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

Regents Geometry June 2026 — Part III Answers & Explanations

Free worked solutions for every Part III question on the June 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 June 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.

Sections

  1. What Part III tests
  2. Part III — Questions 32–34 (click to reveal)
  3. Part III answer key
  4. Big ideas to remember
  5. More Regents Geometry help

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 right-triangle trigonometry with two angles from one point, three-dimensional volume & density with unit conversion, and a formal similar-triangles proof.

Angle of elevation vs. angle of depression

Both are measured from a horizontal line of sight at the observer. If you look up to a point, you have an angle of elevation; if you look down, an angle of depression. When one observer sees both the top and the bottom of something across a horizontal gap, split the target height at the eye-line: the top part is d · tan(elevation) and the bottom part is d · tan(depression).

Volume of a pyramid

V = ⅓ · B · h, where B is the area of the base and h is the perpendicular height (from apex to base). For a square base with side s, B = s². This is on the Geometry Regents reference sheet.

Density & unit conversion

Mass = volume × density. Watch the units: if density is given in g/cm³, convert kilograms to grams (× 1000) before dividing. “Maximum number of pieces” always uses the floor function — you cannot make 31.25 pyramids, only 31.

Isosceles triangle base-angle theorem

In a triangle, if two sides are congruent, the angles opposite those sides are congruent. Any perpendicular dropped from a point on one leg to the base creates a right triangle whose acute angle at the base is the base angle of the isosceles triangle.

AA (Angle-Angle) similarity

Two triangles are similar if two pairs of corresponding angles are congruent. Because triangle angle sums are 180°, the third pair is automatically congruent. Once similar, all pairs of corresponding sides are in the same ratio — the standard way to conclude a proportion like DE / BE = HF / CF.

Part III — Questions 32–34

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

Question 32

Maria wants to determine the height of the building across the street from her position, M. The angle of elevation from M to the top of the building, T, is 36°. From M, the angle of depression to the base of the building, B, is 18°. The buildings are 80 feet apart. Determine and state, to the nearest foot, the height of the building TB across the street.

Step 1 — Draw the horizontal line of sight

Let H be the point on the far building directly across from M. Then MH is horizontal, MH = 80 ft (the given gap between buildings), and MH meets TB at a right angle. The building's height TB splits into two pieces at H: an upper piece TH (above the line of sight) and a lower piece HB (below).

Step 2 — Use the elevation to find TH

In right ▵MHT, the 36° elevation angle sits at M with opposite side TH and adjacent side MH = 80:

tan(36°) = TH / 80 ⇒ TH = 80 · tan(36°) ≈ 58.12 ft.

Step 3 — Use the depression to find HB

In right ▵MHB, the 18° depression angle sits at M with opposite side HB and adjacent side MH = 80:

tan(18°) = HB / 80 ⇒ HB = 80 · tan(18°) ≈ 25.99 ft.

Step 4 — Add the two pieces

TB = TH + HB ≈ 58.12 + 25.99 ≈ 84.12 ft.

TB ≈ 84 feet

Rubric. Full 4 credits requires both trig set-ups shown, both partial heights computed, the sum written, and the final rounded answer. Round only at the very end — rounding TH and HB to whole numbers before adding drops a credit.
Theory. tan(θ) = opposite / adjacent in a right triangle. Angles of elevation and depression are both measured from the same horizontal line, so a single observer looking both up and down at a target creates two right triangles that share the horizontal leg.
Question 33

An artist uses clay to make solid pyramids with a square base whose sides measure 12 cm, and each pyramid has height 8 cm. The density of the clay is 1.25 g/cm³. If the artist has 15 kilograms of clay, determine and state the maximum number of pyramids that can be made.

Step 1 — Volume of one pyramid

V = ⅓ · (side)² · height = ⅓ · 12² · 8 = ⅓ · 144 · 8 = 384 cm³.

Step 2 — Mass of one pyramid

Mass = volume × density = 384 · 1.25 = 480 g per pyramid.

Step 3 — Convert 15 kg to grams

15 kg = 15 · 1000 = 15,000 g.

Step 4 — Divide and take the floor

Number of pyramids = 15,000 ÷ 480 = 31.25.

You cannot make a fraction of a pyramid, and the problem asks for the maximum. Round down.

The artist can make at most 31 pyramids.

Rubric. Full 4 credits requires (a) the volume formula with the ⅓, (b) the density calculation, (c) the kg-to-g conversion, and (d) the floor. Writing “31.25 pyramids” as the final answer costs a credit — the interpretation matters.
Theory. A pyramid’s volume is one-third the volume of the prism with the same base and height. Always look on the reference sheet before writing pyramid or cone formulas — forgetting the ⅓ is one of the most common Regents Geometry errors.
Question 34

Given: ▵ABC, AB ≅ AC, DE ⊥ BC, and HF ⊥ BC. (Points D on AB, E on BC, H on AC, F on BC.) Prove: DE / BE = HF / CF.

Two-column proof

StatementsReasons
1. ▵ABC with AB ≅ AC, DE ⊥ BC, and HF ⊥ BC.1. Given.
2. ∠DEB and ∠HFC are right angles.2. Perpendicular segments form right angles.
3. ∠DEB ≅ ∠HFC.3. All right angles are congruent.
4. ∠B ≅ ∠C.4. Base angles of an isosceles triangle are congruent (from AB ≅ AC).
5. ▵DBE ∼ ▵HCF.5. AA similarity (from steps 3 and 4).
6. DE / BE = HF / CF.6. Corresponding sides of similar triangles are proportional. ■

∴ DE / BE = HF / CF, as required.

Rubric. Full 4 credits requires (a) both right angles established, (b) the base-angle-theorem step (this is the pivot of the proof), (c) AA similarity named, and (d) the concluding proportion. Common credit loss: skipping the “right angles are congruent” step or forgetting to name the similarity criterion.
Theory. Whenever a Geometry proof asks for a proportion of segments, aim for two similar triangles. AA is usually the easiest similarity criterion to reach, and the isosceles-triangle base-angle theorem is often the trick that unlocks the second angle pair.

Part III 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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