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

Regents Geometry January 2026 — Part II Answers & Explanations

Free worked solutions for every Part II short-answer question on the January 2026 New York State Regents Geometry exam. Each 2-credit question is broken down into the exact steps that earn full credit, with the theorem or formula named. Click a question to reveal the answer. This is Part II of a four-post series covering all 35 questions.

📄 Original NYSED exam (PDF)

All 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 Part II is graded. Each question is worth 2 credits. A correct numerical answer with no work shown earns only 1 credit. To earn full credit you must show formula substitutions, name any theorem you cite, and clearly state your final answer with correct units. Write in pen; graphs and drawings in pencil.

Sections

  1. What Part II tests
  2. Part II — Questions 25–31 (click to reveal)
  3. Part II answer key
  4. Big ideas to remember
  5. More Regents Geometry help

What Part II tests

Part II is where the Geometry Regents rewards students who show what they know. The seven questions test whether you can execute a formula cleanly, cite the theorem behind a proof step, or perform a construction with a compass and straightedge. Nothing on this section requires proof-writing at the Part IV level — but the graders are looking for reasoning, not just numbers.

Rigid motions on the coordinate plane

Translations, rotations, and reflections all preserve distance and angle. A reflection over the x-axis sends (x, y) → (x, −y); a translation of “a right and b down” sends (x, y) → (x + a, y − b). To find the image after a composition, apply the transformations in order.

Cylinder volume & density

Cylinder volume is V = πr²h. Weight equals volume times density when density is given as weight per unit volume. Always convert diameter to radius, and keep the full unrounded volume until the final step.

Right-triangle trigonometry (SOH-CAH-TOA)

In a right triangle, sin(θ) = opposite / hypotenuse, cos(θ) = adjacent / hypotenuse, tan(θ) = opposite / adjacent. Identify which side is opposite and adjacent to the given angle before choosing a ratio.

Area of a triangle from two sides & the included angle

Area = ½ a b sin(C), where C is the angle between sides of length a and b. This formula is on the Geometry Regents reference sheet and works for any triangle — acute, right, or obtuse.

Annulus (ring) area

The area between two concentric circles equals the outer-circle area minus the inner-circle area: A = πR² − πr² = π(R² − r²). Convert every diameter to a radius before substituting.

Compass & straightedge constructions

You must be able to construct: an equilateral triangle on a given segment, a perpendicular bisector, an angle bisector, a perpendicular to a line at (or from) a point, and a line parallel to a given line through a point. Leave all construction marks visible — erased arcs cost credit.

Inscribed angles & Thales’ theorem

An inscribed angle is half of the central angle that intercepts the same arc. As a special case (Thales’ theorem): an inscribed angle is a right angle if and only if its intercepted arc is a semicircle — equivalently, its chord is a diameter of the circle.

Part II — Questions 25–31

2 credits each · 7 questions · click any card to reveal the full worked solution.

Question 25

A triangle has vertices with coordinates (2, 1), (0, 3), and (−2, −1). Determine and state the coordinates of the vertices of the image of the triangle after a reflection over the x-axis followed by a translation of 3 units to the right and 2 units down.

Image vertices: (5, −3), (3, −5), (1, −1)

Step-by-step

Step 1 — reflect over the x-axis: (x, y) → (x, −y).

  • (2, 1) → (2, −1)
  • (0, 3) → (0, −3)
  • (−2, −1) → (−2, 1)

Step 2 — translate 3 right, 2 down: (x, y) → (x + 3, y − 2).

  • (2, −1) → (5, −3)
  • (0, −3) → (3, −5)
  • (−2, 1) → (1, −1)
Theory. Rigid-motion compositions are applied one step at a time in the order given. A reflection over the x-axis flips signs of the y-coordinates; a translation shifts each coordinate by a constant.
Question 26

A cylindrical bucket is being used to transport topsoil. The bucket has an inside diameter of 10 inches and a height of 15 inches. If the topsoil weighs 0.0231 pound per cubic inch, determine and state the weight of the topsoil in the bucket when the bucket is full, to the nearest pound.

Weight ≈ 27 pounds

Step-by-step

Radius r = 10 / 2 = 5 in. Cylinder volume: V = πr²h = π(5)²(15) = 375π ≈ 1178.097 in³.

Weight = volume × density = 1178.097 × 0.0231 ≈ 27.213 lb. Rounded to the nearest pound: 27 pounds.

Theory. Cylinder volume is V = πr²h. Weight = volume × density when density is expressed as weight per unit volume. Do not round until the final answer.
Question 27

In right triangle SRT, m∠R = 90°, m∠S = 27°, and ST = 31.8. Determine and state the length of SR, to the nearest tenth.

SR ≈ 28.3

Step-by-step

∠R is the right angle, so the hypotenuse is opposite R — that is ST = 31.8. Relative to ∠S, side SR is the adjacent leg (it shares vertex S) and RT is the opposite leg.

Use cosine: cos(27°) = SR / ST ⇒ SR = 31.8·cos(27°) ≈ 31.8·0.89101 ≈ 28.3.

Theory (SOH-CAH-TOA). cos(θ) = adjacent / hypotenuse. Identify the adjacent leg and hypotenuse relative to the given angle before choosing a ratio.
Question 28

In ▵LET, LE = 7.5, ET = 9.3, and m∠LET = 115°. Determine and state the area of ▵LET, to the nearest tenth.

Area ≈ 31.6 square units

Step-by-step

∠LET is at vertex E, sandwiched between sides LE and ET — so it is the angle included between the two given sides. Use the “two-sides-and-included-angle” area formula:

Area = ½·LE·ET·sin(∠LET) = ½·7.5·9.3·sin(115°).

sin(115°) ≈ 0.90631. Area ≈ 0.5·7.5·9.3·0.90631 ≈ 31.607. Rounded to the nearest tenth: 31.6.

Theory. For any triangle with sides a and b and included angle C, the area is ½·a·b·sin(C). This is on the Regents reference sheet and works for acute, right, and obtuse triangles.
Question 29

A pool owner has a circular deck that surrounds her circular pool. The pool has a diameter of 24 feet, and the distance from the edge of the pool to the outer edge of the deck is 8 feet. Determine and state the number of square feet of the deck, to the nearest square foot.

Deck area ≈ 804 square feet

Step-by-step

Pool radius r = 24 / 2 = 12 ft. Deck extends 8 ft beyond the pool, so the outer radius is R = 12 + 8 = 20 ft.

The deck is an annulus (ring): Area = πR² − πr² = π(20² − 12²) = π(400 − 144) = 256π.

256π ≈ 804.247. Rounded to the nearest square foot: 804 sq ft.

Theory. The area between two concentric circles is A = π(R² − r²). Do not add the two radii together — subtract the squared radii, then multiply by π.
Question 30

Use a compass and straightedge to construct an equilateral triangle with AB, shown in the exam, as one of the sides. [Leave all construction marks.]

Standard equilateral-triangle construction on segment AB.

How to construct it (rubric)

  1. Place the compass point on A. Set the compass width to the length AB. Draw an arc above the segment.
  2. Without changing the compass width, place the compass point on B. Draw a second arc above the segment so it crosses the first arc. Label the intersection point C.
  3. Use the straightedge to draw segments AC and BC.
  4. Triangle ABC is equilateral because AC = AB (both radii of the arc centered at A) and BC = AB (both radii of the arc centered at B). So AB = AC = BC.
  5. Leave all arcs visible — do not erase your construction marks.
Theory. A circle’s radius is constant, so any two arcs drawn with the same compass opening produce congruent segments. Two arcs of length AB, centered at A and B, intersect at the third vertex of the equilateral triangle. This is Euclid’s Proposition I.1.
Question 31

In the exam diagram, right triangle ABC is inscribed in the circle with right angle ABC. Explain why AC must be a diameter of the circle.

AC is a diameter because ∠ABC = 90° is an inscribed angle, so its intercepted arc AC must be a semicircle — and the chord of a semicircle is a diameter.

The written argument (sample credit-earning response)

“∠ABC is an inscribed angle in the circle, and its intercepted arc is arc AC. By the inscribed-angle theorem, the measure of an inscribed angle is half the measure of its intercepted arc, so m(arc AC) = 2·m∠ABC = 2·90° = 180°. An arc of 180° is a semicircle, and the chord that cuts off a semicircle passes through the center of the circle. Therefore AC is a diameter.”

Theory (Thales’ theorem). An inscribed angle in a circle is a right angle if and only if its intercepted arc is a semicircle — equivalently, the chord opposite the right angle is a diameter. Name the inscribed-angle theorem explicitly to earn full credit.

Part II answer key

  • Q25. Image vertices: (5, −3), (3, −5), (1, −1)
  • Q26. Weight ≈ 27 pounds
  • Q27. SR ≈ 28.3
  • Q28. Area ≈ 31.6 square units
  • Q29. Deck area ≈ 804 square feet
  • Q30. Equilateral-triangle construction: two arcs of radius AB from A and B meeting at C, then draw AC and BC.
  • Q31. Inscribed ∠ABC = 90° → arc AC = 180° → AC is a diameter (Thales/inscribed-angle theorem).

Big ideas to remember on test day

  • Show the theorem name. Part II graders want to see “inscribed-angle theorem,” “area = ½ ab sin(C),” “annulus,” and so on — a correct number with no justification typically caps at 1 credit.
  • Never round early. Q26 and Q28 both lose credit if you round the volume or the sine before finishing.
  • Convert diameter to radius first. Q26 (bucket) and Q29 (pool) both start with diameters — halve them before substituting into a formula.
  • Apply rigid motions in the given order. Q25 is “reflect, then translate” — do not reverse the order or the answer will be different.
  • Right-triangle trig starts with side identification. On Q27, name the hypotenuse (opposite the right angle) and the adjacent leg (touching the given angle) before you pick sine, cosine, or tangent.
  • Leave your construction marks. Q30 costs credit if the arcs are erased. The intersection of two equal arcs is what makes the triangle equilateral — that’s the whole proof.
  • Round up, down, or to the nearest? Read each prompt carefully. Q26 wants nearest pound, Q27 nearest tenth, Q28 nearest tenth, Q29 nearest square foot. Different rounding rules on the same page.

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This walkthrough is provided for educational purposes. The January 2026 Geometry Regents exam is publicly released by the New York State Education Department. Questions and figures are the property of NYSED.

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