August 2025 Geometry Regents: Part I answers

Choice (1): cone.

Original SOMATH worked solution

1. Use the altitude as the fixed axis. It runs from a vertex to the midpoint of the opposite side.
2. As the triangle turns, the base sweeps out a circular disk and the sloping sides sweep out the curved surface of a cone.

Key idea: A solid of revolution is determined by the region being rotated and the location of its axis.

Avoid this mistake: A pyramid has flat triangular faces. Continuous rotation produces a circular base and curved surface, not a pyramid.

Choice (3): MQ.

Original SOMATH worked solution

1. Follow the vertex correspondence in the transformation: B → M, D → Q, G → S, and F → P.
2. A segment maps to the segment joining the images of its endpoints. Therefore BD maps to MQ.
3. As a coordinate check, the rotation sends (x, y) to (y, −x); reflecting that result across the y-axis gives (−y, −x).

Key idea: Rigid motions preserve lengths, and corresponding endpoints identify corresponding sides.

Avoid this mistake: Do not match sides only by how they look on the page. Track both endpoints.

Choice (2): 25.1.

Original SOMATH worked solution

1. EJ is the leg adjacent to angle J, while JO is the hypotenuse.
2. cos 38° = EJ / 31.8, so EJ = 31.8 cos 38°.
3. EJ ≈ 25.0587, which rounds to 25.1.

Key idea: Cosine is adjacent ÷ hypotenuse. Use degree mode when the angle is given in degrees.

Avoid this mistake: Sine would find the opposite leg, OE, rather than EJ.

Choice (3): 1072 cm³.

Original SOMATH worked solution

1. A hemisphere is half of a sphere, so V = ½(4πr³/3) = 2πr³/3.
2. Substitute r = 8: V = (2/3)π(8³) = 1024π/3.
3. V ≈ 1072.3303 cm³, so the requested volume is 1072 cm³.

Key idea: Volume uses cubic units and the cube of the radius.

Avoid this mistake: Using the full sphere formula produces approximately 2145 cm³, twice the needed volume.

Choice (4): AC ⊥ BD.

Original SOMATH worked solution

1. The diagonals of every parallelogram bisect each other, so AE = EC and BE = ED.
2. If the diagonals are also perpendicular, triangles AEB and CEB are right triangles with AE = EC and shared leg BE. They are congruent, so AB = BC.
3. Opposite sides of a parallelogram are congruent. Since these adjacent sides are also congruent, all four sides are congruent and ABCD is a rhombus.

Key idea: A parallelogram with perpendicular diagonals is a rhombus.

Avoid this mistake: Congruent diagonals or perpendicular adjacent sides guarantee a rectangle, which need not be a rhombus.

Choice (4): 70°.

Original SOMATH worked solution

1. The parallel bases are HS and JO. Along transversal SO, the same-side interior angles HSO and SOJ are supplementary.
2. m∠HSO = 180° − 30° = 150°.
3. Ray SA splits ∠HSO into ∠HSA and ∠ASO. Thus m∠HSA = 150° − 80° = 70°.

Key idea: Same-side interior angles formed by parallel lines add to 180°.

Avoid this mistake: The 65° and 60° labels are not needed. Do not force every given number into the calculation.

Choice (2): 18.

Original SOMATH worked solution

1. Because DE ∥ AC, corresponding angles are congruent and △BDE ∼ △BAC.
2. The whole side BA = BD + DA = 4 + 8 = 12. The small-to-large scale factor is BD/BA = 4/12 = 1/3.
3. DE/AC = 1/3, so 6/AC = 1/3 and AC = 18.

Key idea: In similar triangles, compare corresponding whole sides in the same order.

Avoid this mistake: AD is only the lower piece of AB. Using 4/8 instead of 4/12 gives the wrong scale factor.

Choice (1): (x − 2)² + (y + 1)² = 25.

Original SOMATH worked solution

1. A circle with center (h, k) and radius r has equation (x − h)² + (y − k)² = r².
2. Substitute the center (2, −1): (x − 2)² + (y + 1)² = r².
3. Count 5 grid units from the center to the circle, so r = 5 and r² = 25.

Key idea: The right side of the circle equation is the radius squared.

Avoid this mistake: Subtracting the center's y-coordinate −1 gives y + 1, not y − 1.

Choice (3): point reflection through the origin.

Original SOMATH worked solution

1. Compare corresponding coordinates: D(2, 2) → D′(−2, −2), E(3, 5) → E′(−3, −5), and F(7, 3) → F′(−7, −3).
2. Every image follows (x, y) → (−x, −y). That is a point reflection through the origin, equivalently a 180° rotation.
3. For example, reflection over y = −x would send E(3, 5) to (−5, −3), not E′(−3, −5).

Key idea: Check the same rule on more than one point to distinguish transformations.

Avoid this mistake: The proposed translation fits D alone. It fails for E and F.

Choice (3): 12.9.

Original SOMATH worked solution

1. On the lower secant, PD is the external portion and DB is the internal portion. The whole secant PB = 10 + 12 = 22.
2. Apply external × whole = external × whole: PC · PA = PD · PB.
3. 17PC = 10(22) = 220, so PC = 220/17 ≈ 12.9412. To the nearest tenth, PC = 12.9.

Key idea: For two secants from one external point, the products of the external lengths and whole lengths are equal.

Avoid this mistake: Do not use BD = 12 as the entire lower secant. PB includes both PD and DB.

Choice (4): △CDB is an isosceles triangle.

Original SOMATH worked solution

1. Since CB bisects ∠ABD, ∠ABC ≅ ∠CBD.
2. Since CD ∥ AB, alternate interior angles ∠ABC and ∠BCD are congruent.
3. Therefore ∠CBD ≅ ∠BCD. The sides opposite these equal angles, CD and BD, are congruent, so △CDB is isosceles.

Key idea: The converse of the isosceles triangle theorem turns equal angles into equal opposite sides.

Avoid this mistake: The diagram does not establish a right angle. Use the stated parallel lines and angle bisector.

Choice (3): y − 1 = (−3/2)(x − 6).

Original SOMATH worked solution

1. The slope of h is 2/3. A perpendicular line has slope −3/2, because (2/3)(−3/2) = −1.
2. Use point-slope form y − y₁ = m(x − x₁) with (x₁, y₁) = (6, 1).
3. This gives y − 1 = (−3/2)(x − 6). Substituting (6, 1) makes both sides zero.

Key idea: Nonvertical perpendicular lines have slopes that are negative reciprocals.

Avoid this mistake: Point-slope form subtracts the given coordinates. The plus signs in choice (4) do not pass through (6, 1).

Choice (1): 300 grams.

Original SOMATH worked solution

1. The square base area is B = 8.2² = 67.24 cm².
2. The pyramid volume is V = Bh/3 = (67.24)(17.4)/3 = 389.992 cm³.
3. Mass = density × volume = 0.77(389.992) = 300.29384 g. Rounded to the nearest gram, the mass is 300 g.

Key idea: First find the volume of the shape, then multiply by its density.

Avoid this mistake: Omitting the pyramid's factor of 1/3 treats the block as a prism and triples the mass.

Choice (2): 14.

Original SOMATH worked solution

1. A triangle's midsegment is half as long as the parallel third side. Thus 2DE = AC.
2. 2(x + 3) = 3x − 5. Expanding gives 2x + 6 = 3x − 5, so x = 11.
3. The question asks for DE, not x: DE = 11 + 3 = 14. Check: AC = 3(11) − 5 = 28, twice 14.

Key idea: A midsegment joins two side midpoints and is parallel to, and half the length of, the third side.

Avoid this mistake: Solving for x is an intermediate step. Substitute back to obtain the requested segment.

Choice (3): 10.

Original SOMATH worked solution

1. The side opposite the right angle, DU, is the hypotenuse.
2. In an isosceles right triangle, hypotenuse = leg · √2.
3. GU = (10√2)/√2 = 10.

Key idea: The side ratio in a 45-45-90 triangle is 1 : 1 : √2.

Avoid this mistake: The hypotenuse is longer than either leg. Do not assign 10√2 to a leg.

Choice (3): 29 cm².

Original SOMATH worked solution

1. The given angle lies between the two given sides, so use A = ½ab sin C.
2. A = ½(9)(8) sin 55° = 36 sin 55° ≈ 29.4895 cm².
3. Rounded to the nearest square centimeter, the area is 29 cm².

Key idea: The sine factor supplies the perpendicular height when the two sides are not perpendicular.

Avoid this mistake: ½(9)(8) = 36 would be correct only if the included angle were 90°.

Choice (2): (2, 1).

Original SOMATH worked solution

1. Read corresponding points from the graph, for example A(1, 5) and R(0, 9).
2. For scale factor 2 with center O, R = O + 2(A − O) = 2A − O. Rearranging gives O = 2A − R.
3. O = (2·1 − 0, 2·5 − 9) = (2, 1). Check with B(3, 3) and S(4, 5): 2B − S = (2, 1) again.

Key idea: A point, its image, and the dilation center lie on one line; the distances from the center scale by the dilation factor.

Avoid this mistake: Multiplying coordinates directly by 2 assumes the center is the origin, which it is not here.

Choice (4): 12 + 4√5.

Original SOMATH worked solution

1. AB is vertical, so AB = |3 − (−1)| = 4. BC is horizontal, so BC = |6 − (−2)| = 8.
2. AC = √[(6 − (−2))² + (−1 − 3)²] = √(64 + 16) = √80 = 4√5.
3. The perimeter is AB + BC + AC = 4 + 8 + 4√5 = 12 + 4√5.

Key idea: Perimeter is the sum of actual side lengths; the distance formula includes a square root.

Avoid this mistake: Adding 4 + 8 + 80 gives 92, but 80 is AC², not AC.

Choice (4): PE/FE = PG/FT.

Original SOMATH worked solution

1. ∠PEG and ∠FET are vertical angles, so they are congruent. Together with the given ∠P ≅ ∠F, this proves △PEG ∼ △FET by AA.
2. The matching vertices are P ↔ F, E ↔ E, and G ↔ T.
3. Thus PE corresponds to FE, PG corresponds to FT, and GE corresponds to ET. Keeping the same order gives PE/FE = PG/FT.

Key idea: Write the similarity correspondence before forming a proportion.

Avoid this mistake: Reversing only one fraction compares small-to-large on one side with large-to-small on the other.

Choice (2): 23 bags.

Original SOMATH worked solution

1. Convert depth to feet: 4 inches = 4/12 foot = 1/3 foot.
2. Volume = length × width × depth = 10·4·(1/3) = 40/3 cubic feet.
3. Bags needed = (40/3)/0.6 = 200/9 ≈ 22.2222. Since 22 bags are insufficient, purchase 23 bags.
4. Check: 22 bags yield 13.2 ft³, less than 13⅓ ft³; 23 bags yield 13.8 ft³, enough.

Key idea: Use consistent units before multiplying. A minimum whole-item purchase requires rounding up.

Avoid this mistake: Rounding to the nearest whole number would give 22 and leave part of the sidewalk unfilled.

Choice (1): y = (2/3)x − 12.

Original SOMATH worked solution

1. Rewrite the original equation as y = (2/3)x − 4.
2. Under the dilation, an original point (x, y) becomes (X, Y) = (3x, 3y). Substitute x = X/3 and y = Y/3.
3. Y/3 = (2/3)(X/3) − 4. Multiply by 3: Y = (2/3)X − 12.
4. The image has the same slope and a y-intercept three times as far from the origin.

Key idea: A dilation sends a line not through its center to a parallel line.

Avoid this mistake: The slope is a ratio of two lengths, both scaled equally. It does not become 2.

Choice (1): a rotation of 180° about the origin.

Original SOMATH worked solution

1. The diagonals meet at (1, 0), the center of the rhombus. A half-turn about that point interchanges opposite vertices.
2. The two diagonal lines are y = (1/2)x − 1/2 and y = −2x + 2. Each is an axis of reflection symmetry for the rhombus.
3. A half-turn about the origin would move its center from (1, 0) to (−1, 0). Since the center changes, the rhombus cannot coincide with its original position.

Key idea: Every rhombus has half-turn symmetry about the intersection of its diagonals and reflection symmetry across each diagonal.

Avoid this mistake: This is a “does not” question. Three choices are valid symmetries; select the exception.

Choice (2): 12.

Original SOMATH worked solution

1. First find the other leg: HA = √(25² − 20²) = √225 = 15.
2. Compute the area using the two perpendicular legs: A = ½(15)(20) = 150.
3. Using HY as the base gives the same area: 150 = ½(25)(AL). Therefore AL = 300/25 = 12.

Key idea: A triangle has the same area regardless of the base-altitude pair used.

Avoid this mistake: HA = 15 is a leg, not the altitude AL. Finish the second area calculation.

Choice (1): 9.

Original SOMATH worked solution

1. A dilation with linear scale factor k multiplies area by k².
2. Image area = 36(1/2)² = 36/4 = 9.
3. As a check, the original side length is 6; the image side is 3, so its area is 3² = 9.

Key idea: Lengths scale by k, areas by k², and volumes by k³.

Avoid this mistake: Halving area directly gives 18. Both dimensions shrink, so the area is quartered.