Geometry Problems of the Day (Geometry Regents, June 2025 Part I)

This exam was adminstered in June 2025.

More Regents problems.

June 2025 Geometry Regents

Part I

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


1. The perimeter of a triangle is 18. What is the perimeter of a similar triangle after a dilation with a scale factor of 3?

(1) 6
(2) 18
(3) 54
(4) 162

Answer: (3) 54

The perimeter of a dilated triangle is the perimeter of the original times the scale factor, so 18 times 3 equals 54, which is Choice (3).

It is not the scale factor squared -- that would be the area, which has two dimensions being expanded.

2. The Washington Monument, shown below, is in Washington, D.C. At a point on the ground 200 feet from the center of the base of the monument, the angle of elevation to the top of the monument is 70.19°.

What is the height of the monument, to the nearest foot?

(1) 188
(2) 213
(3) 555
(4) 590

Answer: (3) 555

If you've ever been there, then you might know that it is 555 feet tall, and they did NOT change this fact for this problem. If you didn't know that, you can calculate it using the tangent ratio, because you have the angle and the adjacent side and you are looking for the opposite side.

tan 70.19 = x / 200
x = 200 tan 70.19 = 555.217...,

which is about 555, which is Choice (3).

3. On the set of axes below, △EQA and △SUL are graphed.

Which sequence of transformations shows that △EQA ≅ △SUL?

(1) Rotate △EQA 90° counterclockwise about the origin and then translate 9 units right and 1 unit down.
(2) Rotate △EQA 90° counterclockwise about the origin and then reflect over the line x = 4.
(3) Reflect △EQA over the x-axis and then rotate 90° clockwise about the origin.
(4) Translate △EQA 10 units right and then reflect over the line x = -1.

Answer: (1) Rotate △EQA 90° counterclockwise about the origin and then translate 9 units right and 1 unit down.

Make sure you are going the correct direction: you want to go from the top left to the bottom right. There are multiple methods of getting there, so check the choices one by one.

In Choice (1), EQA goes to Quadrant III with E'(-5,-2), Q(-1,-2), A'(-1,-5), which is facing the same direction as SUL. A transformation of T_9,-1 brings AEQ to SUL. This is the correct answer.

In Choice (2), EQA goes to Quadrant III with E'(-5,-2), Q(-1,-2), A'(-1,-5), which is facing the same direction as SUL. A reflection would change the orientation and would not map onto SUL. Eliminate Choice (2).

In Choice (3), EQA goes to Quadrant I but changes its orientation. When E'Q'A' is rotated 90 degrees, it will not map ontol SUL because the orientations are different. Eliminate Choice (3).

In Choice (4), EQA goes to Quadrant IV but will be oriented with point E pointing up. A reflection will not rotate the triangle to look like SUL. Eliminate Choice (4).

4. If two sides of a triangle have lengths of 2 and 8, the length of the third side could be

(1) 10
(2) 7
(3) 6
(4) 4

Answer: (2) 7

The Triangle Inequality Theorem states that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side.

If two sides are 2 and 8, then then third side must be less than 10. If two sides are 2 and 8, then the third side must be greater than 6, because 6 + 2 = 8.

This eliminates Choice (1), (3), and (4).

Choice (2) 7 is between 6 and 10, so it is a reasonable length. This is the correct answer.

5. A regular octagon is rotated about its center. Which angle measure will carry the octagon onto itself?

(1) 36°
(2) 90°
(3) 144°
(4) 160°

Answer: (2) 90°

A full rotation is 360°. During that rotation, the octagon will carry onto itself 8 times at increments of 360°/8 = 45°.

Of the four choices, only 90 is a multiple of 45, so the correct answer is Choice (2).

6. The equation of a circle is x² + y² - 6x + 2y = 14. What are the coordinates of the center and the length of the radius of this circle?

(1) (-3,1) and r = 5
(2) (3,-1) and r = 5
(3) (-3,1) and r = √(24)
(4) (3,-1) and r = √(24)

Answer: (4) (3,-1) and r = √(24)

A reminder that old exams not only show you the type of problems you may see but sometimes THE SAME NUMBER QUESTION will have THE SAME TYPE OF QUESTION. This was the same question that was asked in January with slightly different numbers, and it was Question #6 then, too.

To find the center of the circle, you need to complete the squares, twice, and find the square root of the polynomials.

x² + y² - 6x + 2y = 14
x² - 6x + y² + 2y = 14
x² - 6x + (3)² + y² + 2y + (1)² = 14 + 9 + 1
(x - 3)² + (y + 1)² = 24
(x - 3)² + (y + 1)² = (√(24))²

The correct answer is Choice (4) (3,-1) and r = √(24).

7. In △HSF below, m∠S = 90°, HF = 30, and FS = 23.

What is m∠F, to the nearest degree?

(1) 53°
(2) 50°
(3) 40°
(4) 37°

Answer: (3) 40°

You are looking for an angle in a right triangle and you are given the adjacent side and the hypotenuse, so you need to use the cosine ratio.

cos F = 23/30

F = cos^-1(23/30) = 39.94°

Choice (3) is the correct answer.

8. In nCAB below, midsegments DE, EF, and FD are drawn.

If CA = 14, CB = 20, and FB = 9, what is the perimeter of quadrilateral DEFA?

(1) 26
(2) 32
(3) 44
(4) 52

Answer: (2) 32

Midsegments are half the length of the side of the triangle that they are parallel to. Therefore, AD = 7 and EF = 7. Since FB = 9, then AF = 9 and DE = 9.

Thus the perimeter of DEFA is 9 + 7 + 9 + 7 = 32, which is Choice (2).

Not that they asked by DEFA is also a parallelogram.

More to come. Comments and questions welcome.

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