Thursday, April 25, 2024

Geometry Problems of the Day (Geometry Regents, January 2024 Part I)



This exam was adminstered in January 2024.

More Regents problems.

January 2024 Geometry Regents

Part I

Each correct answer will receive 2 credits. No partial credit.


17. In parallelogram ABCD with AC ⟂ BD, AC = 12 and BD = 16. What is the perimeter of ABCD?

(1) 10
(2) 24
(3) 40
(4) 56

Answer: (3) 40


The parallelogram is a rhombus because the diagonals are perpendicular. The diagonals form four right triangles with legs of 6 and 8 (which is half the size of 12 and 16, respectively.)

Using Pythagorean Theorem (or just knowing about 6-8-10 right triangles), you will find that each side of the rhombus has a length of 10. This means that the perimeter is 10 * 4 = 40, which is Choice (3).

I'm surprised that neither 20 nor 80 were listed as incorrect choices.





18.In the diagram of △CAT below, m∠A = 90° and altitude AE is drawn from vertex A.


Which statement is always true?

(1) CE / AE = AE / ET
(2) AE / CE = AE / ET
(3) AC / CE = AT / ET
(4) CE / AC = AC / ET

Answer: (1) CE / AE = AE / ET


When an altitude is drawn from the hypotenuse to the right angle of a right triangle, two more right triangles are formed and all three right triangles are similar to each other. That means that the corresponding sides of any two of the triangles are proportional.

Choice (1) is the Right Triangle Altitude Theorem. It is the correct choice.

Choice (2) shows long leg / short leg = short leg / long leg. This is incorrect.

Choice (3) shows hypotenuse / short leg = hypotenuse / long leg. This is incorrect.

Choice (4) has a ratio that contains two legs from different triangles. This is incorrect.





19. A sandbox in the shape of a rectangular prism has a length of 43 inches and a width of 30 inches. Jack uses bags of sand to fill the sandbox to a depth of 9 inches. Each bag of sand has a volume of 0.5 cubic foot. What is the minimum number of bags of sand that must be purchased to fill the sandbox?

(1) 14
(2) 13
(3) 7
(4) 4

Answer: (1) 14


Find the volume, convert it to cubic feet, and then divide it by 0.5 to find the number of bags.

V = l * w * h = 43 * 30 * 9 = 11610, 11610 / (12^3) = 6.71875 cu. ft.

6.71875 / .5 = 13.4375

You would need 14 bags to fill the sandbox. (Note: 13 bags wouldn't fill the sandbox, so don't round down. Round up.)





20.Parallelogram EATK has diagonals ET and AK. Which information is always sufficient to prove EATK is a rhombus?

(1) EA ⟂ AT
(2) EA ≅ AT
(3) ET ≅ AK
(4) ET ≅ AT

Answer: (2) EA ≅ AT


It is a rhombus is the diagonals are perpedicular or the consecutive sides are congruent.

In Choice (1), EA and AT are not diagonals. This makes the shape a rectangle.

In Choice (2), EA and AT are consecutive sides. This makes the shape a rhombus. Choice (2) is the correct answer.

In Choice (3), ET and AK are congruent diagonals. This makes the shape a rectangle.

In Choice (4), ET is a diagonal and AT is a side. Being congruent means nothing. It's just a parallelogram.





21. In the diagram below, ABCD || EHK, and MBHP and NCHL are drawn such that BC ≅ BH.


If m∠NCD = 62°, what is m∠PHK?

(1) 118°
(2) 68°
(3) 62°
(4) 56°

Answer: (4) 56°


Because BC ≅ BH, then BCD is an isosceles triangle.

This means that ∠BCH = 62, ∠BHC = 62 and ∠CBH = 56.

Since the lines are parallel, ∠CBH and ∠PHK are corresponding angles, which must be congruent.

The measure of angle PHK is 56 degrees.





22. Triangles YEG and POM are two distinct non-right triangles such that ∠G > ∠M. Which statement is sufficient to prove △YEG is always congruent to △POM?

(1) ∠E ≅ ∠O and ∠Y ≅ ∠P
(2) YG ≅ PM and YE ≅ PO
(3) There is a sequence of rigid motions that maps ∠E onto ∠O and YE onto PO.
(4) There is a sequence of rigid motions that maps point Y onto point P and YG onto PM.

Answer: (3) There is a sequence of rigid motions that maps ∠E onto ∠O and YE onto PO.


Make a little sketch, so it'll be easier to compare.

Choice (1) would just show that the two triangles are similar, but not congruent. Eliminate Choice (1).

Choice (2) shows SSA, which is not sufficient for congruency. The question states that they are not right triangles, so it HL Theorem doesn't apply either. Eliminate Choice (2).

Choice (3) shows AAS, which is sufficient to show that two triangles are congruent. Choice (3) is the correct answer.

Choice (4) has one side and one angle. This is insufficient for congruency. Eliminate Choice (4).





23.In the diagram of triangles ABD and CBE below, sides AD and CE intersect at F, and ∠ADB ≅ ∠CEB.


Which statement can not be proven?


(1) △ADB ≅ △CEB
(2) ∠EAF ≅ ∠DCF
(3) △ADB ~ △CEB
(4) △EAF ~ △DCF

Answer: (1) △ADB ≅ △CEB


There is not enough information to show that the two triangles are congruent. Note that no line segements are said to be either congruent or even perpendicular.

Choice (1) is not possible.

Choice (2) can be proven because CEA is congruent to ADC because they are supplementary to the congruent angles. Angles AFE and CFD are congruent because they are vertical angles. If two pairs of angles in two triangles are congruent, then the third pair must be congruent. So angle EAF is congruent to angle DCF.

Choice (3) can be proven because ∠B is congruent to itself by the Reflexive Property, and we were given ∠ADB ≅ ∠CEB. By AA, the two triangles are similar.

Choice (4) can be proven because ∠ADC ≅ ∠CEA because they are supplementary to the given congruent angles. Angles AFE and CFD are congruent because they are vertical angles. By AA, the two triangles are similar.





24. A small town is installing a water storage tank in the shape of a cylinder. The tank must be able to hold at least 100,000 gallons of water. The tank must have a height of exactly 30 feet. [1 cubic foot holds 7.48 gallons of water]
What should the minimum diameter of the tank be, to the nearest foot?

(1) 12
(2) 24
(3) 65
(4) 75

Answer: (2) 24


Figure out the volume from the number of gallons. V = pi r2h. Since you know the height, you can find the radius. Double that to get the diameter.

100,000 gallons * 1 cu ft / 7.48 gallons = 13368.984 cu ft

V = π r2 h

13368.984 = π r2 (30)

r2 = 13368.984 / (30π) = 141.85

r = 11.9

To the nearest foot, the minimum diameter is 24 feet, which is Choice (2).





More to come. Comments and questions welcome.

More Regents problems.

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