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

### January 2020 Geometry, Part III

Each correct answer is worth up to 4 credits. Partial credit is available. Work must be shown. Correct answers without work receive only 1 point.

**32.** *Quadrilateral NATS has coordinates N(-4,-3), A(1,2), T(8,1), and S(3,-4).
Prove quadrilateral NATS is a rhombus.
[The use of the set of axes below is optional.]
*

**Answer:**

Shortest method: If the diagonals are perpendicular, the quadrilateral is a rhombus.

Other method: All four sides have the same length -- distance formula four times.

Other method: Opposite sides are parallel (same slope) and two consecutive sides have the same length -- distance formula two times.

Method 1:

Show that NT is perpendicular to AS:

NT: (1 - -3) / (8 - -4) = 4 / 12 = 1/3

AS: (-4 - 2) / (3 - 1) = -6 / 2 = -3

Diagonal NT is perpendicular to AS, therefore the quadrilateral is a rhombus.
*(You MUST have a concluding statement!)*

Method 2:

NA = SQRT( (1 - -4)^{2} + (2 - -3)^{2}) = SQRT(25 + 25) = SQRT(50)

AT = SQRT( (8 - 1)^{2} + (1 - 2)^{2}) = SQRT(49 + 1) = SQRT(50)

TS = SQRT( (3 - 8)^{2} + (-4 - 1)^{2}) = SQRT(25 + 25) = SQRT(50)

SN = SQRT( (-4 - 3)^{2} + (-3 - -4)^{2}) = SQRT(49 + 1) = SQRT(50)

All four sides are congruent, therefore the quadrilateral is a rhombus.

Method 3:

NA = SQRT( (1 - -4)^{2} + (2 - -3)^{2}) = SQRT(25 + 25) = SQRT(50)

AT = SQRT( (8 - 1)^{2} + (1 - 2)^{2}) = SQRT(49 + 1) = SQRT(50)

(Note that the math involved in finding the slope is the same that was done in Method 2)

Slope of AT: -1/7

Slope of SN: -1/7

Slope of NA: 5/5 = 1

Slope of TS: 5/5 = 1

Opposite sides are parallel and consecutive sides are congruent, therefore the quadrilateral is a rhombus.

**33.** *David has just finished building his treehouse and still needs to buy a ladder to be attached to the ledge of the treehouse and anchored at a point on the ground, as modeled below. David is standing 1.3 meters from the stilt supporting the treehouse. This is the point on the ground where
he has decided to anchor the ladder. The angle of elevation from his eye level to the bottom of
the treehouse is 56 degrees. David’s eye level is 1.5 meters above the ground.
*

Determine and state the minimum length of a ladder, to the nearest tenth of a meter, that David will need to buy for his treehouse.

Determine and state the minimum length of a ladder, to the nearest tenth of a meter, that David will need to buy for his treehouse.

**Answer:**

You need to find the height of the stilt, but you don't have an angle from the ground. Instead you have the angle from David's eye level. So you have to find that height of that portion of the stilt and then ADD 1.5 meters, which is the height of David's eye. That will give you the height of the stilt.

One you have the height of the stilt and the distance along the ground, you have the two legs of a right triangle. The ladder will be the hypotenuse. Use the Pythagorean Theorem, and round to the nearest tenth of an inch.

tan = opp/adj

tan 56 = x / 1.3

x = 1.3 tan 56 = 1.927329. Do NOT round in the middle of the problem.

1.927329 + 1.5 = 3.427329 This is the Stilt height.

a^{2} + b^{2} = c^{2}

1.3^{2} + 3.427329^{2} = 13.436584

c = SQRT(13.436584) = 3.665 = 3.7 meters.

Note: you don't need to keep 7 or 8 decimal places, but if they are already in the calculator's memory, leave them there. If you round, always keep 2 or 3 decimal digits or you will have rounding errors in your final answer.

**34.** *A manufacturer is designing a new container for their chocolate-covered almonds. Their original container was a cylinder with a height of 18 cm and a diameter of 14 cm. The new container can be modeled by a rectangular prism with a square base and will contain the same amount of
chocolate-covered almonds.
*

If the new container’s height is 16 cm, determine and state, to the nearest tenth of a centimeter, the side length of the new container if both containers contain the same amount of almonds.

A store owner who sells the chocolate-covered almonds displays them on a shelf whose dimensions are 80 cm long and 60 cm wide. The shelf can only hold one layer of new containers when each new container sits on its square base. Determine and state the maximum number of new containers the store owner can fit on the shelf

If the new container’s height is 16 cm, determine and state, to the nearest tenth of a centimeter, the side length of the new container if both containers contain the same amount of almonds.

A store owner who sells the chocolate-covered almonds displays them on a shelf whose dimensions are 80 cm long and 60 cm wide. The shelf can only hold one layer of new containers when each new container sits on its square base. Determine and state the maximum number of new containers the store owner can fit on the shelf

**Answer:**

The first part of the problem wants you to compare Volumes. This is a semi-realistic problem. It's realistic because manufacturers change packaging for business reasons. It's less realistic because the Volume itself doesn't take empty space between almonds and the sides of the container into account. These could be different in different-shaped packaging. But this isn't a marketing exam.

Volume = Area of the Base times the Height.

The Area of the cylinder base is *(pi)(r) ^{2}*. The Area of the square prism base is

*s*

^{2}.Use the pi key on your calculator. And remember to cut the diameter in half to get the radius.

Volume of the cylinder: (3.141592)(7)^{2}(18) = 2770.8841...

Volume of prism: s^{2} * 16 = 2770.8841

s^{2} = 173.18

s = 13.15... = 13.2

Area of the base = (6)(10) + (1/2)(6)(2.65) = 67.95

Volume = Area * height = 67.95 * 6.5 = 441.675 = 442 cubic feet

The second part asks you to determine the number of boxes that can fit on the shelf. Do NOT compare Areas because the boxes cannot be split apart to fit into leftover spaces.

Find the maximum number of boxes that can fit on the shelf from front to back, and from side to side. Then multiply those numbers.

80 / 13.2 = 6.06... = 6

60 / 13.2 = 4.54... = 4, round down. There is no space to round up.

6 * 4 = 24 boxes can be displayed on one shelf.

### Part IV

A correct answer is worth up to 6 credits. Partial credit is available.

**35.** * In quadrilateral ABCD, E and F are points on BC and AD, respectively, and BGD and EGF
are drawn such that ∠ABG = ∠CDG, AB = CD, and CE = AF.
*

Prove: FG = EG

Prove: FG = EG

**Answer:**

You can prove that FG = EG if you can prove that triangles BGE = DGF.

First, mark off what you know, so you can figure out how to proceed.

The image *looks* like a parallelogram, but we have to prove that it is a parallelogram (which we can).

Once we know that we'll know that BC and AD are both parallel and congruent. This will make BE and DF congruent through the Subtraction property. Also, BD and EF create pairs of vertical angles, which of congruent, but this needs to be stated as well.

Statement | Reason |

1. ∠ABG = ∠CDG, AB = CD, and CE = AF. | Given |

2. AB || CD | If the Alternate Interior Angles are congruent, the lines are parallel |

3. ABCD is a parallelogram | If an opposite pair of sides of a quadrilateral are both congruent and parallel, the quadrilateral is a parallelogram. |

4. BC = AD | Opposite sides of a parallelogram are congruent. |

5. BE = DF | Subtraction property. |

6. BC || AD | Opposite sides of a parallelogram are parallel. |

7. ∠EBG = ∠ADG | Alternate Interior angles. |

8. ∠BGE = ∠DGF | Vertical angles. |

9. Triangle BGE = Triangle DGF | AAS |

10. FG = EG | CPCTC (Corresponding Parts of Congruent Triangles are Congruent) |

You could also have shown that triangle ABD and CDB were congruent. In that case, CPCTC would give you angles CBG and FDG, and AD = CB. Then you could use the Subtraction property to get the smaller triangles.

**End of Exam**

How did you do?

Comments and corrections welcome. *(I get many of the latter!)*

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