June 2022 Geometry Regents, Part 3

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

June 2022 Geometry, Part III

Each correct answer is worth up to 4 credits. Partial credit can be given. Work must be shown or explained.

32. As modeled below, a projector mounted on a ceiling is 3.74 m from a wall, where a whiteboard is displayed. The vertical distance from the ceiling to the top of the whiteboard is 0.41 m, and the height of the whiteboard is 1.17 m.

Determine and state the projection angle, θ, to the nearest tenth of a degree.

Answer:
Find the angle in the large right triangle, and the angle in the skinny right triangle. Subtract those and you have θ. In both cases, you have the opposite and adjacent sides, so you need to use tangent.

tan x = 0.41/3.74
x = tan^-1(0.41/3.74) = 6.2561...

tan x = (0.41+1.17)/3.74
x = tan^-1 ((0.41+1.17)/3.74) = 22.9021...

θ = 22.9021 - 6. 2561 = 16.646 = 16.6 to the nearest tenth of a degree.

33. Given: Parallelogram PQRS, QT ⟂ PS, SU ⟂ QR

Prove: PT ≅ RU

Answer:

Answer:
You need to make a two-column proof. You can also write a paragraph proof, if you want, but you still need to have all the correct statements and reasons/justifications. The scorer will look to see that all the points are there.

You can prove this by showing that the two triangles are congruent and using CPCTC (corresponding parts of congruent triagnles are congruent). Looking at those triangles, you know that they are right triangles because the lines are perpendicular. And you know that the hypotenuses are congruent because they are the opposite sides of a parallelogram. And you know that angle P is congruent to angle R because its a parallelogram. This is what you need to show.

Note that the above paragraph isn't a paragraph proof because I didn't fully state the reasons and properties. It was meant as a guideline for a two-column proof.

Update: Proof added. Formatting takes time.

Statement	Reason
1. Parallelogram PQRS, QT ⟂ PS, SU ⟂ QR	Given
2. RS ≅ QP	Opposite sides of a parallelogram are congruent.
3. ∠P ≅ ∠R	Opposite angles of a parallelogram are congruent.
4. ∠PTQ and ∠RUS are right angles	Definition of perpendicular lines
5. ∠PTQ ≅ ∠RUS	All right angles are congruent.
6. △PTQ ≅ △RUS	AAS Theorem
7. PT ≅ RU	CPCTC (Corresponding Parts of Congruent Triangles are Congruent)

The first is required but not worth any points. Lines 2 and 3 are each worth 1 point. Lines 4-6 together are worth 1 point. Line 7 is worth the final point. Basically, proving AAS is three points, but you'll lose one point for each pair of "A" or "S" that you didn't show to be congruent, or for not stating AAS.

You can also prove PT = RU by showing that QUST is a parallelgragm (a rectangle even), with opposite sides congruent. Since QR = PS because it's a paralellogram, you can use the subtraction property to show PT = RU.

Given that PQRS is a parallelogram, then RQ ≅ PS because the opposite sides of a parallelogram are congruent. QU || ST because the opposite sides of a parallelogram are parallel. Angle QUS is a right angle because of the definition of perpendicular lines. Angle UST is a right angle because the same-side interior angles of a transversal between parallel lines are supplementary. Angles UST and STQ are supplementary because they add up to 180 degrees. So US || QT because the same-side angles are supplementary. QUST is a parallelogram because its oppisite sides are parallel. QU ≅ ST because the opposite sides of a parallelogram are congruent. Therefore, PT &cong RU by the Subtraction Property.

Okay, I'll admit that this took longer than I thought it would because I misread the given and didn't realize that I still had a couple extra things to prove to get where I wanted to go. HOWEVER, I knew I could get there, and I couldn't get the idea out of my head, so I went with it.

34. A concrete footing is a cylinder that is placed in the ground to support a building structure. The cylinder is 4 feet tall and 12 inches in diameter. A contractor is installing 10 footings.

If a bag of concrete mix makes ; of a cubic foot of concrete, determine and state the minimum number of bags of concrete mix needed to make all 10 footings.

Answer:
First, convert the inches to feet. Then find the radius from the diameter. Then find the volume of ONE footing. Multiply it by 10 to get the volume of TEN footings. Then DIVIDE by 2/3 (which means multiply by 3/2) to find the number of bags of concrete. Then, and this is Important, ROUND UP to the next full bag. If you round down, you won't have enough concrete mix to finish the tenth footing.

d = 12 inches = 1 foot, so r = 1/2 foot

V = π r²h = π (.5)²(4) = π = 3.14159....

10V = 10(3.14159...)

Divide 10(3.14159)/(2/3) = 47.12385

48 bags of concrete mix are needed.

Basically, you got a point for the Volume of one footing, a point for the Volume of 10 footings, a points for the amount of concrete mix needed and a point for the final number of bags. You didn't have to show those numbers to get full credit (the test didn't ask for any in-between answers), but they could give you partial credit if your final answer was incorrect. For example, you could have written one big equation to find the answer in one step. As long as the work was there, you got full credit.

End of Part III

How did you do?

Questions, comments and corrections welcome.

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