June 2023 Geometry Regents — Answer Key & Worked Solutions

Every question from the June 2023 Geometry Regents exam, with the correct answer and a full step-by-step solution. Check your work, learn the method, then convert your raw score.

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These are real questions from the June 2023 Geometry New York State Regents examination, published by NYSED. Worked solutions are written by RegentsLab.

Question 1 · 2 pts

A square pyramid is intersected by a plane passing through the vertex and perpendicular to the base. Which two-dimensional shape describes this cross section?

(A) square
(B) triangle ✓
(C) pentagon
(D) rectangle

Worked solution

$\text{Plane through the apex, perpendicular to the base, slices two opposite lateral faces and the apex}$

$\text{The cross section has 3 sides: two slanted sides meeting at the apex and one side on the base}$

Answer: triangle

A plane through the apex of a square pyramid, perpendicular to the base, cuts through the apex and two opposite base edges, producing a triangular cross section.

Question 2 · 2 pts

Trapezoid $ABCD$ is drawn such that $\overline{AB} \parallel \overline{DC}$. Trapezoid $A'B'C'D'$ is the image of trapezoid $ABCD$ after a rotation of $110°$ counterclockwise about point $P$. Which statement is always true?

(A) $\angle A \cong \angle D'$
(B) $\overline{AC} \cong \overline{B'D'}$
(C) $\overline{A'B'} \parallel \overline{D'C'}$ ✓
(D) $\overline{B'A'} \cong \overline{C'D'}$

Worked solution

$\overline{AB} \parallel \overline{DC} \text{ (given)}$

$\text{Rotations preserve parallelism, so images of parallel segments remain parallel}$

$\overline{A'B'} \parallel \overline{D'C'}$

Answer: $\overline{A'B'} \parallel \overline{D'C'}$

Rotations are rigid motions that preserve parallelism, so since AB is parallel to DC, their images A'B' and D'C' must also be parallel.

Question 3 · 2 pts

What is the volume of a right circular cone that has a height of 7.2 centimeters and a radius of 2.5 centimeters, to the nearest tenth of a cubic centimeter?

(A) 37.7
(B) 47.1 ✓
(C) 113.1
(D) 141.4

Worked solution

$V=\frac{1}{3}\pi r^2 h$

$V=\frac{1}{3}\pi (2.5)^2(7.2) = \frac{1}{3}\pi(6.25)(7.2)$

$V=15\pi \approx 47.1 \text{ cm}^3$

Answer: 47.1

Substitute the radius and height into the cone volume formula and evaluate.

Question 4 · 2 pts

In the diagram below of right triangle $SUN$, where $\angle N$ is a right angle, $SU = 13.6$ and $SN = 12.3$. What is $m\angle S$, to the nearest degree?

(A) 25° ✓
(B) 42°
(C) 48°
(D) 65°

Worked solution

$\cos S = \frac{SN}{SU} = \frac{12.3}{13.6}$

$\cos S \approx 0.9044$

$m\angle S = \cos^{-1}(0.9044) \approx 25°$

Answer: 25°

SN is adjacent to angle S and SU is the hypotenuse, so use cosine and take the inverse cosine.

Question 5 · 2 pts

In the diagram below of circle $O$, diameter $\overline{AOB}$ and chord $\overline{CB}$ are drawn, and $m\angle B = 28°$. What is $m\widehat{BC}$?

(A) 56°
(B) 124° ✓
(C) 152°
(D) 166°

Worked solution

$\angle ACB = 90° \text{ (inscribed angle in a semicircle, since } \overline{AB} \text{ is a diameter)}$

$m\angle A = 180° - 90° - 28° = 62°$

$m\widehat{BC} = 2(m\angle A) = 2(62°) = 124°$

Answer: 124°

Since AB is a diameter, inscribed angle ACB is 90 degrees; angle A is then found from the triangle angle sum, and the inscribed angle at A doubles to give arc BC.

Question 6 · 2 pts

In the diagram below of parallelogram $ABCD$, diagonal $\overline{BED}$ and $\overline{EF}$ are drawn, $\overline{EF} \perp \overline{DFC}$, $m\angle DAB = 111°$, and $m\angle DBC = 39°$. What is $m\angle DEF$?

(A) 30°
(B) 51°
(C) 60° ✓
(D) 120°

Worked solution

$\overline{AD} \parallel \overline{BC} \Rightarrow m\angle ADB = m\angle DBC = 39° \text{ (alternate interior angles)}$

$m\angle ADC = m\angle ABC = 180° - 111° = 69° \text{ (parallelogram consecutive/opposite angles)}$

$m\angle BDC = m\angle ADC - m\angle ADB = 69° - 39° = 30°$

$\text{In right triangle } DFE, \angle DFE = 90°, \angle EDF = 30° \Rightarrow m\angle DEF = 60°$

Answer: 60°

Using parallel sides and parallelogram angle relationships, angle BDC is found to be 30 degrees, and since EF is perpendicular to DC, the remaining angle in the right triangle DEF is 60 degrees.

Question 7 · 2 pts

In the diagram below of $\triangle ACT$, $\overline{ES}$ is drawn parallel to $\overline{AT}$ such that $E$ is on $\overline{CA}$ and $S$ is on $\overline{CT}$. Which statement is always true?

(A) $\frac{CE}{CA} = \frac{CS}{ST}$
(B) $\frac{CE}{ES} = \frac{EA}{AT}$
(C) $\frac{CE}{EA} = \frac{CS}{ST}$ ✓
(D) $\frac{CE}{ST} = \frac{EA}{CS}$

Worked solution

$\overline{ES} \parallel \overline{AT} \Rightarrow \text{Triangle Proportionality Theorem applies}$

$\frac{CE}{EA} = \frac{CS}{ST}$

Answer: $\frac{CE}{EA} = \frac{CS}{ST}$

By the Triangle Proportionality Theorem, a line parallel to one side of a triangle divides the other two sides proportionally.

Question 8 · 2 pts

On the set of axes below, congruent triangles $ABC$ and $DEF$ are drawn. Which sequence of transformations maps $\triangle ABC$ onto $\triangle DEF$?

(A) A counterclockwise rotation of 90 degrees about the origin, followed by a translation 8 units to the right. ✓
(B) A counterclockwise rotation of 90 degrees about the origin, followed by a reflection over the y-axis.
(C) A counterclockwise rotation of 90 degrees about the origin, followed by a translation 4 units down.
(D) A clockwise rotation of 90 degrees about the origin, followed by a reflection over the x-axis.

Worked solution

$\text{A 90° ccw rotation about the origin maps } (x,y) \to (-y,x)$

$\text{Apply this rotation to each vertex of } \triangle ABC, \text{ then check which additional transformation lands the image on } \triangle DEF$

$\text{Testing the translation 8 units right maps the rotated triangle exactly onto } \triangle DEF$

Answer: A counterclockwise rotation of 90 degrees about the origin, followed by a translation 8 units to the right.

Applying the 90 degree counterclockwise rotation rule to the coordinates of triangle ABC and then testing each remaining transformation shows only the translation 8 units right completes the mapping onto triangle DEF.

Question 9 · 2 pts

An equation of circle $M$ is $x^2 + y^2 + 6x - 2y + 1 = 0$. What are the coordinates of the center and the length of the radius of circle $M$?

(A) center (3,-1) and radius 9
(B) center (3,-1) and radius 3
(C) center (-3,1) and radius 9
(D) center (-3,1) and radius 3 ✓

Worked solution

$x^2+6x+y^2-2y=-1$

$(x^2+6x+9)+(y^2-2y+1)=-1+9+1$

$(x+3)^2+(y-1)^2=9$

$\text{center } (-3,1), \text{ radius } \sqrt{9}=3$

Answer: center (-3,1) and radius 3

Completing the square on both x and y terms converts the equation to standard circle form, revealing the center and radius.

Question 10 · 2 pts

Parallelogram $BETH$, with diagonals $\overline{BT}$ and $\overline{HE}$, is drawn below. Which additional statement is sufficient to prove that $BETH$ is a rectangle?

(A) $\overline{BT} \perp \overline{HE}$
(B) $\overline{BE} \parallel \overline{HT}$
(C) $\overline{BT} \cong \overline{HE}$ ✓
(D) $\overline{BE} \cong \overline{ET}$

Worked solution

$\text{A parallelogram is a rectangle if and only if its diagonals are congruent}$

$\overline{BT} \cong \overline{HE} \Rightarrow BETH \text{ is a rectangle}$

Answer: $\overline{BT} \cong \overline{HE}$

A key theorem states a parallelogram is a rectangle exactly when its diagonals are congruent, so proving BT congruent to HE is sufficient.

Question 11 · 2 pts

A gardener wants to buy enough mulch to cover a rectangular garden that is 3 feet by 10 feet. One bag contains 2 cubic feet of mulch and costs $3.66. How much will the minimum number of bags cost to cover the garden with mulch 3 inches deep?

(A) $3.66
(B) $10.98
(C) $14.64 ✓
(D) $29.28

Worked solution

$\text{Area} = 3 \times 10 = 30 \text{ ft}^2$

$\text{Depth} = 3 \text{ in} = 0.25 \text{ ft}$

$\text{Volume} = 30 \times 0.25 = 7.5 \text{ ft}^3$

$\text{Bags needed} = \lceil 7.5/2 \rceil = 4$

$\text{Cost} = 4 \times \$3.66 = \$14.64$

Answer: $14.64

Compute the required volume of mulch, divide by the volume per bag and round up to a whole number of bags, then multiply by the price per bag.

Question 12 · 2 pts

In the diagram below, $\triangle DOG \sim \triangle CAT$, where $\angle G$ and $\angle T$ are right angles. Which expression is always equivalent to $\sin D$?

(A) $\cos A$ ✓
(B) $\sin A$
(C) $\tan A$
(D) $\cos C$

Worked solution

$\triangle DOG \sim \triangle CAT \Rightarrow \angle D \cong \angle C, \angle O \cong \angle A$

$\text{In right triangle } DOG, \angle D + \angle O = 90° \Rightarrow \sin D = \cos O$

$\text{Since } \angle O \cong \angle A, \sin D = \cos A$

Answer: $\cos A$

Since angle D and angle O are complementary in right triangle DOG, sin D equals cos O, and because angle O corresponds to angle A in the similar triangle, sin D equals cos A.

Question 13 · 2 pts

On the set of axes below, $\triangle DEF$ is the image of $\triangle ABC$ after a dilation of scale factor $\frac{1}{3}$. The center of dilation is at

(A) (0,0)
(B) (2,-3) ✓
(C) (0,-2)
(D) (-4,0)

Worked solution

$\text{The center of dilation lies on every line connecting a point to its image, e.g., line } AD, \text{ line } BE, \text{ line } CF$

$\text{These lines all intersect at the single point } (2,-3)$

Answer: (2,-3)

The center of dilation is the common intersection point of the lines through each pair of corresponding vertices, read from the graph as (2,-3).

Question 14 · 2 pts

In the diagram below of isosceles triangle $AHE$ with the vertex angle at $H$, $\overline{CB} \perp \overline{AE}$ and $\overline{FD} \perp \overline{AE}$. Which statement is always true?

(A) $\frac{AH}{AC} = \frac{EH}{EF}$
(B) $\frac{AC}{EF} = \frac{AB}{ED}$ ✓
(C) $\frac{AB}{ED} = \frac{CB}{FE}$
(D) $\frac{AD}{AB} = \frac{BE}{DE}$

Worked solution

$\triangle ABC \text{ and } \triangle EDF \text{ are both right triangles (right angles at } B, D)$

$\angle A \cong \angle E \text{ (base angles of isosceles } \triangle AHE)$

$\triangle ABC \sim \triangle EDF \text{ (AA)} \Rightarrow \frac{AC}{EF} = \frac{AB}{ED}$

Answer: $\frac{AC}{EF} = \frac{AB}{ED}$

Triangle ABC is similar to triangle EDF by AA (equal base angles of the isosceles triangle and equal right angles), so corresponding sides are proportional.

Question 15 · 2 pts

Rectangle $ABCD$ has two vertices at coordinates $A(-1,-3)$ and $B(6,5)$. The slope of $\overline{BC}$ is

(A) $-\frac{7}{8}$ ✓
(B) $\frac{7}{8}$
(C) $-\frac{8}{7}$
(D) $\frac{8}{7}$

Worked solution

$\text{slope of } \overline{AB} = \frac{5-(-3)}{6-(-1)} = \frac{8}{7}$

$\overline{BC} \perp \overline{AB} \text{ (rectangle)} \Rightarrow \text{slope of } \overline{BC} = -\frac{7}{8}$

Answer: $-\frac{7}{8}$

In a rectangle adjacent sides are perpendicular, so the slope of BC is the negative reciprocal of the slope of AB.

Question 16 · 2 pts

In right triangle $ABC$, $m\angle A = 90°$, $m\angle B = 18°$, and $AC = 8$. To the nearest tenth, the length of $\overline{BC}$ is

(A) 2.5
(B) 8.4
(C) 24.6
(D) 25.9 ✓

Worked solution

$\sin B = \frac{AC}{BC}$

$\sin 18° = \frac{8}{BC}$

$BC = \frac{8}{\sin 18°} \approx \frac{8}{0.309} \approx 25.9$

Answer: 25.9

AC is opposite angle B and BC is the hypotenuse, so sine relates them; solve for BC.

Question 17 · 2 pts

The measure of one of the base angles of an isosceles triangle is 42°. The measure of an exterior angle at the vertex of the triangle is

(A) 42°
(B) 84° ✓
(C) 96°
(D) 138°

Worked solution

$\text{Both base angles} = 42°$

$\text{Exterior angle at vertex} = \text{sum of the two remote interior angles} = 42°+42°$

$= 84°$

Answer: 84°

An exterior angle of a triangle equals the sum of the two non-adjacent (remote) interior angles, which here are the two equal base angles.

Question 18 · 2 pts

In the diagram below, $\overline{AFKB} \parallel \overline{CHLM}$, $\overline{FH} \cong \overline{LH}$, $\overline{FL} \cong \overline{KL}$, and $\overline{LF}$ bisects $\angle HFK$. Which statement is always true?

(A) $2(m\angle HLF) = m\angle CHE$
(B) $2(m\angle FLK) = m\angle LKB$
(C) $m\angle AFD = m\angle BKL$
(D) $m\angle DFK = m\angle KLF$ ✓

Worked solution

$\overline{FH} \cong \overline{LH} \Rightarrow \triangle FHL \text{ isosceles} \Rightarrow \angle HFL \cong \angle HLF$

$\overline{LF} \text{ bisects } \angle HFK \Rightarrow \angle HFL \cong \angle LFK$

$\overline{FL} \cong \overline{KL} \Rightarrow \triangle FKL \text{ isosceles} \Rightarrow \angle LFK \cong \angle LKF$

$m\angle FLK = 180° - 2(m\angle LFK) \text{ (angle sum in } \triangle FKL)$

$D,F,H \text{ collinear} \Rightarrow m\angle DFK = 180° - m\angle HFK = 180° - 2(m\angle LFK)$

$\Rightarrow m\angle DFK = m\angle FLK = m\angle KLF$

Answer: $m\angle DFK = m\angle KLF$

Using the isosceles triangles formed by the congruent segments and the angle bisector, the supplementary angle DFK works out algebraically to equal the vertex angle KLF of triangle FKL.

Question 19 · 2 pts

The line whose equation is $6x + 3y = 3$ is dilated by a scale factor of 2 centered at the point (0,0). An equation of its image is

(A) $y = -2x + 1$
(B) $y = -2x + 2$ ✓
(C) $y = -4x + 1$
(D) $y = -4x + 2$

Worked solution

$6x+3y=3 \Rightarrow 2x+y=1$

$\text{Dilation by factor 2 about origin: } (x,y) \to (2x,2y), \text{ so let } x=\frac{x'}{2}, y=\frac{y'}{2}$

$2\left(\frac{x'}{2}\right)+\frac{y'}{2}=1 \Rightarrow x' + \frac{y'}{2} = 1 \Rightarrow 2x'+y'=2$

$y' = -2x' + 2$

Answer: $y = -2x + 2$

Since the line does not pass through the center of dilation, its image is a parallel line (same slope) with the y-intercept and constant scaled appropriately, found by substitution.

Question 20 · 2 pts

Which figure will not carry onto itself after a 120-degree rotation about its center?

(A) equilateral triangle
(B) regular hexagon
(C) regular octagon ✓
(D) regular nonagon

Worked solution

$\text{A regular n-gon carries onto itself under rotations that are multiples of } \frac{360°}{n}$

$\text{Triangle: } 360/3=120° \text{ (works)}; \text{ Hexagon: } 360/6=60°, 120 \text{ is a multiple (works)}$

$\text{Octagon: } 360/8=45°, 120 \text{ is not a multiple of } 45° \text{ (fails)}$

$\text{Nonagon: } 360/9=40°, 120 \text{ is a multiple (works)}$

Answer: regular octagon

A regular octagon's rotational symmetries occur only at multiples of 45 degrees, and 120 is not a multiple of 45, so it does not map onto itself.

Question 21 · 2 pts

Triangle ADF is drawn and $\overline{BC} \parallel \overline{DF}$.

(A) $\frac{AB}{BC} = \frac{BD}{DF}$
(B) $BC = \frac{1}{2}DF$
(C) $AB:AD = AC:CF$
(D) $\angle ACB \cong \angle AFD$ ✓

Worked solution

$\overline{BC}\parallel\overline{DF}$ Given, with C on AF and B on AD

$\overline{AF} \text{ is a transversal crossing the two parallel segments}$

$\angle ACB \cong \angle AFD$ Corresponding angles formed by parallel lines cut by a transversal are congruent

Answer: $\angle ACB \cong \angle AFD$

Since BC is parallel to DF and line AF is a transversal cutting both, the corresponding angles at C and F must be congruent regardless of exact side lengths, so this is the statement that must always be true.

Question 22 · 2 pts

In $\triangle ABC$, M is the midpoint of $\overline{AB}$ and N is the midpoint of $\overline{AC}$. If $MN = x+13$ and $BC = 5x-1$, what is the length of $\overline{MN}$?

(A) 3.5
(B) 9
(C) 16.5
(D) 22 ✓

Worked solution

$MN = \tfrac{1}{2}BC$ Midsegment theorem

$x+13 = \tfrac{1}{2}(5x-1)$

$2x+26 = 5x-1$

$27 = 3x \;\Rightarrow\; x = 9$

$MN = 9+13 = 22$

Answer: 22

MN is a midsegment of triangle ABC connecting the midpoints of two sides, so it equals half of BC; solving the resulting equation gives x = 9 and MN = 22.

Question 23 · 2 pts

In the diagram below of isosceles trapezoid STAR, diagonals $\overline{AS}$ and $\overline{RT}$ intersect at O and $\overline{ST} \parallel \overline{RA}$, with nonparallel sides $\overline{SR}$ and $\overline{TA}$. Which pair of triangles are not always similar?

(A) $\triangle STO$ and $\triangle ARO$
(B) $\triangle SOR$ and $\triangle TOA$
(C) $\triangle SRA$ and $\triangle ATS$ ✓
(D) $\triangle SRT$ and $\triangle TAS$

Worked solution

$\triangle STO \sim \triangle ARO$ AA similarity from ST || RA (alternate interior angles at O)

$\triangle SOR \sim \triangle TOA$ AA similarity, vertical angles at O plus alternate interior angles

$\triangle SRT \cong \triangle TAS$ SSS: ST common, SR \cong TA (legs of isosceles trapezoid), RT \cong SA (diagonals of isosceles trapezoid), so they are similar (in fact congruent)

$\triangle SRA \text{ and } \triangle ATS \text{ share no such forced angle/side correspondence}$ No general AA, SAS, or SSS relationship guarantees similarity in every isosceles trapezoid

Answer: $\triangle SRA$ and $\triangle ATS$

The other three pairs are always similar (two by AA from the parallel sides and vertical angles, and one pair is actually always congruent by SSS using the equal legs and equal diagonals of an isosceles trapezoid), but SRA and ATS do not have a relationship that forces similarity in general.

Question 24 · 2 pts

The endpoints of $\overline{AB}$ are $A(0,4)$ and $B(-4,6)$. Which equation of a line represents the perpendicular bisector of $\overline{AB}$?

(A) $y = -\frac{1}{2}x + 4$
(B) $y = -2x + 1$
(C) $y = 2x + 8$
(D) $y = 2x + 9$ ✓

Worked solution

$\text{midpoint of } AB = \left(\tfrac{0+(-4)}{2}, \tfrac{4+6}{2}\right) = (-2,5)$

$\text{slope of } AB = \frac{6-4}{-4-0} = -\frac{1}{2}$

$\text{perpendicular slope} = 2$

$y - 5 = 2(x+2)$

$y = 2x + 9$

Answer: $y = 2x + 9$

Find the midpoint of AB and the negative reciprocal of AB's slope, then write the point-slope equation of the line through the midpoint with that perpendicular slope.

Question 25 · 2 pts · constructed response

In $\triangle ABC$ below, use a compass and straightedge to construct the altitude from C to $\overline{AB}$. [Leave all construction marks.]

Model answer: A segment from C perpendicular to line AB (extended if necessary), constructed using compass arcs to locate two points equidistant from C on line AB and then bisecting that segment's perpendicular through C.

Worked solution

$\text{Place the compass point at } C \text{ and draw an arc of any radius that intersects line } AB \text{ in two points, call them } P \text{ and } Q.$

$\text{Keeping the same or a larger radius, draw arcs from } P \text{ and from } Q \text{ on the opposite side of } AB \text{ from } C \text{ so the arcs intersect at a point } X.$

$\text{Draw segment } CX; \text{ its intersection with line } AB \text{ is the foot of the altitude, and } CX \perp AB.$

$\text{Leave all arc marks visible as the construction record.}$

Answer: A segment from C perpendicular to line AB (extended if necessary), constructed using compass arcs to locate two points equidistant from C on line AB and then bisecting that segment's perpendicular through C.

The altitude from C is constructed by finding a point equidistant from two points on AB using compass arcs, which produces the perpendicular from C to line AB by the perpendicular-construction method.

Question 26 · 2 pts · constructed response

Triangles ABC and DEF are graphed on the set of axes below. Describe a sequence of transformations that maps $\triangle ABC$ onto $\triangle DEF$.

Model answer: A 90-degree clockwise rotation about a fixed point (such as the origin), followed by a translation that slides the rotated image so its vertices coincide with D, E, and F.

Worked solution

$\triangle ABC \text{ has its apex } B \text{ pointing upward with a roughly horizontal base } AC$ read from the graph

$\triangle DEF \text{ has its apex } E \text{ pointing to the right with a roughly vertical side } DF$ read from the graph

$\text{A } 90^{\circ} \text{ clockwise rotation turns a triangle pointing up into one pointing right, matching the orientation change observed}$

$\text{Rotate } \triangle ABC \; 90^{\circ} \text{ clockwise about a convenient center (e.g. the origin)}$

$\text{Then translate the rotated image by the vector that carries the rotated } A,B,C \text{ onto } D,E,F$

Answer: A 90-degree clockwise rotation about a fixed point (such as the origin), followed by a translation that slides the rotated image so its vertices coincide with D, E, and F.

Comparing the orientation of the two triangles (ABC points up, DEF points right, and both appear the same size, so the map is a rigid motion, not a dilation) shows the sequence is a 90-degree rotation followed by a translation; the exact center and translation vector should be read directly off the grid using the vertices' coordinates.

Question 27 · 2 pts · constructed response

Line segment PQ has endpoints $P(-5,1)$ and $Q(5,6)$, and point R is on $\overline{PQ}$. Determine and state the coordinates of R, such that $PR:RQ = 2:3$.

Model answer: R = (-1, 3)

Worked solution

$PR:RQ = 2:3 \;\Rightarrow\; R \text{ divides } PQ \text{ in the ratio } \frac{2}{2+3} = \frac{2}{5} \text{ of the way from } P \text{ to } Q$

$x_R = x_P + \tfrac{2}{5}(x_Q - x_P) = -5 + \tfrac{2}{5}(5-(-5)) = -5 + \tfrac{2}{5}(10) = -5+4 = -1$

$y_R = y_P + \tfrac{2}{5}(y_Q - y_P) = 1 + \tfrac{2}{5}(6-1) = 1 + \tfrac{2}{5}(5) = 1+2 = 3$

$R = (-1,3)$

Answer: R = (-1, 3)

Using the partition (section) formula, R is located 2/5 of the way from P to Q along the segment, giving R = (-1, 3).

Question 28 · 2 pts · constructed response

A circle has a radius of 6.4 inches. Determine and state, to the nearest square inch, the area of a sector whose arc measures $80°$.

Model answer: 29 square inches

Worked solution

$\text{Area of sector} = \frac{80}{360}\pi r^2$

$= \frac{80}{360}\pi (6.4)^2 = \frac{2}{9}\pi(40.96)$

$= \frac{2}{9}(128.6797) \approx 28.5955$

$\approx 29 \text{ square inches}$

Answer: 29 square inches

The sector area is the fraction of the full circle's area corresponding to its 80-degree central angle out of 360 degrees.

Question 29 · 2 pts · constructed response

A large snowman is made of three spherical snowballs with radii of 1 foot, 2 feet, and 3 feet, respectively. Determine and state the amount of snow, in cubic feet, that is used to make the snowman. [Leave your answer in terms of $\pi$.]

Model answer: $48\pi$ cubic feet

Worked solution

$V = \frac{4}{3}\pi r^3 \text{ for each sphere}$

$r=1: V_1 = \frac{4}{3}\pi(1)^3 = \frac{4}{3}\pi$

$r=2: V_2 = \frac{4}{3}\pi(2)^3 = \frac{32}{3}\pi$

$r=3: V_3 = \frac{4}{3}\pi(3)^3 = 36\pi$

$V_1+V_2+V_3 = \frac{4}{3}\pi + \frac{32}{3}\pi + 36\pi = 12\pi + 36\pi = 48\pi$

Answer: $48\pi$ cubic feet

Add the volumes of the three spheres using the sphere volume formula, keeping the answer in terms of pi as instructed.

Question 30 · 2 pts · constructed response

In the diagram below of right triangle ACB, altitude $\overline{CD}$ is drawn to hypotenuse $\overline{AB}$, $AD = 2$ and $AC = 6$. Determine and state the length of $\overline{AB}$.

Model answer: AB = 18

Worked solution

$AC^2 = AD \cdot AB$ Geometric mean (leg) relationship in a right triangle with an altitude to the hypotenuse

$6^2 = 2 \cdot AB$

$36 = 2 \cdot AB$

$AB = 18$

Answer: AB = 18

The altitude to the hypotenuse creates the geometric mean relationship AC^2 = AD times AB, which is solved directly for AB.

Question 31 · 2 pts · constructed response

Triangle RST has vertices with coordinates $R(-3,-2)$, $S(3,2)$ and $T(4,-4)$. Determine and state an equation of the line parallel to $\overline{RT}$ that passes through point S.

Model answer: $y = -\frac{2}{7}x + \frac{20}{7}$

Worked solution

$\text{slope of } RT = \frac{-4-(-2)}{4-(-3)} = \frac{-2}{7} = -\frac{2}{7}$

$\text{Line through } S(3,2) \text{ with slope } -\tfrac{2}{7}:$

$y - 2 = -\tfrac{2}{7}(x-3)$

$y = -\tfrac{2}{7}x + \tfrac{6}{7} + 2 = -\tfrac{2}{7}x + \tfrac{20}{7}$

Answer: $y = -\frac{2}{7}x + \frac{20}{7}$

Find the slope of RT, then use point-slope form through S with that same slope since parallel lines share slope.

Question 32 · 4 pts · constructed response

Cape Canaveral, Florida is where NASA launches rockets into space. As modeled in the diagram below, a person views the launch of a rocket from observation area A, 3280 feet away from launch pad B. After launch, the rocket was sighted at C with an angle of elevation of $15°$. The rocket was later sighted at D with an angle of elevation of $31°$. Determine and state, to the nearest foot, the distance the rocket traveled between the two sightings, C and D.

Model answer: Approximately 1092 feet

Worked solution

$BC = AB \tan(15°) = 3280\tan(15°) \approx 878.87 \text{ ft}$ height of rocket at first sighting, right angle at B

$BD = AB \tan(31°) = 3280\tan(31°) \approx 1970.82 \text{ ft}$ height of rocket at second sighting

$CD = BD - BC \approx 1970.82 - 878.87 \approx 1091.95$

$CD \approx 1092 \text{ feet}$

Answer: Approximately 1092 feet

Since B, C, and D lie on the same vertical line above the launch pad, the vertical heights BC and BD can be found using the tangent ratio in the two right triangles, and the distance CD is simply the difference of those two heights.

Question 33 · 4 pts · constructed response

A small can of soup is a right circular cylinder with a base diameter of 7 cm and a height of 9 cm. A large container is also a right circular cylinder with a base diameter of 9 cm and a height of 13 cm. Determine and state the volume of the small can and the volume of the large container to the nearest cubic centimeter. What is the minimum number of small cans that must be opened to fill the large container? Justify your answer.

Model answer: Small can volume: 346 cm^3; large container volume: 827 cm^3; minimum number of small cans needed: 3

Worked solution

$\text{small can: } r = 3.5, V = \pi r^2 h = \pi(3.5)^2(9) = 110.25\pi \approx 346.36 \approx 346 \text{ cm}^3$

$\text{large container: } r = 4.5, V = \pi r^2 h = \pi(4.5)^2(13) = 263.25\pi \approx 827.03 \approx 827 \text{ cm}^3$

$2 \text{ cans} = 2(346.36) \approx 692.7 \text{ cm}^3 < 827.03 \text{ cm}^3 \text{ (not enough)}$

$3 \text{ cans} = 3(346.36) \approx 1039.1 \text{ cm}^3 > 827.03 \text{ cm}^3 \text{ (enough)}$

$\text{Minimum number of cans} = 3$

Answer: Small can volume: 346 cm^3; large container volume: 827 cm^3; minimum number of small cans needed: 3

Compute each cylinder's volume with V = pi r^2 h, then compare multiples of the small can's volume to the large container's volume to find the least whole number of cans whose total volume is at least as large.

Question 34 · 4 pts · constructed response

Parallelogram MATH has vertices $M(-7,-2)$, $A(0,4)$, $T(9,2)$, and $H(2,-4)$. Prove that parallelogram MATH is a rhombus. Determine and state the area of MATH.

Model answer: MATH is a rhombus because all four sides equal $\sqrt{85}$; the area of MATH is 68 square units.

Worked solution

$MA = \sqrt{(0-(-7))^2+(4-(-2))^2} = \sqrt{7^2+6^2} = \sqrt{85}$

$AT = \sqrt{(9-0)^2+(2-4)^2} = \sqrt{9^2+(-2)^2} = \sqrt{85}$

$TH = \sqrt{(2-9)^2+(-4-2)^2} = \sqrt{(-7)^2+(-6)^2} = \sqrt{85}$

$HM = \sqrt{(-7-2)^2+(-2-(-4))^2} = \sqrt{(-9)^2+2^2} = \sqrt{85}$

$\text{All four sides} = \sqrt{85}, \text{ so the parallelogram is a rhombus (a parallelogram with all sides congruent)}$

$\text{diagonal } MT = \sqrt{(9-(-7))^2+(2-(-2))^2} = \sqrt{16^2+4^2} = \sqrt{272} = 4\sqrt{17}$

$\text{diagonal } AH = \sqrt{(2-0)^2+(-4-4)^2} = \sqrt{2^2+(-8)^2} = \sqrt{68} = 2\sqrt{17}$

$\text{Area} = \frac{d_1 \cdot d_2}{2} = \frac{4\sqrt{17}\cdot 2\sqrt{17}}{2} = \frac{8(17)}{2} = 68$

Answer: MATH is a rhombus because all four sides equal $\sqrt{85}$; the area of MATH is 68 square units.

Using the distance formula on all four sides shows they are each length sqrt(85), proving the parallelogram is a rhombus; the area then follows from the rhombus diagonal-product formula (or equivalently the cross-product of adjacent side vectors), giving 68 square units.

Question 35 · 6 pts · constructed response

Given: Quadrilateral ABCD, $\overline{AB} \cong \overline{CD}$, $\overline{AB} \parallel \overline{CD}$, diagonal $\overline{AC}$ intersects $\overline{EF}$ at G, and $\overline{DE} \cong \overline{BF}$. Prove: G is the midpoint of $\overline{EF}$.

Model answer: G is the midpoint of EF because triangle AEG is congruent to triangle CFG by AAS, which gives EG congruent to FG.

Worked solution

$\overline{AB}\cong \overline{CD} \text{ and } \overline{AB}\parallel \overline{CD} \;\Rightarrow\; ABCD \text{ is a parallelogram}$ one pair of opposite sides both congruent and parallel

$\text{In parallelogram } ABCD: \overline{AD} \parallel \overline{BC} \text{ and } \overline{AD} \cong \overline{BC}$ opposite sides of a parallelogram are parallel and congruent

$E \text{ lies on } \overline{AD} \text{ and } F \text{ lies on } \overline{BC}$ from the figure

$AE = AD - DE \text{ and } CF = BC - BF$

$\text{Since } AD \cong BC \text{ and } DE \cong BF, \; AE \cong CF$

$\overline{AD}\parallel \overline{BC} \;\Rightarrow\; \angle EAG \cong \angle FCG$ alternate interior angles, transversal AC

$\angle AGE \cong \angle CGF$ vertical angles at G, the intersection of AC and EF

$\triangle AEG \cong \triangle CFG$ AAS: angle EAG \cong angle FCG, angle AGE \cong angle CGF, side AE \cong side CF

$\overline{EG} \cong \overline{FG}$ corresponding parts of congruent triangles (CPCTC)

$\text{Therefore } G \text{ is the midpoint of } \overline{EF}$

Answer: G is the midpoint of EF because triangle AEG is congruent to triangle CFG by AAS, which gives EG congruent to FG.

Because AB is both congruent and parallel to CD, ABCD is a parallelogram, which makes AD parallel and congruent to BC; combined with the given DE congruent to BF, this yields AE congruent to CF, and the resulting AAS congruence of triangles AEG and CFG (using alternate interior angles and vertical angles at G) proves EG congruent to FG, so G bisects EF.

June 2023 Geometry Regents — Answer Key & Worked Solutions

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