Wednesday, March 15, 2023

Geometry Problems of the Day (Geometry Regents, January 2023)



This exam was adminstered in January 2023.

More Regents problems.

January 2023 Geometry Regents

Part I

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


9. Which polygon does not always have congruent diagonals?

(1) square
(2) rectangle
(3) rhombus
(4) isosceles trapezoid

Answer: (3) rhombus


Rectangles have congruent diagonals, so squares do as well. Isosceles trapezoids can be proven to have congruent diagonals using a similar proof to the one for rectangles.

Unless a rhombus is a square, one diagonal is longer than the other, which becomes even more obvious as the rhombus is "squished" with two small angles and two large ones.

The corrent answer is Choice (3).





10. If the circumference of a standard lacrosse ball is 19.9 cm, what is the volume of this ball, to the nearest cubic centimeter?

(1) 42
(2) 133
(3) 415
(4) 1065

Answer: (2) 133


Find the radius from the circumference and then use this to find the volume. Note that since this is a multiple-choice test, the most "obvious" mistakes will use diameter instead of radius. If I were to guess from looking at the choices, I would think that at least one of them will be surface area if you picked the wrong formula.

C = 2πr = 19.9, so r = 19.9/(2π) = 3.167... Leave a few decimal places to avoid rounding errors.

V = 4/3 π r3 = 4/3 π (3.167)3 = 133.055... or 133, which is Choice (2).





11. Which polygon always has a minimum rotation of 180° about its center to carry it onto itself?


Answer: (1) rectangle


The key is that it says "minimum" rotation. A rectangle rquires 180 degrees of rotation to carry onto itself, but a square only requires 90 degrees, even though 180 degrees always works.

An isosceles trapezoid requires a full 360 degrees to carry onto itself.

A regular pentagong carries onto itself with a 72-degree rotation. And 180 degrees would NOT carry it onto itself.

The correct answer is Choice (1).





12. Circle O is drawn below with secant BCD. The length of tangent AD is 24.


If the ratio of DC:CB is 4:5, what is the length of CB?

(1) 36
(2) 20
(3) 16
(4) 4

Answer: (2) 20


Use the formula: AD2 = (CD)(BD). Note that it is (part)(whole), not (part)(part). Since the ratio is 4:5, we have to use 4x and (4+5)x.

(4x)(9x) = 242
36x2 = 576
x2 = 16
x = 4

So CD = 4(4) = 16 and CB = 5(4) = 20, which is Choice (2).

Note the incorrect choices: x = 4, CD = 16, and BD = 36.





13.The equation of a line is 3x - 5y = 8. All lines perpendicular to this line must have a slope of

(1) 3/5
(2) 5/3
(3) -3/5
(4) -5/3

Answer: (4) -5/3


Find the slope of the given line. A perpendicular line will be the inverse reciprocal.

3x - 5y = 8
-5y = -3x + 8
y = 3/5x - 8/5

The slope of the given line is 3/5, so the perpendicular slope is -5/3, which is Choice (4).





14. What are the coordinates of the center and length of the radius of the circle whose equation is x2 + y2 + 2x - 16y + 49 = 0?

(1) center (1,-8) and radius 4
(2) center (-1,8) and radius 4
(3) center (1,-8) and radius 16
(4) center (-1,8) and radius 16

Answer: (2) center (-1,8) and radius 4


You have to rearrange the terms and then complete the squares to find the center and the radius.

x2 + y2 + 2x - 16y + 49 = 0

x2 + 2x + y2 - 16y + 49 = 0

x2 + 2x + y2 - 16y = -49

x2 + 2x + 1 + y2 - 16y + 64 = -49 + 1 + 64

x2 + 2x + 1 + y2 - 16y + 64 = 16

(x + 1)2 + (y - 8)2 = 42

So the center of the circle is (-1, 8) because the signs are flipped, and the radius is 4, not 16.

The correct choice is (2).





15. In the diagram below of right triangle MDL, altitude DG is drawn to hypotenuse ML.


If MG = 3 and GL = 24, what is the length of DG?

(1) 8
(2) 9
(3) √(63)
(4) √(72)

Answer: (4) √(72)


The Right Triangle Altitude Theorem states that (MG)(GL) = (DG)2. Substitute and solve.

(3)(24) = x2
x = √(72)

This is Choice (4).





16. Segment AB is the perpendicular bisector of CD at point M. Which statement is always true?

(1) CB ≅ DB
(2) CD ≅ AB
(3) △ACD ~ △BCD
(4) △ACM ~ △BCM

Answer: (1) CB ≅ DB


If AB is the perpendicular bisector of CD then AC = AD and BC = BD, but AC =/= BC and AD =/= BD.

So Choice (1) is the correct answer.

In Choice (2), just because AB bisects CD, it doesn't mean that the two segments must be congruent.

In Choices (3) and (4), the triangles aren't similar because the AC =/= BC but CD = CD and CM = CM, so the sides of the triangles cannot be corresponding.

So AD/AB = AE/AC = DE/BC

Choice (1): If AD/AB = DE/BC then AD/DE = AB/BC, not DB/BC. Eliminate Choice (1).

Choice (2): If AD/AB = DE/BC then AD/DE = AB/BC. THis is the correct answer.

In Choices (3) and (4): AD/BC is not a proper ratio. It compares one side of the small triangle with a different leg of the larger triangle. These are not corresponding sides. Eliminate these two choices.





More to come. Comments and questions welcome.

More Regents problems.

No comments:

Post a Comment