The following are some of the multiple questions from the recent August 2018 New York State Common Core Geometry Regents exam.
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.
Answer: (4) 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.
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
Answer: (1) congruent and 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
Answer: (4) cone with a diameter of 12
4. In regular hexagon ABCDEF shown below, AD, BE, and CF all intersect at G.
Answer: (1) Triangle FEG
5. A right cylinder is cut perpendicular to its base. The shape of the cross section is a
Answer: (3) 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.
Answer: (4) 28.2
7. In the diagram below of right triangle ABC, altitude BD is drawn to hypotenuse AC.
Answer: (3) 8
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?
Answer: (4) 4 * SQRT(10)
9. In the diagram below of triangle HAR and triangle NTY, angles H and N are right angles, and HAR ~ NTY.
Answer: (1) 23°
10. In the diagram below, AKS, NKC, AN, and SC are drawn such that AN = SC.
Answer: (4) AN || SC
11. Which equation represents a line that is perpendicular to the line represented by y = (2/3)x + 1 ?
Answer: (1) 3x + 2y = 12
12. In the diagram of ABC below, points D and E are on sides AB and CB respectively, such that DE || AC.
Answer: (2) 8
13. Quadrilateral MATH has both pairs of opposite sides congruent and parallel. Which statement about quadrilateral MATH is always true?
Answer: (4) ∠MAT = ∠MHT
14. In the figure shown below, quadrilateral TAEO is circumscribed around circle D. The midpoint of TA is R, and HO = PE .
Answer: (2) 64
AP = 10 so AR =10. R is the midpoint of AT, so RT = 10, which means TH = 10.
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
Answer: (1) (1, -5)
16. In triangle ABC, points D and E are on sides AB and BC, respectively,
such that DE || AC, and AD:DB 3:5.
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.
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).
Answer: (3) a rotation of 180° about the point (6,6)
18. A circle with a diameter of 10 cm and a central angle of 30° is drawn below.
Answer: (2) 6.5
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?
Answer: (2) 58
20. In the accompanying diagram of right triangle ABC, altitude BD is
drawn to hypotenuse AC.
Answer: (2) AD/AB = AB/AC
21. An equation of circle O is x2 + y2 + 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
22. In triangle ABC, BD is the perpendicular bisector of ADC. Based upon
this information, which statements below can be proven?
Answer: (4) I, II, and III
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'?
Answer: (3) area of 54 and perimeter of 36
24. If sin (2x + 7)° = cos (4x - 7)°, what is the value of x?
Answer: (2) 15
End of Part I
How did you do?
Questions, comments and corrections welcome.
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
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°.
So EGF = 137 – 32 = 105 degrees.
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.
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.
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.
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.
Opposite and hypotenuse mean you need to use sine.
sin 16.5 = 8 / x
x = 8 / sin 16.5 = 28.167…
According to the Right Triangle Altitude Theorem, (BD)2 = (AD)(CD)
So 42 = (x – 6)(x)
16 = x2 - 6x. At this point, you could substitute the choices, or solve.
x2 - 6x – 16 = 0
(x – 8)(x + 2) = 0
x = 8 or x = -2. Discard the negative length.
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).
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.
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.
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.
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.
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.
HO = PE means that OZ = ZE, and since OE = 12, then HO = PE = OZ = PE = 6.
4 * 10 + 4 * 6 = 64
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.
So 8/5 = 9.4/x
8x = (5)(9.4)
x = (5)(9.4)/8 = 5.875.
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.
What is the area, to the nearest tenth of a square centimeter, of the sector formed by the 30° angle?
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..
60 inches = 5 feet, 84 inches = 7 feet
Volume = (1/3) (Area of the base) (height)
V = (1/3) (5) (5) (7) = 58.333...
Short leg is to hypotenuse as short leg is to hypotenuse.
The other choices do not use corresponding sides in the correct order.
Rewrite the equation in standard form by grouping the variables and completing the square.
x2 + y2 + 4x - 8y = -16
x2 + 4x + y2 - 8y = -16
x2 + 4x + 4 + y2 - 8y + 16 = -16 + 4 + 16
(x + 2)2 + (y - 4) = 4 = 22
The center of the circle is (-2, 4) and the radius is 2, which would make it tangent to the y-axis.
I. is a median.
II. bisects ∠ABC.
III. ABC is isosceles.
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.
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 32 = 9.
12 * 3 = 36, and 6 * 9 = 54.
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
21. please explain
ReplyDeleteDo you remember how to Complete the Square from Algebra?
ReplyDeleteBriefly, when you put x^2 + 4x together, it isn't the square of a binomial, which you need to rewrite the equation in standard form.
If you take half of 4 (from the 4x), you get 2. The binomial you want is (x + 2)^2. But when you square that, you get x^2 + 4x + 4. So you have to add 4 to both sides of the equation to allow you to do this.
Say thing with the y terms.
This comes from the identity: (x + a)^2 = x^2 + 2ax + a^2.
I hope this helps.
WHEN WILL THE ALGEBRA 1 JUNE 2019 REGENTS ANSWERS COME OUT????!!!
ReplyDeleteProbably not for a week. For whatever reasons, they can't be posted or discussed online for a while.
ReplyDeleteFor question 15 mind explaining how you get 3/5?
ReplyDeleteSure. AB:BC is 3:2. If AB was 3 inches and BC was 2 inches, than AC would be 5 inches.
ReplyDeleteSo the ratio AB/AC is 3/5. B is 3/5th of the way from A to C.