This exam was adminstered in August 2025.
More Regents problems.
August 2025 Geometry Regents
Part I
Each correct answer will receive 2 credits. No partial credit.
1. An equilateral triangle is continuously rotated around one of its altitudes. The three-dimensional object formed is a
(1) cone
(2) sphere
(3) cylinder
(4) pyramid
Answer: (1) cone
Imagine an equilateral triangle taped to a lollipop stick that running from one corner and through the midpoint of the opposite side. If you were to spin the stick, the stuff that the triangle will pass through will form a cone. That is Choice (1).
It is impossible to get any of the others from a triangle.
Side BD will always map onto
2. On the set of axes below, quadrilateral BDGF is rotated 90 degrees clockwise about the origin and then reflected over the y-axis. The image of quadrilateral BDGF is quadrilateral MQSP.
(1) MP
(2) PS
(3) MQ
(4) SQ
Answer: (3) MQ
Follow the movement. When BDGF is rotated 90 degree clockwise, the image appears in Quadrant IV with BD on top, closest to the x-axis. When this image is reflected across the y-axix, B'D' maps onto MQ, which is Choice (3).
3. In right triangle JOE, hypotenuse JO = 31.8 and m∠J = 38°. To the nearest tenth, the length of EJ is
(1) 19.6
(2) 25.1
(3) 40.4
(4) 51.7
Answer: (2) 25.1
Angle E is the right angle. EJ is adjacent to angle J, and you're given the hypotenuse. That means that you need to use cosine.
cos 38 = x / 31.8
x = 31.8 cos 38 = 25.0587... = 25.1
Choice (2) is the correct answer.
4. The hemisphere below has a radius of 8 cm.
To the nearest cubic centimeter, the volume of the hemisphere is
(1) 201
(2) 268
(3) 1072
(4) 2145
Answer: (3) 1072
Use the formula for the Volume of a Sphere and then take half of it. The formula is V = (4/3) π r3.
So V = (1/2) (4/3) π (8)3 = 1072.33...
Choice (3) is the correct choice.
If you'd forgotten the 1/2, you would've gotten choice (4).
If you'd forgotten the 1/2 AND halved the radius to 4, you would've gotten Choice (2).
If you used π r2, you would've gotten Choice (1).
Yes, I needed to figure that one out. I wanted to know!
5. In parallelogram ABCD, diagonals AC and BD intersect at E. Which information is sufficient to prove ABCD is a rhombus?
(1) AE ≅ EC
(2) AC ≅ BD
(3) AB ⊥ BC
(4) AC ⊥ BD
Answer: (4) AC ⊥ BD
In a parallelogram, if the two diagonals are perpendicular, then the parallelogram is a rhombus.
6. Trapezoid JOSH, shown below, has non-parallel sides JH and OS, m∠J = 65°, m∠O = 30°, m∠OSA = 80°, and m∠SHU = 60°.
(1) 55°
(2) 60°
(3) 65°
(4) 70°
Answer: (4) 70°
SAO is a triangle, and the sum of its angles is 180 degrees. Because m∠O = 30° and m∠OSA = 80°, then m∠OAS = 70°. Angle OAS and angle HSA are alternate interior angles. Since JO || SH, then ∠OAS ≅ ∠OSA.
Therefore, m∠OSA = 70°, which is Choice (4).
If AD = 8, DB = 4, and DE = 6, what is the length of AC
7. In △ABC below, points D and E are on such that AC. DE || AC.
(1) 24
(2) 18
(3) 12
(4) 10
Answer: (2) 18
The two triangles are have sides that are proportional because they are similar triangles. If AD = 8 and DB = 4, then AB = 12, and the scale factor is 3.
Therefore, if DE = 6 then AC = 6 * 3 = 18.
Choice (2) is the correct answer.
Which equation represents circle C?
8. On the set of axes below, circle C has a center with coordinates (2,–1).
(1) (x - 2)2 + (y + 1)2 = 25
(2) (x - 2)2 + (y + 1)2 = 16
(3) (x + 2)2 + (y - 1)2 = 25
(4) (x + 2)2 + (y - 1)2 = 16
Answer: (1) (x - 2)2 + (y + 1)2 = 25
The formula for the equation of a circle is (x - h)2 + (y - k)2 = r2, where (h,k) is the coordinates of the center and r is the length of the radius.
The center point C is at (1,-2), so eliminate Choices (3) and (4).
If you count across from C, you'll see that the radius is 5. (5)2 = 25, so the correct answer is Choice (1).
More to come. Comments and questions welcome.
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