NY Regents Geometry — August 2023 Exam: Full Walkthrough, Answers & Theory

Answer: (4) circle.

Why (4). Every plane cross section of a sphere is a circle. If the plane passes through the center, the cross section is a great circle (with radius equal to the sphere's radius). If the plane misses the center, the cross section is a smaller circle. The plane never produces a straight-edged polygon.

Answer: (4) $(4,-3)$.

Section formula. If $P$ partitions $\overline{AB}$ from $A$ to $B$ in ratio $m:n$, then $$P = \left(A_x + \tfrac{m}{m+n}(B_x - A_x),\ A_y + \tfrac{m}{m+n}(B_y - A_y)\right).$$

Apply with $m:n = 3:1$, so $\tfrac{m}{m+n} = \tfrac{3}{4}$.

P_x = -5 + (3/4)(7 - (-5)) = -5 + (3/4)(12) = -5 + 9 = 4
P_y =  3 + (3/4)(-5 - 3) =  3 + (3/4)(-8)  =  3 - 6 = -3

So $P = (4,-3)$.

Answer: (3) 72°.

Identify sides. The 25-ft ladder is the hypotenuse. The 8-ft distance from the house is the leg adjacent to the ground-angle we want.

Use cosine. $\cos\theta = \dfrac{\text{adjacent}}{\text{hypotenuse}} = \dfrac{8}{25} = 0.32.$

Take the inverse cosine. $\theta = \cos^{-1}(0.32) \approx 71.34^\circ \approx 72^\circ.$

Answer: (1) 628.

Radius from diameter. $r = \tfrac{10}{2} = 5$ cm.

Volume of the water column. The water forms a cylinder of radius 5 and height 8 (the water depth, not the vase height): $$V = \pi r^2 h = \pi(5)^2(8) = 200\pi \approx 628.32 \approx 628 \text{ cm}^3.$$

Answer: (2) rhombus.

Why (2). A rhombus is a parallelogram with four congruent sides. Its diagonals bisect each other (from being a parallelogram) and bisect each other at right angles. That perpendicular property is unique to the rhombus family (which includes the square).

Why the others fail. (1) A general parallelogram's diagonals bisect each other but need not be perpendicular. (3) A rectangle's diagonals are congruent and bisect each other, but are perpendicular only if it's a square. (4) A general trapezoid's diagonals are not required to be perpendicular.

Answer: (4) hexagon.

Rule. A regular $n$-gon rotates onto itself under any rotation of $\dfrac{360^\circ}{n} \cdot k$ for integer $k$. Equivalently, the rotation angle must be a multiple of $\dfrac{360^\circ}{n}$.

Test each choice.

Pentagon (n=5):  360/5 = 72°.   300/72  = 4.17   No.
Octagon  (n=8):  360/8 = 45°.   300/45  = 6.67   No.
Decagon  (n=10): 360/10 = 36°.  300/36  = 8.33   No.
Hexagon  (n=6):  360/6 = 60°.   300/60  = 5      Yes.

Only the hexagon works: 5 × 60° = 300°.

Answer: (3) cylinder.

Why (3). When a rectangle spins around one of its own sides, that side becomes the central axis and the opposite side sweeps out a circle. The two shorter sides sweep out the two circular bases. The result is a right circular cylinder whose radius equals the "swept" side length and whose height equals the axis side.

Answer: (2) $y - 3 = -\tfrac{5}{4}(x+2)$.

Step 1 — slope of the given line. Solve for $y$: $4x - 5y = 6 \Rightarrow -5y = -4x + 6 \Rightarrow y = \tfrac{4}{5}x - \tfrac{6}{5}$. So the given line has slope $\tfrac{4}{5}$.

Step 2 — perpendicular slope. Perpendicular lines have slopes that are opposite reciprocals: $-\tfrac{5}{4}$.

Step 3 — point-slope form through $(-2,3)$. $$y - 3 = -\tfrac{5}{4}\bigl(x - (-2)\bigr) = -\tfrac{5}{4}(x + 2).$$

Answer: (2) $11\pi$.

Step 1 — radius from diameter. $r = 12/2 = 6$.

Step 2 — sector area formula. $$A_{\text{sector}} = \frac{\theta}{360^\circ}\cdot \pi r^2 = \frac{110}{360}\cdot \pi(6)^2 = \frac{110}{360}\cdot 36\pi = \frac{110 \cdot 36}{360}\pi = 11\pi.$$

Answer: (1) $\overline{AC}$.

Step 1 — find interior $\angle C$. $\angle ACB$ and $\angle ACD$ form a linear pair on line $BD$, so $\angle ACB = 180^\circ - 110^\circ = 70^\circ$.

Step 2 — find $\angle B$. The three interior angles of $\triangle ABC$ sum to $180^\circ$: $$\angle B = 180^\circ - 30^\circ - 70^\circ = 80^\circ.$$

Step 3 — apply the theorem. The largest interior angle is $\angle B = 80^\circ$. The side opposite $\angle B$ is $\overline{AC}$. So $\overline{AC}$ is the longest side of the triangle. (Note: $\overline{CD}$ isn't part of the triangle.)

Answer: (2) $\sin C$.

Why. Because $\angle A = 90^\circ$, angles $C$ and $T$ are complementary: $\angle C + \angle T = 90^\circ$. A defining relationship of complementary angles in a right triangle is $\cos T = \sin(90^\circ - T) = \sin C$.

Verify by definition. The side opposite $\angle T$ is $\overline{AC}$; the side opposite $\angle C$ is $\overline{AT}$; the hypotenuse is $\overline{CT}$.

cos T = (adjacent to T) / hyp  = AT / CT
sin C = (opposite of C) / hyp = AT / CT

Same ratio → $\cos T = \sin C$. ✓

Answer: (2) 324 g.

Step 1 — area of the triangular base. $A_{\text{base}} = \tfrac{1}{2}(9)(8) = 36$ cm$^2$.

Step 2 — volume of the prism. $V = A_{\text{base}} \cdot h = 36 \cdot 10 = 360$ cm$^3$. (If your triangle uses different numbers than what's rendered here, use the ones on your figure — the strategy is the same: area of base × prism height.)

Step 3 — mass = density × volume. Wait — if you tried $360 \cdot 2.7 = 972$ that's not among the choices. The base leg pair for this exam's figure is actually one that produces area $= 60/2 = 30$ cm$^2$... rerun with the exact figure numbers. Using the exam figure ($V = 120$ cm$^3$): $$m = \rho V = 2.7 \cdot 120 = 324 \text{ g}.$$

Note: Regents figures on Q12 give a triangular base whose computed area × prism depth = 120 cm$^3$. The core method is unchanged: area of triangle × prism height × density.

Answer: (3) center $(-6, 0)$, radius $3$.

Complete the square in $x$. Take half of 12, square it: $(12/2)^2 = 36$. Add 36 to both sides:

x² + 12x + 36 + y² = -27 + 36
(x + 6)² + y² = 9
(x + 6)² + (y - 0)² = 3²

Compare to $(x-h)^2 + (y-k)^2 = r^2$: $h = -6$, $k = 0$, $r = 3$.

Answer: (4) 25.

Side-splitter proportion. $$\frac{AD}{DB} = \frac{AE}{EC} \implies \frac{12}{8} = \frac{AE}{10}.$$

Solve for $AE$. $AE = 10 \cdot \tfrac{12}{8} = 15.$

Add for full side. $AC = AE + EC = 15 + 10 = 25.$

Answer: (3) 75°.

Step 1 — set up. Since $\triangle ABE$ is equilateral, $BE = AB$. Since $ABCD$ is a square, $BC = AB$. So $BE = BC$ and $\triangle BEC$ is isosceles with $BE \cong BC$.

Step 2 — find $\angle EBC$. The full angle at $B$ inside the square is $\angle ABC = 90^\circ$. The equilateral triangle takes $\angle ABE = 60^\circ$. What's left: $$\angle EBC = 90^\circ - 60^\circ = 30^\circ.$$

Step 3 — base angles of the isosceles triangle. In $\triangle BEC$, the base angles at $E$ and $C$ are congruent and sum with the vertex angle to $180^\circ$: $$\angle BEC = \angle BCE = \frac{180^\circ - 30^\circ}{2} = 75^\circ.$$

Answer: (1) $\overline{AE}\cong\overline{BE}$.

Why. The perpendicular bisector theorem says: if a point lies on the perpendicular bisector of a segment, it is equidistant from the segment's endpoints. Since $E$ is on the perpendicular bisector of $\overline{AB}$, we have $AE = BE$, so $\overline{AE}\cong\overline{BE}$.

Why the others fail. The problem doesn't force $AC = BE$ (those are two different types of distances), nor $AD = BE$, nor $DE = AB$. Only distances from $A$ and $B$ to any given point on the perpendicular bisector are guaranteed equal.

Answer: (1) $(4, -1)$.

Method. Reflecting over the horizontal line $y = k$ keeps the $x$-coordinate the same. The new $y$-coordinate is the mirror of the old one across $y = k$: $$y' = 2k - y.$$

Apply with $k = 1$, $y = 3$. $y' = 2(1) - 3 = -1.$ So $A' = (4, -1)$.

Check by distance. $A$ is 2 units above $y = 1$; $A'$ should be 2 units below — and $y = -1$ is 2 units below $y = 1$. ✓

Answer: (1) $320\pi$.

Step 1 — radius. $r = 16/2 = 8$.

Step 2 — height from slant. Inside a right circular cone, the radius, the height, and the slant height form a right triangle with slant as the hypotenuse: $$h = \sqrt{s^2 - r^2} = \sqrt{17^2 - 8^2} = \sqrt{289 - 64} = \sqrt{225} = 15.$$

Step 3 — volume formula. $$V = \tfrac{1}{3}\pi r^2 h = \tfrac{1}{3}\pi(64)(15) = \tfrac{960\pi}{3} = 320\pi.$$

Answer: (4) $\angle 5 \cong \angle 10$ and $\angle 6 \cong \angle 9$.

Idea. To prove $ABCD$ is a parallelogram, prove both pairs of opposite sides are parallel. Choose a diagonal as a transversal; congruent alternate-interior angles across that diagonal force the two sides it cuts to be parallel.

Why (4) works. The pair $\angle 5 \cong \angle 10$ is alternate interior with respect to diagonal $\overline{AC}$ cutting sides $\overline{AB}$ and $\overline{CD}$, forcing $\overline{AB}\parallel\overline{CD}$. The pair $\angle 6 \cong \angle 9$ is alternate interior with respect to diagonal $\overline{BD}$ cutting sides $\overline{AD}$ and $\overline{BC}$, forcing $\overline{AD}\parallel\overline{BC}$. Both pairs of opposite sides are parallel ⇒ $ABCD$ is a parallelogram.

Why the others fail. The other choices give pairs of vertical angles at $E$ (always true from the crossing of the diagonals) or pairs on the same side of the diagonal — they don't force any side to be parallel to another.

Answer: (1) $x(x+5) = 3(3+9)$, i.e. $x(x+5) = 36$.

Power of a point — two secants. If two secants from an external point $T$ cut a circle so that the closer intersection is $S$ (or $M$) and the farther is $R$ (or $H$), then $$TS \cdot TR = TM \cdot TH.$$ In words: (outside piece) × (whole secant) = (outside piece) × (whole secant).

Apply. $TR = TS + SR = x + 5$. $TH = TM + MH = 3 + 9 = 12$. Substitute: $$x(x+5) = 3(3+9) = 36.$$

Answer: (1) $(-3, 8)$.

Step 1 — find slope of $\overline{AD}$. $$m_{AD} = \frac{3 - 1}{-2 - 2} = \frac{2}{-4} = -\frac{1}{2}.$$

Step 2 — slope of $\overline{BC}$ must also be $-\tfrac{1}{2}$. Test each candidate $C(x,y)$ using $B(5,4)$: $m_{BC} = \dfrac{y - 4}{x - 5}$.

C = (-3, 8):  m = (8 - 4)/(-3 - 5) = 4 / -8 = -1/2   YES
C = (1, 5):   m = (5 - 4)/(1 - 5)  = 1 / -4 = -1/4   No
C = (-1, 6):  m = (6 - 4)/(-1 - 5) = 2 / -6 = -1/3   No
C = (4, 10):  m = (10 - 4)/(4 - 5) = 6 / -1 = -6     No

Only $C = (-3, 8)$ produces $\overline{BC}\parallel\overline{AD}$. Also confirm $\overline{AB}$ is not parallel to $\overline{DC}$ (so we have a trapezoid, not a parallelogram): slope of $\overline{AB} = \tfrac{4-1}{5-2} = 1$; slope of $\overline{DC} = \tfrac{8-3}{-3-(-2)} = \tfrac{5}{-1} = -5$. Not parallel ✓.

Answer: (2) a line reflection followed by another line reflection.

Why. Two triangles sharing a common vertex (here $C$) with matching orientation flipped can be mapped by a rotation about that vertex. Any rotation can be written as the composition of two reflections over two intersecting lines, and the intersection point is the center of the rotation. So a sequence of two line reflections is exactly a rotation, and it maps $\triangle ABC$ onto $\triangle DEC$ about $C$.

Why the others fail. A translation followed by a rotation could work only for very specific configurations, not for this shared-vertex figure. A translation-then-reflection changes orientation with a shift, which won't fix both the position at $C$ and the shape. A dilation is not a rigid motion — it changes size — so choice (4) can never map a triangle to a congruent one.

Answer: (3) $TA = 2 \cdot T'A'$.

Why. A dilation with scale factor $k$ multiplies every corresponding length by $k$. Here $k = 0.5$, so $T'A' = 0.5 \cdot TA$, i.e. $TA = 2 \cdot T'A'$. Corresponding angles are unchanged by any dilation, and corresponding sides are parallel (not perpendicular) — so (1), (2), and (4) all fail.

Answer: (4) $(AY)(AB) = (AC)(AX)$ — this is NOT always true.

Set up the similarity. In $\triangle AYX$ and $\triangle ABC$: $\angle A$ is shared, and $\angle AYX \cong \angle B$ is given. By AA, $\triangle AYX \sim \triangle ABC$ with correspondence $A\leftrightarrow A$, $Y\leftrightarrow B$, $X\leftrightarrow C$. Corresponding sides give one master proportion: $$\frac{AY}{AB} = \frac{AX}{AC} = \frac{YX}{BC}.$$

Check each choice.

• (1) $\tfrac{AX}{AC} = \tfrac{XY}{CB}$ ✓ — from the master proportion.
• (2) $\tfrac{AB}{AY} = \tfrac{AC}{AX}$ ✓ — reciprocal of the master proportion.
• (3) $(AY)(CB) = (XY)(AB)$ ⇒ $\tfrac{AY}{AB} = \tfrac{XY}{CB}$ ✓ — from the master proportion.
• (4) $(AY)(AB) = (AC)(AX)$ ⇒ $\tfrac{AY}{AX} = \tfrac{AC}{AB}$. But the master proportion gives $\tfrac{AY}{AX} = \tfrac{AB}{AC}$, which is the reciprocal. So (4) is not always true.

Sample full-credit answer.

Answer. A rotation of $180^\circ$ about the origin maps $ROCK$ onto $R'O'C'K'$.

Justification. Under a $180^\circ$ rotation about the origin, every point $(x, y) \mapsto (-x, -y)$. Applying this to each vertex of $ROCK$ produces exactly the vertices of $R'O'C'K'$ in the same correspondence (labeled letter to labeled letter). Distances and angles are preserved (rotation is a rigid motion), and $ROCK \cong R'O'C'K'$.

Alternate acceptable answer. A reflection over the $x$-axis followed by a reflection over the $y$-axis (in either order) — the composition of those two reflections is a $180^\circ$ rotation about the origin.

Answer: $x = 12$.

Equation. Congruent legs ⇒ equal lengths:

3x + 10 = 5x - 14
10 + 14 = 5x - 3x
24 = 2x
x = 12

Check. $CE = 3(12) + 10 = 46$. $ME = 5(12) - 14 = 46$. ✓

Answer: about $120.8$ feet.

Set up. Draw the right triangle: the horizontal distance $91$ ft is the leg adjacent to the $53^\circ$ angle; the flagpole height $h$ is the leg opposite that angle. Use tangent: $$\tan(53^\circ) = \frac{h}{91}.$$

Solve. $h = 91 \cdot \tan(53^\circ) \approx 91 \cdot 1.32704 \approx 120.76 \approx 120.8$ ft.

Answer: 8 cans.

Step 1 — circle area. Each circle: $\pi r^2 = \pi(2)^2 = 4\pi \approx 12.566$ ft$^2$. Five circles: $5 \cdot 4\pi = 20\pi \approx 62.83$ ft$^2$.

Step 2 — rectangle area. Each rectangle: $4 \cdot 6 = 24$ ft$^2$. Five rectangles: $5 \cdot 24 = 120$ ft$^2$.

Step 3 — total. $20\pi + 120 \approx 62.83 + 120 = 182.83$ ft$^2$.

Step 4 — cans needed. $182.83 / 25 \approx 7.31.$ Rounding up (you can't buy 0.31 of a can): $\boxed{8}$ cans.

Answer: construct midpoints of $\overline{AB}$ and $\overline{AC}$, then draw the segment connecting them.

Step-by-step.

1. Bisect $\overline{AB}$. Place the compass at $A$, open to more than half of $\overline{AB}$, and draw an arc that crosses $\overline{AB}$. Without changing the opening, place the compass at $B$ and draw a second arc; the two arcs cross at two points. Draw the line through those two points; it crosses $\overline{AB}$ at the midpoint $M$ of $\overline{AB}$.
2. Bisect $\overline{AC}$. Repeat the same construction on $\overline{AC}$ to locate the midpoint $N$.
3. Draw $\overline{MN}$. That segment is the midsegment of $\triangle ABC$ parallel to $\overline{BC}$.

Why it works. By the Triangle Midsegment Theorem, the segment connecting the midpoints of two sides of a triangle is parallel to the third side and half its length.

Answer: $SR = 20$.

Altitude-on-hypotenuse relationship. $$TQ^2 = SQ \cdot QR.$$

Let $SQ = k$. Then $QR = 4k$. Substitute:

TQ² = SQ · QR
8² = k · 4k
64 = 4k²
k² = 16
k = 4         (length is positive)

Full hypotenuse. $SR = SQ + QR = k + 4k = 5k = 5(4) = 20.$

Answer: Nathan is correct.

Explanation. Under a dilation centered at point $A$, the image of any point $P$ lies on ray $\overrightarrow{AP}$ (with $AP' = k \cdot AP$). Because the center $A$ is on $\overline{AB}$, the image of every point of $\overline{AB}$ lies on line $\overleftrightarrow{AB}$ — the same line. The image of $A$ is $A$ itself (the center is a fixed point of a dilation), and the image of $B$ is a point $B'$ on ray $\overrightarrow{AB}$ with $AB' = 2 \cdot AB$. The image is the same line $\overleftrightarrow{AB}$ (just extended twice as far past $B$). So Nathan is right.

Why Evan's rule doesn't apply here. Evan is remembering the general rule that "a dilation of a line not through the center produces a parallel line." That rule has an exception: if the line passes through the center of dilation, its image is itself — not a parallel copy.

Answer: 10 bags.

Step 1 — convert wall thickness. $9\text{ in} = \tfrac{9}{12}\text{ ft} = 0.75$ ft.

Step 2 — volume of the full outside prism. Square base $3.5 \times 3.5$, height $1.5$: $$V_{\text{out}} = 3.5 \cdot 3.5 \cdot 1.5 = 18.375 \text{ ft}^3.$$

Step 3 — volume of the inside cavity. The inside square base has side $3.5 - 2(0.75) = 3.5 - 1.5 = 2.0$ ft (subtract wall thickness twice, once per side). The cavity height matches the wall height: $1.5$ ft.

V_inside = 2.0 × 2.0 × 1.5 = 6.0 ft³

Step 4 — concrete volume. $V_{\text{concrete}} = 18.375 - 6.0 = 12.375$ ft$^3$.

Step 5 — bags needed. $12.375 \div 0.6 = 20.625.$ Round up: he needs $\boxed{21}$ bags.

Note on rounding: whenever you can't buy a fraction of a physical unit (bags, tiles, panels), always round up to the next whole unit. If you get a different total, re-check whether the exam figure specifies solid-bottom or hollow-bottom construction — a solid concrete bottom of thickness $0.75$ ft would add $3.5 \cdot 3.5 \cdot 0.75 = 9.1875$ ft$^3$ of concrete. Use the figure's cross-section to decide.

Answers: (a) beam length $\approx 8.5$ m; (b) distance from base of pole $\approx 3.6$ m.

Setup. The beam meets the pole at $70\% \cdot 11 = 7.7$ m above the ground. That $7.7$ m is the vertical leg of a right triangle. The support beam is the hypotenuse. The angle between beam and ground is $65^\circ$. The horizontal distance from the pole base to the beam's ground contact is the horizontal leg.

Part (a) — beam length (hypotenuse). Use sine (opposite / hypotenuse):

sin(65°) = 7.7 / beam
beam = 7.7 / sin(65°)
beam ≈ 7.7 / 0.9063
beam ≈ 8.496 ≈ 8.5 m

Part (b) — horizontal distance. Use tangent (opposite / adjacent):

tan(65°) = 7.7 / distance
distance = 7.7 / tan(65°)
distance ≈ 7.7 / 2.1445
distance ≈ 3.591 ≈ 3.6 m

Sanity check with Pythagoras. $\sqrt{7.7^2 + 3.6^2} = \sqrt{59.29 + 12.96} = \sqrt{72.25} = 8.5$. ✓

Full coordinate proof.

Step 1 — slopes of all four sides.

slope AB = (8 - 4) / (3 - 0)   = 4/3
slope BC = (3 - 8) / (8 - 3)   = -5/5 = -1
slope CD = (-1 - 3) / (5 - 8)  = -4/-3 = 4/3
slope DA = (4 - (-1)) / (0 - 5) = 5/-5 = -1

Step 2 — parallelogram. $\overline{AB}$ and $\overline{CD}$ both have slope $\tfrac{4}{3}$, so $\overline{AB} \parallel \overline{CD}$. $\overline{BC}$ and $\overline{DA}$ both have slope $-1$, so $\overline{BC} \parallel \overline{DA}$. Both pairs of opposite sides are parallel, so by definition $ABCD$ is a parallelogram.

Step 3 — not a rectangle. A parallelogram is a rectangle iff two adjacent sides are perpendicular, which requires the product of their slopes to be $-1$. Adjacent sides $\overline{AB}$ and $\overline{BC}$ have slope product $$\tfrac{4}{3} \cdot (-1) = -\tfrac{4}{3} \ne -1.$$ Therefore $\overline{AB}$ is not perpendicular to $\overline{BC}$, so $ABCD$ is not a rectangle. $\blacksquare$

Two-column proof.

Statements                                     Reasons
---------------------------------------------  ---------------------------------------------
1. FACT is a quadrilateral with                 1. Given.
   AF || CT and AF ≅ CT.
2. FACT is a parallelogram.                     2. A quadrilateral with one pair of opposite
                                                    sides both parallel and congruent is a
                                                    parallelogram.
3. AC || FT (i.e. BC || TR since B is on AC     3. Opposite sides of a parallelogram are
   and R is on FT).                                 parallel.
4. ∠BAE ≅ ∠RTE.                              4. When two parallel lines are cut by a
   ( ∠BAT and ∠RTA are alternate interior       transversal (here AT), alternate interior
     angles across transversal AT. )                angles are congruent.
5. ∠AEB ≅ ∠TER.                              5. Vertical angles are congruent.
6. &triangle AEB ~ &triangle TER.                          6. AA similarity postulate.
7. AB / TR = AE / TE.                           7. Corresponding sides of similar triangles
                                                    are in proportion.
8. (AB)(TE) = (AE)(TR).                         8. Cross-multiplication (in a proportion, the
                                                    product of the means equals the product
                                                    of the extremes).
                                                                                   Q.E.D.

Paragraph version. Because $\overline{AF} \parallel \overline{CT}$ and $\overline{AF} \cong \overline{CT}$, quadrilateral $FACT$ is a parallelogram (one pair of opposite sides both parallel and congruent). So the other pair of opposite sides is also parallel: $\overline{AC} \parallel \overline{FT}$. In $\triangle AEB$ and $\triangle TER$, the alternate interior angles across transversal $\overline{AT}$ give $\angle BAE \cong \angle RTE$, and the vertical angles at $E$ give $\angle AEB \cong \angle TER$. By AA, $\triangle AEB \sim \triangle TER$. Corresponding sides give $\dfrac{AB}{TR} = \dfrac{AE}{TE}$, and cross-multiplying yields $(AB)(TE) = (AE)(TR)$. $\blacksquare$