SOMATH Journal · Regents Geometry · Part IV
August 2026 Geometry Regents: Part IV
This August 2026 Geometry Regents Part IV walkthrough covers a coordinate proof that SHED is a parallelogram but neither a rectangle nor a rhombus. Find the original questions and diagrams below, then select the blue Answer button for each original SOMATH solution: identify the idea, show the steps, and check the result.
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Questions and original diagrams are reproduced from the NYSED August 2026 Geometry exam, with web formatting and mathematical notation normalized. The explanations are newly written by SOMATH, not NYSED solutions or an official scoring rubric; results were checked against the NYSED scoring key and rating guide.
Open the original exam PDF for the full booklet and reference sheet.
Show the reasoning, substitutions, and conclusions, not just an answer. The NYSED rating guide awards credit for the required work; for a numerical question, a correct answer alone receives only 1 credit. Keep construction marks visible and close each proof with an explicit conclusion.
Coordinate proof: classifying SHEDQuestion 35
The coordinates of the vertices of quadrilateral \(SHED\) are \(S(0,0)\), \(H(3,4)\), \(E(8,0)\), and \(D(5,-4)\).
Prove \(SHED\) is a parallelogram.
[The use of the set of axes on the next page is optional.]
Prove \(SHED\) is not a rectangle.
[The use of the set of axes on the next page is optional.]
Prove \(SHED\) is not a rhombus.
[The use of the set of axes below is optional.]
Answer
Answer: \(SHED\) is a parallelogram, but it is neither a rectangle nor a rhombus.
- Prove it is a parallelogram. Compute the first pair of opposite-side slopes: \(m_{SH}=\frac{4-0}{3-0}=\frac43\) and \(m_{ED}=\frac{-4-0}{5-8}=\frac43\). Therefore \(SH\parallel ED\).
- For the other pair, \(m_{HE}=\frac{0-4}{8-3}=-\frac45\) and \(m_{DS}=\frac{0-(-4)}{0-5}=-\frac45\). Therefore \(HE\parallel DS\). Both pairs of opposite sides are parallel, so \(SHED\) is a parallelogram.
- Prove it is not a rectangle. Adjacent sides \(SH\) and \(HE\) have slope product \(\frac43(-\frac45)=-\frac{16}{15}\), not \(-1\). They are not perpendicular, so the angle at \(H\) is not a right angle. A rectangle requires four right angles; therefore \(SHED\) is not a rectangle.
- Prove it is not a rhombus. Compare adjacent side lengths: \(SH=\sqrt{3^2+4^2}=5\), while \(HE=\sqrt{(8-3)^2+(0-4)^2}=\sqrt{41}\).
- Because \(5\ne\sqrt{41}\), the quadrilateral does not have four equal sides. Therefore \(SHED\) is not a rhombus.
Another way to check your classification
The diagonals have the same midpoint: both \(\overline{SE}\) and \(\overline{HD}\) have midpoint \((4,0)\). Diagonals that bisect each other confirm a parallelogram. Their lengths are \(SE=8\) and \(HD=\sqrt{68}\), which are unequal, confirming it is not a rectangle. Since \(SE\) is horizontal and \(HD\) is not vertical, the diagonals are not perpendicular, also confirming that this parallelogram is not a rhombus. These are alternative checks, not extra requirements once the first proof is complete.
SOMATH takeaway: This question needs three conclusions supported by three tests. A correct graph is useful, but visual appearance alone is not a coordinate proof.
Answer key
Keep the key closed while you practice. Each question above includes the reasoning behind its answer.
Show answer key
Q35: \(SHED\) is a parallelogram, but it is neither a rectangle nor a rhombus.
Class review: what to remember
Coordinate classification uses different tools for different properties. Equal opposite slopes establish parallel sides. Negative reciprocal adjacent slopes establish perpendicular sides. The distance formula establishes equal or unequal side lengths.
To disprove a special classification, one failed required property is enough. One non-right angle rules out a rectangle; one unequal adjacent pair rules out a rhombus. Still prove the positive classification separately: being neither a rectangle nor a rhombus does not by itself prove a parallelogram.
Quick questions and answers
What is the answer to August 2026 Geometry Regents Question 35?
SHED is a parallelogram, but it is not a rectangle and not a rhombus. Equal opposite-side slopes prove parallelism; adjacent slopes are not negative reciprocals, and adjacent sides have unequal lengths.
How do slopes prove SHED is a parallelogram?
SH and ED both have slope 4/3. HE and DS both have slope −4/5. Therefore both pairs of opposite sides are parallel.
Why is SHED not a rectangle?
Adjacent slopes 4/3 and −4/5 have product −16/15, not −1. Those adjacent sides are not perpendicular.
Why is SHED not a rhombus?
Adjacent side lengths are 5 and square root of 41. They are unequal, so the figure does not have four equal sides.
Is the coordinate graph required?
No. The exam makes the graph optional. Correct slope and distance calculations with explicit conclusions supply a coordinate proof.