← SOMATH Blog · Regents Prep NYC · Updated August 4, 2026

Regents Geometry January 2026 — Part I Answers & Explanations

Free worked solutions for every Part I question on the January 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 January 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 January 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

On the set of axes shown in the official NYSED PDF, ▵RST and its image ▵R′S′T′ are graphed. Which rigid motion is sufficient to prove ▵RST ≅ ▵R′S′T′?

  1. a rotation of 90° clockwise about the origin
  2. a translation 4 units to the right
  3. a reflection over the x-axis
  4. a reflection over the y-axis

Answer: (4) a reflection over the y-axis

Why

Compare corresponding vertices in the figure: T and T′ lie on the same horizontal line, at equal distances on opposite sides of the y-axis. The same is true for the other paired vertices. A reflection over the y-axis sends every point (x, y) to (−x, y), which matches the figure. A rotation of 90° would swap the coordinates, and a translation would not flip orientation. A reflection over the x-axis would send positive y‑values to negative ones, which does not match.

Theory. A rigid motion is enough to prove congruence. Rigid motions preserve length and angle; reflections reverse orientation, while rotations and translations preserve it.
Question 2

Which regular polygon would carry onto itself after a rotation of 60° about its center?

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

Answer: (2) hexagon

Why

A regular n‑gon carries onto itself under a rotation of 360°/n and any multiple of it. Set 360°/n = 60°, so n = 6. A regular hexagon has six-fold rotational symmetry.

Theory. The smallest rotation that maps a regular n‑gon onto itself is 360°/n.
Question 3

A right triangle with legs 4 cm and 7 cm is continuously rotated about the 4 cm side. The solid formed is

  1. a cone with a height of 4 cm and a radius of 7 cm
  2. a cone with a height of 4 cm and a radius of 14 cm
  3. a pyramid with a height of 4 cm and a base length of 7 cm
  4. a pyramid with a height of 4 cm and a base length of 14 cm

Answer: (1) a cone with a height of 4 cm and a radius of 7 cm

Why

Rotating a right triangle about one of its legs sweeps out a cone. The axis leg becomes the height, and the perpendicular leg becomes the radius. Here the 4 cm leg is the axis, so height = 4 cm and radius = 7 cm.

Theory. Rotating a right triangle about a leg → cone. The rotation axis is the height; the perpendicular leg is the radius.
Question 4

In isosceles triangle AHP, AH ≅ PH, and AH is extended through H to C. If m∠A = (2x + 12)° and m∠P = (3x − 8)°, what is the measure of ∠CHP?

  1. 52°
  2. 76°
  3. 104°
  4. 128°

Answer: (3) 104°

Why

Because AH ≅ PH, ∠A ≅ ∠P (base angles of an isosceles triangle). Set them equal: 2x + 12 = 3x − 8, so x = 20. Then m∠A = m∠P = 52°. ∠CHP is the exterior angle of ▵AHP at vertex H, so it equals the sum of the two remote interior angles: m∠CHP = 52° + 52° = 104°.

Theory (exterior-angle theorem). An exterior angle of a triangle equals the sum of the two non-adjacent interior angles.
Question 5

In the diagram in the official NYSED PDF, ▵XYZ is the image of ▵ABC after a dilation of scale factor ½. Which point must be the center of dilation?

  1. (2, −1)
  2. (8, 3)
  3. (5, 1)
  4. (0, 0)

Answer: (4) (0, 0)

Why

A dilation of scale factor ½ centered at point P sends every vertex V to a point V′ with PV′ = ½·PV, on ray PV. From the figure, each image vertex is exactly halfway from the origin to its preimage: A→X, B→Y, C→Z. The three lines AX, BY, CZ all pass through the origin, so the center is (0, 0).

Theory. Under a dilation with center P, every line through the center maps to itself. To locate the center: draw the line through each preimage-image pair and find where they concur.
Question 6

In a right triangle, the acute angles have the relationship cos(5x + 7)° = sin(3x + 3)°. Which equation is always true?

  1. 5x + 7 = 3x + 3
  2. 5x + 7 + 3x + 3 = 90
  3. 5x + 7 + 3x + 3 = 180
  4. 5x + 7 + 3x + 3 + x = 180

Answer: (2) 5x + 7 + 3x + 3 = 90

Why

The co-function identity says cos(θ) = sin(90° − θ). If cos(A) = sin(B), the two angles must be complementary: A + B = 90°. So (5x + 7) + (3x + 3) = 90.

Theory (co-function). sin A = cos(90° − A). In a right triangle the two acute angles are always complementary.
Question 7

In circle O, radius = 8 cm and central angle AOB measures 140°. What is the length of arc AB, to the nearest centimeter?

  1. 10
  2. 20
  3. 25
  4. 78

Answer: (2) 20

Why

Arc length = r·θ when θ is in radians. Convert 140° to radians: 140π/180 = 7π/9. Arc length = 8·(7π/9) = 56π/9 ≈ 19.55, which rounds to 20 cm.

Theory. Arc length s = r θ (radians) = (θ°/360)·2πr (degrees).
Question 8

Lucy models a pile of sand as a cone with height 3.5 ft and radius 3.5 ft. Her wagon holds 5 cubic feet of sand. What is the fewest number of trips she needs to move the entire pile?

  1. 27
  2. 14
  3. 3
  4. 9

Answer: (4) 9

Why

Cone volume: V = ⅓πr²h = ⅓π(3.5)²(3.5) = ⅓π(42.875) ≈ 44.9 ft³. Number of full trips = 44.9 / 5 = 8.98, and you cannot make a partial trip — round up to 9.

Theory. Cone volume is V = ⅓πr²h. Real-world “how many trips” problems require ceiling, not rounding.
Question 9

Segments BD and AE intersect at C, and AB and DE are drawn with AB ∥ DE. Which statement is not always true?

  1. ∠ABC ≅ ∠EDC
  2. ∠ACB ≅ ∠ECD
  3. ▵ABC ∼ ▵EDC
  4. ▵ABC ≅ ▵EDC

Answer: (4) ▵ABC ≅ ▵EDC

Why

∠ABC and ∠EDC are alternate interior angles (parallel lines cut by transversal BD), so (1) is true. ∠ACB and ∠ECD are vertical angles, so (2) is true. Two pairs of congruent angles ⇒ AA similarity, so (3) is true. But the triangles are only congruent if AB = DE, which is not given — so (4) is not always true.

Theory. Parallel lines cut by a transversal generate similar triangles at the intersection (AA). Congruence requires an extra side-length condition.
Question 10

The face of a shed is a 10 ft-wide rectangle BFGK topped by isosceles triangle FEG with vertex angle at E and height 6 ft. What is m∠EGD, to the nearest degree?

  1. 34°
  2. 40°
  3. 50°
  4. 56°

Answer: (3) 50°

Why

The triangle is isosceles with base 10 ft (same width as the rectangle) and altitude 6 ft to the base. The altitude drops to the midpoint of the base, creating a right triangle with legs 5 (half-base) and 6 (height). ∠EGD is the base angle of the isosceles triangle, so tan(∠EGD) = opposite / adjacent = 6/5. ∠EGD = arctan(6/5) ≈ 50.19°, rounding to 50°.

Theory. In an isosceles triangle, the altitude from the vertex angle bisects the base. Use SOH-CAH-TOA on the resulting right triangle.
Question 11

Triangles ABC, A′B′C′, and A″B″C″ are graphed on the set of axes in the official NYSED PDF. Which sequence maps ▵ABC onto ▵A′B′C′, and then ▵A′B′C′ onto ▵A″B″C″?

  1. a translation followed by a rotation
  2. a rotation followed by a translation
  3. a line reflection followed by a rotation
  4. a translation followed by a line reflection

Answer: (1) a translation followed by a rotation

Why

From the figure, ▵ABC (below the x-axis, left of the y-axis) has the same orientation as ▵A′B′C′ (below the x-axis, right of the y-axis). Same orientation ⇒ a translation. The second image ▵A″B″C″ (above the x-axis) is rotated relative to the first; the orientation is still preserved (no mirroring), so it is a rotation, not a reflection.

Theory. Reflections reverse orientation. Rotations and translations preserve it. Check orientation first, then decide which pair of rigid motions fits.
Question 12

A line contains the points (−1, −4) and (3, −1). An equation of a line perpendicular to this line is

  1. y + 4 = ¾(x + 1)
  2. y − 4 = &frac43;(x − 1)
  3. y − 1 = −¾(x + 3)
  4. y + 1 = −&frac43;(x − 3)

Answer: (4) y + 1 = −&frac43;(x − 3)

Why

Slope of the given line: (−1 − (−4)) / (3 − (−1)) = 3/4. A perpendicular line has slope equal to the negative reciprocal: −4/3. Only option (4) has slope −4/3.

Theory. Two non-vertical lines are perpendicular iff their slopes multiply to −1 (are negative reciprocals).
Question 13

In the diagram of right triangles DAY and NIT, AD = 6, DY = 6, IT = 16, and ▵DAY ∼ ▵NIT. The length of TN is

  1. 8
  2. 8√2
  3. 8√3
  4. 16√2

Answer: (2) 8√2

Why

In ▵DAY, the right angle is at D, so the hypotenuse is AY. By the Pythagorean theorem AY² = 6² + 6² = 72, so AY = 6√2. Similarity ▵DAY ∼ ▵NIT gives the correspondence D ↔ N, A ↔ I, Y ↔ T, so AY ↔ IT and DY ↔ NT. Ratio of similarity = IT / AY = 16 / (6√2) = 8 / (3√2) = 4√2 / 3. Then NT = DY·(4√2 / 3) = 6·(4√2 / 3) = 8√2.

Theory. Similar figures have all corresponding sides in the same ratio. Use the correspondence given by the similarity statement to line up matching sides.
Question 14

The volume of a sphere is 333 cm³. To the nearest tenth of a centimeter, the diameter of the sphere is

  1. 4.3
  2. 5.2
  3. 8.6
  4. 10.4

Answer: (3) 8.6

Why

Sphere volume: V = &frac43;πr³. Solve for r: r³ = 3V / (4π) = 3·333 / (4π) ≈ 79.5. Then r ≈ √3{79.5} ≈ 4.30 cm, so the diameter is 2r ≈ 8.6 cm.

Theory. Sphere volume V = &frac43;πr³. Diameter = 2r.
Question 15

Line BTS is parallel to line MAVR, and AE ⊥ TV. If m∠STE = 38°, what is the measure of ∠VAE?

  1. 38°
  2. 52°
  3. 128°
  4. 142°

Answer: (2) 52°

Why

Because BTS ∥ MAVR with transversal TV, the alternate interior angles are equal, so m∠TVA = m∠STE = 38°. In ▵AVE the segment AE is perpendicular to TV, giving a right angle at E. The three interior angles must sum to 180°, so m∠VAE = 180° − 90° − 38° = 52°.

Theory. Parallel lines ⇒ alternate interior angles are congruent. Then use the angle-sum of a triangle (or the fact that acute angles of a right triangle are complementary).
Question 16

Segment RAZ has endpoints R(6, 6) and Z(−12, −3). If A divides RZ such that RA : AZ = 5 : 4, then the coordinates of A are

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

Answer: (4) (−4, 1)

Why

The section formula for point A that partitions RZ in ratio 5 : 4 from R to Z: A = R + (5 / 9)(Z − R). Compute Z − R = (−18, −9). Then (5 / 9)(−18, −9) = (−10, −5). Finally A = (6, 6) + (−10, −5) = (−4, 1).

Theory (section formula). If P divides AB in ratio m : n from A, then P = A + [m / (m + n)](B − A).
Question 17

In ▵ABC, points D and E are on AB and CB respectively, with DE ∥ AC. If BD = 9, DA = 3, and EC = 4, what is the length of BC?

  1. 10
  2. 12
  3. 14
  4. 16

Answer: (4) 16

Why

Because DE ∥ AC, ▵BDE ∼ ▵BAC (AA). Corresponding-side ratio: BD / BA = BE / BC. With BA = BD + DA = 12, the ratio is 9 / 12 = 3 / 4. Let BC = x. Then BE = x − EC = x − 4, and (x − 4) / x = 3 / 4. Cross-multiply: 4(x − 4) = 3x ⇒ x = 16.

Theory (side-splitter theorem). A line parallel to one side of a triangle cuts the other two sides proportionally.
Question 18

Triangle ABC is mapped onto ▵A′B′C′ after a sequence of rigid motions. Which statement is always true?

  1. Segment AB is parallel to segment A′B′.
  2. Segment AB is congruent to segment A′B′.
  3. The measure of ∠A is the same as the measure of ∠B′.
  4. The orientation of ▵ABC is the same as the orientation of ▵A′B′C′.

Answer: (2) Segment AB ≅ segment A′B′

Why

Rigid motions preserve distance (length) and angle measure, so AB = A′B′. Rigid motions do not guarantee parallelism (rotations tilt segments), matching corresponding vertices to different letters (m∠A = m∠A′, not B′), or the same orientation (reflections reverse it).

Theory. Rigid motions preserve length and angle measure. Orientation is preserved by rotations and translations but reversed by reflections.
Question 19

What are the coordinates of the center and the length of the radius of the circle whose equation is x² − 16x + y² + 20y = −155?

  1. center (8, −10) and radius 9
  2. center (−8, 10) and radius 9
  3. center (8, −10) and radius 3
  4. center (−8, 10) and radius 3

Answer: (3) center (8, −10) and radius 3

Why

Complete the square on x and y: x² − 16x = (x − 8)² − 64, and y² + 20y = (y + 10)² − 100. Substitute: (x − 8)² + (y + 10)² − 164 = −155, so (x − 8)² + (y + 10)² = 9. Center (8, −10), radius √9 = 3.

Theory. To convert circle general form x² + y² + Dx + Ey + F = 0 to standard form, complete the square in x and y separately.
Question 20

The 2020 US Census populations and land areas of Connecticut, New Jersey, New York, and Pennsylvania are shown in the table in the official NYSED PDF. Which list shows the state population densities, in order from smallest to largest?

  1. Pennsylvania, New York, Connecticut, New Jersey
  2. Connecticut, New Jersey, Pennsylvania, New York
  3. New York, Pennsylvania, New Jersey, Connecticut
  4. New Jersey, Connecticut, New York, Pennsylvania

Answer: (1) Pennsylvania, New York, Connecticut, New Jersey

Why

Population density = population ÷ land area.

  • Pennsylvania: 13,002,700 ÷ 44,743 ≈ 290.6 /mi²
  • New York: 20,201,249 ÷ 47,126 ≈ 428.7 /mi²
  • Connecticut: 3,605,944 ÷ 4,842 ≈ 744.7 /mi²
  • New Jersey: 9,288,994 ÷ 7,354 ≈ 1,262.9 /mi²

Smallest to largest: PA, NY, CT, NJ.

Theory. Density is a rate: quantity per unit area. Divide, then sort.
Question 21

Line t is represented by the equation y = 2x − 1. If the line is dilated by a scale factor of 3 centered at the origin, which equation represents the image of line t after the dilation?

  1. y = 2x − 3
  2. y = 6x − 3
  3. y = 2x − 1
  4. y = 6x − 1

Answer: (1) y = 2x − 3

Why

A dilation centered at the origin maps a line that does not pass through the origin to a parallel line (same slope, different intercept). Line t has slope 2, so the image also has slope 2. The y-intercept (0, −1) is dilated by factor 3 to (0, −3). Image equation: y = 2x − 3.

Theory. Dilations preserve slope. A line through the center of dilation maps to itself; any other line maps to a parallel line whose intercept is scaled by the dilation factor.
Question 22

Quadrilateral ABCD is a parallelogram. Which additional statement is sufficient to prove ABCD is a rhombus?

  1. AB ≅ CD
  2. AD ∥ BC
  3. AD ≅ DC
  4. ∠ADC ≅ ∠ABC

Answer: (3) AD ≅ DC

Why

In a parallelogram opposite sides are already congruent (so AB ≅ CD and AD ≅ BC automatically) and opposite angles are already congruent. To upgrade a parallelogram to a rhombus you need two adjacent sides to be congruent, which forces all four sides equal. AD and DC are adjacent sides, so (3) is sufficient.

Theory. A parallelogram is a rhombus iff two adjacent sides are congruent (equivalently, all four sides congruent, or the diagonals are perpendicular).
Question 23

In right triangle ABC, m∠ABC = 90°, and BD ⊥ ADC. If AD = 3 and CD = 12, the length of AB is

  1. 6
  2. 9
  3. 3√5
  4. 5√3

Answer: (3) 3√5

Why

When an altitude is drawn from the right angle to the hypotenuse of a right triangle, the geometric-mean (leg) relation says: (leg)² = (adjacent segment of hypotenuse)·(entire hypotenuse). Here AB² = AD·AC = 3·(3 + 12) = 3·15 = 45, so AB = √45 = 3√5.

Theory (altitude-on-hypotenuse). If h is the altitude to the hypotenuse of a right triangle, then each leg is the geometric mean between its adjacent hypotenuse segment and the full hypotenuse: (leg)² = (adjacent piece)·(full hypotenuse). Also h² = (segment 1)·(segment 2).
Question 24

In ▵GBT, segments GXM, BXR, and TXE are drawn such that point X is the centroid. Which statement is always true?

  1. MX / GX = 1 / 3
  2. TX / EX = 2 / 1
  3. BX ≅ RX
  4. TM ≅ TR

Answer: (2) TX / EX = 2 / 1

Why

A centroid divides every median into a 2 : 1 ratio, measured from the vertex. Median TXE starts at vertex T and ends at the midpoint E of the opposite side, so TX : XE = 2 : 1. Option (1) is wrong (the correct ratio is MX : GX = 1 : 2, not 1 : 3). Options (3) and (4) are not required by the centroid theorem.

Theory (centroid theorem). The three medians of a triangle meet at the centroid, which divides each median in a 2 : 1 ratio, with the longer segment adjacent to the vertex.

Part I answer key

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

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

Big ideas to remember before test day

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