This exam was adminstered in June 2023.
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__June 2023 Geometry Regents__

__Part I __

Each correct answer will receive 2 credits. No partial credit.

*17. The measure of one of the base angles of an isosceles triangle is 42°.
The measure of an exterior angle at the vertex of the triangle is
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(1) 42°

(2) 84°

(3) 96°

(4) 138°

**Answer: (2) 84° **

If one base angle is 42° then the second base angle is also 42°. According to the Remote Angle Theorem, the measure of the exterior angle is equal to the sum of the two remote angles, which in the case of the vertex angle, the remote angles would be the two base angles. That sum is 84°, so **Choice (2) is the correct answer**.
If you forgot about the Remote Angle Teheorem, you could have used the sum of 84° to find the size of the vertex angle, which is 180 - 84 = 96°. Then the exterior angle would be supplementary to the vertex angle, so 180 - 96 = 84°. You not only got the same answer, but you just demonstrated the Remote Angle Theorem.

Two supplementary angles, A and B, add up to 180 degrees. The three angles of a triangle. B, C, and D, add up to 180 degrees. This means that A + B = B + C + D, so if you remove B (which in this problem is the vertex angle), you are left with A = C + D. This is Remote Angle Theorem.

*18. In the diagram below, AFKB || CHLM, FH ≅ LH, FL ≅ KL, and LF bisects ∠HFK.
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Which statement is always true?

(1) 2(m∠HLF) = m∠CHE

(2) 2(m∠FLK) = m∠LKB

(3) m∠AFD = m∠BKL

(4) m∠DFK = m∠KLF

**Answer: (4) m∠DFK = m∠KLF **

Look at the choices one at a time.
Angle HLF is one of two base angles and is congruent to angle HFL. Angle FHL (and therefore angle CHE) will equal 180 - 2(m∠HLF), not 2(m∠HLF). Eliminate Choice (1).

Angle FLK is the vertex angle of isosceles triangle FLK. LKB is an exterior angle. That means that, according to the Remote Angle Theorem that the sum of KFL and FLK equals the measure of LKB. For Choice (2) to be true, angle KFL must be congruent to angle FLK and there is no reason to suggest that they always will be. Eliminate Choice (2).

Angles AFD and BKL have no relationship between them DE and KL are not given to be parallel. Eliminate Choice (3).

Angles DFK and KLF look like they have no relationship, but they do. You may want to mark the illustration with this information. Angles HFL and LFK are congruent becuase LF bisects HFK. Angles LFP and LKF are congruent because triangle FKL is isosceles (with FL ≅ LK). This means that 2(LFK) + DFK = 180 degrees because that's the measure of a striaght line. And 2(LFK) + KLF = 180 because that's the measure of a traingle. Therefore, DFK must be congurnet to KLF. A little convoluted for a multiple-choice queston, but **the correct answer is Choice (4)**.

*19. The line whose equation is 6x + 3y = 3 is dilated by a scale factor of
2 centered at the point (0,0). An equation of its image is
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(1) y = -2x + 1

(2) y = -2x + 2

(3) y = -4x + 1

(4) y = -4x + 2

**Answer: (2) y = -2x + 2 **

When you dilate a line, the image will either be the same line or a line parallel to the original line. The slope will not change. The y-intercept will only change if the center of dilation is a point that is not on the pre-image.
First, find the slope of the original line. Since it is in Standard Form, Ax + By = C, the slope is given by the formula m = -A/B = -6/3 = -2.

If you didn't remember how to find the slope that way, then you have to rewrite it in slope-intercept form.

6x + 3y = 3

3y = -6x + 3

y = -2x + 1

Elimnate Choice (3) and (4).

So (0,1) is a point on the pre-image. If that point is dilated by a factor of 2, centered on the origin, the image will be at (0,2), which will be the y-interecept of the image. Then **the correct answer is Choice (2)**.

*20. Which figure will not carry onto itself after a 120-degree rotation
about its center?
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(1) equilateral triangle

(2) regular hexagon

(3) regular octagon

(4) regular nonagon

**Answer: (3) regular octagon **

Consider that 120 degrees is 1/3 of a full rotation.

If an equilateral triangle rotates 1/3 of the way, it will carry onto itself because it has 3 sides.

If a regular hexagon rotates 1/3 of the way, two sides will have rotated around and it will still look the same.

If a regular nonagon (which has nine sides) rotates 1/3 of the way, three sides will have rotated and it will still look the same.

An octagon, however, has eight sides, so if it rotates 1/3 of the way, it will not look the same because 1/3 * 8 is not a whole number.

The correct choice is Choice (3).

*21.Triangle ADF is drawn and BC || DF.
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Which statement must be true?

(1) AB/BC = BD/DF

(2) BC = 1/2 DF

(3) AB:AD = AC:CF

(4) ∠ACB ≅ ∠AFD

**Answer: (4) ∠ACB ≅ ∠AFD **

Because BC || DF, the small triangle is similar to the larger triangle and the corresponding sides are proportional.

Choice (1) and (3) show incorrect proportions because in each case there is are non-corresponding sides. In Choice (1), if BD said AD, or in Choice (3), if CF said AF, the proportions would have been correct. Eliminate Choice (1) and (3).

Choice (2) looks like a good answer EXCEPT it is only true if B and C are the midpoints of AD and AF, respectively, which would make BC a midsegment. We are not given this piece of information and we cannot assume it. Eliminate Choice (2).

Since the two triangles have the same shape, their corresponding angles are congruent. **The correct answer is Choice (4)**.

*22. In △ABC, M is the midpoint of AB and N is the midpoint of AC.
If MN = x + 13 and BC = 5x - 1, what is the length of MN?
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Which statement is always true?

(1) 3.5

(2) 9

(3) 16.5

(4) 22

**Answer: (4) 22 **

The thing to remember here is that BC is NOT congruent to MN. MN is a midsegment, so it is HALF the length of BC. That is, BC is TWICE the length of MN.

Write that as an equation. Then substitute and solve.

BC = 2(MN)

5x - 1 = 2(x + 13)

5x - 1 = 2x + 26

3x = 27

x = 9

If x = 9, which is NOT the final answer, then MN = x + 13 = 9 + 13 = 22. That is **Choice (4), which is the correct answer.**

If you forgot to multiply by 2, then you would have gotten 3.5 for x and 16.5 for MN. Incorrect.

*23. In the diagram below of isosceles trapezoid STAR, diagonals AS and RT intersect at O and ST || RA, with nonparallel sides SR and TA.
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Which pair of triangles are not always similar?

(1) △STO and △ARO

(2) △SOR and △TOA

(3) △SRA and △ATS

(4) △SRT and △TAS

**Answer: (3) △SRA and △ATS **

The legs of the isosceles trapezoid are congruent and the diagonals are congruent. The bases are not congruent. The bases are parallel, which creates alternate interior angles that are congruent. And the diagonals form vertical angles that are congruent.

In Choice (1), the top and bottom triangles are always similar because of the alternate interior angles and the vertical angles.

In Choice (2), the left and right can be show to be not only similar but congruent using AAS.

In Choice (3), the left and the bottom together is not similar to the top and right together. They cannot be. Because SR ≅ TA, then ST must be congruent to AR for the corresponding sides to be proportional. However, we know that ST is NOT congruent to AR, because then we would have a parallelogram instead of a trapezoid.** Choice (3) is the correct answer.**

In Choice (4), these two triangles can be shown to be congruent (and therefore similar) through SAS.

*24. The endpoints of AB are A(0,4) and B(–4,6). Which equation of a line represents the perpendicular bisector of AB?
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(1) y = -1/2 x + 4

(2) y = -2x + 1

(3) y = 2x + 8

(4) y = 2x + 9

**Answer: (4) y = 2x + 9 **

The perpendicular bisector has a slope that is the inverse reciprocal to the line connecting (0,4) and (-4,6). It also must contain the midpoint of AB, which is (-2,5), which is half the distance from A to B.

(6 - 4) / (-4 - 0) = 2/-4 = -1/2

Slope of AB = -1/2

Perpendicular slope = 2

The slope of the bisector is 2, so y = mx + b, and 5 = (2)(-2) + b.

5 = -4 + b, so b = 9.

End of Part I. How did you do?

Comments and questions welcome.

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