June 2026 Geometry Regents: All 35 Worked Solutions

Question 1: Identify a transformation that changes area

1. Reflections, translations and rotations are rigid motions. They preserve lengths and therefore preserve the rectangle’s area.
2. A vertical stretch by a factor of 3 changes each point’s distance from the x-axis by that factor. Horizontal lengths remain unchanged while vertical lengths triple.
3. Consequently the image area is A′=3A, so the area changes.

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Question 2: Apply the triangle midsegment theorem

1. D and E are the midpoints of two sides of triangle ABC. The segment joining them is a midsegment.
2. A triangle’s midsegment is parallel to the third side and half its length. Here the third side is AB:
   DE=12AB

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Question 3: Find the hypotenuse using cosine

1. The right angle is at R, so MJ is the hypotenuse. Relative to the 35° angle at M, MR is the adjacent leg.
2. Set up cosine and solve for the hypotenuse:
   cos(35°)=7.25MJ
   MJ=7.25cos(35°)≈8.8506
3. Round to the nearest hundredth of an inch.

Exam detail: Use degree mode and divide by cosine when the unknown is the hypotenuse.

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Question 4: Convert the water volume to liters

1. The water stops 3 cm below the top, so its height is 25−3=22 cm.
2. Calculate the volume of the water, using this height rather than the tank’s full height:
   V=35(17)(22)=13090
   cubic centimeters.
3. Since 1 liter is 1,000 cubic centimeters, divide by 1,000:
   130901000=13.09
   liters. Round to the nearest liter.

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Question 5: Track a side through two reflections

1. Under the first reflection, triangle ACE maps to triangle AXE. The vertex correspondence is A to A, C to X, and E to E. Thus side AC maps to side AX.
2. Under the second reflection, triangle AXE maps to triangle LXE. A maps to L, while X and E stay fixed on the reflecting line XE.
3. Therefore the image of the original side AC is side LX:
   AC→AX→LX

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Question 6: Compare slopes and intercepts

1. The first line already has slope-intercept form:
   y=−12x+2
2. Rewrite the second equation:
   x+2y=8
   y=−12x+4
3. Both slopes are −12, but the y-intercepts are different. The lines are distinct and parallel.

Exam detail: Equal slopes alone do not distinguish parallel lines from the same line; compare the intercepts too.

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Question 7: Find the cone formed by rotation

1. Side AB is the fixed axis, so it becomes the cone’s height: h=6. The perpendicular leg BC sweeps out the circular base, so r=8.
2. Use the cone-volume formula:
   V=13πr²h
   V=13π(8)²(6)=128π

Exam detail: The rotating leg gives the radius, not the diameter.

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Question 8: Use the perpendicular-bisector theorem

1. CE is perpendicular to AB and passes through its midpoint E. Therefore the line CE is the perpendicular bisector of AB.
2. Every point on a segment’s perpendicular bisector is the same distance from its endpoints. Since C lies on this line, CA=CB.
3. Equal lengths mean the corresponding segments are congruent.

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Question 9: Choose the correct smaller right triangle

1. The altitude BD is perpendicular to AC, so triangle BCD is a right triangle with right angle at D.
2. In this smaller triangle, BC is the hypotenuse and has length 12. For the angle at C, BD is the opposite leg.
3. Sine is opposite divided by hypotenuse:
   sin(C)=BDBC=BD12

Exam detail: Use the hypotenuse of triangle BCD, not the hypotenuse of the original triangle.

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Question 10: Test both rotational symmetries

1. A regular n-sided polygon has a basic rotational symmetry of 360°n. Other symmetry rotations are whole-number multiples of this angle.
2. A regular hexagon has a basic rotation of 60°. Both 120°=2(60°) and 180°=3(60°) carry it onto itself.
3. A triangle fails the 180° test; a square and an octagon fail the 120° test.

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Question 11: Divide a segment into five equal parts

1. The ratio PC:CA=1:4 means that C lies one-fifth of the way from P to A.
2. Apply one-fifth of each coordinate change:
   x_C=3+15(−2−3)=2
   y_C=−6+15(9−(−6))=−3

Exam detail: The total number of equal parts is five, not four.

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Question 12: Find the scale factor and dilation center

1. Read corresponding points from the grid: R(1,3)→B(1,9), E(−2,1)→L(−8,3), and D(2,−3)→U(4,−9).
2. Test the center (1,0). Relative to this center, R is 3 units above it and B is 9 units above it, giving scale factor 93=3.
3. The resulting rule is (x,y)→(1+3(x−1),3y). It sends E to (−8,3) and D to (4,−9), agreeing with the other image vertices.

Exam detail: Measure each displacement from the center of dilation, rather than automatically multiplying coordinates from the origin.

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Question 13: Complete the square for a circle

1. Move the x-term to the left:
   x²−4x+y²=45
2. Complete the square by adding 4 to both sides:
   (x−2)²+y²=49
3. Compare with (x−h)²+(y−k)²=r². The center is (2,0) and the radius is √49=7.

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Question 14: Use the right-triangle leg theorem

1. An altitude to the hypotenuse creates similar right triangles. A leg squared equals the full hypotenuse times the segment of the hypotenuse next to that leg.
2. For leg EF, the adjacent hypotenuse segment is FT:
   EF²=FT×FG
   17²=15×FG
   FG=28915
3. Subtract FT to obtain only TG:
   TG=28915−15=6415≈4.2667

Exam detail: The intermediate result is the entire hypotenuse FG. The question asks for its remaining segment TG.

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Question 15: Use supplementary angles on a trapezoid leg

1. ER and TJ are parallel, and RJ is a transversal. The interior angles at R and J are on the same side of that transversal, so they add to 180°.
2. Solve for x:
   (2x+15)+(3x−40)=180
   5x=205
   x=41
3. Substitute into the expression for angle J:
   m∠J=(3(41)−40)°=83°

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Question 16: Find a leg of an isosceles right triangle

1. The two legs have the same length. Let that length be x.
2. Apply the Pythagorean theorem:
   x²+x²=14²
   2x²=196
   x=√98=7√2≈9.8995
3. Round to the nearest tenth.

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Question 17: Apply the tangent-secant theorem

1. The full secant from S to E includes both segments:
   SE=SL+LE=8+10=18
2. The tangent length squared equals the external secant length times the full secant length:
   AS²=SL×SE=8(18)=144
3. A length is positive, so AS=√144=12.

Exam detail: Use the full secant length 18 in the product, not just the inside portion 10.

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Question 18: Recognize the rhombus diagonal condition

1. The vertices T and M lie on different diagonals, and I is their intersection. Thus m∠TIM=90° says the diagonals are perpendicular.
2. A parallelogram with perpendicular diagonals is a rhombus. One way to see this is that its diagonals bisect each other, creating right triangles with equal corresponding legs; adjacent sides are therefore equal.

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Question 19: Find the smaller complementary angle

1. For acute angles, sin(α)=cos(β) implies that α+β=90°.
2. Set the given angle expressions equal to this sum:
   3x+(x+10)=90
   x=20
3. The acute angles are 3x=60° and x+10=30°. The third angle is 90°, so the smallest is 30°.

Exam detail: The value of x is 20; the question asks for an angle measure, not x itself.

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Question 20: Bound the number of sides in a plane section

1. A rectangular prism has six faces. A slicing plane can contribute at most one straight boundary segment from each face.
2. Therefore a nondegenerate polygonal cross-section has at most six sides. An octagon has eight sides, so it cannot occur.

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Question 21: Calculate the length of the minor arc

1. D is the center, so DA gives radius 12. The 150° central angle subtends the minor arc AB.
2. Take the corresponding fraction of the circumference:
   ℓ=150360×2π(12)=10π

Exam detail: The curved arc symbol over AB asks for arc length, not the straight chord length.

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Question 22: Apply both triangle inequalities

1. If the third side has length x, it must be larger than the difference of the known sides and smaller than their sum:
   30−12<x<30+12
   18<x<42
2. Among the listed values, only 28 lies strictly inside this interval. Values 18 and 42 would give a straight, degenerate figure rather than a triangle.

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Question 23: Use similarity to find the perimeter

1. The angles at B are vertical angles. Since AE is parallel to CD, a second pair of angles is congruent by the alternate-interior-angle theorem. Hence △ABE∼△CBD by AA.
2. The scale factor from the smaller triangle CBD to ABE is ABCB=62=3. The corresponding sides give AE=3(3)=9 and BE=3(4)=12.
3. Add all three sides of triangle ABE:
   P=AB+BE+EA=6+12+9=27

Exam detail: The correspondence is A to C, B to B, and E to D.

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Question 24: Dilate a line through the center

1. Under a dilation, a point and its image lie on the same line through the center of dilation.
2. Because the center is already on the given line, all its image points remain on that line. The line maps onto itself, so its slope and y-intercept are unchanged.

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Question 25: Describe and verify two rigid motions

1. Read the original vertices as D(−6,9), A(−1,8), and N(−3,4). The target correspondence is D to M, A to I, and N to K.
2. First reflect across the y-axis, using (x,y)→(−x,y). Then translate 11 units downward, using (x,y)→(x,y−11).
3. Check every vertex:

   Verification of the two motions

   Original	Reflect in y-axis	Move 11 down
   D(−6,9)	(6,9)	M(6,−2)
   A(−1,8)	(1,8)	I(1,−3)
   N(−3,4)	(3,4)	K(3,−7)

4. Both transformations are rigid motions, and the complete rule (x,y)→(−x,y−11) maps all three original vertices to their required images.

Exam detail: Other valid sequences may also work. State the reflecting line and the translation direction and distance explicitly.

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Question 26: Find the obtuse angle beside the perpendicular

1. Opposite sides of a parallelogram are parallel, so LE∥DA. The diagonal DE is a transversal, giving m∠CDS=m∠LED=21°.
2. Since SC is perpendicular to DE, triangle DCS has a right angle at C:
   m∠CSD=180°−90°−21°=69°
3. D, S and A are collinear, so the angles CSD and CSA form a linear pair:
   m∠CSA=180°−69°=111°

Exam detail: The requested angle is CSA, which is obtuse. The 69° angle is the adjacent angle CSD.

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Question 27: Find weight from cylindrical volume

1. The diameter is 1.5 feet, so the radius is r=1.52=0.75 feet.
2. Find the cylinder’s volume:
   V=πr²h=π(0.75)²(8)=4.5π
   cubic feet.
3. Multiply by the stated 25 pounds per cubic foot:
   W=25(4.5π)=112.5π≈353.4292
   pounds. Round only the final weight to the nearest pound.

Exam detail: Halve the diameter before using the cylinder-volume formula.

Explore the radius and height relationship with the cylinder calculator.

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Question 28: Use the converse of proportional division

1. For BF to be parallel to CD, it must divide sides AC and AD proportionally. Use corresponding parts:
   ABBC=AFFD
2. Substitute and cross-multiply:
   4030=64FD
   40×FD=30(64)
   FD=48
3. With this length, the two ratios are equal, so the converse of the triangle proportionality theorem establishes BF∥CD.

Exam detail: The requested length is FD, not the full side AD, which would be 112.

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Question 29: Use two sides and their included angle for area

1. The 33° angle lies between the sides of lengths 8 and 15. Therefore the triangle area formula is A=12absin(C).
2. Substitute the two sides and included angle:
   A=12(8)(15)sin(33°)=60sin(33°)
   A≈32.6783
3. Round to the nearest tenth.

Exam detail: Use degree mode and the angle included between the two known sides.

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Question 30: Construct the angle bisector at B

1. Place the compass point at B. Draw an arc crossing both rays BA and BC. Label those intersection points P and Q; the single compass setting ensures BP=BQ.
2. Choose a compass opening large enough for arcs centered at P and Q to meet inside angle ABC. Draw both arcs with the same opening, and label their interior intersection R.
3. Use a straightedge to draw the ray from B through R. Keep the original arc and both intersecting arcs visible.
4. Why it works: BP=BQ, PR=QR, and BR is common. Thus △BPR≅△BQR by SSS, so the angles at B are equal and ray BR bisects angle ABC.

Exam detail: A measured or freehand line without the appropriate construction arcs is not an acceptable compass-and-straightedge construction. The bisector must start at B.

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Question 31: Use congruent diagonals to prove a rectangle

1. The diagonals of parallelogram GRAM are GA and RM. Diagonals of a parallelogram bisect each other, so P is the midpoint of RM.
2. Therefore PM=RP=12 and RM=12+12=24. Since GA=24, the two diagonals are congruent.
3. A parallelogram whose diagonals are congruent is a rectangle. Therefore GRAM is a rectangle.

Exam detail: Explain why RM is twice RP; RP is only half a diagonal.

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Question 32: Add the heights above and below the observer

1. H is level with M, so the horizontal distance MH is 80 feet. Split the required building height into TB=TH+HB.
2. Use the angle of elevation for the portion above eye level:
   tan(36°)=TH80
   TH=80tan(36°)≈58.1234
   feet.
3. Use the angle of depression for the portion below eye level:
   tan(18°)=HB80
   HB=80tan(18°)≈25.9936
   feet.
4. Add using full calculator precision, then round to the nearest foot:
   TB=80tan(36°)+80tan(18°)≈84.1170

Exam detail: Finding only the portion above Maria’s eye level does not give the full building height. Add both vertical portions.

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Question 33: Determine the maximum number of complete pyramids

1. Each square base has area 12²=144 square centimeters. The volume of one pyramid is:
   V=13(144)(8)=384
   cubic centimeters.
2. Multiply volume by density to find the mass of one pyramid:
   m=384(1.25)=480
   grams.
3. Convert the available clay into the same mass unit:
   15×1000=15000
   grams. Divide by the mass per pyramid:
   15000480=31.25
4. Only whole pyramids count, so take the whole-number part: 31. As a check, 31 pyramids use 14,880 grams, leaving 120 grams. Another pyramid needs 480 grams, which exceeds the amount left.

Exam detail: A maximum number of complete objects must be rounded down, even when the division gives a fraction.

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Question 34: Prove the ratio using similar right triangles

1. Because AB≅AC, triangle ABC is isosceles and its base angles at B and C are congruent. With D on AB, H on AC, and E and F on BC, this gives ∠DBE≅∠HCF.
2. The perpendicular givens make m∠BED=m∠CFH=90°. These right angles are congruent.
3. Two angle pairs are congruent, so △BED∼△CFH by AA. The correspondence is B to C, E to F, and D to H.
4. Corresponding sides of similar triangles are proportional:
   DEHF=BECF
   Cross-multiply and divide by the positive product of BE and CF:
   DE×CF=HF×BE
   DEBE=HFCF
   This is the required relationship.

Exam detail: Do not stop after proving similarity. Include the proportional-side step and the requested concluding equation.

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Question 35: Complete the parallelogram and prove an isosceles triangle

1. Find D. The move from B to C is (−1−(−5),8−2)=(4,6). Opposite sides of a parallelogram have the same directed displacement, so apply this move to A:
   D=(−3+4,−1+6)=(1,5)
2. Prove ABCD is a parallelogram by comparing both pairs of opposite slopes:
   m_AB=2−(−1)−5−(−3)=−32
   m_CD=5−81−(−1)=−32
   m_BC=8−2−1−(−5)=32
   m_AD=5−(−1)1−(−3)=32
   Thus AB∥CD and BC∥AD. Both pairs of opposite sides are parallel, so ABCD is a parallelogram.
3. Find the midpoint E of BC by averaging coordinates:
   E=(−5+(−1)2,2+82)=(−3,5)
4. Prove triangle ABE is isosceles. Calculate two of its side lengths:
   AB=√(−5−(−3))²+(2−(−1))²=√13
   BE=√(−3−(−5))²+(5−2)²=√13
   Since AB=BE, triangle ABE has two congruent sides and is isosceles.

Exam detail: There are four tasks across pages 22–23. State both coordinate answers and give both proofs with explicit concluding statements. The graph is optional.

Check another coordinate example in the 2D Points mode of the distance calculator.

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