## Saturday, February 10, 2024

### January 2024 Geometry Regents Part II

This exam was adminstered in January 2024 .

### January 2024 Geometry, Part II

Each correct answer is worth up to 2 credits. Partial credit can be given. Work must be shown or explained.

25. In isosceles triangle ABC shown below, AB ≅ AC, and altitude AD is drawn.

The length of AD is 12 cm and the length of BC is 10 cm.
Determine and state, to the nearest cubic centimeter, the volume of the solid formed by continuously rotating triangle ABC about AD.

Rotating triangle ABC around AD would create a cone with a height equal to AD (which is 12 cm) and a radius equal to BD (which is half of 10, or 5).

Use the formula for the Volume of a cylinder.

V = (1/3) π r2 h = (1/3) π (5)2 (12) = 314.159 ... = 314

Amusingly, 12 * 25 = 300, and 1/3 of 300 is 100, so the answer is 100 pi, rounded.

(Okay, that might only be amusing to me, but I'm amused and you may choose to be as well.

26. The diagram below models the projection of light from a lighthouse, L. The sector has a radius of 38 miles and spans 102°.

Determine and state the area of the sector, to the nearest square mile.

PLEASE, PLEASE, PLEASE!!! Make sure you round correctly. It hurts to see students do all the work to answer this question and then lose half of the points because they didn't round correctly.

You find the area of a sector in the same way that you find the area of an entire circle. However, in the same way that a cone is only 1/3 of a cylinder of the same dimensions, a sector is only a fraction of the entire circle, and that fraction will be the central angle divided by 360 degrees.

Use the formula for the Area of a Circle and multiply by the fraction.

A = (102/360) π r2 = (102/360) π (38)2 = 1285.33... = 1285 square miles.

27. A Segment CA is drawn below. Using a compass and straightedge, construct isosceles right triangle CAT where CA is perpendicular to CT and CA ≅ CT. [Leave all construction marks.]

Extend line CA past point C (to the left). Draw a semicircle with radius AC, mark the new point on line AC and call it point D.

Construct a perpendicular bisector of DA through point C. Put the compass and point D and open it wider than the length of DC. Make arcs above and below line DA. Move the compass to point A without changing the width of the compass. Make marks above and below DA that intersect the arcs you just make.

Draw the vertical line between the two intersections and through point C.

Label one of the intersections point T. Use a straightedge to draw CT and TA.

You have constructed isosceles triangle CAT. Done.

28. A On the set of axes below, congruent triangles ABC and DEF are graphed.

Describe a sequence of rigid motions that maps triangle ABC onto triangle DEF.

This is a "mean" question for two reasons. First, there is no reason to write a "sequence" because it can be done in one rotation. Second, it's set up in such a way to encourage a student to map ABC onto FED instead of DEF. Yes, that matters.

A rotation of 90 degrees counterclockwise around the origin would map ABC onto DEF.

A reflection over the y-axis and a translation of +1,+1 would map to the FED and receive half credit.

Under the new curriculum in NYS, you could also state, "Translate triangle ABC along ray AD until A comes to D. Then rotate ABC until point B coincides with point E."
(I'm not kidding.)

29. In triangle ADC below, EB is drawn such that AB = 4.1, AE = 5.6, BC = 8.22, and ED = 3.42.
Is triangle ABE similar to trianlge ADC? Explain why.

There's a little bit of a hint when they didn't state "or why not".

Set up a proportion of corresponding sides. Remember to add the lines segments to get the lengths of AD and AC. We don't know anything about the lengths of BE or CD, but we do know that both triangles share angle A, which, of course, is congruent to itself because of the Reflexive Property.

Is 4.1 / (5.6 + 3.42) = 5.6 / (4.1 + 8.22) ?

Dividing, we get 0.4545... = 0.4545...

Or by cross-multiplying:

(5.6 + 3.42) (5.6) = 50.512; (4.1) (4.1 + 8.22) = 50.512

Since the corresponding sides are proportional, and because the two triangles share angle A, the triangles are similar by SAS.

If you didn't show the sides were propotional or you didn't mention angle A and SAS, you couldn't receive full credit.

30. Determine and state the coordinates of the center and the length of the radius of the circle represented by the equation x2 + 16x + y2 + 12y - 44 = 0.

You have to complete the squares for both the x terms and the y terms to get the equation into the form (x - h)2 + (y - k)2 = r2, where (h,k) is the center of the circle and r is the radius of the circle. (Not r2 -- don't make that mistake!)

Has of 16 is 8, and 8 squared is 64. Half of 12 is 6, and 6 squared is 36. Add 64 and 36 to both sides of the equation. This means that (x + 8) and (y + 6) will both be in the final answer.

x2 + 16x + y2 + 12y - 44 = 0

x2 + 16x + 64 + y2 + 12y + 36 - 44 = 64 + 36

x2 + 16x + 64 + y2 + 12y + 36 = 100 + 44

(x + 8)2 + (y + 6)2 = 144

(x + 8)2 + (y + 6)2 = 122

The center of the circle is (-8,-6), and the radius is 12.

Both forget to "flip the signs" of the coordinates and to take the square root to find the radius.

31. In the diagram below, traingle SBC ~ triangle CMJ and cos J = 3/5.
Determine and state m∠S, to the nearest degree.

This is another question that I do not like BECAUSE they hide important information. Since they stated that △SBC ~ △CMJ, then ∠S corresponds to ∠C, not ∠J.

This is necessary information. The problem is that in my years of teaching, many educators are a little lax in their naming conventions when stating the order of the vertices in the triangles. This could confuse many students. What's more since triangle SBC is the same as triangle BCS and triangle CBS, it isn't WRONG, per se, to write the letters in the incorrect order because it's still true.

So the entire question hinges on convention and whether or not the instructor followed it.

If you know enough about right triangles, you know enough to remember that if cos J = 3/5, then sin J = 4/5, and sin JCM = 3/5 and cos JCM = 4/5, because it is a multiple of a 3-4-5 triangle.

You can use trig ratios to find angles or Pythogean Theorem but that's a simple fact.

Since angle S correspond to angle C, then sin S = 3/5, and S = sin -1 (3/5) = 37 degrees.

End of Part II

How did you do?

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