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Regents Geometry January 2026 — Part III Answers & Explanations

Free worked solutions for every Part III question on the January 2026 New York State Regents Geometry exam. Each 4-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 III 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 III is graded. Each question is worth 4 credits. A correct numerical answer with no work receives only 1 credit. Show formula substitutions, name the theorem or definition you use, and give the final answer with correct units. Proofs may be written in two-column, paragraph, or flow-chart form — the grader looks for a valid reason beside every statement.

What Part III tests

Part III is where the Regents shifts from short answers to multi-step reasoning. The three questions each ask you to apply several ideas in the same problem: identify the geometric setup, choose the right formula or theorem, execute the algebra or the proof cleanly, and finish with a correctly stated answer. On this exam Part III covers the altitude of an isosceles triangle with right-triangle trigonometry, a formal congruent-triangles proof inside a parallelogram, and the volume of a composite prism-plus-pyramid solid with density.

Altitude of an isosceles triangle

In an isosceles triangle, the altitude drawn from the vertex angle to the base is also the perpendicular bisector of the base and the angle bisector of the vertex angle. That single segment cuts the isosceles triangle into two congruent right triangles — the fastest way to set up trigonometry when the vertex angle and altitude are known.

SOH-CAH-TOA in a right triangle

sin(θ) = opposite / hypotenuse, cos(θ) = adjacent / hypotenuse, tan(θ) = opposite / adjacent. On Q32, the half-triangle at the base has the half-vertex angle at A, its opposite leg is half the base, and its adjacent leg is the altitude — so tan gives the half-base and cos gives the slant leg.

Parallelogram from one pair of sides

If one pair of opposite sides of a quadrilateral is both parallel and congruent, the quadrilateral is a parallelogram. Once you have a parallelogram, both pairs of opposite sides are parallel and congruent, and — crucially for Q33 — the diagonals bisect each other.

SAS triangle congruence

Two triangles are congruent if two sides and the included angle of one triangle are congruent to two sides and the included angle of the other. Vertical angles at the intersection of two lines are always congruent, which supplies the included angle whenever a proof lives at the crossing of two segments.

Volume of a composite solid + density

For a solid glued together from simpler shapes, add the individual volumes. Rectangular prism: V = l · w · h. Rectangular pyramid: V = ⅓ · l · w · h. Then mass = total volume × density. Both formulas are on the reference sheet — do not write them from memory.

Part III — Questions 32–34

4 credits each · 3 questions · click any card to reveal the full worked solution.

Question 32

In isosceles ▵ABC, AD is an altitude drawn to base BC. If m∠BAC = 80° and AD = 8, determine and state the perimeter of ▵ABC, to the nearest tenth.

Step 1 — Use the altitude to split the triangle in half

Because ▵ABC is isosceles with ABAC, the altitude AD from the vertex angle is also the perpendicular bisector of BC and the bisector of ∠BAC. So ▵ABD and ▵ACD are congruent right triangles, and ∠BAD = ∠CAD = 80° / 2 = 40°.

Step 2 — Solve for half the base using tangent

In right ▵ABD, the 40° angle at A has opposite side BD and adjacent side AD = 8:

tan(40°) = BD / 8 ⇒ BD = 8 · tan(40°) ≈ 6.7128.

By symmetry, DC = BD, so BC = 2 · BD ≈ 13.4256.

Step 3 — Solve for a leg using cosine

In the same right triangle, the 40° angle has adjacent side AD = 8 and hypotenuse AB:

cos(40°) = 8 / ABAB = 8 / cos(40°) ≈ 10.4433.

Because ABAC, we also have AC ≈ 10.4433.

Step 4 — Add all three sides and round at the end

Perimeter = AB + AC + BC ≈ 10.4433 + 10.4433 + 13.4256 ≈ 34.3122.

Perimeter of ▵ABC ≈ 34.3

Rubric. Full 4 credits requires (a) the half-angle statement 80° ÷ 2 = 40° (or an equivalent argument), (b) a trig equation with substitution to find half the base or a leg, (c) both BC and AB computed, and (d) the perimeter rounded correctly at the very end. Rounding BD to 6.7 before doubling can drop a credit for a rounding-mid-work error.
Theory. The altitude from the vertex angle of an isosceles triangle is a triple-duty segment: perpendicular bisector of the base, angle bisector of the vertex angle, and median to the base. This is what lets you attack an isosceles-triangle problem with plain right-triangle trigonometry.
Question 33

In quadrilateral SMIL, diagonals IS and ML intersect at point E, MSIL, and MSIL. Prove: ▵MIE ≅ ▵LSE.

Plan

The given information — one pair of opposite sides both parallel and congruent — is the standard shortcut for a parallelogram. Once SMIL is a parallelogram, its diagonals bisect each other, which supplies two pairs of congruent sides at E. Vertical angles at E then close the proof by SAS.

Two-column proof

StatementsReasons
1. Quadrilateral SMIL with diagonals IS and ML intersecting at E, MSIL, and MSIL.1. Given.
2. SMIL is a parallelogram.2. A quadrilateral is a parallelogram if one pair of opposite sides is both parallel and congruent.
3. MELE and IESE.3. The diagonals of a parallelogram bisect each other.
4. ∠MEI ≅ ∠LES.4. Vertical angles are congruent.
5. ▵MIE ≅ ▵LSE.5. SAS (from steps 3 and 4: MELE, ∠MEI ≅ ∠LES, IESE). ■

∴ ▵MIE ≅ ▵LSE, as required.

Rubric. Full 4 credits requires (a) naming the “parallelogram from one pair” theorem, (b) using “diagonals bisect each other” to get two pairs of congruent segments, (c) citing vertical angles, and (d) naming SAS. Jumping straight to SAS without justifying why MELE and IESE drops credit.
Theory. The five “is-a-parallelogram” theorems (both pairs parallel, both pairs congruent, one pair parallel and congruent, diagonals bisect each other, both pairs of opposite angles congruent) are worth memorizing verbatim — they unlock most Regents parallelogram proofs in one or two lines.
Question 34

A solid glass trophy is composed of a rectangular prism and a rectangular pyramid. The prism has a length of 12 cm, a width of 6 cm, and a height of 3 cm. The pyramid sits on top of the prism and shares its 12 cm × 6 cm face; the pyramid’s height is 10 cm. If the density of glass is 2.5 g/cm³, determine and state the mass of the trophy, in grams.

Step 1 — Volume of the rectangular prism

Vprism = l · w · h = 12 · 6 · 3 = 216 cm³.

Step 2 — Volume of the rectangular pyramid

Vpyramid = ⅓ · (base area) · height = ⅓ · (12 · 6) · 10 = ⅓ · 72 · 10 = 240 cm³.

Step 3 — Total volume of the trophy

Vtotal = Vprism + Vpyramid = 216 + 240 = 456 cm³.

Step 4 — Mass from density

Mass = volume × density = 456 · 2.5 = 1140 g.

Mass of the trophy = 1140 grams

Rubric. Full 4 credits requires (a) the prism formula with substituted numbers, (b) the pyramid formula with the ⅓, (c) the two volumes added, and (d) the density multiplication with correct units. Forgetting the ⅓ on the pyramid is the single most common Regents mistake on solid-volume problems.
Theory. A rectangular pyramid’s volume is exactly one-third that of a rectangular prism with the same base and height. For any composite solid, add the piece volumes first and multiply by density last — never re-do the density calculation on each piece unless different materials are used.

Part III answer key

Big ideas to remember on test day

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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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