The following are some of the multiple questions from the recent January 2020 New York State Common Core Geometry Regents exam.

### January 2020 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, FAD || EHC, and ABH and BC are drawn.
*

If m∠FAB = 48° and m∠ECB = l8°, what is m∠ABC?

If m∠FAB = 48° and m∠ECB = l8°, what is m∠ABC?

**Answer: (3) 66° **

Angle ABC is exterior to triangle CHB. Angles BCH and BHC are the Remote Angles of the triangle, and ABC is equal to their sum.

Angle BHC = FAB = 48 because of Alternate Interior Angles.

48 + 18 = 66.

**2.** *A cone has a volume of 108(π) and a base diameter of 12. What is the height of the cone?
*

**Answer: (2) 9 **

Volume = 1/3 (π) r^{2} h.

The radius is half of the diameter, so r = 6.

108 (π) = 1/3 (π) (6)^{2} h

h = [108 (π)] / [ 1/3 (π) (36) ]

h = 108 / 12 = 9

**3.** *Triangle JGR is similar to triangle MST. Which statement is not always true?
*

**Answer: (2) ∠G = ∠T **

G and T are not corresponding angles. J goes with M, G goes with S, R goes with T.

**4.** *In paralellogram ABCD, diagonals AC and BD intersect at E. Which statement proves ABCD is a rectangle?
*

**Answer: (1) AC = BD **

A parallelogram with congruent diagonals is a rectangle.

Choice (2) makes no sense, having one side perpendicular to the diagonal.

Choices (3) and (4) would prove that ABCD is a square, which would make it a rectangle, but it could be a rectangle without those being true.

**5.** *The endpoints of directed line segment PQ have coordinates of P(-7, -5) and Q(5, 3). What are the coordinates of point A, on PQ, that divide PQ into a ratio of 1:3?
*

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

A ratio of 1:3 means 4 parts, and we want 1 of them. That is, 1/4.

From -7 to 5 is a distance of 12, 1/4 of 12 = 3. -7 + 3 = -4. (That's enough to give you the answer.)

From -5 to 3 is a distance of 8, 1/4 of 8 is 2. -5 + 2 = -3. (-4, -3).

**6.** *In the trapezoid below, AB || CD.
*

If AE = 5.2, AC = 11.7, and CD = 10.5, what is the length of AB, to the nearest tenth?

If AE = 5.2, AC = 11.7, and CD = 10.5, what is the length of AB, to the nearest tenth?

**Answer: (3) 8.4 **

Triangles ABE and CDE are similar because of Alternate Interior and Vertical Angles.

That makes the corresponding sides proportional.

EC = AC - AE = 11.7 - 5.2 = 6.5

6.5 / 10.5 = 5.2 / x

6.5 x = (10.5)(5.2)

x = (10.5)(5.2) / 6.5 = 8.4

^{2}+ y

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

x

^{2}- 8x + y

^{2}+ 6y = 39

x

^{2}- 8x + 16 + y

^{2}+ 6y + 9 = 39 + 16 + 9

(x - 4)

^{2}+ (y + 3)

^{2}= 64

Center (4, -3), Radius 8

Quick answer: Move the x and y term to the left, you get -8x and +6y.

This tells you that the center is going to (-1/2) of those coefficients: (4, -3)

The radius is the square root of the number on the right, and 64^2 is a really big number, so it doesn't make sense.

**7.** *Kayla was cutting right triangles from wood to use for an art project. Two of the right triangles she cut are shown below.
*

If triangle ABC ~ triangle DEF, with right angles B and E, BC = 15 cm, and AC = 17 cm, what is the measure of ∠F, to the nearest degree?

If triangle ABC ~ triangle DEF, with right angles B and E, BC = 15 cm, and AC = 17 cm, what is the measure of ∠F, to the nearest degree?

**Answer: (1) 28° **

The measure of angle F is the same as the measure of angle C.

BC is adjacent and AC is the hypotenuse. That means, use cosine.

Cos C = 15/17

C = cos^{-1}(15/17) = 28.07... or 28.

**8.** *The line represented by 2y = x + 8 is dilated by a scale factor of k centered at the origin, such that the image of the line has an equation of y - 1/2 x = 2. What is the scale factor?
*

**Answer: (1) k = 1/2 **

Divide the first equation by two: y = 1/2 x + 4.

Add 1/2x to both sides of the second equation: y = 1/2x + 2.

The new y-intercept is 1/2 the distance from the origin as the original line.

**9.** *In quadrilateral ABCD, below, AB || CD, and E, H, and F are the midpoints of AD, AC, and BC respectively.
*

If AB = 24, CD = 18, and AH = 10, then FH is

If AB = 24, CD = 18, and AH = 10, then FH is

**Answer: (3) 12 **

This may be confusing because of the extra information which is not needed.

Triangle ABC has a midsegment HF, because both H and F are midpoints. The midsegment is parallel to and half the size of the third side, AB. AB is 24, so HF = 12.

**10.** *Jaden is comparing two cones. The radius of the base of cone A is twice as large as the radius of the base of cone B. The height of cone B is twice the height of cone A. The volume of cone A is
*

**Answer: (1) twice the volume of cone B.**

V = 1/3 (π) r^{2} h.

If you double the radius, you get (2)^{2} = 4 times the volume.

If you halve the height, you get (1/2) the volume.

(4)*(1/2) = 2

**11.** *A regular hexagon is rotated about its center. Which degree measure will carry the regular hexagon onto itself?
*

**Answer: (3) 120° **

The sum of the interior angles in a hexagon is 720°. Each individual angle in a regular hexagon is 1/6 of that, or 120 degrees.

(6 - 2) * 180 / 6 = 120

**12.** *In triangle MAH below, MT is the perpendicular bisector of AH.
*

Which statement is not always true?

Which statement is not always true?

**Answer: (2) Triangle MAT is isosceles **

If MT is a perpendicular bisector, then MAT = MHT because AT = AH, the two angles are right angles and they shared side MT.

This means that MA = MH, so triangle MAH is isosceles, Choice (1).

Angles AMT = HMT, so MT is a bisector of angle AMH, Choice (3).

Since MTA is a right angle, that makes angles A and TMH complementary, Choice (4).

There is no need for AT = MT (although it could in some cases), so it is not always true that triangle MAT is isosceles.

**13.** *In circle B below, diameter RT, radius BE and chord RE are drawn.
*

If m∠TRE = 15° and BE = 9, then the area of sector EBR is

If m∠TRE = 15° and BE = 9, then the area of sector EBR is

**Answer: (3) 33.75π**

Triangle BRE is an isosceles triangle because two sides are radii, which are equal.

This means that angle E is also 15 degrees, and the central angle EBR is 150 degrees.

The area of the sector of the circle is (150 / 360)*(π)*(9)^{2} = 33.75 π.

Note that they wanted the answer in terms of π, so don't hit the π key when multiplying.

**14.** *Lou has a solid clay brick in the shape of a rectangular prism with a
length of 8 inches, a width of 3.5 inches, and a height of 2.25 inches.
If the clay weighs 1.055 oz/in ^{3}, how much does Lou's brick weigh, to
the nearest ounce?
*

**Answer: (1) 66**

Find the Volume and multiply it by the weight: L * W * H * Wt.

8 * 3.5 * 2.25 * 1.055 = 66.465, or 66 ounces.

**15.** *Rhombus ABCD can be mapped onto rhombus KLMN by a rotation about point P, as shown below.
*

What is the measure of ∠KNM if the measure of ∠CAD = 35°?

What is the measure of ∠KNM if the measure of ∠CAD = 35°?

**Answer: (4) 110° **

There are four right triangles. The complement of 35 degrees is 90 - 35 = 55 degrees.

Angle KNM is the sum of two 55 degree angles, which is 110 degrees.

**16.** * In right triangle RST below, altitude SV is drawn to hypotenuse RT.
*

If RV = 4.1 and TV = 10.2, what is the length of ST, to the nearest tenth?

If RV = 4.1 and TV = 10.2, what is the length of ST, to the nearest tenth?

**Answer: (4) 12.1**

TV / ST = ST / RT. RT = RV + VT = 4.1 + 10.2 = 14.3

ST ^{2} = (10.2)(14.3) = 145.86

ST = SQRT(145.86) = 12.077... = 12.1

**17.** *On the set of axes below, pentagon ABCDE is congruent to
A"B"C"D"E".
*

Which describes a sequence of rigid motions that maps ABCDE onto A"B"C"D"E"?

Which describes a sequence of rigid motions that maps ABCDE onto A"B"C"D"E"?

**Answer: (2) a rotation of 90° counterclockwise about the origin followed by a translation down 7 units **

In choice (1), a reflection over the x-axes would leave A" at (-2,-2) and E" at (-2,-5), instead of the other way around.

In choice (3), two reflections would have the image pointing downward. That is, C" would be the lowest point on the graph.

In choice (4), the image would wind up in Quadrant 1.

**18.** *On the set of axes below, rhombus ABCD has vertices whose
coordinates are A(l,2), B(4,6), C(7,2), and D(4,-2).
*

What is the area of rhombus ABCD?

What is the area of rhombus ABCD?

**Answer: (2) 24**

The rhombus can be divided into 4 congruent right triangles with base 3 and height 4.

So A = (4) * (1/2) * (3) * (4) = 24

There is another formula: Area of a rhombus is one half the product of the diagonals.

**19.** *Which figure(s) below can have a triangle as a two-dimensional cross
section?
*

I. cone

II. cylinder

III. cube

IV. square pyramid

I. cone

II. cylinder

III. cube

IV. square pyramid

**Answer: (4) I, III, and IV, only **

If you slice a cone vertically, the cross section is a triangle.

If you slice a cube on its corner so that you go through 3 of its 6 sides, you will get a triangle.

If you slice a square pyramid through its apex, or in any place where you cut through only 3 sides, you will get a triangle.

You can't get a triangle from a cylinder. You will either get a rectangle, circle, ellipse or something else, but not a triangle.

**20.** *What is an equation of a circle whose center is at (2, -4) and is tangent to the line x = -2?
*

**Answer: (2) (x - 2) ^{2} + (y + 4)^{2} = 16 **

Start with (x - h)

^{2}+ (y - k)

^{2}= r

^{2}

You know h, and k, but you need r.

If x = -2 (a vertical line) is tangent, then a radius goes from (2, -4) to (-2, -4), which is a distance of 4, and 4

^{2}= 16.

**21.** *For the acute angles in a right triangle, sin (4x)° = cos (3x + 13)°.
What is the number of degrees in the measure of the smaller angle?
*

**Answer: (3) 44°**

If the sin of one angle equals the cosine of a second angle, then the two angles are complementary, adding up to 90 degrees.

So 4x + 3x + 13 = 90

7x + 13 = 90

7x = 77

x = 11

The answer is NOT 11°.

Substitute 11 for x and find the smaller angle.

4x = 4(11) = 44°. *Note:* Since it is less than 45°, we already know this has to be smaller.

3(11) + 13 = 33 + 13 = 46°.

44° is the smaller angle.

**22.** *Triangle PQR is shown on the set of axes below.
*

Which quadrant will contain point R", the image of point R, after a 90° clockwise rotation centered at (O,O) followed by a reflection over the x-axis?

Which quadrant will contain point R", the image of point R, after a 90° clockwise rotation centered at (O,O) followed by a reflection over the x-axis?

**Answer: (1) I**

The clockwise rotation puts R' in Quadrant IV. The reflection puts it back in Quadrant I.

**23.** *In the diagram below of right triangle ABC, altitude BD is drawn.
*

Which ratio is always equivalent to cos A?

Which ratio is always equivalent to cos A?

**Answer: (2) BD/BC **

Whichever triangle it is, COS A means the adjacent over the hypotenuse.

Choice (1) uses the two legs of the bigger triangle, so it is tangent.

Choice (3) uses the opposite side and hypotenuse of the small triangle, which is sine.

Choice (4) uses the opposite side and hypotenuse of the big triangle, which is sine.

Choice (2) doesn't use point A at all. That means that they are using properties of similar triangles.

In triangle BCD, angle A is congruent to angle CDB, which is where the short leg meets the hypotenuse. BD is adjacent and BC is the hypotenuse.

**24.** *In the diagram below of triangle RST, L is a point on RS, and M is a pointon RT, such that LM || ST.
*

If RL = 2, LS = 6, LM = 4, and ST = x + 2, what is the length of ST?

If RL = 2, LS = 6, LM = 4, and ST = x + 2, what is the length of ST?

**Answer: (4) 16**

Since LM || ST, that means that triangle RLM ~ RST (corresponding angles are congruent).

RL/LM = RS/ST

2/4 = (2+6)/(x + 2)

4*8 = 2(x + 2)

32 = 2x + 4

28 = 2x

14 = x

ST = x + 2 = 16.

Note that you could have saved a step if you just used ST instead of x.

**End of Part I**

How did you do?

Questions, comments and corrections welcome.

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## 1 comment:

Great explanations! Thank you!

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