The following are some of the multiple questions from the recent August 2018 New York State Common Core Geometry Regents exam.

The answers to Part II can be found here

The answers to Parts III and IV can be found here

### August 2018 Geometry, Part I

Each correct answer is worth up to 2 credits. No partial credit. Work need not be shown.

**1.** *In the diagram below, AEFB || CGD, and GE and GF drawn.
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If m∠EFG = 32° and m∠AEG 137°, what is m∠EGF?
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**Answer: (4) 105°**

If ∠AEG = 137, then ∠FEG = 180 – 137 = 43.

The sum of the angles of triangle EFG is 180.

So 180 – (43 + 32) = 105. Angle EGF is 105°.

Alternatively, Angle AEG is an exterior angle to triangle EFG. The Remote Angle Theorem says that it must be the sum of the two remote angles EFG and and EGF.

So EGF = 137 – 32 = 105 degrees.

**2.** *If triangle ABC is mapped onto triangle DEF after a line reflection and triangle DEF is mapped onto triangle XYZ after a translation, the relationship between triangle ABC and triangle XYZ is that they are always
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**Answer: (1) congruent and similar**

Reflections and translations are rigid motions which preserve the shape of the object. Also, choice (2) congruent but not similar is not possible. If two triangles are congruent, they are automatically similar.

**3.** *An isosceles right triangle whose legs measure 6 is continuously rotated about one of its legs to form a three-dimensional object. The three-dimensional object is a
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**Answer: (4) cone with a diameter of 12**

When a right triangle is rotated about a leg, the resulting 3-D shape will be a cone. Because the base of the triangle is 6, the radius will be 6, so the diameter is twice that, or 12.

**4.** *In regular hexagon ABCDEF shown below, AD, BE, and CF all intersect at G.
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When triangle ABG is reflected over BG and then rotated 180° about point G, triangle ABG is mapped onto
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**Answer: (1) Triangle FEG**

When it is reflected, it is mapped onto triangle BGC, the top center of the hexagon. When it is rotated 180 degrees, it maps onto FEG, the bottom center of the hexagon.

**5.** *A right cylinder is cut perpendicular to its base. The shape of the cross section is a
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**Answer: (3) rectangle**

A horizontal (parallel to base) cut would give a circle. A vertical (perpendicular to base) cut gives a rectangle. If you look at a can on a shelf, it appears to be a rectangle.

**6.** *Yolanda is making a springboard to use for gymnastics. She has 8-inch-tall springs and wants to form a 16.5° angle with the base, as modeled in the diagram below.
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To the nearest tenth of an inch, what will be the length of the springboard, x?
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**Answer: (4) 28.2 **

Opposite and hypotenuse mean you need to use sine.

sin 16.5 = 8 / x

x = 8 / sin 16.5 = 28.167…

**7.** *In the diagram below of right triangle ABC, altitude BD is drawn to hypotenuse AC.
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If BD = 4, AD = x – 6, and CD = x what is the length of CD?
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**Answer: (3) 8 **

According to the Right Triangle Altitude Theorem, (BD)^{2} = (AD)(CD)

So 4^{2} = (x – 6)(x)

16 = x^{2} - 6x. At this point, you could substitute the choices, or solve.

x^{2} - 6x – 16 = 0

(x – 8)(x + 2) = 0

x = 8 or x = -2. Discard the negative length.

**8.** *Rhombus STAR has vertices S(–1,2), T(2,3), A(3,0), and R(0,–1). What is the perimeter of rhombus STAR?
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**Answer: (4) 4 * SQRT(10) **

Each side of the rhombus has the same length, so calculate the length of one side using the distance formula, and multiply the result by 4.

SQRT( (2 - -1)^{2} + (3-2)^{2} ) = SQRT( (3)^{2} + (1)^{2} )

= SQRT(9 + 1) = SQRT(10) per side.

Perimeter is 4 * SQRT(10).

**9.** *In the diagram below of triangle HAR and triangle NTY, angles H and N are right angles, and HAR ~ NTY.
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If AR = 13 and HR = 12, what is the measure of angle Y, to the nearest degree?
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**Answer: (1) 23° **

Angle Y corresponds to angle R. AR is the hypotenuse, and HR is adjacent to R. This means use cosine to find the size of angle R, which will also be the size of angle Y.

cos R = 12/13

r = cos^{-1}(12/13) = 22.61 = 23 degrees.
Note that choice (4) is the angle if you either used sine, or if you solved for angle A. You would get choices (2) and (3) if you used tangent, and if you used tangent and the wrong angle.

**10.** *In the diagram below, AKS, NKC, AN, and SC are drawn such that AN = SC.
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Which additional statement is sufficient to prove triangle KAN = triangle KSC by AAS?
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**Answer: (4) AN || SC**

If AN is parallel to SC, then you would get alternate interior angles, along with the vertical angles. That is enough to prove congruency by either ASA or AAS. Choice (1) would prove congruency by SSS, but wouldn’t give any more information about the angles. Choice (2) would yield SSA, which is not sufficient to prove congruency. Choice (3) would establish that the vertical angles are also right angles, but we already know that those two angles are congruent.

**11.** *Which equation represents a line that is perpendicular to the line represented by y = (2/3)x + 1 ?
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**Answer: (1) 3x + 2y = 12 **

The negative reciprocal of 2/3 is -3/2, so choices (3) and (4) are both incorrect.

y = -(3/2)x + b

3/2x + y = b, multiply by 2 to get rid of the fraction

3x + 2y = 2b, which gives us the first equation, if 2b is replaced by 12.

**12.** *In the diagram of ABC below, points D and E are on sides AB and CB respectively, such that DE || AC.
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If EB is 3 more than DB, AB = 14, and CD = 21, what is the length of AD?
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**Answer: (2) 8 **

Be careful because you need to find the length of DB first, but they are asking for the length of AD, so you have to subtract DB from AB.

The sides are proportional, so you can set up an equation:

x / 14 = (x + 3) / 21. You can substitute the choices from this point if you want.

21x = 14x + 42

7x = 42

x = 6

AD = 14 – 6 = 8.

Notice that the four choices represent the lengths of DB, AD, BE, and CE. You might have reasoned this one out from the diagram and then checked your work to see if you were correct.

**13.** *Quadrilateral MATH has both pairs of opposite sides congruent and parallel. Which statement about quadrilateral MATH is always true?
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**Answer: (4) ∠MAT = ∠MHT**

Check Part IV of this exam for a coordinate geometry problem involving a quadrilateral with vertices M, A, T, and H.

Opposite sides congruent and parallel mean that this is a parallelogram. That means that the opposite angles are also congruent.

Choice (1) says that the diagonals must be congruent, which is only true in rectangles. (You could use this in Part IV)

Choice (2) says that the consecutive angles are right angles, which is only true in rectangles. (Again, Part IV)

Choice (3) would occur in rectangles as well because the two diagonals would create four isosceles triangles.

**14.** *In the figure shown below, quadrilateral TAEO is circumscribed around circle D. The midpoint of TA is R, and HO = PE .
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If AP = 10 and EO = 12, what is the perimeter of quadrilateral TAEO?
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**Answer: (2) 64**

AP = 10 so AR =10. R is the midpoint of AT, so RT = 10, which means TH = 10.

HO = PE means that OZ = ZE, and since OE = 12, then HO = PE = OZ = PE = 6.

4 * 10 + 4 * 6 = 64

**15.** *The coordinates of the endpoints of directed line segment ABC are A(-8,7) and C(7,-13). If AB:BC = 3:2, the coordinates of B are
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**Answer: (1) (1, -5) **

B is 3/5 of the distance from A to C. The x-coordinates of A and C are 7 – (-8) = 15 units apart.

3/5(15) = 9, and -8 + 9 = 1, so the x-coordinate of B is 1, which is choice (1).

To check, -13 – 7 = -20, and (3/5)(-20) = -12, and 7 – 12 = -5, which is the y-coordinate of B.

**16.** *In triangle ABC, points D and E are on sides AB and BC, respectively,
such that DE || AC, and AD:DB 3:5.
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If DB = 6.3 and AC = 9.4, what is the length of DE, to the nearest tenth?
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**Answer: (3) 5.9**

If AD:DB = 3:5, then AB:DB = 8:5, which is the ratio of the larger triangle to the smaller triangle.

So 8/5 = 9.4/x

8x = (5)(9.4)

x = (5)(9.4)/8 = 5.875.

**17.** *In the diagram below, rectangle ABCD has vertices whose coordinates
are A(7,1), B(9,3), C(3,9), and D(1,7).
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Which transformation will not carry the rectangle onto itself?
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**Answer: (3) a rotation of 180° about the point (6,6) **

The rectangle will carry onto itself in a rotation about the rectangle's center. (5, 5) is the center of the rectangle. (6, 6) is not. If you rotate about (6, 6), the image will be adjacent to the original rectangle.

**18.** *A circle with a diameter of 10 cm and a central angle of 30° is drawn below.
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What is the area, to the nearest tenth of a square centimeter, of the sector formed by the 30° angle?
**Answer: (2) 6.5**

The radius is 5 cm. The central angle is 30 degrees, which is 1/12 of the total circle. (30/360 = 1/12)

V = (1/12)(pi)(5)^{2} = 6.54..

**19.** *A child’s tent can be modeled as a pyramid with a square base whose
sides measure 60 inches and whose height measures 84 inches. What is the volume of the tent, to the nearest cubic foot?
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**Answer: (2) 58**

60 inches = 5 feet, 84 inches = 7 feet

Volume = (1/3) (Area of the base) (height)

V = (1/3) (5) (5) (7) = 58.333...

**20.** *In the accompanying diagram of right triangle ABC, altitude BD is
drawn to hypotenuse AC.
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Which statement must always be true?
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**Answer: (2) AD/AB = AB/AC**

Short leg is to hypotenuse as short leg is to hypotenuse.

The other choices do not use corresponding sides in the correct order.

**21.** *An equation of circle O is x*^{2} + y^{2} + 4x - 8y = -16. The statement that best describes circle O is the

**Answer: (4) center is (-2,4) and is tangent to the y-axis **

Rewrite the equation in standard form by grouping the variables and completing the square.

x^{2} + y^{2} + 4x - 8y = -16

x^{2} + 4x + y^{2} - 8y = -16

x^{2} + 4x **+ 4** + y^{2} - 8y **+ 16** = -16 **+ 4 + 16**

(x + 2)^{2} + (y - 4) = 4 = 2^{2}

The center of the circle is (-2, 4) and the radius is 2, which would make it tangent to the y-axis.

**22.** *In triangle ABC, BD is the perpendicular bisector of ADC. Based upon
this information, which statements below can be proven?
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I. is a median.

II. bisects ∠ABC.

III. ABC is isosceles.*
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**Answer: (4) I, II, and III**

If BD is a bisector of ADC, it's also a median.

If BD is a perpendicular bisector, than AB and AC must be congruent. (This comes in handy to know in Part II!)

IF AB = AC, the triangle is isosceles. In an isosceles triangle, the median bisects the vertex angle.

**23.** *Triangle RJM has an area of 6 and a perimeter of 12. If the triangle
is dilated by a scale factor of 3 centered at the origin, what are the area and perimeter of its image, triangle R'J'M'?
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**Answer: (3) area of 54 and perimeter of 36**

If the image is dilated by a scale factor of 3, then the perimeter is increased by a factor of 3, but the area is increased by a factor of 3^{2} = 9.

12 * 3 = 36, and 6 * 9 = 54.

**24.** *If sin (2x + 7)° = cos (4x - 7)°, what is the value of x?
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**Answer: (2) 15**

The Sine of any angle is equal to the Cosine of the complementary angle. And the sum of the two complementary angles is 90 degrees

2x + 7 + 4x - 7 = 90

6x = 90

x = 15

**End of Part I**

How did you do?

Questions, comments and corrections welcome.