June 2022 Geometry Regents, Part I (multiple choice)

This exam was adminstered in June 2022.

More Regents problems.

June 2022 Geometry Regents

Part I

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


1. Triangle A'B'C' is the image of △ABC after a dilation centered at the origin. The coordinates of the vertices of △ABC are A(-2,1), B(2,4), and C(2,-3).

If the coordinates of A' are (-4,2), the coordinates of B' are

(1) (8,4)
(2) (4,8)
(3) (4,-6)
(4) (1,2)

Answer: (2) (4,8)

If A is at (-2,1) and A' is at (-4,2), then the scale factor of the dilation is 2.

If B is at (2,4), then B' must be at (2*2,4*2) or (4,8), which is Choice (2).

2. In the diagram below, a plane intersects a square pyramid parallel to its base.

Which two-dimensional shape describes this cross section?

(1) circle
(2) square
(3) triangle
(4) pentagon

Answer: (2) square

If it is parallel to the square base then the cross section will also be a square.

You would have a similar square pyramid which proportional sides.

Note that Choice (1) is silly. There is no way that you could slice that pyramid with a plane with the result being a circle or an oval (ellipse). Choice (4) pentagon isn't much better. There are only four sides along with the base that the plane is parallel to, so the cross section couldn't have five sides to it.

If the plane was parallel to one of the sides, cutting off one of the base corners, then the cross section would have been a triangle.

3. In the diagram below, △CDE is the image of △CAB after a dilation of DE/AB centered at C.

(1) sin A = CE/CD
(2) cos A - CD/CE
(3) sin A = DE/CD
(4) cos A = DE/CE

Answer: (1) sin A = CE/CD

The two triangles are similar, so they have the same shape. That is to say, the corresponding angles are congruent. That means that the m∠A = m∠CDE. This also means that sin A = sin CDE and cos A = cos CDE.

The hypotenuse of △CDE is CD, not CE. Eliminate Choice (2) and (4).

Sine is the opposite over the hypotenuse. The opposite of CDE is CE, so the sine is CE/CD, which is Choice (1).

4. A regular pentagon is rotated about its center. What is the minimum number of degrees needed to carry the pentagon onto itself?

(1) 72°
(2) 108°
(3) 144°
(4) 360°

Answer: (1) 72°

A regular pentagon has five sides. A rotation will carry it onto itself with every 1/5 of a complete circle it is turned.

One-fifth of 360° is 72°, which is Choice (1).

Choice (2) is the measure of one interior angle of a regular pentagon. However, turning the pentagon by that amount will not carry it onto itself.

Choices (3) and (4) will carry the pentagon onto itself (they are both multiples of 72°) but they are not the minimum.

5. On the set of axes below, △ABC ≅ △A'B'C'.

Triangle ABC maps onto triangle A'B'C' after a

(1) reflection over the line y = -x
(2) reflection over the line y = -x + 2
(3) rotation of 180° centered at (1,1)
(4) rotation of 180° centered at the origin

Answer: (3) rotation of 180° centered at (1,1)

The transformation cannot be a reflection because A' and B' would have to switch places. (Also, C' wouldn't line up either.) Eliminate Choice (1) and (2).

Neither A nor B are on the x-axis or y-axis, but A' and B' are. That means that the rotation cannot be centered on the origin. Eliminate Choice (4).

Point A is at (6,2) which is (1+5,1+1). Point A' is at (-4,0) which is (1-5,1-1). Both points are equidistant from the center of rotation at (1,1). If you check B and B', and C and C', you will get the same results. Choice (3) is the answer.

6. Right triangle TMR is a scalene triangle with the right angle at M. Which equation is true?

(1) sin M = cos T
(2) sin R = cos R
(3) sin T = cos R
(4) sin T = cos M

Answer: (3) sin T = cos R

You don't want the sin or cos of a right angle. (They're 1 and 0, by the way.) So eliminate Choices (1) and (4).

In any right triangle, the sin of one acute angle is equal to the cos of the other (complementary) acute angle. So the correct answer is Choice (3).

Choice (2) is only true in a right isosceles triangle where both acute angles are 45 degrees. The question states that it is a scalene triangle, so Choice (2) is incorrect.

7. In the diagram below of △AED and line segment ABCD, AE ≅ DE.

Which statement is always true?

(1) EB ≅ EC
(2) AC ≅ DB
(3) ∠EBA ≅ ∠ECD
(4) ∠EAC ≅ ∠EDB

Answer: (4) ∠EAC ≅ ∠EDB

If you ignore the part about line segment ABCD, the only information you are given tells you that this is an isosceles triangle. Noting that point B and C are on side AD doesn't actually tell you anything. We can deduce nothing about the lengths of the BE or CE or about the size of the angles these lnes create.

However, we know that angle A and angle D must be congruent because they are the base angles of the isosceles triangle. Choice (4) names these angle EAC and EDB. Many names could have been used.

Note that if Choice (1) had been true, then Choices (2) and (3) would have to be true as well. This isn't allowed in an exam, so we know that all three must be incorrect.

8. As shown in the diagram below, right triangle ABC has side lengths of 8 and 15.

If the triangle is continuously rotated about AC, the resulting figure will be

(1) a right cone with a radius of 15 and a height of 8
(2) a right cone with a radius of 8 and a height of 15
(3) a right cylinder with a radius of 15 and a height of 8
(4) a right cylinder with a radius of 8 and a height of 15

Answer: (1) a right cone with a radius of 15 and a height of 8

Imagine this triangle is taped to a pencil or a straw. If you spin the pencil, the area of space that the triangle will pass through will be cone-shaped, not cylinder-shaped. Eliminate Choices (3) and (4).

Rotating around AC means that AC is where the triangle is taped to the pencil. The height of the cone will be the 8 and the radius is 15. This is Choice (1).

Be careful with questions of this type! Sometimes they will list a diameter instead of a radius.

More to come. Comments and questions welcome.

More Regents problems.