Thursday, October 28, 2021

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

35. Given: AD bisects BC at E, AB ⊥ BC, DC ⊥ BC.

Prove: AB ≅ DC


To prove that AB ≅ DC, you first need to show that the two triangles are congruent. We know that both triangles have right angles and vertical angles and that E is a midpoint of BC. That means that you can use ASA to show the triangles are congruent.

Start with all the Given information. Your last line MUST be what you are proving.

1. AD bisects BC at E, AB ⊥ BC, DC ⊥ BC1. Given
2. Angle BEA ≅ Angle CED2. Vertical angles are congruent.
3. BE ≅ CE3. Definition of midpoint.
4. Angle B ≅ = Angle C4. All right angles are congruent.
5. Triangle ABE ≅ Triangle DCE5. ASA
6. AB ≅ DC6. CPCTC

36. TThe coordinates of trapezoid ABCD are A(-4,5), B(1,5), C(1,2), and D(-6,2). Trapezoid A"B"C"D" is the image after the composition rx-axis ° ry = x is performed on trapezoid ABCD. State the coordinates of trapezoid A"B"C"D".
[The use of the set of axes below is optional.


In a composition of transformations, you work from right to left. You are doing a reflection over the x-axis OF THE IMAGE of the reflection over y = x.

Without using the grid, here are the rules for reflections:

Reflection over the line y = x: (x, y) --> (y, x)

Reflection over x-axis (x, y) --> (x, -y)

You could conceivably do this in one step. DON'T. Show the middle step.

A(-4,5), B(1,5), C(1,2), and D(-6,2)

A'(5,-4), B'(5,1), C'(2,1), and D'(2,-6)

A"(5,4), B"(5,-1), C"(2,-1), and D"(2,6)

37. In the diagram below of circle O, chords RT and QS intersect at M. Secant PTR and tangent PS are drawn to circle O. The length of RM is two more than the length of TM, QM = 2, SM = 12, and PT = 8.

Find the length of RT.
Find the length of PS.


Rules you need to know:
(RM)(MT) = (QM)(MS)
(PT)(PR) = (PS)2

We know that QM = 2 and SM = 12. We also know that RM = TM + 2. (Be careful -- I first read that as "two times" not "two more than"!)


(x)(x + 2) = (2)(12)
x2 + 2x = 24
x2 + 2x - 24 = 0
(x + 6)(x - 4) = 0
x + 6 = 0 or x - 4 = 0
x = -6 or x = 4
Discard the negative answer.

Since x = 4, the TM = 4 and RM = 4 + 2 = 6, so RT = 10

PT = 8 and RT = 10, so PR = 18. I used x, so let y = PS

y2 = (8)(18)
y2 = 144
y = 12

So PS = 12.

End of Part III.

More to come. Comments and questions welcome.

More Regents problems.

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