is the concurrence point for three angle bisectors of a triangle.

The orthocenter is the concurrence point for three altitudes of a triangle.

15. What is the equation of the line that passes through the point (-9,6) and is perpendicular to the line y = 3x - 5?

1) y = 3x + 21
2) y = -1/3 x - 3
3) y = 3x + 33
4) y = -1/3 x + 3

Answer: 4) y = -1/3 x + 3

Perpendicular lines have slopes that are inverse reciprocals. Parallel lines have the same slopes.

The given line has a slope of 3. Choices (1) and (3) are lines with a slope of 3, so they are parallel to the given line. Eliminate them.

The slope of the perpendicular line must be -1/3. Check (-9,6) to see if it is a point on either line.

y = -1/3(-9) - 3 = 3 - 3 = 0. (-9, 6) is not on this line.

y = -1/3(-9) + 3 = 3 + 3 = 6. (-9, 6) IS a point on this line.

More to come. Comments and questions welcome.

More Regents problems.

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