August 2025 Geometry Regents: Part II answers

△D′A′N′ ≅ △DAN because a translation is a rigid motion.

Original SOMATH worked solution

1. A translation moves every point the same distance in the same direction.
2. It preserves distances and angle measures. In particular, D′A′ = DA, A′N′ = AN, and D′N′ = DN.
3. The corresponding sides are congruent, so the triangles are congruent by SSS. Equivalently, a rigid motion mapping one triangle onto another establishes congruence.

Key idea: Congruence describes figures with the same shape and size, even when their positions differ.

Avoid this mistake: “They look the same” is not a geometric explanation. Name the translation's distance-preserving property.

Iron.

Original SOMATH worked solution

1. A cube with edge length 5 cm has volume V = 5³ = 125 cm³.
2. Density = mass / volume = 982.5/125 = 7.86 g/cm³.
3. The table lists iron at 7.86 g/cm³, so the cube is made of iron.

Key idea: Density is mass per unit volume, not mass per unit length.

Avoid this mistake: Dividing by 5 or by 25 uses an edge length or face area instead of the cube's volume.

A = (3, 4).

Original SOMATH worked solution

1. The ratio 3:2 divides CS into 5 equal parts. A is 3/5 of the way from C toward S.
2. The displacement from C to S is (7 − (−3), 6 − 1) = (10, 5). Three fifths of that displacement is (6, 3).
3. Add it to C: A = (−3 + 6, 1 + 3) = (3, 4).
4. Check: CA has displacement (6, 3) and AS has displacement (4, 2). Their lengths are 3√5 and 2√5, giving the required 3:2 ratio.

Key idea: For CA:AS = m:n, start at C and take m/(m + n) of the displacement toward S.

Avoid this mistake: The first number of the ratio measures the part from C to A. Starting from S without reversing the fraction gives the wrong point.

7.2 meters.

Original SOMATH worked solution

1. Let L be the ramp length. The vertical rise of 0.6 m is opposite the 4.8° angle, and the ramp is the hypotenuse.
2. sin 4.8° = 0.6/L, so L = 0.6/sin 4.8°.
3. In degree mode, L ≈ 7.17036 m. To the nearest tenth, the ramp is 7.2 m long.

Key idea: Use sine when the known side is opposite the angle and the unknown is the hypotenuse.

Avoid this mistake: Tangent would find the horizontal run, not the slanted ramp.

m∠KMN = 30°.

Original SOMATH worked solution

1. Because the vertex angle is at M, the base angles are at K and L. Thus m∠L = m∠K = 15°.
2. The interior angle at M is 180° − 15° − 15° = 150°.
3. Since MN extends ML in the opposite direction, ∠KMN and ∠KML form a linear pair. Therefore m∠KMN = 180° − 150° = 30°.
4. A shorter check uses the exterior angle theorem: 15° + 15° = 30°.

Key idea: An exterior angle equals the sum of the two nonadjacent interior angles.

Avoid this mistake: The vertex angle is at M, so it is not one of the two equal base angles.

m∠KLM = 108°.

Original SOMATH worked solution

1. The entire circle has area πr² = π(5²) = 25π.
2. The shaded sector is 7.5π/(25π) = 0.3 of the circle.
3. Its central angle is the same fraction of a full turn: 0.3(360°) = 108°.
4. Check: (108/360)π(5²) = 7.5π, the given sector area.

Key idea: Sector area / circle area = central angle / 360°.

Avoid this mistake: This is an area problem. Do not substitute the arc-length formula using 2πr.

A′ is the second intersection of the two compass arcs, on the opposite side of BC from A.

Original SOMATH worked solution

1. Place the compass point at B and open it to A. Without changing that opening, draw an arc on the other side of BC where the reflected point will lie.
2. Place the compass point at C and open it to A. Draw a second arc on that same side of BC, crossing the first arc.
3. Label their intersection A′. The two circles centered at B and C meet at A and A′; choose the intersection different from A.
4. Leave both intersecting arcs visible. Use the straightedge to show AA′ if desired; do not erase the construction marks.
5. Why it works: BA = BA′ and CA = CA′ are compass radii. Both B and C are equidistant from A and A′, so line BC is the perpendicular bisector of AA′. That is exactly the defining condition for reflection.

Key idea: Reflection across a line places a point and its image at equal perpendicular distances on opposite sides of that line.

Avoid this mistake: A point placed by eye, with no compass arcs, is not a compass-and-straightedge construction.