Geometry Problems of the Day (Geometry Regents, August 2025 Part I)

This exam was adminstered in August 2025.

More Regents problems.

August 2025 Geometry Regents

Part I

Each correct answer will receive 2 credits. No partial credit.


9. On the set of axes below, △D'E'F' is the image of △DEF.

A transformation that maps △DEF onto △D'E'F' is a

(1) reflection over the line y = x
(2) reflection over the line y = - x
(3) point reflection through the origin
(4) translation 4 units left and 4 units down

Answer: (3) point reflection through the origin

A point reflection through the origin is the same as a 180 degree rotation. Choice (3) is the correct answer.

In Choice (1), a reflection over y = x would have the image in Quadrant I not III. Eliminate Choice (1).

In Choice (2), a reflection over y = -x would have E' closer to the x-axis and F' closer to the y-axis. The orientation is not correct for a reflection. Eliminate Choice (2).

In Choice (4), a translation doesn't change the orientation of the original shape, but D'E'F' has a different orientation. Eliminate Choice (4).

10. In circle O below, secants PCA and PDB are drawn from external point P.

If PA = 17, PD = 10, and BD = 12, what is the length of PC, to the nearest tenth?

(1) 7.1
(2) 7.7
(3) 12.9
(4) 14.2

Answer: (3) 12.9

The product of the length of the secant line times the length of the line that is outside the circle (PC, PD) will be equal for all of the secant lines.

That is to say: (PA)(PC) = (PB)(PD)

(17)(PC) = (10 + 12)(10)

17 (PC) = 220

PC = 220 / 17 = 12.94..., or 12.9

Choice (3) is the correct answer.

11. In the diagram below, CD || AB, and CB bisects ∠ABD.

Which statement must be true?

(1) CD ≅ AB
(2) AB ≅ BD
(3) △CDB is a right triangle
(4) △SNA is an isosceles triangle

Answer: (4) △SNA is an isosceles triangle

BC and BD are transversals crossing parallel lines. The means that ∠ABC ≅ ∠BCD. CB bisects ∠ABD, so ∠ABC ≅ ∠CBD. Since two angles in the triangle are congruent, then the triangle is an isosceles triangle.

Choice (4) is the correct answer.

12. Line h is represented by the equation y = 2/3 x - 4. Which equation represents the line that is perpendicular to line h and passes through the point (6,1)?

(1) y - 1 = 2/3(x - 6)
(2) y + 1 = 2/3(x + 6)
(3) y - 1 = -3/2(x - 6)
(4) y + 1 = -3/2(x + 6)

Answer: (3) y - 1 = -3/2(x - 6)

A line perpendicular to a line with a slope of 2/3 would have a slope of -3/2. Eliminate Choice (1) and (2).

Point slope form is y - h = m(x - k), with two minus signs. Those signs will flip to plus if the coordinate is negative. The slope of the line is m, and (h,k) is a point on that line.

Choice (3) is the correct answer.

Choice (4) goes through the point (-6,-1).

13. A wooden toy block can be modeled by a pyramid with a square base, as shown below. The height of the block is 17.4 cm and the square base has a side length of 8.2 cm.

The block is made of solid oak, which has a density of 0.77 g/cm³. What is the mass of the block, to the nearest gram?

(1) 300
(2) 506
(3) 637
(4) 901

Answer: (1) 300

The Volume of a pyramid is 1/3 the Area of the Base times the height. The Area of the square base is s². Density is mass divided by the Volume, so mass is equal to Density times Volume.

V = 1/3 (8.2)(8.2)(17.4) = 389.992

389.992 * 0.77 = 300.29..., which rounds to 300. Choice (1) is the correct answer.

If you forgot the (1/3), you would've gotten 901, which is Choice (4).

If you squared 17.4 instead of 8.2, you would've gotten 637, which is Choice (3).

If you found the correct Volume but divided by the Density, then you would've gotten 506, which is Choice (2).

14. In △ABC below, midsegment DE is drawn.

If DE = x + 3 and AC = 3x - 5, what is the length of DE?

(1) 28
(2) 14
(3) 7
(4) 4

Answer: (2) 14

Since DE is a midsegment, it is half the size of AC. That is, AC is twice the size of DE.

AC = 2 DE
3x - 5 = 2(x + 3)
3x - 5 = 2x + 6
x = 11

If x = 11, the DE = x + 3 = 11 + 3 = 14, which is Choice (2).

15. Triangle DUG is an isosceles right triangle with the right angle at G. If DU = 10√(2), what is the length of GU?

(1) 5
(2) 5√(2)
(3) 10
(4) 10√(2)

Answer: (3) 10

Side DU is the hypotenuse, not a leg. The two legs are congruent. The acute angles are each 45 degrees.

If you forgot the rules for Special Right Triangles, such as 45-45-90 (right isosceles), you can figure it out using Pythagorean Theorem.

x² + x² = (10√(2))²
2x² = 200
x² = 100
x = 10

Choice (3) is the correct answer.

16. In △RST below, RS = 9 cm, RT = 8 cm, and m∠TRS = 55°.

What is the area of △RST, to the nearest square centimeter?

(1) 59
(2) 36
(3) 29
(4) 21

Answer: (3) 29

Use the Law of Sines, A = 1/2 ab sin(C) to find the area, where a and b are sides of the triangle (not base and altitude) and C is the included angle between them.

A = 1/2 (8) (9) sin(55) = 29.4..., which rounds to 29.

Choice (3) is the correct answer.

More to come. Comments and questions welcome.

MY NEWEST BOOK IS OUT Burke's Lore Briefs: Yesterday's Villains, the following to Tomorrow's Heroes is now available on Amazon and Kindle Unlimited. If Heroes who don't die live long enough to become the villain, what happens to Villains who live long enough? When do schemes of global conquest become dreams of a quiet place away from all those annoying people you once wanted to subjugate? And does anyone really want to rule over the world's ashes if it means we can't have nice things?
My older books include three more books in my Burke's Lore Briefs series, and the anthologies A Bucket Full of Moonlight and In A Flash 2020. Vampires, werewolves, angels, demons, used-car salesmen, fairies, superheroes, space and time travel, and little gray aliens talking to rock creatures and living plants. Plus pirates, spies, horror, and kindergarten noir! If you enjoy my books, please consider leaving a rating or review on Amazon or on Good Reads. Thank you!

Geometry Problems of the Day (Geometry Regents, August 2025 Part I)

This exam was adminstered in August 2025.

More Regents problems.

August 2025 Geometry Regents

Part I

Each correct answer will receive 2 credits. No partial credit.


1. An equilateral triangle is continuously rotated around one of its altitudes. The three-dimensional object formed is a

(1) cone
(2) sphere
(3) cylinder
(4) pyramid

Answer: (1) cone

Imagine an equilateral triangle taped to a lollipop stick that running from one corner and through the midpoint of the opposite side. If you were to spin the stick, the stuff that the triangle will pass through will form a cone. That is Choice (1).

It is impossible to get any of the others from a triangle.

2. On the set of axes below, quadrilateral BDGF is rotated 90 degrees clockwise about the origin and then reflected over the y-axis. The image of quadrilateral BDGF is quadrilateral MQSP.

Side BD will always map onto

(1) MP
(2) PS
(3) MQ
(4) SQ

Answer: (3) MQ

Follow the movement. When BDGF is rotated 90 degree clockwise, the image appears in Quadrant IV with BD on top, closest to the x-axis. When this image is reflected across the y-axix, B'D' maps onto MQ, which is Choice (3).

3. In right triangle JOE, hypotenuse JO = 31.8 and m∠J = 38°. To the nearest tenth, the length of EJ is

(1) 19.6
(2) 25.1
(3) 40.4
(4) 51.7

Answer: (2) 25.1

Angle E is the right angle. EJ is adjacent to angle J, and you're given the hypotenuse. That means that you need to use cosine.

cos 38 = x / 31.8
x = 31.8 cos 38 = 25.0587... = 25.1

Choice (2) is the correct answer.

4. The hemisphere below has a radius of 8 cm.

To the nearest cubic centimeter, the volume of the hemisphere is

(1) 201
(2) 268
(3) 1072
(4) 2145

Answer: (3) 1072

Use the formula for the Volume of a Sphere and then take half of it. The formula is V = (4/3) π r³.

So V = (1/2) (4/3) π (8)³ = 1072.33...

Choice (3) is the correct choice.

If you'd forgotten the 1/2, you would've gotten choice (4).

If you'd forgotten the 1/2 AND halved the radius to 4, you would've gotten Choice (2).

If you used π r², you would've gotten Choice (1).
Yes, I needed to figure that one out. I wanted to know!

5. In parallelogram ABCD, diagonals AC and BD intersect at E. Which information is sufficient to prove ABCD is a rhombus?

(1) AE ≅ EC
(2) AC ≅ BD
(3) AB ⊥ BC
(4) AC ⊥ BD

Answer: (4) AC ⊥ BD

In a parallelogram, if the two diagonals are perpendicular, then the parallelogram is a rhombus.

6. Trapezoid JOSH, shown below, has non-parallel sides JH and OS, m∠J = 65°, m∠O = 30°, m∠OSA = 80°, and m∠SHU = 60°.

(1) 55°
(2) 60°
(3) 65°
(4) 70°

Answer: (4) 70°

SAO is a triangle, and the sum of its angles is 180 degrees. Because m∠O = 30° and m∠OSA = 80°, then m∠OAS = 70°. Angle OAS and angle HSA are alternate interior angles. Since JO || SH, then ∠OAS ≅ ∠OSA.

Therefore, m∠OSA = 70°, which is Choice (4).

7. In △ABC below, points D and E are on such that AC. DE || AC.

If AD = 8, DB = 4, and DE = 6, what is the length of AC

(1) 24
(2) 18
(3) 12
(4) 10

Answer: (2) 18

The two triangles are have sides that are proportional because they are similar triangles. If AD = 8 and DB = 4, then AB = 12, and the scale factor is 3.

Therefore, if DE = 6 then AC = 6 * 3 = 18.

Choice (2) is the correct answer.

8. On the set of axes below, circle C has a center with coordinates (2,–1).

Which equation represents circle C?

(1) (x - 2)² + (y + 1)² = 25
(2) (x - 2)² + (y + 1)² = 16
(3) (x + 2)² + (y - 1)² = 25
(4) (x + 2)² + (y - 1)² = 16

Answer: (1) (x - 2)² + (y + 1)² = 25

The formula for the equation of a circle is (x - h)² + (y - k)² = r², where (h,k) is the coordinates of the center and r is the length of the radius.

The center point C is at (1,-2), so eliminate Choices (3) and (4).

If you count across from C, you'll see that the radius is 5. (5)² = 25, so the correct answer is Choice (1).

More to come. Comments and questions welcome.

MY NEWEST BOOK IS OUT Burke's Lore Briefs: Yesterday's Villains, the following to Tomorrow's Heroes is now available on Amazon and Kindle Unlimited. If Heroes who don't die live long enough to become the villain, what happens to Villains who live long enough? When do schemes of global conquest become dreams of a quiet place away from all those annoying people you once wanted to subjugate? And does anyone really want to rule over the world's ashes if it means we can't have nice things?
My older books include three more books in my Burke's Lore Briefs series, and the anthologies A Bucket Full of Moonlight and In A Flash 2020. Vampires, werewolves, angels, demons, used-car salesmen, fairies, superheroes, space and time travel, and little gray aliens talking to rock creatures and living plants. Plus pirates, spies, horror, and kindergarten noir! If you enjoy my books, please consider leaving a rating or review on Amazon or on Good Reads. Thank you!

Geometry Problems of the Day (Geometry Regents, June 2025 Part I)

This exam was adminstered in June 2025.

More Regents problems.

June 2025 Geometry Regents

Part I

Each correct answer will receive 2 credits. No partial credit.


17. Point O divides COA such that CO:OA = 1:4. If C has coordinates (–2, –9) and A has coordinates (3,6), the coordinates of O are

(1) (2,3)
(2) (1,0)
(3) (0,-3)
(4) (-1,-6)

Answer: (4) (-1,-6)

Dividing 1:4 means that there are 5 parts, with CO being 1/5 of the length COA and OA being 4/5 of the length. Find 1/5 of the change in x values and 1/5 of the change of y values and add those values to (-2,-9).

1/5(3 - -2) = 1/5(5) = 1. 1/5(6 - -9) = 3.

(-2 + 1, -9 + 3) = (-1, -6), which is choice (4).

Another formula you can use is this:

4/5(-2,-9) + 1/5(3,6)
(-8/5, -36/5) + (3/5, 6/5)
(-5/5, -30/5)
(-1, -6)

Remember to put the 4/5 in front of the first coordinate and the 1/5 in front of the second.

This always works.

18. 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?

(1) $2.06
(2) $2.42
(3) $3.22
(4) $16.46

Answer: (1) $2.06

Find the volume of the sphere and then multiply by 0.80.

The volume of a sphere is V = 4/3 π r³

V = 4/3 (3.141592...)(.85)³ = 2.57244...

Multiply that by $0.80 = 2.0579..., which rounds to $2.06.

11. In right triangle ABD below, altitude BC is drawn to hypotenuse AD, AC = 2.5, and CD = 4.3.

What is the length of BA, to the nearest tenth?

(1) 3.3
(2) 3.4
(3) 4.1
(4) 5.4

Answer: (3) 4.1

The three right triangles are all similar and their sides are proportional. That means that AC / AB = AB / AD (short leg / hypotenuse).

2.5 / x = x / (2.5 + 4.3)
x² = 17
x = 4.123...

This rounds to 4.1, which is Choice (3).

20. Trapezoid ZOYD has parallel sides ZO and DY. If m∠Z = 141° and m∠Y = 73°, what is m/D?

(1) 39°
(2) 73°
(3) 107°
(4) 141°

Answer: (1) 39°

If ZO and DY are parallel sides, then OY and DZ are the other two sides. The sum of the measures of angles O and Y must be 180 degrees, and the sum of the measures of angles D and Z must be 180 degrees.

180 - 141 = 39.

Choice (1) is the correct answer.

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

(1) △PQR ≅ △ABC
(2) ∠A ≅ ∠Q
(3) BQ = √(29)
(4) RQ = CB

Answer: (2) ∠A ≅ ∠Q

In a translation, size and shape are preserved. The corresponding angles will be congruent. However, in ABC and PQR, angles A and Q are NOT corresponding angles. They do not have to be congruent. They could be if the triangle is isosceles or equilateral.

In Choice (1), the triangles are always congruent. Eliminate Choice (1).

In Choice (3), point B moves 5 left and 2 up to point Q, so line segment BQ would be √(5^2 + 2^2), which is √(29). Eliminate Choice (3).

In Choice (4), RQ corresponds to CB, so length is preserved. Eliminate Choice (4).

22. In the diagram below, congruent triangles PEN and PAL are drawn.

Which rigid motion maps nPEN onto nPAL?

(1) a point reflection of △PEN through P
(2) a reflection of △PEN over the angle bisector of ∠EPA
(3) a rotation of △PEN about point P, mapping PE onto PA
(4) a translation of △PEN along EA, mapping point E onto A

Answer: (2) a reflection of △PEN over the angle bisector of ∠EPA

The triangles do not have the same orientation. There is a reflection involved.

A point reflection would be 180 degrees about point P. Eliminate Choice (1).

A reflection over the a line bisect the image will carry PEN onto PAL. Choice (2) is the correct answer.

A rotation mapping PE to PA would not move N to point L. The result would be a kite. Eliminate Choice (3).

A translation would not change the orientation. Moving E to A would create a bowtie-like image. Eliminate Choice (4).

23. 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?

(1) 9
(2) 2
(3) 3
(4) 6

Answer: (4) 6

Use the formula for the volume of a cone, substitute what you know, and solve for the unknown.

The formula for the volume of a cone is V = 1/3 π r² h.

V = 1/3 π r² h
75.4 = 1/3 π r² (8)
r² = (3) 75.4 / (8 π)
r² = 9.00...
r = 3

So the DIAMETER is 2 * 3 = 6, which is Choice (4).

24. 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

(1) is the original line
(2) passes through the point (4,0)
(3) passes through the point (0,–10)
(4) is perpendicular to the original line

Answer: (4) are congruent

A dilation of a straight line will be a parallel line or the same line if the center of dilation is on the line itself. The image cannot be have a different slope, so it cannot be perpendicular. Eliminate Choice (4).

Substitute (2,0) into the original question: 5(2) - 2(0) = 10. This is a true statement, so (2,0) is a point on the original line. Therefore, the image will be the same line. Choice (1) is the correct answer.

End of Part I. Comments and questions welcome.

MY NEWEST BOOK IS OUT Burke's Lore Briefs: Yesterday's Villains, the following to Tomorrow's Heroes is now available on Amazon and Kindle Unlimited. If Heroes who don't die live long enough to become the villain, what happens to Villains who live long enough? When do schemes of global conquest become dreams of a quiet place away from all those annoying people you once wanted to subjugate? And does anyone really want to rule over the world's ashes if it means we can't have nice things?
My older books include three more books in my Burke's Lore Briefs series, and the anthologies A Bucket Full of Moonlight and In A Flash 2020. Vampires, werewolves, angels, demons, used-car salesmen, fairies, superheroes, space and time travel, and little gray aliens talking to rock creatures and living plants. Plus pirates, spies, horror, and kindergarten noir! If you enjoy my books, please consider leaving a rating or review on Amazon or on Good Reads. Thank you!

Geometry Problems of the Day (Geometry Regents, June 2025 Part I)

This exam was adminstered in June 2025.

More Regents problems.

June 2025 Geometry Regents

Part I

Each correct answer will receive 2 credits. No partial credit.


9. 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?

(1) 103
(2) 51
(3) 34
(4) 11

Answer: (3) 54

A two-part problem because first you need to know how many cubic centimeters are in the candle before you can answer the question.

The volume of a square pyramid is 1/3 (side)² times height, which is (1/3) (6)² (10) = 120 cm³.

If it burns at a rate of 3.5 cm³/hr, then divide 120 / 3.5 = 34.2857..., or about 34.

Choice (3) is the correct answer.

If you forgot the (1/3), you would've gotten choice (1). Oops.

10. In the diagram below, tangent SR and secant STU are drawn to circle O from external point S.

If TU ≅ RU and mTR = 68°, what is m∠S?

(1) 22°
(2) 34°
(3) 39°
(4) 78°

Answer: (3) 39°

The measure of angle S is one half of the difference of arc RU and arc TR. You need to find the size of RU.

Since TU ≅ RU, and TU + RU + TR = 360, then 2(RU) + 68 = 360, and 2(RU) = 292, and RU = 146.

One half of the difference of RU and TR is 1/2(146 - 68) = 39, which is Choice (3).

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

(1) 22.9
(2) 27.3
(3) 35.2
(4) 45.7

Answer: (1) 22.9

Use the Law of Sines: A = 1/2 ab sin C = (1/2)(7)(12)sin(33) = 22.87...

12. 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 nABC. The transformation that maps △ABC onto △A'B'C' is a

(1) dilation
(2) translation
(3) vertical stretch
(4) horizontal stretch

Answer: (4) horizontal stretch

It is not a dilation because the y-coordinates don't change. On the other hand, the x-coordinates of the image are double the original.

Choice (4) is the correct answer.

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

(1) y - 2 = 4(x + 3)
(2) y + 3 = 4(x - 2)
(3) y - 2 = -1/4(x + 3)
(4) y + 3 = -1/4(x - 2)

Answer: (1) y - 2 = 4(x + 3)

Parallel lines have the same slope so the answer must have a slope of 4. Eliminate choices (3) and (4).

The answers are given in point-slope form. The x-coordinate of the given point goes with the x variable, and the y-coordinate goes with the y variable.

Choice (1) is the correct answer.

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

(1) circle
(2) triangle
(3) rhombus
(4) rectangle

Answer: (4) rectangle

If you slice a round cake down the center and separate the two pieces, you will see two rectangles, composed of the top, bottom and the sides of the cake.

Choice (4) is the correct answer.

15. In △KMP below, CE is drawn parallel to MP.

If KC = 8, CM = 3, and CE = 12, what is the length of MP?

(1) 24
(2) 16.5
(3) 15
(4) 4.5

Answer: (2) 16.5

The sides of the two triangles are proportional, but first you need to find KM = KC + CM = 8 + 3 = 11.

So 8/12 = 11/MP.

Then MP = 11 * 12 / 8 = 16.5

The correct answer is Choice (2).

16. A parallelogram must be a rectangle if its diagonals

(1) are perpendicular
(2) bisect each other
(3) bisect its angles
(4) are congruent

Answer: (4) are congruent

What do you know about parallelograms and rectangles?

The diagonals of parallelograms and rectangles are not perpendicular unless they are also rhombuses. Eliminate Choice (1).

The diagonals of both parallelograms and rectangles are bisect each other, so that is not enough information. Eliminate Choice (2).

The diagonals of parallelograms and rectangles only are bisect each other is they are also rhombuses. Eliminate Choice (3).

The diagonals of parallelograms are only congruent if the parallelogram is a rectangle. Choice (4) is the answer.

More to come. Comments and questions welcome.

MY NEWEST BOOK IS OUT Burke's Lore Briefs: Yesterday's Villains, the following to Tomorrow's Heroes is now available on Amazon and Kindle Unlimited. If Heroes who don't die live long enough to become the villain, what happens to Villains who live long enough? When do schemes of global conquest become dreams of a quiet place away from all those annoying people you once wanted to subjugate? And does anyone really want to rule over the world's ashes if it means we can't have nice things?
My older books include three more books in my Burke's Lore Briefs series, and the anthologies A Bucket Full of Moonlight and In A Flash 2020. Vampires, werewolves, angels, demons, used-car salesmen, fairies, superheroes, space and time travel, and little gray aliens talking to rock creatures and living plants. Plus pirates, spies, horror, and kindergarten noir! If you enjoy my books, please consider leaving a rating or review on Amazon or on Good Reads. Thank you!

Geometry Problems of the Day (Geometry Regents, June 2025 Part I)

This exam was adminstered in June 2025.

More Regents problems.

June 2025 Geometry Regents

Part I

Each correct answer will receive 2 credits. No partial credit.


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

(1) 6
(2) 18
(3) 54
(4) 162

Answer: (3) 54

The perimeter of a dilated triangle is the perimeter of the original times the scale factor, so 18 times 3 equals 54, which is Choice (3).

It is not the scale factor squared -- that would be the area, which has two dimensions being expanded.

2. 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?

(1) 188
(2) 213
(3) 555
(4) 590

Answer: (3) 555

If you've ever been there, then you might know that it is 555 feet tall, and they did NOT change this fact for this problem. If you didn't know that, you can calculate it using the tangent ratio, because you have the angle and the adjacent side and you are looking for the opposite side.

tan 70.19 = x / 200
x = 200 tan 70.19 = 555.217...,

which is about 555, which is Choice (3).

3. On the set of axes below, △EQA and △SUL are graphed.

Which sequence of transformations shows that △EQA ≅ △SUL?

(1) Rotate △EQA 90° counterclockwise about the origin and then translate 9 units right and 1 unit down.
(2) Rotate △EQA 90° counterclockwise about the origin and then reflect over the line x = 4.
(3) Reflect △EQA over the x-axis and then rotate 90° clockwise about the origin.
(4) Translate △EQA 10 units right and then reflect over the line x = -1.

Answer: (1) Rotate △EQA 90° counterclockwise about the origin and then translate 9 units right and 1 unit down.

Make sure you are going the correct direction: you want to go from the top left to the bottom right. There are multiple methods of getting there, so check the choices one by one.

In Choice (1), EQA goes to Quadrant III with E'(-5,-2), Q(-1,-2), A'(-1,-5), which is facing the same direction as SUL. A transformation of T_9,-1 brings AEQ to SUL. This is the correct answer.

In Choice (2), EQA goes to Quadrant III with E'(-5,-2), Q(-1,-2), A'(-1,-5), which is facing the same direction as SUL. A reflection would change the orientation and would not map onto SUL. Eliminate Choice (2).

In Choice (3), EQA goes to Quadrant I but changes its orientation. When E'Q'A' is rotated 90 degrees, it will not map ontol SUL because the orientations are different. Eliminate Choice (3).

In Choice (4), EQA goes to Quadrant IV but will be oriented with point E pointing up. A reflection will not rotate the triangle to look like SUL. Eliminate Choice (4).

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

(1) 10
(2) 7
(3) 6
(4) 4

Answer: (2) 7

The Triangle Inequality Theorem states that the sum of the lengths of any two sides of a triangle must be greater than the length of the third side.

If two sides are 2 and 8, then then third side must be less than 10. If two sides are 2 and 8, then the third side must be greater than 6, because 6 + 2 = 8.

This eliminates Choice (1), (3), and (4).

Choice (2) 7 is between 6 and 10, so it is a reasonable length. This is the correct answer.

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

(1) 36°
(2) 90°
(3) 144°
(4) 160°

Answer: (2) 90°

A full rotation is 360°. During that rotation, the octagon will carry onto itself 8 times at increments of 360°/8 = 45°.

Of the four choices, only 90 is a multiple of 45, so the correct answer is Choice (2).

6. The equation of a circle is x² + y² - 6x + 2y = 14. What are the coordinates of the center and the length of the radius of this circle?

(1) (-3,1) and r = 5
(2) (3,-1) and r = 5
(3) (-3,1) and r = √(24)
(4) (3,-1) and r = √(24)

Answer: (4) (3,-1) and r = √(24)

A reminder that old exams not only show you the type of problems you may see but sometimes THE SAME NUMBER QUESTION will have THE SAME TYPE OF QUESTION. This was the same question that was asked in January with slightly different numbers, and it was Question #6 then, too.

To find the center of the circle, you need to complete the squares, twice, and find the square root of the polynomials.

x² + y² - 6x + 2y = 14
x² - 6x + y² + 2y = 14
x² - 6x + (3)² + y² + 2y + (1)² = 14 + 9 + 1
(x - 3)² + (y + 1)² = 24
(x - 3)² + (y + 1)² = (√(24))²

The correct answer is Choice (4) (3,-1) and r = √(24).

7. In △HSF below, m∠S = 90°, HF = 30, and FS = 23.

What is m∠F, to the nearest degree?

(1) 53°
(2) 50°
(3) 40°
(4) 37°

Answer: (3) 40°

You are looking for an angle in a right triangle and you are given the adjacent side and the hypotenuse, so you need to use the cosine ratio.

cos F = 23/30

F = cos^-1(23/30) = 39.94°

Choice (3) is the correct answer.

8. In △CAB below, midsegments DE, EF, and FD are drawn.

If CA = 14, CB = 20, and FB = 9, what is the perimeter of quadrilateral DEFA?

(1) 26
(2) 32
(3) 44
(4) 52

Answer: (2) 32

Midsegments are half the length of the side of the triangle that they are parallel to. Therefore, AD = 7 and EF = 7. Since FB = 9, then AF = 9 and DE = 9.

Thus the perimeter of DEFA is 9 + 7 + 9 + 7 = 32, which is Choice (2).

Not that they asked by DEFA is also a parallelogram.

More to come. Comments and questions welcome.

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