August 2026 Geometry Regents: Part I

Answer: (3), a translation.

1. Congruent triangles have equal corresponding side lengths and angle measures. A rigid motion preserves both.
2. A translation slides every point the same distance in the same direction. Here the rule is \((x,y)\mapsto(x+3,y+4)\).
3. The stretches change lengths, and a dilation with scale factor 3 triples every side. Those are not congruence-preserving transformations.

SOMATH takeaway: Ask whether the transformation changes size. A slide, turn, or reflection preserves it.

Answer: (2), \(44^\circ\).

1. Look from angle \(C\): \(AB\) is opposite and \(AC\) is the hypotenuse.
2. Use sine: \(\sin C=\frac{10.3}{14.8}\).
3. In degree mode, \(C=\sin^{-1}(10.3/14.8)\approx44.10^\circ\). Round to \(44^\circ\).

SOMATH takeaway: Name the sides relative to the requested angle before choosing sine, cosine, or tangent.

Answer: (4), \(\angle B\cong\angle Y\).

1. Read corresponding vertices in order: \(A\leftrightarrow K\), \(B\leftrightarrow L\), and \(C\leftrightarrow M\).
2. For the other mapping, \(X\leftrightarrow K\), \(Y\leftrightarrow L\), and \(Z\leftrightarrow M\).
3. Both rotations and translations preserve angle measures. Thus \(\angle B\cong\angle L\) and \(\angle Y\cong\angle L\), giving \(\angle B\cong\angle Y\).

SOMATH takeaway: Matching the same target vertex is more reliable than matching letters that look familiar.

Answer: (3), 235.7 grams.

1. The radius is half the diameter: \(r=3.5/2=1.75\) cm.
2. The sphere's volume is \(V=\frac43\pi r^3=\frac43\pi(1.75)^3\).
3. Mass equals density times volume: \(m=10.5\left[\frac43\pi(1.75)^3\right]\approx235.717\) grams.
4. Round only the final result: 235.7 grams.

SOMATH takeaway: Using the diameter as the radius makes the volume eight times too large.

Answer: (1), \(AE=3(FE)\).

1. A segment from a vertex to the midpoint of the opposite side is a median. Therefore \(AE\) and \(BD\) are medians and \(F\) is the centroid.
2. A centroid divides each median in a \(2:1\) ratio, with the longer piece next to the vertex. Thus \(AF=2(FE)\).
3. The whole median is \(AE=AF+FE=3(FE)\). On the other median, \(BF=2(FD)\), not \(3(FD)\).

SOMATH takeaway: Distinguish a whole median, which is three short pieces, from the vertex-to-centroid piece, which is two.

Answer: (3), cylinder.

1. The fixed side is the axis of rotation. Every point on the opposite side travels around that axis in a circle.
2. Those circles have the same radius all along the axis, so the swept solid has two circular bases and a constant width.
3. That solid is a cylinder. Its height and radius each equal the side length of the square.

SOMATH takeaway: Imagine the shape swept out during the rotation, not just the square's starting position.

Answer: (3), \(80^\circ\).

1. The exterior angle at \(U\) equals the sum of the two remote interior angles: \(m\angle R+m\angle T=120^\circ\).
2. Let \(m\angle T=x\). Then \(m\angle R=2x\), so \(3x=120\) and \(x=40\).
3. The question asks for \(\angle R\), so double 40: \(m\angle R=80^\circ\).

SOMATH takeaway: After solving for a variable, check whether it represents the requested angle.

Answer: (1), \(y=-1\).

1. The upper horizontal edge lies at \(y=4\), and the lower one lies at \(y=-6\). Their halfway height is \(\frac{4+(-6)}2=-1\).
2. Reflection across \(y=-1\) pairs the upper and lower vertices while keeping their x-coordinates unchanged.
3. The two middle vertices lie on \(y=-1\), so they stay fixed. Every edge lands on a matching edge.

SOMATH takeaway: A symmetry line must work for the entire shape. It does not have to be a coordinate axis.

Answer: (2), \(\angle TAH\).

1. Focus on the smaller triangle \(AHT\), because it contains the requested angle.
2. Since \(AH=TH\), this triangle is isosceles with base \(AT\). The base angles at \(A\) and \(T\) are congruent.
3. Those angles are \(\angle TAH\) and \(\angle HTA\). The larger triangle being scalene does not change this fact.

SOMATH takeaway: Equal sides face equal angles. Trace each angle's middle letter to locate its vertex.

Answer: (1), 24.

1. The rectangular base area is \(B=16(12)=192\).
2. A pyramid has volume \(V=\frac13Bh\), so \(1536=\frac13(192)h=64h\).
3. Divide by 64: \(h=1536/64=24\). Substitution gives \(\frac13(192)(24)=1536\), as required.

SOMATH takeaway: A pyramid uses one third of base area times height. Omitting that factor gives the prism answer.

Answer: (3), 19 inches.

1. Relative to angle \(B\), \(BC\) is the adjacent leg and \(AB\) is the hypotenuse.
2. Use \(\cos37^\circ=\frac{BC}{24}\). Multiply by 24 to get \(BC=24\cos37^\circ\).
3. This is approximately \(19.167\), which rounds to 19 inches. It is shorter than the 24-inch hypotenuse, as it should be.

SOMATH takeaway: Cosine connects the adjacent leg and hypotenuse; keep the calculator in degree mode.

Answer: (4), \((1,3)\).

1. The ratio \(3:1\) divides the whole segment into four equal parts. Point \(S\) is three fourths of the way from \(R\) to \(T\).
2. The displacement from \(R\) to \(T\) is \((3-(-5),2-6)=(8,-4)\).
3. Three fourths of that displacement is \((6,-3)\). Add it to \(R\): \((-5+6,6-3)=(1,3)\).

SOMATH takeaway: The fraction from the first endpoint is 3/(3+1), not 3/1. The point should be closer to T.

Answer: (2), reflect across the x-axis, then translate 8 units right.

1. The combined rule for choice (2) is \((x,y)\mapsto(x+8,-y)\). Reflection changes the sign of y; the translation then adds 8 to x.
2. Read the vertices: \(A(-6,6)\), \(B(-3,6)\), \(C(-1,2)\), and \(D(-7,2)\).
3. They map to \(E(2,-6)\), \(F(5,-6)\), \(G(7,-2)\), and \(H(1,-2)\), respectively. All four points match the graph.

SOMATH takeaway: Check the labeled correspondence, not just whether the transformed shape reaches the same quadrant.

Answer: (2), \(102^\circ\).

1. In the order \(P,A,R,K\), angles \(P\) and \(K\) are adjacent. Adjacent parallelogram angles are supplementary.
2. Write \((3x+42)+(5x-22)=180\). Then \(8x+20=180\), so \(x=20\).
3. Angle \(P\) measures \(3(20)+42=102^\circ\). Angle \(R\) is opposite \(P\), so \(m\angle R=102^\circ\).

SOMATH takeaway: Adjacent angles add to 180°; opposite angles are equal. Identify which relationship you have before writing an equation.

Answer: (1), \(10+5\sqrt2\).

1. The vertical leg \(CE\) has length \(4-(-1)=5\). The horizontal leg \(EM\) has length \(3-(-2)=5\).
2. By the Pythagorean theorem, \(CM=\sqrt{5^2+5^2}=\sqrt{50}=5\sqrt2\).
3. Add all three side lengths: \(P=5+5+5\sqrt2=10+5\sqrt2\).

SOMATH takeaway: Perimeter adds lengths. The answer 12.5 is the triangle's area, not its perimeter.

Answer: (2), \(-\frac13\).

1. First find the slope of \(MT\): \(m=\frac{8-(-7)}{3-(-2)}=\frac{15}{5}=3\).
2. A perpendicular nonvertical line has the negative reciprocal slope: \(-1/3\).
3. The word “bisector” tells us where the line passes, but no midpoint calculation is needed when only its slope is requested.

SOMATH takeaway: A negative reciprocal changes the sign and flips the fraction. Simply changing 3 to −3 is not enough.

Answer: (3), \(9\sqrt3\).

1. Triangle \(SLW\) is a \(45^\circ\)-\(45^\circ\)-\(90^\circ\) triangle with hypotenuse \(SW=18\sqrt2\). Each leg, including \(LW\), is 18.
2. In triangle \(LOW\), \(LW=18\) is now the hypotenuse. Side \(OW\) is adjacent to the \(30^\circ\) angle at \(W\).
3. Thus \(OW=18\cos30^\circ=18(\sqrt3/2)=9\sqrt3\).

SOMATH takeaway: A shared side can be a leg in one triangle and the hypotenuse in another. Relabel its role when switching triangles.

Answer: (2), \(x^2+(y+4)^2=225\).

1. The standard form is \((x-h)^2+(y-k)^2=r^2\), where \((h,k)\) is the center.
2. Here \(h=0\), \(k=-4\), and \(r=30/2=15\).
3. Substitute: \(x^2+(y-(-4))^2=15^2\), or \(x^2+(y+4)^2=225\).

SOMATH takeaway: The center's y-coordinate is −4 even though the equation contains y+4. Also square the radius, not the diameter.

Answer: (1), approximately 7 cups.

1. The batter fills a rectangular prism with dimensions \(8\times8\times1.5\), not the full 2-inch depth.
2. Its volume is \(V=8(8)(1.5)=96\) cubic inches.
3. Divide by cubic inches per cup: \(96/14.4375\approx6.65\) cups. The closest choice is 7.

SOMATH takeaway: Use the actual filled height, then divide by the volume of one cup.

Answer: (4), 8.5.

1. Start with right triangle \(SDA\). Its hypotenuse is 34 and one leg is 16, so \(AD=\sqrt{34^2-16^2}=\sqrt{900}=30\).
2. An altitude to a right triangle's hypotenuse satisfies \(SD^2=WD\cdot AD\). This follows from the similarity of the two smaller triangles.
3. Substitute: \(16^2=WD(30)\). Thus \(WD=256/30\approx8.533\), which rounds to 8.5.

SOMATH takeaway: Find the missing hypotenuse piece first. The altitude is the geometric mean of the two pieces, not their arithmetic mean.

Answer: (4), the image's perimeter is half the original perimeter.

1. A scale factor \(k=\frac12\) halves every corresponding side length, so adding the image's sides gives half the original perimeter.
2. Angles are unchanged under dilation, so choice (2) is not correct.
3. Area scales by \(k^2=\frac14\), not \(\frac12\). Also \(WX=\frac12 AB\), so choice (1) reverses the length relationship.

SOMATH takeaway: Lengths and perimeter scale by k, area by k², and angle measures do not change.

Answer: (1), 39 square units.

1. The given angle \(H\) is between the two given sides \(AH\) and \(HL\). Use \(A=\frac12 ab\sin C\).
2. Substitute: \(A=\frac12(10)(12)\sin40^\circ=60\sin40^\circ\).
3. The result is approximately 38.567, which rounds to 39 square units.

SOMATH takeaway: The sine factor supplies the perpendicular height. Using ½(10)(12) alone assumes a right angle that was not given.

Answer: (3), \(m\angle DEC=90^\circ\).

1. A parallelogram already has diagonals that bisect one another, so \(AE=EC\) adds nothing new.
2. If \(\angle DEC=90^\circ\), the diagonals are perpendicular. A parallelogram with perpendicular diagonals is a rhombus.
3. Why? The bisected diagonal pieces form right triangles with matching legs, so adjacent outer sides are congruent. Together with the parallelogram's opposite-side equality, all four sides are equal.
4. Congruent diagonals or a right vertex angle would instead establish a rectangle, which need not be a rhombus.

SOMATH takeaway: Use the condition that distinguishes equal sides, rather than the condition that distinguishes right angles.

Answer: (4), \(y=-2x-6\).

1. Check whether the dilation center lies on the line. At \(x=-1\), the equation gives \(y=-2(-1)-6=-4\). It does.
2. A line through the center of a nonzero dilation maps onto itself. Its points move along that same line, rather than onto a new parallel line.
3. For example, \((0,-6)\) maps halfway toward \((-1,-4)\), landing at \((-0.5,-5)\). This point still satisfies \(y=-2x-6\).

SOMATH takeaway: Do not halve a slope or intercept automatically. First locate the center relative to the line.