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

Regents Geometry June 2026 — Part II Answers & Explanations

Free worked solutions for every Part II short-answer question on the June 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 June 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. Two congruent figures always have a rigid-motion sequence between them. If the two figures have the same orientation (both clockwise or both counterclockwise), you can map one to the other with a translation, rotation, or their composition. If orientations differ, your sequence must include at least one reflection.

Parallel lines & parallelogram angle relationships

When a transversal cuts two parallel lines, corresponding, alternate interior, and alternate exterior angles are congruent, while same-side (co-interior) angles are supplementary. Opposite sides of a parallelogram are parallel, so any segment inside a parallelogram can act as a transversal.

3-D volume & density

Cylinder volume is V = πr²h. Weight (or mass) equals volume times density, when density is given as weight per unit volume. Always convert diameter to radius before substituting.

Side-splitter / basic proportionality theorem

If a segment inside a triangle is parallel to one side, it divides the other two sides in the same ratio. This is how “prove parallel by proportion” questions like Q28 are set up.

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 is on the Geometry Regents reference sheet.

Compass & straightedge constructions

You must be able to construct: perpendicular bisector of a segment, angle bisector, perpendicular to a line at (or from) a point, and a line parallel to a given line through a point. All construction marks must be left visible — do not erase your arcs.

Proving a parallelogram is a special quadrilateral

A parallelogram is a rectangle iff its diagonals are congruent. A parallelogram is a rhombus iff its diagonals are perpendicular or its diagonals bisect its vertex angles. A parallelogram is a square iff both conditions hold.

Part II — Questions 25–31

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

Question 25

On the set of axes in the exam, ▵DAN ≅ ▵MIK. Describe a sequence of rigid motions that maps ▵DAN onto ▵MIK.

Coordinates (from the figure in the official NYSED PDF): D(−6, 5), A(−2, 5), N(−4, 2); M(5, −3), I(1, −3), K(3, −6).

Step 1 — Check orientation

Going D → A → N, we move right and then down: clockwise. Going M → I → K, we move left and then down: counterclockwise. Orientations differ, so a reflection must be part of the sequence.

Step 2 — Try a reflection over the y-axis

Reflecting (x, y) → (−x, y):

  • D(−6, 5) → D′(6, 5)
  • A(−2, 5) → A′(2, 5)
  • N(−4, 2) → N′(4, 2)

Step 3 — Translate to land on MIK

Translate D′(6, 5) to M(5, −3): shift = ⟨−1, −8⟩. Verify the other two:

  • A′(2, 5) + ⟨−1, −8⟩ = (1, −3) = I ✓
  • N′(4, 2) + ⟨−1, −8⟩ = (3, −6) = K ✓

A sequence that works: reflect ▵DAN over the y-axis, then translate the image 1 unit left and 8 units down.

Rubric. Any valid rigid-motion sequence receives full 2 credits. Other correct answers include composition of two reflections whose axes are related, or a rotation + a reflection. Just make sure the two orientations differ and that your final image lands exactly on MIK.
Theory. Fundamental Theorem of Isometries: every isometry of the plane is a composition of at most 3 reflections. Same-orientation congruent figures are related by 0 or 2 reflections (i.e. a translation or rotation). Opposite-orientation figures need exactly 1 or 3 reflections.
Question 26

In parallelogram LEAD, C is on diagonal DE and S is on side DA such that SC ⊥ DE. If m∠LED = 21°, determine and state the measure of ∠CSA.

Step 1 — Use the parallel-side property of a parallelogram

In parallelogram LEAD, opposite sides LE and DA are parallel. Diagonal DE is a transversal crossing both.

Step 2 — Identify alternate interior angles

On the transversal DE, angles ∠LED (at E) and ∠EDA (at D) are alternate interior angles between the parallels LE and DA. Alternate interior angles are congruent:

m∠EDA = m∠LED = 21°.

Because C lies on DE and S lies on DA, this is the same angle as ∠CDS. So m∠CDS = 21°.

Step 3 — Solve the small right triangle SCD

Given SC ⊥ DE, so m∠SCD = 90°. The three angles of ▵SCD sum to 180°:

m∠DSC = 180° − 90° − 21° = 69°.

Step 4 — Finish with a linear pair

Points D, S, A are collinear (S is on DA), so ∠DSC and ∠CSA form a linear pair at S:

m∠CSA = 180° − 69° = 111°.

m∠CSA = 111°

Theory. Parallel-line angles (alternate interior, corresponding, co-interior), the triangle-angle-sum theorem, and the linear-pair postulate together handle almost every parallelogram angle problem on the Regents.
Question 27

A section of a white pine tree trunk is modeled by a cylinder. The diameter of the trunk is 1.5 feet, and the height of this section is 8 feet. If white pine weighs 25 pounds per cubic foot, determine and state the weight of this section, to the nearest pound.

Step 1 — Find the radius

r = ½ · 1.5 = 0.75 ft.

Step 2 — Compute the cylinder volume

V = πr²h = π(0.75)²(8) = π(0.5625)(8) = 4.5π ≈ 14.137 ft³.

Step 3 — Multiply by the density

Weight = volume × density = 14.137 · 25 ≈ 353.43 lb.

Weight ≈ 353 pounds

Rubric. Show V = πr²h with numbers substituted, then multiply by 25. Rounding must happen at the very end — do not round the volume prematurely, or you may lose the credit for the final answer.
Theory. Density = mass / volume, so mass (or weight, in a uniform gravitational field) = density × volume. When density is given per unit volume, this collapses to a one-step multiplication once the volume is known.
Question 28

In ▵ADC, points B and F are on AC and AD, respectively, such that AB = 40, BC = 30, and BF is drawn. If AF = 64, determine and state the length of FD that would prove BF ∥ CD.

Step 1 — Recall the side-splitter theorem

If a segment inside a triangle is parallel to one side of the triangle, it divides the other two sides proportionally. Conversely, if that proportion holds, the segment must be parallel to the third side.

Step 2 — Write the proportion for BF ∥ CD

BF would be parallel to CD exactly when it splits sides AC and AD in the same ratio:

AB / BC = AF / FD

40 / 30 = 64 / FD.

Step 3 — Solve for FD

Cross-multiply: 40 · FD = 30 · 64 = 1920.

FD = 1920 / 40 = 48.

FD = 48

Theory (converse of the side-splitter). If B is on AC, F is on AD, and AB / BC = AF / FD, then BF ∥ CD. This is the “converse” that lets you construct parallels from ratios.
Question 29

In ▵CLM, m∠C = 33°, CL = 8, and CM = 15. Determine and state, to the nearest tenth, the area of ▵CLM.

Step 1 — Identify the “SAS” setup

You are given two sides and the included angle: CL = 8, CM = 15, and ∠C is the angle between those two sides. This is the exact setup for the SAS area formula.

Step 2 — Apply the trig area formula

Area = ½ · (CL) · (CM) · sin(C).

= ½ · 8 · 15 · sin(33°)

= 60 · sin(33°)

≈ 60 · 0.5446

≈ 32.678.

Area ≈ 32.7 square units

Rubric. Show the formula, substitute numbers, and give the rounded answer. If you drop the ½, you will double the area — a common Regents error.
Theory (SAS area formula). Area = ½ a b sin(C). Requires the angle to be between the two given sides. This formula lives on the Geometry Regents reference sheet.
Question 30

In the diagram of ▵ABC, use a compass and straightedge to construct the angle bisector of ∠ABC. [Leave all construction marks.]

Step-by-step construction (of the angle bisector of ∠ABC)

  1. Arc from the vertex. Place the compass point on B. Open it to any convenient radius and draw an arc that crosses both ray BA and ray BC. Label the crossing points P (on BA) and Q (on BC).
  2. Two arcs from P and Q. Keep the compass open to any radius greater than half of PQ. Place the compass point on P and draw an arc in the interior of ∠ABC. Without changing the compass width, place the point on Q and draw a second arc that crosses the first. Label the intersection R.
  3. Draw the bisector. Use the straightedge to draw ray BR. This is the angle bisector of ∠ABC.
  4. Leave every construction mark visible. Do not erase your arcs. The graders need to see them to award credit.

Ray BR bisects ∠ABC because BP = BQ (same first-arc radius) and PR = QR (same second-arc radius), so ▵BPR ≅ ▵BQR by SSS, forcing ∠PBR ≅ ∠QBR.

Rubric. Full credit requires (a) the vertex arc that produces P and Q, (b) two equal-radius arcs from P and Q meeting at R, and (c) the ray from B through R. Erasing construction marks is a common credit loss.
Theory. The angle-bisector construction is really an SSS congruence proof in disguise. Equal radii create equal segments; equal segments create congruent triangles; congruent triangles have congruent corresponding angles.
Question 31

The diagonals of parallelogram GRAM intersect at P. If RP = 12 and GA = 24, explain why GRAM is a rectangle.

Step 1 — Use the parallelogram diagonal property

In any parallelogram, the diagonals bisect each other. Point P is therefore the midpoint of both diagonals GA and RM.

Step 2 — Compute RM

Because P is the midpoint of RM, we have RM = 2 · RP = 2 · 12 = 24.

Step 3 — Compare the diagonals

The problem gives GA = 24. So GA = RM = 24 — the two diagonals are congruent.

Step 4 — Apply the rectangle test

Theorem: a parallelogram is a rectangle if and only if its diagonals are congruent. GRAM is a parallelogram (given) with congruent diagonals (just shown), so GRAM is a rectangle.

Therefore GRAM is a rectangle.

Rubric. Full 2 credits requires (a) stating that diagonals of a parallelogram bisect each other, (b) computing RM = 24 from that, (c) naming the theorem “a parallelogram with congruent diagonals is a rectangle” and (d) concluding.
Theory (rectangle test). Parallelogram + congruent diagonals ⇔ rectangle. Parallelogram + perpendicular diagonals ⇔ rhombus. Both ⇔ square.

Part II answer key

Big ideas to remember on test day

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This walkthrough is provided for educational purposes. The June 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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