What follows is a portion of the Common Core Geometry exam. Other parts will be posted on other days. Illustrations will be added at a later time when they become available.

### August 2016 Geometry Regents, Part I

**1. ***In the diagram below, lines l, m, n and p intersect line r.
*

Which statement is true?

Which statement is true?

(2) l || p. If two exterior angles on the same side of a transversal are supplementary, then the lines are parallel. The sum of 112 and 68 is 180 degrees.

**2. ***Which transformation would not always proudce an image that would be congruent to the original figure?
*

(2) dilation. A dilation will change the size of an image, so it will not be congruent. The others only change the orientation.

**3. ***If an equilateral triangle is continuously rotated around ones of its medians, which 3-dimensional object is generated? *

(1) cone. One point will stay in play and the other two will form a circle.

**4. ***In the diagram below, m<BDC = 100, m<A = 50, and m<DBC = 30.
*

Which statement is true?

Which statement is true?

(2) Triangle ABC is isosceles. BDC = 100, DBC = 30, so <C = 50, because 100 + 30 + 50 = 180. Angle A is also 50. Beacuse angle A = angle C, it is an isosceles triangle.

**5. ***Which point shown in the graph below is the image of point P after a counterclockwise rotation of 90 ^{o} about the angle?*

(1) A. Counterclockwise from Quadrant IV brings you to Quadrant I. Point P is close to the x-axis, so the image will be close to the y-axis, which is where A is. Point B looks like a *reflection* across the x-axis.

**6. ***In triangle ABC, where <C is a right angle, cos A = SQRT(21) / 5. What is sin B?
*

(1) SQRT(21) / 5. Because Sin B = Cos A

**7. ***Quadrilateral ABCD with diagonals AC and BD is shown in the diagram below.
*

Which information is not enough to prove ABCD is a parallelogram?

Which information is not enough to prove ABCD is a parallelogram?

(3) AB = CD and BC || AD. Either two pairs of parallel sides are needed, or two pair of congruent sides, or one pair that is *both* parallel and congruent would prove it. However, one pair of parallel sides and a different pair congruent is not sufficient. That information could be true is ABCD was, for example, an isosceles trapezoid.

**8. ***An equilateral triangle has sides of length 20. To the nearest tenth, what is the height of the equilateral triangle?*

(3) 17.3. You could guess this one. Draw a triangle and an altitude. Choice (1) 10 is the length of the base of the right triangle you just created. Choice (4) 23.1 is longer than the hypotenuse. It's either (2) or (3), but you know that it will be much bigger than the base. In fact, it will be the base TIMES SQRT (3), which is approximately 1.732. So 10(1.732) = 17.32.

Using the **Pythagorean Theorem**: x^{2} + 10^{2} = 20^{2}

so x^{2} + 100 = 400

so x^{2} = 300

x = 17.3205..., which is 17.3 to the nearest tenth.

**9. ***Given: Triangle AEC, triangle DEF, and FE perpendicular to CE
*

What is a correct sequence of similarity transformations that shows triangle AEC ~ triangle DEF?

What is a correct sequence of similarity transformations that shows triangle AEC ~ triangle DEF?

(4) a counterclockwise rotation of 90 degrees about point E followed by a dilation with a scale factor 2 centered at point E. You can see that the figure rotated only 90 degrees, not 180 -- it would be upside down then -- so eliminate choices (1) and (3). Next, a translation would move DEF away from ACE, which didn't occur. However, the size did change, which indicates a dilation.

**10. ***In the diagram of right triangle ABC, CD intersects hypotenuse AB at D.
*

*If AD = 4 and DB = 6, which length of AC makes CD perpendicular to AB?*

(1) 2*SQRT(6). If CD is perpendicular to AB, then CD is an altitude. The Right Triangle Altitude Theorem says that the product of AD times DB must equal the length of CD^{2}. Since 4 * 6 = 24, CD must be SQRT(24), which reduces to 2*SQRT(6) in simplest form.

**11. ***Segment CD is the perpendicular bisector of AB at E. Which pair of segments does not have to be congruent? *

(4) DE, CE. AB is not said to bisect CD, so you cannot assume that DE and CE are congruent. AE and BE are congruent by definition. The other points would form congruent right triangles, with the corresponding hypotenuses being congruent.

**12. ***In triangle CHR, O is on HR, and D is on CR so that <H = <RDO
*

If RD = 4, RO = 6 and OH = 4, what is the length of CD?

If RD = 4, RO = 6 and OH = 4, what is the length of CD?

(3) 11. It may not seem obvious at first that the two triangles are similar because DO is not parallel to CH. However, you are given one pair of angles congruent, and both triangles contain angle R. Therefore, they are similar.

The proper proportion to set up is: RD / HR = RO / CR. And HR = 4 + 6 = 10.

So 4 / 6 = 10 / CR

4*CR = 60

CR = 15

CD = CR - DR = 15 - 4 = 11.

Don't forget to subtract the length of RD from the side of the triangle because they are looking for CD.

**13. ***The cross section of a regular pyramid contains the altitude of the pyramid. The shape of this cross section is a *
(3) triangle. If the cross section contains the altitude, then imagine a vertical sheet of paper slicing through the top of a pyramid. No matter how you line it up, the sides of the pyramid will form a triangle on the paper.

**14. ***The diagonals of rhombus TEAM intersect at P(2, 10). If the equation of the line that contains diagonal TA is y = -x + 3, what is the equation of a line that contains diagonal EM?*

(1) y = x - 1. The lines must be perpendicular, and that means the slope of EM must be 1 because the slope of TA is -1. This eliminates choices (3) and (4). Given P(2, 1), substitute 2 for x in the first equation and you get y = 1, which is correct.

**15. ***The coordinates of vertices A and B of triangle ABC are A(3, 4) and B(3, 12). If the area of triangle ABC is 24 square units, what could be the coordinates of point C?*

(3) (-3, 8) Area of a triangle is ** 1/2 b h**. The base has a length of 12 - 4 = 8.

(1/2)(8)h = 24, so 4h = 24 and h = 6.

The third coordinate has to have an x value that is six more or six less than 3. 3 - 6 = -3, so choice (3) works.

**16. ***What are the coordinates of the center and the length of the radius of the circle represented by the equation x ^{2} + y^{2} - 4x + 8y + 11 = 0?*

(1) center (2, -4) and the radius 3.
They are making these more difficult. You need to rewrite this in standard form (x - h)^{2} + (y - k)^{2} = r^{2}.

To do this, we need to **complete the square** ... twice.

^{2}+ y

^{2}- 4x + 8y + 11 = 0

regroup: x

^{2}-4x + y

^{2}+ 8y + 11 = 0

complete the squares: x

^{2}-4x + 4 + y

^{2}+ 8y + 16 + 11 = 0 + 4 + 16

simplify: (x - 2)

^{2}+ (y + 4)

^{2}+ 11 = 20

simplify: (x - 2)

^{2}+ (y + 4)

^{2}= 9

If you didn't remember how to complete the square, you could have started by writing an equation for choice (1), expanding it, and seeing if you got the same equation. If it was wrong, it would have been a matter of signs, so if you started with an incorrect choice, you could have deduced the correct one.

**17. ***The density of the American white oak tree is 752 kilograms per cubic meter. If the trunk of an American white oak tree has a circumference of 4.5 meters and the height of the trunk is 8 meters, what is the approximate number of kilograms of the trunk?*

(2) 9694.
d = m / V or d = m / (pi * r^{2} * h). C = 2*pi*r so r = C / (2 * pi) = 0.7162

752 = m / (pi*0.71612^{2}*8)

m = 752 * (pi*0.71612^{2}*8) = 9692.355

which is approximately 9694, with errors for rounding in the middle of the problem.

Yes, this was an unnecessarily complicated problem for a multiple-choice question. They could have give a *diameter* instead of the circumference, or even just given the radius itself.

(1) 150(0.85)_{m}. The hot chocolate is getting cooler, and the temperature is getting lower. Therefore, the base must be less than one (**exponential decay**), so choices (2) and (4) are out. If m = 0, then 150(0.85)_{0} = 150(1) = 150, which fits the table. The correct choice is (1). In choice (3), m - 1 would give an exponent of -1, which would actually increase the temperature at m = 0.

**18. ***Point P is on the directed line segment from point X(-6, -2) to point Y(6, 7) and divides the segment in the ratio 1:5. What are the coordinates of point P?*

(4) (-4, -1/2). The ratio 1:5 represents 1x + 5x which is 6x. If the calculate the distance between the two x-coordinates, the x-coordinate of P is 1/6th to the right of the x coordinate of point X. If the calculate the distance between the two y-coordinates, the y-coordinate of P is 1/6th to the above of the y coordinate of point X (not point Y).

From -6 to 6 is 12 spaces, and 1/6(12) = 2; -6 + 2 = -4. The answer is choice (4). You can verify with y.

From -2 to 7 is 9 spaces, and 1/6(9) = 1.5; -2 + 1.5 = -1/2, which is choice (4).

**19. **Due to a typographical error in the test booklet, there is no correct answer for question 19. I'm repeating it here just in case a teacher accidentally assigns it and a student puts it into a search engine.

*In circle O, diameter AB, chord BC and radius OC are drawn, and the measure of arc BC is 108 ^{o}. Some students wrote these formulas to find the area of sector COB (omitted)
Which students wrote correct formulas?*

No answer.

**20. ***Tennis balls are sold in cylindrical cans with the balls stacked one on top of the other. A tennis ball has a diameter of 6.7 cm. To the nearest cubic centimeter, what is the minimum volume of the can that holds a stack of 4 tennis balls?*

(4) 945. Volume = (pi)r^{2}h, r = 6.7/2 = 3.35, h = 4(6.7) = 26.8

V = (pi)(3.35)^2(26.8) = 944.87... = 945

**21. ***Line segment A'B', whose endpoints are (4, -2) and (16, 14) is the image of AB after a dilation of 1/2 centered at the origin. What is the length of AB?*

(4) 40. You could either find the length and double it (the image is 1/2 the original) or you can double the x- and y-values and then find the length. Your choice. The distance between the x values is 16 - 4 = 12. The distance between the y-values is 14 - (-2) = 16. Using either the distance formula, or Pythagorean theorem, you will find that the distance between the two is 20. (You know that 12-16-20 is just a 3-4-5 right triangle times 4, right?)

Since the image is half the length of the original, the length of AB is 40.

**22. ***Given: Triangle ABE and triangle CBD shown in the diagram below with DB = BE
*

Which statement is needed to prove triangle ABE = ACBD using only SAS = SAS?

Which statement is needed to prove triangle ABE = ACBD using only SAS = SAS?

(3) AD = CE. You are given DB = BE. Angle B is congruent to itself by the *Reflexive Property*. You need to know that AB = BC, but since you already know that DB = BE, then if AD = CE, you can conclude that AB = BC by the *Addition Postulate*. That will give you SAS = SAS.

**23. ***In the diagram below, BC is the diameter of circle A.
*

Point D, which is unique from points D and C, is plotted on circle A. Which statement must always be true?

Point D, which is unique from points D and C, is plotted on circle A. Which statement must always be true?

(1) Triangle BCD is a right triangle. Angle D will be an inscribed angle and the arc it intercepts is a semi-circle, which is 180 degrees. Therefore, angle D must be half of that, or 90 degrees. So BCD must be a right triangle, always.

**24. ***In the diagram below, ABCD is a parallelogram, AB is extended through B to E, and CE is drawn.
*

*If CE = BE and m<D = 112*

^{o}, what is m<E?(1) 44^{o}.
If angle D = 112, then angle A is supplementary, or 180 - 112 = 68. Angle CBE is congruent to A, so it is also 68 degrees. Because BE = CE, triangle BCE is an isosceles triangle and the base angles are congruent, so angle BCE is also 68 degrees.
Angle E = 180 - 68 - 68 = 112 - 68 = 44 degrees.

**END OF PART I**

How did you do? Any questions? (I appreciate pointing out any "typos" in my problems. Thank you.)

More Geometry problems.

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