The following are some of the multiple questions from the recent June 2019 New York State Common Core Geometry Regents exam.

### June 2019 Geometry, Part I

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

**1.** *On the set of axes below, AB is dilated by a scale factor of 5/2
centered at point P.
*

Which statement is always true?

Which statement is always true?

**Answer: (2) AB | A'B' **

Shape (and therefore slope) is preserved in a dilation, so the line segments are parallel.

Choice (4) is backward. If it had been 5/2(AB) = A'B', that would have been correct.

**2.** *The coordinates of the vertices of parallelogram CDEH are C(-5,5),
D(2,5), E(-1,-1), and H(-8,-1). What are the coordinates of P, the point of intersection of diagonals CE
and DH?
*

**Answer: (3) ( 3,2) **

In a parallelogram, the diagonals bisect each other, so the point of intersection must be the midpoint of both lines.

The midpoint of CE is at ( (-5+-1)/2, (5+-1)/2 ), which is (-3, 2).

Checking the other line, just to be sure:

The midpoint of DH is at ( (2+-8)/2, (5+-1)/2 ), which is (-3, 2).

**3.** *The coordinates of the endpoints of QS are Q(-9,8) and S(9,-4).
Point R is on QS such that QR:RS is in the ratio of 1:2. What are the coordinates of point R?
*

**Answer: (3) (-3,4) **

A ratio of 1:2 means one-third of the way along he line.

The difference in the x-coordinates is 9 - (-9) = 18. One-third of 18 is 6, and -9 + 6 = -3.

The eliminates all but one choice.

Check the y-coordinates to be sure:

The difference in the y-coordinates is -4 - 8 = -12. One-third of -12 is -4, and 8 + -4 = 4.

**4.** *If the altitudes of a triangle meet at one of the triangle’s vertices, then the triangle is
*

**Answer: (1) a right triangle **

There's no way for this to happen unless one of the angles is a right angle, which makes it a right triangle.

**5.** *In the diagram below of triangle ACD, DB is a median to AC, and AB = DB
If m∠DAB = 32°, what is m∠BDC?
*

**Answer: (3) 58°**

If m∠DAB = 32°, then m∠ADB = 32° because it is an isosceles triangle.

This makes m∠DBC = 64° by the Remote Angle Theorem.

Since B is a median, AB = BC, but since AB = DB, then BC = DB, making DBC an isosceles triangle.

This makes BDC = BCD, and the pair of base angles have a sum equal to 180 - 64 = 116.

Half of 116 is 58, so m∠BDC = 58°

**6.** *What are the coordinates of the center and the length of the radius of the circle whose equation is x ^{2} + y^{2} = 8x - 6y + 39?
*

**Answer: (2) 128 **

To put is into standard form (x - h)^{2} + (y - k)^{2} = r^{2}, you need to move the x and y terms to the left side and then Complete the Square, twice.

Remember, you need to take half of the coefficient of x (or y) and then square it to find out what to add.

^{2}+ y

^{2}= 8x - 6y + 39

x

^{2}- 8x + y

^{2}+ 6y = 39

**7.** *The line 3x 4y 8 is transformed by a dilation centered at the
origin. Which linear equation could represent its image?
*

**Answer: (2) y = 3/4 x + 8 **

A line that has been dilated would keep its slope and be parallel to the original (unless it was the same line).

The slope of the original line can be found by isolating y.

-3x + 4y = 8

4y = 3x + 8

y = 3/4 x + 8

The slope must be 3/4, so the choice is (2).

Also, when written in Standard Form Ax + By = C, the slope of the line is -A/B, which in this case is -3/-4 = 3/4.

**8.** *In the diagram below, AC and BD intersect at E.
Which information is always sufficient to prove triangle ABE = triangle CDE?
*

**Answer: (4) BD and AC bisect each other. **

Bisecting each other means that the two triangles will be congruent because of SAS, usually the vertical angles between the sides.

Choice (1) is enough to prove that they are similar by AAA because of Alternate Interior Angles, but not that they are congruent.

Choice (2) gives you SSA, which isn't a postulate or theorem. (And don't read it backward!)

Choice (3) isn't enough information. You'll have one side and one angle.

**9.** *The expression sin 57° is equal to
*

**Answer: (2) cos 33° **

The sine of an angle (n) is equal to the cosine of the complementary angle (90 - n).

**10.** *What is the volume of a hemisphere that has a diameter of 12.6 cm,
to the nearest tenth of a cubic centimeter?
*

**Answer: (1) 523.7**

Volume of a sphere is V = (4/3) pi * r^{3}

A hemisphere is half of that. The radius is half of the 12.6 diameter, or 6.3

So V = (1/2)(4/3)(3.141592)(6.3)^{3} = 523.6971...

A quick guess of the wrong answers would be (2) forget the 1/2, (3) took 1/2 the volume but used the diameter, (4) made both mistakes.

**11.** *In the diagram below of triangle ABC, D is a point on BA, E is a point on BC, and DE is drawn.
If BD = 5, DA = 12, and BE = 7, what is the length of BC so that AC || DE?
*

**Answer: (1) 23.8 **

For the lines to be parallel, the following proportion needs to be true. Notice that it says EC and NOT BC.

BD / DA = BE / EC

5 / 12 = 7 / EC

5 EC = 84

EC = 16.8

BC = BE + EC = 7 + 16.8 = 23.8

You could have done BD/BA = BE/BC if you preferred.

**12.** *A quadrilateral must be a parallelogram if
*

**Answer: (3) one pair of sides is both parallel and congruent **

If the lines are parallel and congruent, then the quadrilateral must be a parallelogram.

Choices (1), (2) and (4) could be an isosceles trapezoid.

**13.** *In the diagram below of circle O, chords JT and ER intersect at M.
*

If EM = 8 and RM = 15, the lengths of JM and TM could be

If EM = 8 and RM = 15, the lengths of JM and TM could be

**Answer: (3) 16 and 7.5**

If two chords of a circle intersect, then the products of their segments will be equal.

So (JM)(MT) = (RM)(ME) = (15)(8) = 120

Only Choice (3) has two factors with a product of 120.

**14.** *Triangles JOE and SAM are drawn such that ∠E = ∠M and
EJ = MS. Which mapping would not always lead to triangle JOE = triangle SAM?
*

**Answer: (4) JO maps onto SA**

Choice (1) gives you ASA. Choice (2) gives you AAS. Choice (3) gives you SAS. Choice (4) gives you SSA, which is not a theorem nor postulate (as stated in an earlier question).

**15.** *5 In triangle ABC shown below, ∠ACB is a right angle, E is a point on AC, and ED is drawn perpendicular to hypotenuse AB.
If AB = 9, BC = 6, and DE = 4, what is the length of AE?
*

**Answer: (2) 6 **

The triangles are similar because they each have a right angle and they share angle A. That means that the corresponding sides are proportional. However, be careful not to mix up the sides because of the way it is drawn. AE is the hypotenuse of the smaller triangle.

AB / BC = AE / DE

9 / 6 = AE / 4

6 AE = (9)(4) = 36

AE = 6

**16.** *Which equation represents a line parallel to the line whose equation
is -2x + 3y = -4 and passes through the point (1,3)?
*

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

The choices are all in *Point-slope form*, which means y - 3 = m(x - 1), where m is the same slope as the original equation.

Immediately, cross off choices (3) and (4).

Spoiler alert: the slope is going to be positive, so (1) is wrong, too, but let's continue.

-2x + 3y = -4

3y = 2x - 4

y = 2/3x - 4, slope = 2/3.

As stated in an earlier question, when the equation is in Standard Form Ax + By = C, the slope is -A/B.

So the slope was -(-2)/3 = 2/3.

**17.** *In rhombus TIGE, diagonals TG and IE intersect at R. The perimeter
of TIGE is 68, and TG = 16.
*

What is the length of diagonal IE?

What is the length of diagonal IE?

**Answer: (2) 30 **

They chose the numbers carefully. There are four congruent right triangles making up that rhombus.

The perimeter is 68, meaning that each side is 17, which is the hypotenuse of the triangles. TG = 16, but since TG and IE bisect each other, TR and RG are each 8. This is one leg of the right triangles.

So 8^{2} + ER^{2} = 17^{2}

If you've followed my advice before this, you've learned from Pythagorean Triples and know that these are 8-15-17 triangles.

If you didn't ... sigh, okay, let's do it:

^{2}+ ER

^{2}= 17

^{2}

64 + ER

^{2}= 289

ER

^{2}= 225

ER = 15

ER = 15 = IR, so IE = 30.

**18.** *In circle O two secants, ABP and CDP, are drawn to external
point P. If mAC = 72°, and mBD = 34°, what is the measure of ∠P?
*

**Answer: (1) 19°**

Angle P will be one-half of the difference between the two arcs.

So (72° - 34°) / 2 = 19°.

**19.** *What are the coordinates of point C on the directed segment
from A(-8,4) to B(10,-2) that partitions the segment such that AC:CB is 2:1?
*

**Answer: (4) (4,0) **

To get a 2:1 ratio, C has to be 2/3 of the way along the line from A to B.

To get from -8 to 10, you have to move 18 units. Multiply 2/3 * 18 = 12. Add 12 to -8 to get 4.

To get from 4 to -2, you have to move 6 units. Multiply 2/3 * 6 = 4. Subtract 4 from 4 to get 0.

Subtract in the second case because you are moving down.

**20.** *The equation of a circle is x ^{2} + 8x + y^{2} - 12y = 144. What are the
coordinates of the center and the length of the radius of the circle?
*

**Answer: (4) center (-4,6) and radius 14**

You need to *Complete the Square*. (Yes, that's back from Algebra).

Half of 8 is 4, and 4^{2} is 16

Half of -12 is -6, and (-6)^{2} is 36.

x^{2} + 8x + y^{2} - 12y = 144

x^{2} + 8x + 16 + y^{2} - 12y + 36 = 144 + 16 + 36

(x + 4)^{2} + (y - 6)^{2} = 196

The center is (-4, 6) and the radius is SQRT(196) = 14.

**21.** *In parallelogram PQRS, QP is extended to point T and ST is drawn.
*

If ST = SP and m∠R = 130°, what is m∠PST?

If ST = SP and m∠R = 130°, what is m∠PST?

**Answer: (2) 80°**

There is a parallelogram and an isosceles triangle.

Angle QPS = angle R = 130 degrees. Angle SPT is supplementary to QPS, so it is 50 degrees. Since PST is an isosceles triangle, the base angles are equal, so angle T is also 50 degrees. The vertex angle PST is 180 - 50 - 50 = 80 degrees.

**22.** *A 12-foot ladder leans against a building and reaches a window
10 feet above ground. What is the measure of the angle, to the nearest degree, that the ladder forms with the ground?
*

**Answer: (4) 56**

Before starting, realize that the angle with the ground must be bigger than the angle with the building, which means it has to be over 45 degrees. So you can eliminate choices 1 and 2.

The wall is opposite the angle, and the ladder is the hypotenuse, so you need to use Sine to find the angle.

Sin x = opp/hyp

Sin x = (10/12)

x = sin^{-1}(10/12) = 56.44...

If you got a ridiculously low decimal, then your calculator is in Radians mode.

**23.** *In the diagram of equilateral triangle ABC shown below, E and F
are the midpoints of AC and BC, respectively.
*

If EF = 2x + 8 and AB = 7x - 2, what is the perimeter of trapezoid ABFE?

If EF = 2x + 8 and AB = 7x - 2, what is the perimeter of trapezoid ABFE?

**Answer: (3) 100 **

Since EF is a midsegment, then it is half the length of AB.

Since ABC is an equilateral triangle, EA and FB are half the length of AB.

So first we need to find x and then EF.

4x + 16 = 7x - 2

16 = 3x - 2

18 = 3x

x = 6

Since x = 6, then EF = 2(6) + 8 = 12 + 8 = 20.

The perimeter is equal to FIVE TIMES the length of EF, so 20 * 5 = 100.

Where did I get FIVE TIMES from? EF = EA = FB = 1/2(AB), so AB = 2 EF.

**24.** *Which information is not sufficient to prove that a parallelogram is a square?
*

**Answer: (3) The diagonals are perpendicular and one pair of adjacent sides
are congruent. **

Choice (3) describes a rhombus which does not have to be a square. Each of the other choices have one condition that makes it a rhombus and another that makes it a rectangle. If it is both a rhombus and a rectangle, it must be a square.

**End of Part I**

How did you do?

Questions, comments and corrections welcome.

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