## Friday, July 05, 2024

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

This exam was adminstered in June 2024.

More Regents problems.

### Part I

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

1. In the diagram below, △BRI is the image of △JOE after a translation. Triangle CAT is the image of △BRI after a line reflection.

Which statement is always true?

(1) ∠R ≅ ∠T
(2) ∠J ≅ ∠A
(3) JE ≅ RI
(4) OE ≅ AT

Answer: (4) OE ≅ AT

You can follow the transformations, and if you wish to, you can mark up the book to show which points are moving to which new poisitions. Then you can see which angles are corresponding (and therefore congruent) and which sides are corresponding (and congruent).

But here's the little secret for state exams. Keep in mind that this doesn't always work in your classroom because teachers are sloppy (and I can be guilty of this myself sometimes if I'm not paying attention or if I'm just tired or overworked). The order of the letter is the triangle name is important. They will always correspond on state exams, and they should in textbook examples. (Textbook writers aren't always 100% at the top of their game.).

So B goes to J goes to C, R goes to O goes to A, and I goes to E goes to T.

You can eliminate Choices (1) and (2) immediately. Careful checking shows that Choice (3) is wrong and Choice (4) is the correct answer because JOE goes to CAT.

2. right cylinder is cut parallel to its base. The shape of this cross section is a

(1) cone
(2) circle
(3) triangle
(4) rectangle

If it is cut parallel to the base, then the cross-section has to be a circle, which is the same shape as the base of the cylinder. The correct answer is Choice (2).

A rectangle could occur if the cross section was perpendicular to the base.

A triangle would not happen, no matter how you slice it. And a cross section is a 2-D shape, so it could not be a cone, which is 3-D.

3. What is the minimum number of degrees that a regular hexagon must rotate about its center to carry it onto itself?

(1) 45°
(2) 72°
(3) 60°
(4) 120°

Any figure will carry onto itself if you rotate it 360 degrees. Because regular polygons have symmetry, you only have to rotate them a minimum of 360 divided by the number of sides that the polygon has, which in this case is 6.

360 / 6 = 60 degrees, so the correct answer is Choise (3).

120 degrees is the size of each internal angle, but that's not the question that's being asked.

4. In the diagram below, a sphere is inscribed inside a cube. The cube has edge lengths of 18.

What is the volume of the sphere, in terms of π?

(1) 108π
(2) 432π
(3) 972π
(4) 7776π

Use the formula for the volume of a sphere. The length of the box is the diameter of the ball. You need to cut that in half to get the radius.

Note that one of the wrong choices will always be the result if a student uses the diameter instead of the radius. They expect some students to do this. As a result, you can be fairly certain that the largest choise will not be the correct answer.

V = 4/3 π r3 = 4/3 π (9)3 = 972 π, which is Choice (3).

5. In the diagram below, EM intersects HA at J, EA ⟂ HA, and EM ⟂ HM.
If EA = 7, EJ = 9, AJ = 5.4, and HM = 3.29, what is the length of MJ, to the nearest hundredth?

(1) 2.47
(2) 2.63
(3) 4.11
(4) 4.39

You can prove that the two traingles are similar in shape and therefore their corresponding sides are proportional. You don't have to prove anything though because if this wasn't true, there wouldn't be any way to solve this question. So in this particular problem if you just assumed that they were similar, you would be okay. Because they are.

Briefly, each triangle has a right angle, which are congruent to each other, and the vertical angles are congruent. Therefore, by AA, the triangles must be similar. (They are not congruent.)

Write down the lengths that you know and put an x next to the side you're looking for, MJ.

Comparing sides, you will see that

5.4 / x = 7.2 / 3.29

so 7.2 x = (5.4)(3.29)

and x = (5.4)(3.29)/7.2 = 2.4675

To the nearest hundredth, MJ = 2.47, which is Choice (1).

6. Which equation represents the line that passes through the point (2,-7) and is perpendicular to the line whose equation is y = 3/4x + 4?

(1) y + 7 = 3/4(x - 2)
(2) y - 7 = 3/4(x + 2)
(3) y + 7 = -4/3(x - 2)
(4) y - 7 = -4/3(x + 2)

Answer: (3) y + 7 = -4/3(x - 2)

Parallel lines have the same slope. Perpendicular lines have slopes that are NOT the same but are inverse reciprocals -- the product of the slopes is -1.

Since they want perpendicular, you can cross out Choices (1) and (2).

The question uses slope-intercept form, which gives you the slopes and the y-intercept. The choices are given using point-slope form, which gives you the slope and a point on the line.

The thing to remember is that the formula has minus signs in it (as do many, many formulas in Algebra and Geometry): y - k = m(x - h).

You may have learned that formula as y - y0 = m(x - x0), which is the same thing. However, if you get used to seeing h and k, you'll notice that (h,k) is all over the place.

Since the point is (2,-7) and the signs are "flipped" in the equation, then the correct answer is Choice (3).

7. In △RHM below, m∠R = 110° and m∠M = 40°.

If △RHM os reflected over side HM to form quadrilateral RHR'M, which statement is always true?

(1) Quadrilateral RHR'M is a parallelogram.
(2) m∠MHR' = 40°
(3) m∠HMR' = 40°
(4) MR ≅ HR'

Answer: (3) m∠HMR' = 40°

You can make a sketch in the book underneath the original image and mark it up.

You can see that the correct answer is Choice (3) because HMR and HMR' are corresponding angles of congruent triangles.

The figure is not a parallelogram because the opposite sides are not parallel. It's a kite. Eliminate Choice (1).

The measure of angle MHR' is the same as angle MHR, both of which can be calculated to be 30 degrees. (180 - 110 - 40 = 30). Choice (2) is incorrect.

Since the figure is a kite and not a parallelogram, the opposite sides are not congruent, so MR is not congruent to HR'. This eliminates Choice (4).

8. The funnel shown below can be used to decorate cookies with melted chocolate. The funnel can be modeled by a cone whose radius is 6 cm and height is 13 cm.
The baker uses 2 cubic centimeters of chocolate to decorate each cookie. When the funnel is completely filled, what is the maximum number of cookies that can be decorated with the melted chocolate?

(1) 78
(2) 245
(3) 490
(4) 735

Find the volume of the funnel using the formula for the volume of a cone: V = 1/3 π r2 h.

V = (1/3) π (6)2 (13) = 490.08845396...

If the baker uses 2 cm3 per cookie, then divide the result by 2: 490 / 2 = 245.

The correct answer is Choice (2).

More to come. Comments and questions welcome.

More Regents problems.

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