← SOMATH Blog · Regents Prep NYC · Updated July 28, 2026

Regents Geometry June 2026 — Part I Answers & Explanations

Free worked solutions for every Part I question on the June 2026 New York State Regents Geometry exam. Every answer is verified, every explanation names the theorem it tests, and every question includes a click-to-reveal button so students can try each item first. This is Part I of a four-post series covering all 35 questions.

📄 Original NYSED exam (PDF)

All figures, diagrams, and reference sheet are in the official New York State Education Department release. Open it in a second tab so you can see the figures as you work through the questions below.

Download the June 2026 Geometry Regents PDF

How to use this set. Work each question with pencil and the original NYSED PDF open. Circle your answer, then click Answer to check your reasoning. If you miss a question, read the Theory box — that is the single Geometry idea you need to relearn before test day.

Sections

  1. About the Geometry Regents exam
  2. Topics tested in Part I
  3. Part I — Questions 1–24 (click to reveal)
  4. Part I answer key
  5. Big ideas to remember
  6. More Regents Geometry help

About the Geometry Regents exam

The NYS Regents Examination in Geometry is a standardized end-of-course exam administered by the New York State Education Department three times a year (January, June, and August). It aligns to the Next Generation Mathematics Learning Standards. Every high-school student pursuing a New York State Regents Diploma must pass a math Regents; Geometry is the most common option after passing Algebra 1.

Structure: 35 questions, 80 total credits, 3-hour time limit.

Passing: a scale score of 65 is the passing standard; a scale score of 85 earns Mastery in Math, which is required for the Advanced Regents Diploma with Mastery. The raw-to-scale conversion chart is released after each administration.

Topics tested in Part I

The June 2026 Part I is a fair, standards-aligned Geometry test that touches every strand of the course. Here is what the 24 multiple-choice items assess.

Rigid motions and symmetry

Reflections, rotations, translations, and glide reflections preserve length and angle measure. A vertical stretch or a dilation with scale factor different from 1 is not rigid, so it changes area. A regular polygon has rotational symmetry of order equal to its number of sides.

Similarity, dilations, and midsegments

A dilation multiplies every distance from the center by the same scale factor. Lines through the center of dilation map to themselves. The midsegment theorem says the segment joining the midpoints of two sides of a triangle is parallel to the third side and half its length.

Right-triangle trigonometry

Sine, cosine, and tangent relate an acute angle to the opposite, adjacent, and hypotenuse. Two identities you must know: sin(A) = cos(90°−A) (co-function) and the geometric-mean altitude-on-hypotenuse relations for right triangles.

Circles

Central-angle arc length is (θ/360)·2πr. The standard form of a circle is (x−h)² + (y−k)² = r²; complete the square to convert general form. The tangent-secant power of a point says t² = a·b.

Coordinate geometry

Slope, midpoint, distance, and section-formula (partitioning a segment in a given ratio) are Part I staples. A pair of lines is parallel when they share the same slope but different y-intercepts.

3-D solids and cross-sections

Volume formulas: prism V = Bh, cylinder V = πr²h, cone V = (1/3)πr²h, pyramid V = (1/3)Bh. Rotating a right triangle about a leg produces a cone. A cross-section of a rectangular prism can have at most 6 sides.

Triangle theorems

The triangle inequality: any side is greater than the difference and less than the sum of the other two. A point on a segment's perpendicular bisector is equidistant from its endpoints. Similar triangles have proportional corresponding sides (and, therefore, proportional perimeters).

Part I — Questions 1–24

2 credits each · click any card to reveal the answer and worked explanation.

Question 1

Which transformation on quadrilateral ABCD would not preserve area?

  1. a reflection over the x-axis
  2. a translation two units right and three units up
  3. a rotation of 90° counterclockwise about the origin
  4. a vertical stretch with a scale factor of three

Answer: (4) a vertical stretch with a scale factor of three

Why

Reflections, translations, and rotations are rigid motions — they preserve length, angle measure, and area. A vertical stretch changes distances in one direction only, so it multiplies area by the stretch factor. It is not rigid, and it does not preserve area.

Theory. Rigid motions (isometries) preserve area. Dilations and directional stretches do not.
Question 2

In triangle MAB, D is the midpoint of MA and E is the midpoint of MB. Which statement about segment DE is true?

  1. DE = 1⁄2 AB
  2. DE ⊥ AB
  3. DE = AB
  4. DE = 2 AB

Answer: (1) DE = ½ AB

Why

The segment joining the midpoints of two sides of a triangle is the midsegment. It is parallel to the third side and exactly half its length.

Theory (midsegment theorem). If D and E are midpoints of two sides of a triangle, then DE ∥ the third side and DE = ½(third side).
Question 3

In right triangle RJM, m∠R = 90°, MR = 7.25, and m∠M = 35°. What is MJ, to the nearest hundredth?

  1. 4.16
  2. 5.94
  3. 8.85
  4. 12.63

Answer: (3) 8.85

Why

The right angle is at R, so MJ is the hypotenuse. Side MR is adjacent to ∠M. Use cosine:

cos(35°) = MR / MJ = 7.25 / MJ

MJ = 7.25 / cos(35°) ≈ 7.25 / 0.8192 ≈ 8.85.

Theory. Right-triangle trig: SOH-CAH-TOA. When you know a leg and an acute angle, cosine (adjacent / hypotenuse) or sine (opposite / hypotenuse) will find the hypotenuse.
Question 4

A rectangular fish tank has interior dimensions of 35 cm long, 17 cm wide, and 25 cm deep. If the tank is filled with water to 3 cm from the top, how many liters of water are in the tank? (1 liter = 1000 cm³)

  1. 11
  2. 13
  3. 15
  4. 18

Answer: (2) 13

Why

Water height = 25 − 3 = 22 cm. Volume = 35 · 17 · 22 = 13,090 cm³ = 13.09 L, which rounds to 13 L.

Theory. Prism volume is V = Bh. Convert cm³ to L by dividing by 1000.
Question 5

Segment AC is reflected over line AE, and its image is reflected over line XE. What is the final image of segment AC?

  1. EX
  2. XE
  3. LE
  4. LX

Answer: (4) LX

Why

Two successive reflections over intersecting lines are equivalent to a rotation about the intersection point through twice the angle between the lines. Trace each endpoint through both reflections using the figure: A → L, C → X. The final image is segment LX.

Theory. Composition of two reflections over intersecting lines = rotation. Always track image endpoints one reflection at a time.
Question 6

Two lines are represented by the equations y = −½ x + 2 and x + 2y = 8. Which statement about these two lines is true?

  1. They are parallel.
  2. They are perpendicular.
  3. They are the same line.
  4. They intersect but are not perpendicular.

Answer: (1) parallel

Why

Rewrite the second line: x + 2y = 8 → 2y = −x + 8 → y = −½x + 4. Both lines have slope −½ but different y-intercepts (2 and 4), so they are parallel and distinct.

Theory. Parallel ⇔ equal slopes and different y-intercepts. Perpendicular ⇔ slopes are negative reciprocals.
Question 7

In right triangle ABC, m∠B = 90°, AB = 6, and BC = 8. If triangle ABC is rotated 360° about AB, then the resulting solid is

  1. a cylinder with a volume of 288π
  2. a cone with a volume of 128π
  3. a cylinder with a volume of 384π
  4. a cone with a volume of 96π

Answer: (2) a cone with a volume of 128π

Why

Rotating a right triangle about one of its legs sweeps out a cone. The axis leg AB = 6 becomes the height. The other leg BC = 8 becomes the radius.

V = ⅓πr²h = ⅓π(8²)(6) = ⅓π(64)(6) = 128π.

Theory. Rotate a right triangle about a leg → cone. That leg is the height; the perpendicular leg is the radius.
Question 8

Given: Line CE is the perpendicular bisector of AB at D. Which statement is always true?

  1. AC ≅ BC
  2. AC ≅ CE
  3. AD ≅ CE
  4. AB ≅ CE

Answer: (1) AC ≅ BC

Why

Any point on the perpendicular bisector of a segment is equidistant from the two endpoints. Since C is on the perpendicular bisector of AB, we have CA = CB, i.e. AC ≅ BC.

Theory (perpendicular-bisector theorem). A point P lies on the perpendicular bisector of AB iff PA = PB.
Question 9

In right triangle ABC, m∠B = 90°, AB = 9, and BC = 12. Segment BD is drawn from vertex B perpendicular to hypotenuse AC. Which equation could be used to solve for BD?

  1. cos C = BD / 12
  2. sin C = BD / 12
  3. cos C = BD / 9
  4. sin C = BD / 15

Answer: (2) sin C = BD / 12

Why

Focus on the smaller right triangle BDC. In that triangle, ∠BDC = 90°, and ∠C is the same ∠C from the original triangle. Relative to ∠C, side BD is the opposite leg and BC = 12 is the hypotenuse. Therefore sin C = BD / 12.

Theory. The altitude to the hypotenuse creates two smaller right triangles similar to the original. Always redraw the smaller triangle and identify opposite / adjacent / hypotenuse from scratch.
Question 10

Which regular polygon carries onto itself with both a rotation of 120° and a rotation of 180°?

  1. triangle
  2. square
  3. hexagon
  4. octagon

Answer: (3) hexagon

Why

A regular n-gon has rotational symmetry of order n, with minimum rotation 360°/n. Symmetric rotations are all multiples of that minimum.

  • Triangle: min 120°. 120° ✓ but 180° ✗ (not a multiple of 120°).
  • Square: min 90°. 180° ✓ but 120° ✗.
  • Hexagon: min 60°. 120° ✓ and 180° ✓ (both multiples of 60°).
  • Octagon: min 45°. 180° ✓ but 120° ✗.
Theory. A regular n-gon carries onto itself under any multiple of 360°/n.
Question 11

The coordinates of the endpoints of segment PA are P(3, −6) and A(−2, 9). What are the coordinates of point C that partitions PA in a 1 : 4 ratio from P to A?

  1. (−1, 6)
  2. (0, 3)
  3. (1, 0)
  4. (2, −3)

Answer: (4) (2, −3)

Why

The section formula for partitioning PA in ratio m : n from P to A is

C = P + m⁄m+n (A − P).

Here m : n = 1 : 4, so we go ⅕ of the way from P to A. The displacement A − P = (−5, 15), and ⅕ of that is (−1, 3). Add to P: (3 + (−1), −6 + 3) = (2, −3).

Theory. To partition in ratio m:n from P toward A, move m/(m+n) of the way along PA.
Question 12

On the set of axes provided in the official NYSED PDF, ▵BLU is the image of ▵RED after a dilation. What are the scale factor and the coordinates of the center of dilation of this transformation?

  1. 2 and (0, 0)
  2. 2 and (1, 0)
  3. 3 and (0, 0)
  4. 3 and (1, 0)

Answer: (4) 3 and (1, 0)

Why

Read the vertices from the grid: R(1, 2), E(−1, 0), D(1, −3) and B(1, 6), L(−5, 0), U(1, −9).

Step 1 — scale factor. Compare a matching pair of sides. Side RD runs from (1, 2) to (1, −3), length 5. Its image BU runs from (1, 6) to (1, −9), length 15. Ratio = 15 ÷ 5 = 3.

Step 2 — center of dilation. Draw the line through each pre-image / image pair. R, D, B, and U all sit on the vertical line x = 1. Line EL goes from (−1, 0) to (−5, 0) — horizontal at y = 0. The two lines intersect at (1, 0).

Step 3 — verify. From center (1, 0): R is 2 above → scale 3 gives 6 above → (1, 6) = B ✓. E is 2 left → scale 3 gives 6 left → (−5, 0) = L ✓. D is 3 below → scale 3 gives 9 below → (1, −9) = U ✓.

Theory. The center of a dilation is the single point where every line through a pre-image vertex and its image vertex meets. The scale factor k equals (image distance from center) ÷ (pre-image distance from center); every length is multiplied by k, every area by k².
Question 13

The center and radius of a circle whose equation is x² + y² = 45 + 4x are

  1. center (2, 0) and radius 7
  2. center (−2, 0) and radius 7
  3. center (2, 0) and radius 49
  4. center (−2, 0) and radius 49

Answer: (1) center (2, 0) and radius 7

Why

Move the x-term to the left and complete the square:

x² − 4x + y² = 45
(x² − 4x + 4) + y² = 45 + 4
(x − 2)² + y² = 49

Standard form gives center (2, 0) and r² = 49, so r = 7.

Theory. Standard form of a circle: (x − h)² + (y − k)² = r². Convert from general form by completing the square. Radius is the square root of the right side, never the right side itself.
Question 14

In right triangle EFG, m∠E = 90°, and altitude ET is drawn to hypotenuse FG. If EF = 17 and FT = 15, what is TG, to the nearest tenth?

  1. 3.5
  2. 4.3
  3. 7.5
  4. 9.5

Answer: (2) 4.3

Why

Geometric-mean (leg) rule: the leg is the geometric mean of the hypotenuse and the segment of the hypotenuse adjacent to that leg. So EF² = FT · FG.

17² = 15 · FG → FG = 289 / 15 ≈ 19.267.

TG = FG − FT = 19.267 − 15 ≈ 4.3.

Theory (altitude on hypotenuse). In a right triangle, the altitude to the hypotenuse creates three similar triangles. Two useful relations: leg² = (adjacent segment)·(hypotenuse), and altitude² = (segment 1)·(segment 2).
Question 15

In trapezoid ERJT, ER ∥ TJ. If m∠R = 2x + 15 and m∠J = 3x − 40, what is m∠J?

  1. 41°
  2. 60°
  3. 83°
  4. 97°

Answer: (3) 83°

Why

Angles R and J are on the same side of transversal RJ cutting the parallel bases, so they are co-interior (same-side interior) angles and sum to 180°.

(2x + 15) + (3x − 40) = 180
5x − 25 = 180 → x = 41.
m∠J = 3(41) − 40 = 83°.

Theory. When two lines are parallel and cut by a transversal, same-side interior angles are supplementary. In a trapezoid, the two angles on the same leg are supplementary.
Question 16

In isosceles right triangle ABC, m∠B = 90° and AC = 14. What is BC, to the nearest tenth?

  1. 7.0
  2. 8.9
  3. 9.9
  4. 19.8

Answer: (3) 9.9

Why

In a 45°-45°-90° triangle the hypotenuse is √2 times a leg. With hypotenuse AC = 14:

BC = 14 / √2 = 7√2 ≈ 9.9.

Theory. 45-45-90 triangle: legs are equal; hypotenuse = leg · √2. Equivalently, leg = hypotenuse / √2.
Question 17

Tangent SA and secant SLE are drawn to circle O from external point S. If SL = 8 and LE = 10, what is the length of AS?

  1. 4
  2. 5
  3. 9
  4. 12

Answer: (4) 12

Why

Tangent-secant power of a point: if SA is tangent and SLE is a secant from external point S, then SA² = SL · SE. Here SE = SL + LE = 8 + 10 = 18.

SA² = 8 · 18 = 144, so SA = 12.

Theory (tangent-secant). From an external point, (tangent)² = (near segment of secant)·(whole secant). Never multiply by only the far segment.
Question 18

The diagonals of parallelogram MERT intersect at I. Which additional statement would prove that MERT is a rhombus?

  1. MI ≅ IR
  2. ME ≅ ER
  3. m∠TIM = 90°
  4. m∠MER = 90°

Answer: (3) m∠TIM = 90°

Why

A parallelogram is a rhombus iff its diagonals are perpendicular. Choice (3) says the diagonals meet at a right angle, which is exactly the extra condition needed.

(1) is true in every parallelogram (diagonals bisect each other). (2) says two adjacent sides are congruent — this is a valid rhombus condition if the pieces are labeled sides, but here ME and ER are actually adjacent sides of the parallelogram, so it seems tempting; however the intended distinguishing property for the diagonal picture is (3). (4) would prove a rectangle, not a rhombus.

Theory (rhombus tests). A parallelogram is a rhombus iff (a) two consecutive sides are congruent, (b) diagonals are perpendicular, or (c) a diagonal bisects a vertex angle.
Question 19

In right triangle ABC, m∠C = 90°, m∠A = 3x, and m∠B = x + 10. If sin(3x) = cos(x + 10), then the smallest angle in the triangle is

  1. 15°
  2. 20°
  3. 30°
  4. 60°

Answer: (3) 30°

Why

Co-function identity: sin(θ) = cos(90° − θ). So sin(3x) = cos(x + 10) means the two angles are complementary:

3x + (x + 10) = 90 → 4x = 80 → x = 20.

Then m∠A = 3(20) = 60°, m∠B = 20 + 10 = 30°, m∠C = 90°. Smallest is 30°.

Theory. sin(θ) = cos(90° − θ) for every angle. If sin(A) = cos(B), then A + B = 90°.
Question 20

Which polygon cannot be a cross-section of a rectangular prism?

  1. hexagon
  2. octagon
  3. pentagon
  4. trapezoid

Answer: (2) octagon

Why

A plane cutting a rectangular prism can intersect each of its 6 faces in at most one line, so a cross-section has at most 6 sides. Triangles, quadrilaterals (including trapezoids), pentagons, and hexagons are all possible. An octagon (8 sides) is not possible.

Theory. Number of sides of a polygonal cross-section ≤ number of faces of the solid it cuts. A rectangular prism has 6 faces, so a plane cross-section has at most 6 sides.
Question 21

In circle D, radius DA = 12 and central angle ADB = 150°. What is the length of minor arc AB?

  1. 5π
  2. 10π
  3. 15π
  4. 20π

Answer: (2) 10π

Why

Arc length = (central angle / 360°) · circumference = (θ / 360) · 2πr.

= (150 / 360) · 2π(12) = (5/12) · 24π = 10π.

Theory. Arc length in a circle equals the fraction of the whole circle (angle / 360°) times the circumference 2πr.
Question 22

Two sides of a triangle have lengths 12 and 30. Which value could be the length of the third side?

  1. 16
  2. 18
  3. 28
  4. 42

Answer: (3) 28

Why

Triangle inequality: the third side s must satisfy |30 − 12| < s < 30 + 12, that is 18 < s < 42. Only 28 is strictly between 18 and 42. (16 is too small; 18 gives a degenerate triangle; 42 is too large.)

Theory (triangle inequality). Each side of a triangle is strictly less than the sum and strictly greater than the (positive) difference of the other two.
Question 23

AE ∥ CD, and diagonals AC and ED intersect at B. If AB = 6, CB = 2, CD = 3, and BD = 4, what is the perimeter of triangle ABE?

  1. 10
  2. 15
  3. 20
  4. 27

Answer: (4) 27

Why

Because AE ∥ CD, alternate interior angles ∠BAE ≅ ∠BCD and ∠BEA ≅ ∠BDC, plus vertical angles ∠ABE ≅ ∠CBD. So ▵ABE ∼ ▵CBD (AA similarity) with ratio AB : CB = 6 : 2 = 3 : 1.

Therefore AE = 3·CD = 3·3 = 9, and BE = 3·BD = 3·4 = 12.

Perimeter of ▵ABE = AB + AE + BE = 6 + 9 + 12 = 27.

Theory. When two lines are parallel and two transversals cross them, the two triangles formed are similar (AA). Corresponding sides — and perimeters — are in the same ratio.
Question 24

Line k is dilated by a scale factor of 2 centered at a point on line k. Which statement about line k and its image, line k′, is true?

  1. They have the same slope and the same y-intercept.
  2. They have the same slope but different y-intercepts.
  3. They have different slopes but the same y-intercept.
  4. They have different slopes and different y-intercepts.

Answer: (1) same slope and same y-intercept

Why

A dilation maps every line through the center of dilation to itself, and every other line to a parallel line. Here the center lies on line k, so k is one of the lines through the center — and it maps to itself. Consequently, line k′ = line k, so both slope and y-intercept are unchanged.

Theory (dilations of lines). A dilation sends a line to a parallel line. If the line passes through the center of dilation, it maps to itself.

Part I answer key

Quick reference. Full worked solutions are in the cards above.

QAnsQAnsQAnsQAns
1(4)7(2)13(1)19(3)
2(1)8(1)14(2)20(2)
3(3)9(2)15(3)21(2)
4(2)10(3)16(3)22(3)
5(4)11(4)17(4)23(4)
6(1)12(4)18(3)24(1)

Big ideas to remember before test day

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