This exam was adminstered in June 2023.
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June 2023 Geometry Regents
Part I
Each correct answer will receive 2 credits. No partial credit.
1. A square pyramid is intersected by a plane passing through the vertex
and perpendicular to the base.
Which two-dimensional shape describes this cross section?
(1) square
(2) triangle
(3) pentagon
(4) rectangle
Answer: (2) triangle
If this were a cake and you were to slice it straight down the middle so that you cut through the highest point (the vertex) and sliced straight down, when you separated the two pieces, you would have a triangle slice, which is Choice (2).
If you missed the vertex, any other vertical cut would result in a trapezoid (which is not one of the choices).
If the plane was parallel to the base, the cross section would have been a square. If the was not parallel but passed through all four sides, it would result is a rectangle. Note that since a square is a rectangle, you know that square MUST BE incorrect because rectangle would also be correct.
There is no way for a place to intersect the pyramid and result in a pentagon.
2. Trapezoid ABCD is drawn such that AB||DC. Trapezoid A'B'C'D' is the image of trapezoid ABCD after a rotation of 110o counterclockwise about point P.
Which statement is always true?
(1) ∠A ≅ ∠D'
(2) AC ≅ B'D'
(3) A'B' || D'C'
(4) B'A' ≅ C'D'
Answer:(3) A'B' || D'C'
The iamge after the rotation will have the same shape. That means that parallel lines are still parallel. The image is still a trapezoid. Therefore, A'B' || D'C'. Choice (3) is the correct choice.
∠A doesn't correspond to ∠D', so there is no reason for the two angles to be congruent. Eliminate Choice (1).
Since it is not stated that this is an isosceles trapezoid, the diagonals (not drawn) are not necessarily congruent. Eliminate Choice (2).
In a trapezoid, the bases do not have to be congruent. In fact, in standard HS Geometry, the bases cannot be congruent, because that would make it a parallelogram instead of a trapezoid. (When you get to advanced mathematics, parallelograms will become special cases of trapezoids the way that equilateral triangles are also isosceles.)
3. What is the volume of a right circular cone that has a height of
7.2 centimeters and a radius of 2.5 centimeters, to the nearest tenth of
a cubic centimeter?
(1) 37.7
(2) 47.1
(3) 113.1
(4) 141.4
Answer: (2) 47.1
The Volume of a cone is V = 1/3 π r2 h.
V = 1/3 π (2.5)2 (7.2) = 47.1239..., which is approximately 47.1. Choice (2) is the correct answer.
I think the other two choices come from using (2.5)(2) instead of (2.5)2, with and without the 1/3 in front.
4. In the diagram below of right triangle SUN, where ∠N is a right angle,
SU = 13.6 and SN = 12.3.
What is m∠S, to the nearest degree?
(1) 25o
(2) 42o
(3) 48o
(4) 65o
Answer: (1) 25o
You are given the hypotenuse and the side adjacent to the angle, so you need to use the cosine ratio.
Cos x = 12.3/13.6, so x = cos-1 (12.3/13.6) = 25.26 degrees, so Choice (1) is the correct answer.
The other choices are for sine and tangent (using the hypotenuse instead of the opposite side) and tangent flipping the order of the numbers.
5. In the diagram below of circle O, diameter AOB and chord CB are
drawn, and m/B = 28°
What is mBC (arc BC)?
(1) 56°
(2) 124°
(3) 152°
(4) 166°
Answer: (2) 124°
Angle B is an inscribed angle, so it intercepts an arc that is equal to twice the size of its angle. AOB is a diameter which marks off two semicircles, each having arcs of 180 degrees. Arc BC is therefore 180 - 2(28), which is 124 degrees, which is Choice (2).
Choice (1) is the size of arc AC, not BC. Choice (3) is 180 - 28, instead of 2(28). Choice (4) is 180 - 1/2(28).
6. In the diagram below of parallelogram ABCD, diagonal BED and EF are drawn, EF ⊥ DFC, m∠DAB = 111°, and m∠DBC = 39°.
What is m∠DEF?
(1) 11°
(2) 51°
(3) 60°
(4) 120°
Answer: (3) 60°
Work your way around the parallelogram. Angle C = 111°. Angle EFC = 90°. So m∠BEF = 360 - (39 + 111 + 90) = 120°.
This makes m∠DEF = 180 - 120 = 60°.
7. In the diagram below of nACT, ES is drawn parallel to AT such that E is on CA and S is on CT.
Which statement is always true?
(1) CE/CA = CS/ST
(2) CE/ED = EA/AT
(3) CE/EA = CS/ST
(4) CE/ST = EA/CS
Answer: (3) CE/EA = CS/ST
Triangle CES is similar to triangle CAT, so the corresponding sides are proportional. Look for the correct proportion from the choices.
Choice (1) is incorrect because it uses ST instead of CT.
Choice (2) is incorrect because it uses EA instead of CA.
Choice (3) compares segments split by parallel lines. Choice (3) is the correct answer.
Choice (4) is inconsisent with the triangles, comparing a part of one triangle with a part of the a different triangle.
8. On the set of axes below, congruent triangles ABC and DEF are drawn.
Which sequence of transformations maps △ABC onto △DEF?
(1) A counterclockwise rotation of 90 degrees about the origin,
followed by a translation 8 units to the right.
(2)A counterclockwise rotation of 90 degrees about the origin,
followed by a reflection over the y-axis.
(3) A counterclockwise rotation of 90 degrees about the origin,
followed by a translation 4 units down.
(4) A clockwise rotation of 90 degrees about the origin, followed by a
reflection over the x-axis.
Answer: (1) A counterclockwise rotation of 90 degrees about the origin, followed by a translation 8 units to the right.
Look at each of the choices, one at a time.
Choice (1) moves point A to (-6,-2) and B to (-2,-2). If you translate the triangle 8 units to the right, it will line up with DE. Likewise, you can follow point C and it will move onto point F. Choice (1) is the correct answer.
Choice (2) is incorrect because a reflection over the y-axis would reverse the positions of D and E. Furthermore, the image of point C would be must closer to the y-axis (at (2,-5)).
Choice (3) has a translation down, instead of to the right.
Choice (4) has a clockwise rotation, which will not line up with DEF after a reflection over the x-axis.
More to come. Comments and questions welcome.
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