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June 2025 Geometry Regents — Answer Key & Worked Solutions

Every question from the June 2025 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 2025 Geometry New York State Regents examination, published by NYSED. Worked solutions are written by RegentsLab.

Question 1 · 2 pts

The perimeter of a triangle is 18. What is the perimeter of a similar triangle after a dilation with a scale factor of 3?

Worked solution
$P' = k \cdot P$ Dilation scales all linear measures (including perimeter) by the scale factor k
$P' = 3(18) = 54$
Answer: 54

A dilation multiplies every linear dimension, including perimeter, by the scale factor.

Question 2 · 2 pts

The Washington Monument, shown below, is in Washington, D.C. At a point on the ground 200 feet from the center of the base of the monument, the angle of elevation to the top of the monument is 70.19°. What is the height of the monument, to the nearest foot?

geo question 2 figure
Worked solution
$\tan(70.19^\circ) = \frac{h}{200}$
$h = 200\tan(70.19^\circ) \approx 200(2.776) \approx 555 \text{ ft}$
Answer: 555

The height is the side opposite the angle of elevation, and the 200 ft distance is adjacent, so tangent relates them.

Question 3 · 2 pts

On the set of axes below, △EQA and △SUL are graphed. Which sequence of transformations shows that △EQA ≅ △SUL?

geo question 3 figure
Worked solution
$R_{90^\circ,\,O}: (x,y) \to (-y,x)$ Apply the rotation to each vertex of △EQA
$T_{(9,-1)}$ Translate the rotated image to land exactly on △SUL
Answer: Rotate △EQA 90° counterclockwise about the origin and then translate 9 units right and 1 unit down.

Tracing the vertices of △EQA through a 90° counterclockwise rotation about the origin followed by the given translation lands exactly on the corresponding vertices of △SUL, and both are rigid motions so congruence is preserved.

Question 4 · 2 pts

If two sides of a triangle have lengths of 2 and 8, the length of the third side could be

Worked solution
$|8-2| < x < 8+2$
$6 < x < 10$
Answer: 7

By the Triangle Inequality Theorem the third side must be strictly between the difference and sum of the other two sides; only 7 satisfies 6 < x < 10.

Question 5 · 2 pts

A regular octagon is rotated about its center. Which angle measure will carry the octagon onto itself?

Worked solution
$\theta = \frac{360^\circ}{8} = 45^\circ$ Smallest rotation that carries a regular octagon onto itself
$\text{Valid angles are multiples of } 45^\circ: 45,90,135,180,\dots$
Answer: 90°

A regular n-gon maps onto itself under rotations that are multiples of 360°/n; for an octagon that is multiples of 45°, and 90° is the only such multiple among the choices.

Question 6 · 2 pts

An equation of a circle is $x^2 + y^2 - 6x + 2y = 14$. What are the coordinates of the center and the length of the radius of this circle?

Worked solution
$x^2-6x+9 + y^2+2y+1 = 14+9+1$ Complete the square in x and y
$(x-3)^2+(y+1)^2 = 24$
Answer: $(3,-1)$ and $r = \sqrt{24}$

Completing the square puts the equation in standard form, revealing center (3,-1) and radius √24.

Question 7 · 2 pts

In △HSF below, m∠S = 90°, HF = 30, and FS = 23. What is m∠F, to the nearest degree?

geo question 7 figure
Worked solution
$\cos(F) = \frac{FS}{HF} = \frac{23}{30}$ HF is the hypotenuse (opposite the right angle at S); FS is adjacent to angle F
$F = \cos^{-1}\left(\frac{23}{30}\right) \approx 39.96^\circ \approx 40^\circ$
Answer: 40°

Since angle S is the right angle, HF is the hypotenuse and FS is adjacent to angle F, so cosine relates them.

Question 8 · 2 pts

In △CAB below, midsegments $\overline{DE}$, $\overline{EF}$, and $\overline{FD}$ are drawn. If CA = 14, CB = 20, and FB = 9, what is the perimeter of quadrilateral DEFA?

geo question 8 figure
Worked solution
$D,E,F \text{ are midpoints of } CA,CB,AB$ Midsegment theorem
$AD = \tfrac{1}{2}CA = 7,\quad FA = FB = 9 \text{ (F is midpoint of AB)}$
$DE = \tfrac{1}{2}AB,\ EF = \tfrac{1}{2}CA = 7,\ \text{and } DE = \tfrac{1}{2}AB \text{ but here DE is parallel to } AB, DE = \tfrac{1}{2}(AB)$
$\text{Since } AF=FB=9,\ AB=18,\ DE=\tfrac12(18)=9$
$P_{DEFA} = DE+EF+FA+AD = 9+7+9+7 = 32$
Answer: 32

Each midsegment is half the length of the side it is parallel to, so DE = 9, EF = 7, and with AD = 7 and FA = 9 the quadrilateral's perimeter is 32.

Question 9 · 2 pts

A candle can be modeled by a pyramid with a square base, as shown below. The height of the candle is 10 cm, and each side of the base measures 6 cm. If the candle wax burns at a rate of 3.5 cubic centimeters per hour, what is the approximate number of hours this candle could burn?

geo question 9 figure
Worked solution
$V = \frac{1}{3}s^2h = \frac{1}{3}(6)^2(10) = 120 \text{ cm}^3$
$t = \frac{120}{3.5} \approx 34.3 \text{ hours}$
Answer: 34

Compute the pyramid's volume, then divide by the burn rate to find approximate burn time.

Question 10 · 2 pts

In the diagram below, tangent $\overline{SR}$ and secant $\overline{STU}$ are drawn to circle O from external point S. If $\widehat{TU} \cong \widehat{RU}$ and $m\widehat{TR} = 68°$, what is m∠S?

geo question 10 figure
Worked solution
$\widehat{TR}+\widehat{RU}+\widehat{UT} = 360^\circ,\quad \widehat{RU}=\widehat{UT}=x$
$68 + 2x = 360 \Rightarrow x = 146^\circ$
$m\angle S = \frac{1}{2}\left(\widehat{RU} - \widehat{TR}\right) = \frac{1}{2}(146-68) = 39^\circ$
Answer: 39°

The tangent-secant angle equals half the difference of the intercepted arcs; using the arc relationships given, the far arc RU is 146° and the near arc TR is 68°.

Question 11 · 2 pts

Triangle RST has m∠S = 33°, RS = 7, and ST = 12. What is the area of △RST, to the nearest tenth?

Worked solution
$A = \frac{1}{2}(RS)(ST)\sin(S)$
$A = \frac{1}{2}(7)(12)\sin(33^\circ) \approx 42(0.5446) \approx 22.9$
Answer: 22.9

Use the SAS area formula, which requires two sides and the included angle.

Question 12 · 2 pts

Triangle ABC, with vertices whose coordinates are A(-3,-2), B(-1,2), and C(4,1), is graphed on the set of axes below. Triangle A'B'C', whose vertices have coordinates A'(-6,-2), B'(-2,2), and C'(8,1), is the image of △ABC. The transformation that maps △ABC onto △A'B'C' is a

geo question 12 figure
Worked solution
$A(-3,-2)\to A'(-6,-2):\ x\text{-coord doubles, }y\text{-coord unchanged}$
$B(-1,2)\to B'(-2,2),\quad C(4,1)\to C'(8,1)$ Same pattern for all vertices
$(x,y)\to(2x,y)$ This maps every x-coordinate to twice its value while y stays fixed, a horizontal stretch by factor 2
Answer: horizontal stretch

Every x-coordinate doubles while every y-coordinate stays the same, which is a horizontal stretch (not a uniform dilation, since x and y scale differently).

Question 13 · 2 pts

Which equation represents a line parallel to the line represented by $y = 4x + 6$ and passing through the point (-3,2)?

Worked solution
$\text{Parallel lines have equal slopes: } m = 4$
$y - y_1 = m(x-x_1) \Rightarrow y - 2 = 4(x-(-3)) = 4(x+3)$
Answer: $y - 2 = 4(x + 3)$

Parallel lines share the same slope, so using point-slope form with slope 4 and point (-3,2) gives the correct equation.

Question 14 · 2 pts

Which two-dimensional figure is always formed when a plane intersects a right cylinder perpendicular to its base?

Worked solution
$\text{A plane perpendicular to the circular base slices vertically through the cylinder}$
$\text{The cross section has two straight vertical sides (cylinder height) and two straight horizontal sides (chords of the base circles), forming a rectangle}$
Answer: rectangle

A vertical cross-section (perpendicular to the base) of a right cylinder always produces a rectangle, bounded by two parallel vertical segments equal to the height and two parallel chords of the circular bases.

Question 15 · 2 pts

In △KMP below, $\overline{CE}$ is drawn parallel to $\overline{MP}$. If KC = 8, CM = 3, and CE = 12, what is the length of $\overline{MP}$?

geo question 15 figure
Worked solution
$KM = KC + CM = 8+3 = 11$
$\triangle KCE \sim \triangle KMP \text{ (since } CE \parallel MP\text{)}$
$\frac{KC}{KM} = \frac{CE}{MP} \Rightarrow \frac{8}{11} = \frac{12}{MP}$
$MP = \frac{12 \times 11}{8} = 16.5$
Answer: 16.5

Since CE is parallel to MP, triangles KCE and KMP are similar, giving a proportion between corresponding sides.

Question 16 · 2 pts

A parallelogram must be a rectangle if its diagonals

Worked solution
$\text{In a parallelogram, diagonals always bisect each other}$
$\text{Congruent diagonals is the additional condition that forces right angles, making it a rectangle}$
Answer: are congruent

All parallelograms have diagonals that bisect each other, but only a rectangle's diagonals are also congruent.

Question 17 · 2 pts

Point O divides $\overline{COA}$ such that CO:OA = 1:4. If C has coordinates (-2,-9) and A has coordinates (3,6), the coordinates of O are

Worked solution
$O = C + \frac{1}{5}(A-C)$ CO:OA = 1:4 means O is 1/5 of the way from C to A
$O_x = -2 + \frac{1}{5}(3-(-2)) = -2+1 = -1$
$O_y = -9 + \frac{1}{5}(6-(-9)) = -9+3 = -6$
Answer: (-1,-6)

Since CO:OA = 1:4, point O partitions segment CA in a 1:4 ratio, located 1/5 of the way from C to A.

Question 18 · 2 pts

A spherical balloon is fully inflated with helium to a diameter of 1.7 feet. If helium costs $0.80 per cubic foot, what is the cost to completely fill the balloon with helium?

Worked solution
$r = \frac{1.7}{2} = 0.85 \text{ ft}$
$V = \frac{4}{3}\pi r^3 = \frac{4}{3}\pi(0.85)^3 \approx 2.572 \text{ ft}^3$
$\text{Cost} = 2.572 \times 0.80 \approx \$2.06$
Answer: $2.06

Compute the sphere's volume from its radius, then multiply by the cost per cubic foot.

Question 19 · 2 pts

In right triangle ABD below, altitude $\overline{BC}$ is drawn to hypotenuse $\overline{AD}$, AC = 2.5, and CD = 4.3. What is the length of $\overline{BA}$, to the nearest tenth?

geo question 19 figure
Worked solution
$AD = AC + CD = 2.5+4.3 = 6.8$
$BA^2 = AC \cdot AD$ Geometric mean (altitude-on-hypotenuse) relation
$BA = \sqrt{2.5 \times 6.8} = \sqrt{17} \approx 4.1$
Answer: 4.1

The leg adjacent to the smaller segment equals the geometric mean of that segment and the whole hypotenuse.

Question 20 · 2 pts

Trapezoid ZOYD has parallel sides $\overline{ZO}$ and $\overline{DY}$. If m∠Z = 141° and m∠Y = 73°, what is m∠D?

Worked solution
$\overline{ZD} \text{ is a transversal cutting parallel segments } ZO \parallel DY$
$m\angle Z + m\angle D = 180^\circ$ Co-interior (same-side interior) angles are supplementary
$m\angle D = 180-141 = 39^\circ$
Answer: 39°

Angles Z and D lie on the same leg (transversal) between the two parallel sides, so they are supplementary co-interior angles.

Question 21 · 2 pts

Triangle ABC is translated 5 units to the left and 2 units up to map onto $\triangle PQR$. Which statement is not always true?

Worked solution
$A \to P,\ B \to Q,\ C \to R \text{ under the translation}$
$\text{A translation is a rigid motion, so } \triangle PQR \cong \triangle ABC \text{ always}$
$\text{Corresponding angles are } \angle A \cong \angle P,\ \angle B \cong \angle Q,\ \angle C \cong \angle R$
$\angle A \cong \angle Q \text{ is not a valid correspondence, so it is not always true}$
$BQ = \sqrt{5^2+2^2} = \sqrt{29} \text{ always (distance moved by translation)}$
$RQ = CB \text{ always, since translations preserve corresponding side lengths}$
Answer: \angle A \cong \angle Q

A translation is a rigid motion, so it preserves distances and maps corresponding parts (A to P, B to Q, C to R) onto each other; angle A corresponds to angle P, not angle Q, so that statement fails.

Question 22 · 2 pts

In the diagram below, congruent triangles PEN and PAL are drawn. Which rigid motion maps $\triangle PEN$ onto $\triangle PAL$?

geo question 22 figure
Worked solution
$P \text{ is the common vertex of the two congruent, symmetric triangles } PEN \text{ and } PAL$
$\text{The two triangles are mirror images of each other about the bisector of } \angle EPA$
$\text{Reflecting } \triangle PEN \text{ over this bisector sends } E \to A,\ N \to L,\ P \to P$
Answer: a reflection of \triangle PEN over the angle bisector of \angle EPA

Since PE = PA and PN = PL with the triangles arranged symmetrically about the vertex P, the line that maps one onto the other is the angle bisector of angle EPA, which is a line of reflection, not a rotation or point reflection center.

Question 23 · 2 pts

A cone has a height of 8 inches and volume of 75.4 cubic inches. What is the diameter of the cone, to the nearest inch?

Worked solution
$V = \frac{1}{3}\pi r^2 h$
$75.4 = \frac{1}{3}\pi r^2 (8)$
$r^2 = \frac{75.4 \times 3}{8\pi} \approx 9.00$
$r \approx 3.0 \text{ in} \Rightarrow d = 2r \approx 6 \text{ in}$
Answer: 6

Solving the cone volume formula for radius gives r approximately 3 inches, so the diameter is about 6 inches.

Question 24 · 2 pts

The line represented by the equation 5x - 2y = 10 is transformed by a dilation centered at (2,0) with a scale factor of 2. The image of the line

Worked solution
$\text{Check whether the center of dilation } (2,0) \text{ lies on the line}$
$5(2) - 2(0) = 10 - 0 = 10 \checkmark$
$\text{Since the center of dilation lies on the line, the dilation maps the line onto itself}$
Answer: is the original line

A dilation centered at a point that lies on a line maps that line onto itself, regardless of scale factor, since every point on the line dilates to another point on the same line through the center.

Question 25 · 2 pts · constructed response

In the year 2020, the village of Depew, New York had an area of 5.1 square miles and a population of 15,069. In the same year, the village of Lancaster, New York had an area of 2.7 square miles and a population of 10,087. Which village had the larger population density in 2020? Justify your answer.

Model answer: Lancaster had the larger population density (about 3735.9 people per square mile vs. about 2954.7 for Depew).
Worked solution
$\text{Depew density} = \frac{15069}{5.1} \approx 2954.7 \text{ people/mi}^2$
$\text{Lancaster density} = \frac{10087}{2.7} \approx 3735.9 \text{ people/mi}^2$
$3735.9 > 2954.7$
Answer: Lancaster had the larger population density (about 3735.9 people per square mile vs. about 2954.7 for Depew).

Population density equals population divided by area; comparing the two computed densities shows Lancaster is denser than Depew.

Question 26 · 2 pts · constructed response

In $\triangle ABC$ below, $\overline{AC}$ is extended through C to D, $m\angle A = (3x-22)^{\circ}$, $m\angle B = (4x-18)^{\circ}$, and $m\angle BCD = (6x-23)^{\circ}$. Determine and state $m\angle ACB$.

geo question 26 figure
Model answer: m\angle ACB = 101^{\circ}
Worked solution
$\text{Exterior angle theorem: } m\angle BCD = m\angle A + m\angle B$
$6x - 23 = (3x-22) + (4x-18)$
$6x - 23 = 7x - 40$
$17 = x$
$m\angle BCD = 6(17) - 23 = 79^{\circ}$
$m\angle ACB = 180^{\circ} - 79^{\circ} = 101^{\circ} \text{ (linear pair with } \angle BCD\text{)}$
Answer: m\angle ACB = 101^{\circ}

The exterior angle of a triangle equals the sum of the two remote interior angles, which gives x = 17; angle ACB is supplementary to exterior angle BCD since they form a linear pair along line ACD.

Question 27 · 2 pts · constructed response

Parallelogram ABCD is shown below. Using a compass and straightedge, construct the altitude from point A to side $\overline{DC}$. Leave all construction marks.

geo question 27 figure
Model answer: Construct a line through A perpendicular to line DC (extending DC if necessary), meeting DC (or its extension) at a point, say F; segment AF is the required altitude.
Worked solution
$\text{Place the compass point at } A \text{ and draw an arc that intersects line } DC \text{ (extended if needed) at two points, } X \text{ and } Y$
$\text{Open the compass wider than half } XY \text{ and draw intersecting arcs above and below line } DC \text{ from } X \text{ and } Y$
$\text{Draw a line through } A \text{ and the intersection point of these arcs; this line is perpendicular to } DC$
$\text{Label the point where this perpendicular line meets } DC \text{ (or its extension) as } F; \ \overline{AF} \text{ is the altitude from } A \text{ to } \overline{DC}$
Answer: Construct a line through A perpendicular to line DC (extending DC if necessary), meeting DC (or its extension) at a point, say F; segment AF is the required altitude.

This is a classic compass-and-straightedge construction of a perpendicular from an external point to a line, which produces the altitude from vertex A to side DC of the parallelogram.

Question 28 · 2 pts · constructed response

Quadrilateral QUAD is graphed on the set of axes below. Determine and state the area of quadrilateral QUAD.

geo question 28 figure
Model answer: Area of QUAD = 24 square units
Worked solution
$\text{From the graph, } Q(-4,0),\ U(0,3),\ A(4,0),\ D(0,-3)$
$QUAD \text{ is a rhombus (kite) whose diagonals are } \overline{QA} \text{ (horizontal) and } \overline{UD} \text{ (vertical)}$
$QA = 8,\ UD = 6$
$\text{Area} = \frac{d_1 \cdot d_2}{2} = \frac{8 \cdot 6}{2} = 24$
Answer: Area of QUAD = 24 square units

QUAD's diagonals lie on the x- and y-axes; reading their lengths from the grid and applying the diagonals-of-a-rhombus area formula gives the area.

Question 29 · 2 pts · constructed response

In a right triangle, the acute angles have the relationship $\sin(3x-7)^{\circ} = \cos(x+1)^{\circ}$. Determine and state the value of $x$.

Model answer: x = 24
Worked solution
$\text{Since the acute angles of a right triangle are complementary, } \sin\theta = \cos(90-\theta)$
$(3x-7) + (x+1) = 90$
$4x - 6 = 90$
$4x = 96$
$x = 24$
Answer: x = 24

Sine and cosine of complementary angles are equal, so the two given angle expressions must sum to 90 degrees; solving that linear equation gives x = 24.

Question 30 · 2 pts · constructed response

In circle A, $m\angle BAM = 36^{\circ}$. If $AB = 20$, determine and state the length of $\widehat{MB}$. Leave your answer in terms of $\pi$.

Model answer: \widehat{MB} = 4\pi
Worked solution
$\text{Radius } r = AB = 20$
$\text{Circumference} = 2\pi r = 40\pi$
$\text{Arc } \widehat{MB} \text{ corresponds to central angle } 36^{\circ}$
$\text{arc length} = \frac{36}{360} \times 40\pi = \frac{1}{10} \times 40\pi = 4\pi$
Answer: \widehat{MB} = 4\pi

The central angle BAM equals the measure of its intercepted arc MB; the arc length is that fraction of the full circumference (2 pi r).

Question 31 · 2 pts · constructed response

In triangles ANT and ELM below, AN = 6, NT = 5.6, TA = 4, EL = 9, LM = 8.4, and ME = 6. Explain why $\triangle ANT \sim \triangle ELM$.

geo question 31 figure
Model answer: All three pairs of corresponding sides are in the same ratio (1.5), so by the SSS similarity theorem, \triangle ANT \sim \triangle ELM.
Worked solution
$\frac{EL}{AN} = \frac{9}{6} = 1.5$
$\frac{LM}{NT} = \frac{8.4}{5.6} = 1.5$
$\frac{ME}{TA} = \frac{6}{4} = 1.5$
$\text{All three ratios of corresponding sides are equal to } 1.5$
Answer: All three pairs of corresponding sides are in the same ratio (1.5), so by the SSS similarity theorem, \triangle ANT \sim \triangle ELM.

Since the ratios of all three pairs of corresponding sides (AN to EL, NT to LM, TA to ME) are equal, the SSS similarity theorem guarantees the triangles are similar.

Question 32 · 4 pts · constructed response

A store sells colored craft sand in the three different containers below. Container 1: A square prism with a base length of 4 inches and a height of 7.5 inches. Container 2: A cylinder with a diameter of 5 inches and a height of 6 inches. Container 3: A cone with a diameter of 7.5 inches and a height of 8.5 inches. If the containers are filled to the top, which container will hold the most sand? Justify your answer.

Model answer: Container 3 (the cone) holds the most sand, with a volume of approximately 125.2 cubic inches.
Worked solution
$\text{Container 1: } V = 4^2 \times 7.5 = 16 \times 7.5 = 120 \text{ in}^3$
$\text{Container 2: } r = 2.5,\ V = \pi r^2 h = \pi (2.5)^2 (6) = 37.5\pi \approx 117.8 \text{ in}^3$
$\text{Container 3: } r = 3.75,\ V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi (3.75)^2 (8.5) \approx 125.2 \text{ in}^3$
$125.2 > 120 > 117.8$
Answer: Container 3 (the cone) holds the most sand, with a volume of approximately 125.2 cubic inches.

Computing each solid's volume with the appropriate formula (square prism, cylinder, cone) shows the cone (Container 3) has the greatest volume and therefore holds the most sand.

Question 33 · 4 pts · constructed response

Quadrilateral MIKE has vertices with coordinates M(-1,-3), I(-3,3), K(5,4), and E(7,-2). Prove MIKE is a parallelogram, and prove MIKE is not a rhombus.

Model answer: MIKE is a parallelogram because its diagonals bisect each other (both have midpoint (2, 0.5)); it is not a rhombus because adjacent sides MI = sqrt(40) and IK = sqrt(65) are not equal in length.
Worked solution
$\text{Midpoint of } \overline{MK} = \left(\frac{-1+5}{2}, \frac{-3+4}{2}\right) = (2, 0.5)$
$\text{Midpoint of } \overline{IE} = \left(\frac{-3+7}{2}, \frac{3-2}{2}\right) = (2, 0.5)$
$\text{Since the diagonals share the same midpoint, they bisect each other, so } MIKE \text{ is a parallelogram}$
$MI = \sqrt{(-3-(-1))^2 + (3-(-3))^2} = \sqrt{(-2)^2+6^2} = \sqrt{40}$
$IK = \sqrt{(5-(-3))^2 + (4-3)^2} = \sqrt{8^2+1^2} = \sqrt{65}$
$\sqrt{40} \neq \sqrt{65}, \text{ so adjacent sides are not congruent and } MIKE \text{ is not a rhombus}$
Answer: MIKE is a parallelogram because its diagonals bisect each other (both have midpoint (2, 0.5)); it is not a rhombus because adjacent sides MI = sqrt(40) and IK = sqrt(65) are not equal in length.

A quadrilateral is a parallelogram if its diagonals bisect each other, which is shown by the equal midpoints of MK and IE; since a rhombus requires all sides congruent, showing two adjacent sides have different lengths proves MIKE is not a rhombus.

Question 34 · 4 pts · constructed response

A park ranger needs to secure two different trees with wire. A wire is to be attached from a stake in the ground to each tree. The wire is attached at two different heights and two different angles of elevation, as indicated in the model: one tree has height 4 ft with the wire making a 24 degree angle of elevation from the stake, and the other tree has height 3 ft with the wire making an 11.6 degree angle of elevation from the stake. The park ranger has 20 feet of wire. Does the park ranger have enough wire to secure both trees? Justify your answer.

Model answer: No; the two wires require a total of about 24.75 feet, which is more than the 20 feet of wire the park ranger has.
Worked solution
$\text{Wire}_1 = \frac{4}{\sin(24^{\circ})} \approx \frac{4}{0.4067} \approx 9.83 \text{ ft}$
$\text{Wire}_2 = \frac{3}{\sin(11.6^{\circ})} \approx \frac{3}{0.2011} \approx 14.92 \text{ ft}$
$\text{Total wire needed} \approx 9.83 + 14.92 \approx 24.75 \text{ ft}$
$24.75 \text{ ft} > 20 \text{ ft, so there is not enough wire}$
Answer: No; the two wires require a total of about 24.75 feet, which is more than the 20 feet of wire the park ranger has.

Each wire is the hypotenuse of a right triangle where the tree height is opposite the angle of elevation at the stake, so wire length = height divided by sine of the angle; summing both required lengths and comparing to the available 20 feet shows the ranger falls short.

Question 35 · 6 pts · constructed response

Given: Quadrilateral CARI with $\overline{CA} \cong \overline{RI}$ and $\overline{CI} \cong \overline{RA}$, and $\overline{AEI}$ and $\overline{LEH}$ are drawn. Prove: $HA \cdot EL = LI \cdot EH$

geo question 35 figure
Model answer: HA \cdot EL = LI \cdot EH, proven via similar triangles AEH and IEL.
Worked solution
$\text{Given } \overline{CA} \cong \overline{RI} \text{ and } \overline{CI} \cong \overline{RA}$
$\text{Both pairs of opposite sides of quadrilateral } CARI \text{ are congruent, so } CARI \text{ is a parallelogram}$
$\text{Opposite sides of a parallelogram are parallel, so } \overline{CA} \parallel \overline{IR}, \text{ hence } \overline{HA} \parallel \overline{LI} \text{ (H on } CA\text{, L on } IR\text{)}$
$\overline{AI} \text{ is a transversal cutting parallel segments } HA \text{ and } LI$
$\angle HAE \cong \angle LIE \text{ (alternate interior angles, } HA \parallel LI\text{)}$
$\angle HEA \cong \angle LEI \text{ (vertical angles, since } \overline{AEI} \text{ and } \overline{LEH} \text{ intersect at } E\text{)}$
$\triangle AEH \sim \triangle IEL \text{ by AA similarity}$
$\text{Corresponding sides are proportional: } \frac{HA}{LI} = \frac{EH}{EL}$
$\text{Cross-multiplying: } HA \cdot EL = LI \cdot EH$
Answer: HA \cdot EL = LI \cdot EH, proven via similar triangles AEH and IEL.

Because both pairs of opposite sides are congruent, CARI is a parallelogram, making CA parallel to IR (and therefore HA parallel to LI); this parallelism, together with the vertical angles at E where diagonals AI and LH cross, establishes triangle AEH similar to triangle IEL by AA, whose proportional sides directly yield the desired product equality.

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