Geometry Regents High School Examination Study Notes

Examination Overview and Administrative Regulations

• Exam Identity: The University of the State of New York Regents High School Examination in Geometry, held on Wednesday, January 21, 2026, from $9:15\,a.m.$ to $12:15\,p.m.$

• Prohibited Items: The possession or use of any communications device is strictly prohibited. Use of such a device, even briefly, results in exam invalidation and a zero score.

• Required Tools: Students must have access to a graphing calculator, a straightedge (ruler), and a compass.

• Exam Structure: The test consists of four parts with a total of $35$ questions. All questions must be answered.

  • Part I: $24$ multiple-choice questions ($2$ credits each, no partial credit).

  • Part II: $7$ constructed-response questions ($2$ credits each).

  • Part III: $3$ constructed-response questions ($4$ credits each).

  • Part IV: $1$ multi-step proof/constructed-response question ($6$ credits).

• Recording Standards: Answers for Part I go on a separate sheet. Parts II, III, and IV are written in the booklet. Use pen for text and pencil for graphs/drawings. Work must show all steps, including formula substitutions and diagrams.

• Materials: Scrap paper is not permitted; the booklet and a provided perforated leaf of graph paper should be used instead.

• Integrity: A student declaration must be signed after completion to certify no unlawful knowledge or assistance was exchanged.

Geometric Transformations and Rigid Motions

• Reflection Identifying Rigid Motion: For triangle $\Delta RST$ and its image $\Delta R'S'T'$, if the coordinates show a flip across the vertical axis, a reflection over the $y$-axis is the sufficient rigid motion to prove congruence.

• Rotational Symmetry: A regular polygon carries onto itself when rotated by an angle of $\frac{360^{\circ}}{n}$, where $n$ is the number of sides. For a rotation of $60^{\circ}$, the polygon must be a regular hexagon since $\frac{360^{\circ}}{6} = 60^{\circ}$.

• Dilation and Centers: Dilations involve a scale factor ($k$) and a center of dilation. If $\Delta XYZ$ is an image of $\Delta ABC$ with a scale factor of $k = \frac{1}{2}$, the center is found by connecting corresponding vertices (e.g., $A$ to $X$) and finding the intersection of those lines.

• Sequences of Transformations: Multiple transformations can map a figure. In one example, $\Delta ABC$ is mapped to $\Delta A'B'C'$ via translation, and then to $\Delta A''B''C''$ via a line reflection.

• Properties of Rigid Motions: Rigid motions (translations, reflections, rotations) preserve segment length and angle measure.

  • Statement: Segment $AB$ is always congruent to segment $A'B'$ after a rigid motion.

  • Rigid motions do not necessarily preserve parallelism between the original and the image (unless it is a specific translation or $180^{\circ}$ rotation).

• Dilation of Lines: If a line $y = 2x - 1$ is dilated by a scale factor of $3$ centered at the origin, the slope remains the same ($m = 2$), but the $y$-intercept is multiplied by the scale factor. The new equation becomes $y = 2x - 3$.

• Coordinate Transformation Procedures: To perform a reflection over the $x$-axis followed by a translation of $3$ units right ($+3$ to $x$) and $2$ units down ($-2$ to $y$) on vertices $(2, 1)$, $(0, 3)$, and $(-2, -1)$, the steps are:

  1. Reflect over $x$-axis: $(x, -y) \rightarrow (2, -1), (0, -3), (-2, 1)$.

  2. Translate: $(x+3, y-2) \rightarrow (5, -3), (3, -5), (1, -1)$.

Three-Dimensional Geometry and Volumetric Analysis

• Solids of Revolution: Rotating a right triangle about one of its legs creates a cone.

  • Example: Rotating a triangle with height $4\,cm$ and base $7\,cm$ about the $4\,cm$ side results in a cone with height $h = 4\,cm$ and radius $r = 7\,cm$.

• Volume Formulas and Practical Application:

  • Cone Volume: $V = \frac{1}{3} \pi r^2 h$.

  • Sphere Volume: $V = \frac{4}{3} \pi r^3$.

  • Rectangular Prism Volume: $V = l \times w \times h$.

  • Pyramid Volume: $V = \frac{1}{3} B h$ (where $B$ is the area of the base).

• Sandpile Logic (Lucy's Wagon):

  • Sandpile modeled as a cone ($r = 3.5\,ft, h = 3.5\,ft$).

  • $V = \frac{1}{3} \pi (3.5)^2 (3.5) \approx 44.899\,ft^3$.

  • Wagon capacity = $5\,ft^3$.

  • Trips required: $\frac{44.899}{5} \approx 8.97$, which rounds up to $9$ trips.

• Density and Mass:

  • Topsoil Calculation: Cylinder with $d = 10\,in$ ($r = 5$) and $h = 15\,in$. Volume $V = \pi (5)^2 (15) \approx 1178.097\,in^3$. With density $0.0231\,lb/in^3$, weight is $1178.097 \times 0.0231 \approx 27\,lb$.

  • Trophy Mass: Composite solid of a prism ($12 \times 6 \times 3 = 216\,cm^3$) and a pyramid ($\frac{1}{3} \times (12 \times 6) \times 10 = 240\,cm^3$). Total Volume = $456\,cm^3$. Mass = $456 \times 2.5\,g/cm^3 = 1140\,g$.

• Sphere Diameter: For $V = 333\,cm^3$, solve for $r$ using $333 = \frac{4}{3} \pi r^3$. $r^3 = \frac{333 \times 3}{4\pi} \approx 79.497$. $r \approx 4.30\,cm$. Diameter $d = 2r \approx 8.6\,cm$.

Triangle Properties, Similarity, and Proofs

• Isosceles Triangle Angles: In $\Delta AHP$, if $AH \cong PH$, the base angles $\angle A$ and $\angle P$ are congruent. If $m\angle A = (2x + 12)^{\circ}$ and $m\angle P = (3x - 8)^{\circ}$, then $2x + 12 = 3x - 8$, so $x = 20$. Base angles are $(2(20) + 12) = 52^{\circ}$. The exterior angle $\angle CHP = m\angle A + m\angle P = 52 + 52 = 104^{\circ}$.

• Parallel Lines and Similarity: In a triangle where a line segment is parallel to the base, similar triangles are formed ($\Delta ABC \sim \Delta EDC$). Vertical angles and alternate interior angles are congruent. However, congruence ($\Delta ABC \cong \Delta EDC$) is NOT always true unless a specific side length equality is given.

• Side Splitter Theorem: In $\Delta ABC$ with $DE \parallel AC$, the segments are proportional: $\frac{BD}{DA} = \frac{BE}{EC}$. If $BD=9, DA=3, EC=4$, then $\frac{9}{3} = \frac{BE}{4} \Rightarrow 3 = \frac{BE}{4} \Rightarrow BE = 12$. Total length $BC = BE + EC = 12 + 4 = 15$.

• Right Triangle Relationships:

  • Cofunctions: $\cos(A) = \sin(B)$ if $A + B = 90^{\circ}$. Thus, if $\cos(5x + 7)^{\circ} = \sin(3x + 3)^{\circ}$, then $(5x + 7) + (3x + 3) = 90$.

  • Geometric Mean (Altitude): In $\Delta ABC$ with altitude $BD$ to hypotenuse $AC$, $BD^2 = AD \times CD$. Also, legs follow $AB^2 = AD \times AC$. If $AD = 3$ and $CD = 12$, then $AC = 15$. $AB^2 = 3 \times 15 = 45$, so $AB = \sqrt{45} = 3\sqrt{5}$.

• Centroids: The centroid $X$ divides medians in a $2:1$ ratio. For segment $GXM$ (median), $GX = 2(XM)$, meaning $\frac{MX}{GX} = \frac{1}{2}$. Also, for median $TXE$, $TX = 2(EX)$, so $\frac{TX}{EX} = \frac{2}{1}$.

• Perimeter and Trigonometry: In isosceles $\Delta ABC$ with altitude $AD=8$ and vertex angle $\angle BAC=80^{\circ}$, the altitude bisects the angle into two $40^{\circ}$ angles. Using $\tan(40^{\circ}) = \frac{BD}{8}$, $BD \approx 6.71$. $BC = 2 \times 6.71 = 13.42$. Using $\cos(40^{\circ}) = \frac{8}{AB}$, $AB \approx 10.44$. Perimeter $= 10.44 + 10.44 + 13.42 \approx 34.3$.

Circle Geometry

• Arc Length: Formula $s = r\theta$ where $\theta$ is in radians, or $s = \frac{\text{Angle}}{360} \times 2\pi r$. For radius $8\,cm$ and angle $140^{\circ}$, length $AB = \frac{140}{360} \times 2\pi(8) \approx 19.55$, rounded to $20\,cm$.

• Circle Equation: Prepared in the form $(x - h)^2 + (y - k)^2 = r^2$. For $x^2 - 16x + y^2 + 20y = -155$:

  1. $(x^2 - 16x + 64) + (y^2 + 20y + 100) = -155 + 64 + 100$.

  2. $(x - 8)^2 + (y + 10)^2 = 9$.

  3. Center is $(8, -10)$, radius is $\sqrt{9} = 3$.

• Annulus (Deck Area): A pool ($d = 24\,ft, r = 12\,ft$) is surrounded by an $8\,ft$ deck. Outer radius $R = 12 + 8 = 20\,ft$. Area of deck $= \pi R^2 - \pi r^2 = \pi(20^2) - \pi(12^2) = 400\pi - 144\pi = 256\pi \approx 804\,ft^2$.

• Inscribed Angles: If $\angle ABC$ is a right angle inscribed in a circle, it must intercept a semicircle because an inscribed angle is half the measure of its intercepted arc ($90^{\circ} \times 2 = 180^{\circ}$). Therefore, the hypotenuse $AC$ must be the diameter.

Trigonometry and Analytic Geometry

• Building a Shed Face (Trig): Face consists of a rectangle and isosceles triangle. Height of triangle $= 6\,ft$, base $= 10\,ft$ (half base $= 5$). $\tan(\angle EGD) = \frac{\text{opposite}}{\text{adjacent}}$. In the right triangle forming the roof half, the angle at the base is $\arctan(\frac{6}{5}) \approx 50.19^{\circ}$. Thus $m\angle EGD \approx 50^{\circ}$.

• Lines and Slopes: A line through $(-1, -4)$ and $(3, -1)$ has a slope $m = \frac{-1 - (-4)}{3 - (-1)} = \frac{3}{4}$. A perpendicular line must have a slope that is the negative reciprocal, $m_{\perp} = -\frac{4}{3}$. The equation uses point-slope form: $y - y_1 = m(x - x_1)$, resulting in $y + 1 = -\frac{4}{3}(x - 3)$.

• Population Density: Formula is $\text{Density} = \frac{\text{Population}}{\text{Land Area}}$.

  • NY: $\frac{20,201,249}{47,126} \approx 428.6$

  • NJ: $\frac{9,288,994}{7,354} \approx 1,263.1$

  • CT: $\frac{3,605,944}{4,842} \approx 744.7$

  • PA: $\frac{13,002,700}{44,743} \approx 290.6$

  • Order (Smallest to Largest): PA, NY, CT, NJ.

• Segment Partitioning: Point $A$ divides segment $RZ$ with $R(6, 6)$ and $Z(-12, -3)$ in ratio $5:4$.

  • Total parts = $9$.

  • $x = 6 + \frac{5}{9}(-12 - 6) = 6 + \frac{5}{9}(-18) = 6 - 10 = -4$.

  • $y = 6 + \frac{5}{9}(-3 - 6) = 6 + \frac{5}{9}(-9) = 6 - 5 = 1$.

  • Coordinates are $(-4, 1)$.

Quadrilaterals and Coordinate Proofs

• Rhombus Proof: A parallelogram is a rhombus if it has at least one of the following:

  • Perpendicular diagonals.

  • Diagonals that bisect the vertex angles.

  • Two consecutive sides are congruent ($AD \cong DC$).

• Trapezoid and Midsegment Proof (Question 35):

  • Trapezoid ABCD: Vertices $A(-3, 1), B(-3, -7), C(6, 5), D(0, 5)$.

  • Slope of $AD = \frac{5 - 1}{0 - (-3)} = \frac{4}{3}$.

  • Slope of $BC = \frac{5 - (-7)}{6 - (-3)} = \frac{12}{9} = \frac{4}{3}$.

  • Since slopes are equal, $AD \parallel BC$. One pair of parallel sides proves it is a trapezoid.

  • Line EF: Endpoints $E(-3, -3)$ and $F(3, 5)$. Slope $m_{EF} = \frac{5 - (-3)}{3 - (-3)} = \frac{8}{6} = \frac{4}{3}$. This confirms $EF \parallel AD$ and $EF \parallel BC$.

  • Midsegment Property Check: Length $AD = \sqrt{(0 - (-3))^2 + (5 - 1)^2} = \sqrt{3^2 + 4^2} = 5$. Length $BC = \sqrt{(6 - (-3))^2 + (5 - (-7))^2} = \sqrt{9^2 + 12^2} = 15$. Length $EF = \sqrt{(3 - (-3))^2 + (5 - (-3))^2} = \sqrt{6^2 + 8^2} = 10$.

  • Verification: $10 = \frac{1}{2}(5 + 15) \Rightarrow 10 = 10$. This proves $EF$ is the midsegment.

Questions & Discussion

• Question 2: Which regular polygon would carry onto itself after a rotation of $60^{\circ}$?

  • Answer: (2) Hexagon.

• Question 10: What is $m\angle EGD$ to the nearest degree for the shed face?

  • Answer: (1) $50^{\circ}$.

• Question 14: Volume of a sphere is $333\,cm^3$. What is the diameter to the nearest tenth?

  • Answer: (3) $8.6\,cm$.

• Question 31: Explain why $AC$ must be a diameter of the circle if $\angle ABC$ is a right angle.

  • Explanation: An inscribed angle measures half its intercepted arc. A $90^{\circ}$ angle intercepts a $180^{\circ}$ arc. A $180^{\circ}$ arc is a semicircle, and the chord that defines a semicircle is the diameter.