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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 PDFHow 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.
Sections
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 32In 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 AB ≅ AC, 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 / AB ⇒ AB = 8 / cos(40°) ≈ 10.4433.
Because AB ≅ AC, 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
Question 33In quadrilateral SMIL, diagonals IS and ML intersect at point E, MS ∥ IL, and MS ≅ IL. 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
| Statements | Reasons |
|---|---|
| 1. Quadrilateral SMIL with diagonals IS and ML intersecting at E, MS ∥ IL, and MS ≅ IL. | 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. ME ≅ LE and IE ≅ SE. | 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: ME ≅ LE, ∠MEI ≅ ∠LES, IE ≅ SE). ■ |
∴ ▵MIE ≅ ▵LSE, as required.
Question 34A 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
Part III answer key
- Q32. Perimeter = 2 · (8 / cos 40°) + 2 · (8 tan 40°) ≈ 20.887 + 13.426 ≈ 34.3
- Q33. One pair of opposite sides parallel and congruent → parallelogram → diagonals bisect each other → SAS → ▵MIE ≅ ▵LSE. ■
- Q34. (12 · 6 · 3) + (⅓ · 12 · 6 · 10) = 216 + 240 = 456 cm³; 456 · 2.5 = 1140 grams
Big ideas to remember on test day
- The altitude of an isosceles triangle is triple-duty. It is the perpendicular bisector of the base, the angle bisector of the vertex angle, and the median to the base — the single fact that unlocks Q32.
- Never round in the middle. Q32 loses a credit if you round BD or AB before summing to get the perimeter.
- Memorize the five parallelogram tests. One pair of opposite sides parallel and congruent → parallelogram was the pivot on Q33.
- Diagonals of a parallelogram bisect each other. Two free pairs of congruent segments — use them.
- Vertical angles are always available. Whenever two segments cross, the two angles opposite each other at the crossing are congruent. Costs nothing to cite, often finishes a proof.
- Reference sheet before every solid. Prism V = lwh and pyramid V = ⅓ lwh live on the sheet — the ⅓ on the pyramid is the most-missed detail on the exam.
- Composite solids: add volumes, then multiply by density once. Never re-apply density to each piece unless the materials are different.
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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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