Wednesday, February 22, 2023

January 2023 Geometry Regents, Part II


This exam was adminstered in January 2023. These answers were not posted until they were unlocked on the NY Regents website or were posted elsewhere on the web.

January 2023 Geometry, Part II

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


25. Using a compass and straightedge, construct the angle bisector of /ABC. [Leave all construction marks.]

Answer:
Angle cisectors are simple. The first thing you want to do is make sure that you start at point B.

To bisect an angle, put the compass at point B and make an arc that crosses AB and BC. Then without changing the size of the compass, go to the intersections on AB and BC and make another arc in the middle of the triangle. The point where these two arcs intersect will lie on the angle bisector. Use your straightedge to draw the bisector from B through this line.




26. On the set of axes below, ∆ABC and ∆DEF are graphed.

Describe a sequence of rigid motions that would map ∆ABC onto ∆DEF.

Answer:
A rotation of 90 degrees clockwise centered on the origin, followed by a translation of 1 unit to the right and four units down would map ABC onto DEF.

You don't need to write an equation, but you could. If you did, make sure you write it in the correct order.

There are other possibilities, depending upon where you center the rotation. You MUST mention the center of the rotation. You can ASSUME it's the origin. The person scoring you will not.




27. As shown in the diagram below, a symmetrical roof frame rises 4 feet above a house and has a width of 24 feet.

Determine and state, to the nearest degree, the angle of elevation of the roof frame.

Answer:
Draw the altitude of the triangle. Now you have two right triangles with legs of 4 and 12.

The angle of elevation is the bottom angle and it can be found using the tangent ratio.

tan x = 4/12, so x = tan-1(4/12) = 18.4, which is 18 degrees.




28. Directed line segment AB has endpoints whose coordinates are A(-2,5) and B(8,-1). Determine and state the coordinates of P, the point which divides the segment in the ratio 3:2. [The use of the set of axes below is optional.]

Answer:
Segmenting (or partitioning) AB into a ratio of 3:2 means that you need two cut it into five pieces. You want AP to be 3 of those pieces long and PB to be 2 of them.

(8 - -2)/5 = 2 and (-1 - 5)/5 = -6/5 = -1.2

2 * 3 = 6 and -1.2 * 3 = -3.6

P is at (-2 + 6, 5 - 3.6), which (4, 1.4).

Don't let the decimal throw you off. Decimals get used in real life, too.




29. In ∆ABC, AB = 5 5, AC 5 12, and m∠A = 90°. In ∆DEF, m∠D = 90°, DF = 12, and EF = 13. Brett claims ∆ABC ≅ ∆DEF and ∆ABC ~ ∆DEF. Is Brett correct? Explain why.

Answer:
Both triangles are right triangles. Both triangles have two sides listed. You can use Pythagorean Theorem to show that both triangles are 5-12-13 triangles. Or you can pretty much just state that because they are common triangles, used all the time.

Since the two triangles are have congruent corresponding sides then by SSS the triangles are congruent. If two triangles are congruent, they must also be similar.




30. The volume of a triangular prism is 70 in3. The base of the prism is a right triangle with one leg whose measure is 5 inches. If the height of the prism is 4 inches, determine and state the length, in inches, of the other leg of the triangle.

Answer:
If the prism has a height of 4 and a volume of 70, then the area of the base is 70/4 = 17.5 in2.

The area of a triangle = 1/2 (5) b = 17.5


So b = 17.5 (2) (1/5) = 7


31. Triangle ABC with coordinates A(-2,5), B(4,2), and C(-8,-1) is graphed on the set of axes below.
Determine and state the area of ∆ABC.

Answer:
There isn't a shortcut. The best method is to make a rectangle around the triangle and calculate the negative space and subtract it from the area of the rectangle until only the triangle remains.

Make rectangle CDEF so that A is on DE and B is on EF.

Area of CDA = 1/2(6)(6) = 18. Area of AEB = 1/2(6)(3) = 9. Area of BFC = 1/2(12)(3)= 18.

Area of CDEF = (6)(12) = 72, and 72 - 18 - 9 - 18 = 27

End of Part II

How did you do?

Questions, comments and corrections welcome.

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