Geometry Problems of the Day (Geometry Regents, June 2011)

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 2011

Part I: Each correct answer will receive 2 credits.


16. In rhombus ABCD, the diagonals AC and BD intersect at E. If AE = 5 and BE = 12, what is the length of AB?

1) 7
2) 10
3) 13
4) 17

Answer: 3) 13

The diagonals of a rhombus are perpendicular bisectors of each other. Drawing both of these will divide the rhombus into four congruent right triangles.

If the two legs of the right triangle (AE and BE) are 5 and 12, then using the Pythagorean Theorem, we know that

AB = SQRT( 5² + 12² ) = SQRT(25 + 144) = SQRT(169) = 13.

You should also recognize 5-12-13 as one of the most commonly-used Pythagorean Triple.

17. In the diagram below of circle O, PA is tangent to circle O at A, and PBC is a secant with points B and C in the circle.

If PA = 8 and PB = 4, what is the length of BC?

1) 20
2) 16
3) 15
4) 12

Answer: 4) 12

THe Tangent-Secant Theorem tells us that PA squared is equal to PB times PC.

PA is 8, and 8² = 64.

That means that (PB)(PC) = 64, so PC = 64/4 = 16.

BC = PC - PC = 16 - 4 = 12.

If you answered 16, you were incorrect because you didn't answer what the question asked, even though you did all the work.

18. Lines m and n intersect at point A. Line k is perpendicular to both lines m and n at point A. Which statement must be true?

1) Lines m, n, and k are in the same plane
2) Lines m and n are in two different planes.
3) Lines m and n are perpendicular to each other
4) Line k is perpendicular to the plane containing lines m and n.

Answer: 4) Line k is perpendicular to the plane containing lines m and n.

The only way that k can be perpendicular to two intersecting lines is if k is perpendicular to the plane that contains those two lines. Like a streetlight at a corner.

Choice (1) is impossible. For k to be perpendicular to two lines in one plane, those lines would have to be parallel. However, we are told that the lines intersect. Eliminate Choice (1).

Choice (2) is impossible. If lines m and n intersect, they lie on the same plane. Only skew lines lie on different planes. Since the lines intersect, they are NOT skew.

Choice (3) is not necessarily true. Using the streetlight example, the streets meeting at the intersection do not have to be perpendicular to each other although they may be.

Choice (4) is the answer.

19. In △DEF, m∠D = 3x + 5, m∠E = 4x − 15, and m∠F = 2x + 10. Which statement is true?

1) DF = FE
2) DE = FE
3) m∠E = m∠F
4) m∠D = m∠F

Answer: 1) DF = FE

The three angles must add up to 180 degrees. Find the size of each of the angles. From the choices given, you know that this will be an isosceles triangle. Two angles will be conruent and that makes their opposite sides congruent as well.

Notice that for DE to be equal to FE, then angle F would have to be congruent to angle D. That means that Choices (2) and (4) are either BOTH true (can't happen) or Both False. Eliminate (2) and (4).

3x + 5 + 4x − 15 + 2x + 10 = 180

9x + 0 = 180

x = 20

m∠D = 3(20) + 5 = 65, m∠E = 4(20) − 15 = 65, and m∠F = 2(20) + 10 = 50. 65 + 65 + 50 = 180 (check).

m∠D = m∠E (which is not a choice), so EF = DF, which is Choice (1).

20. As shown in the diagram below, △ABC ∼ △DEF, AB = 7x, BC = 4, DE = 7, and EF = x.

What is the length of AB?

1) 28
2) 2
3) 14
4) 4

Answer: 3) 14

The scale factor is 7x/7 or simply x.

This means that 4 is x².

You could also write a proportion:

7x / 4 = 7 / x

7x² = 28

x² = 4

So x = 2. That makes AB = 7x = 7(2) = 14.

More to come. Comments and questions welcome.

More Regents problems.

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