## Monday, July 01, 2024

### June 2024 Geometry Regents Part II

This exam was adminstered in June 2024 .

### June 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 △ABC below, m∠C = 90o, AC = 11, and AB = 18.

Determine and state the measure of angle A, to the nearest degree.

It's a right triangle. For angle A, you have the adjacent side and the hypotenuse. This means that you can find the size of the angle using the cosine ratio.

Cos A = 11/18

A = cos-1 (11/18) = 52.3

Angle A is 52o to the nearest degree. Make sure the calculator is in DEGREE Mode.

26. Use a compass and straightedge to construct an equilateral triangle inscribed in circle A below. [Leave all construction marks.]

There are multiple methods that can be used. If you recall that a hexagon inscribed in a circle can be split up into 6 congruent equilateral triangles.

[IMAGE OMITTED]

Here's a simple solution, and it really is simple.

Use your straightedge to draw a diameter (or a radius). Use the compass to measure the distance from the point on the circle to the center of the circle. Make arcs that cross the circle above and below. Move the compass to the new points and make another arc and then another. When you are finished, you will have six points on the circle.

Use the straightedge to draw three lines connecting every other point. The result will be an equilateral triangle.

27. Quadirlateral DEAR and its image, quadrilateral D'E'A'R', are graphed on the set on axes below.

Described a sequence of transformations that maps quadrilateral DEAR onto quadrilateral D'E'A'R'.

There are several possibilities but each of them will involve a reflection and a translation. The reflection must include the line of reflection and the translation must state how far is each direction.

Note that under the new curriculum, you could draw a line, say EE', and say something like "translate DEAR along directed line segement EE'". I won't do that here. I'm just saying that it is valid.

One valid answer is to translate DEAR two spaces to the left and seven spaces down, followed by a reflection over the y-axis.

Or you could reflect over the y-axis and then translate two spaces to the right and seven spaces down.

28. In circle P below, tangent AL and secant AKE are drawn.

If AK = 12 and KE = 36, determine and state the length of AL.

This was a little bit of an "irrelevant" problem. By this, I mean, when grading these problems, the scorer is instructed to give 1 credit to a correct answer without any work at all. However, a correct answer that was found through an obviously incorrect, incoherent or irrelevant manner is supposed to be scored with 0 credits.

In this problem, if you subtracted 36 - 12 = 24, you got the correct answer through an incorrect method. You shouldn't have gotten any points. In reality, scorers are probably looking for excuses to give you points if you did something so desparate.

The correct method involves using the following formula: (AL)2 = (AK)(AE)

x2 = (12)(12 + 36)
x2 = 576
x = 24

AL has a length of 24.

29. The equation of a circle is x2 + y2 + 8x - 6y + 7 = 0. Determine and state the coordinates of the center and the length of the radius of the circle.

You need to Complete the Square, twice. Note that you can find the center without completing the square, but you need to complete them to find the radius.

The formula for a circle is (x - h)2 + (y - k)2 = r2, where (h,k) is the center and r is the radius.

Reminder: (x + 5)2 is x2 + 10x + 25. So if I had x2 + 10x, I could factor the expression if I add 25 to it first.

x2 + y2 + 8x - 6y + 7 = 0
x2 + 8x + y2 - 6y = -7
x2 + 8x + 16 + y2 - 6y + 9 = -7 + 16 + 9
(x + 4)2 + (y - 3)2 = 18

The center of the circle is (-4,3) and the radius is &SQRT;(18).

30. On the set of axes below, △ABC is drawn with vertices that have coordinates A(2,-3), B(4,5) and c(-5,1)

Determine and state the area of △ABC.

This problem is deceptively simple. Draw a rectangle about the triangle, so that that C and A are points on the rectangle and B is a vertex. The rectangle has a length of 9 and a width of 8, and an Area of (9)(8) = 72.

This creates three triangles. Find the areas of the three triangles and subtract those from 72.

1/2(4)(7) = 14, 1/2(4)(9) = 18, 1/2(2)(8) = 8.

72 - 14 - 18 - 8 = 32.

The area of ABC is 32.

31. In the diagram below, AE = 15, EB = 27, AF = 20, and FC = 36.

Explain why EF || BC.

If the proportion AE/AB = AF/AC is true, then the two triangles are similar by SAS since they share angle A. This means that corresponding angles AEF and AEB are congruent. That makes the EF || BC.

Check: 15/27 = 5/9, and 20/36 = 5/9

Therefore, AE/AB = AF/AC, so the triangles are similar and the lines are parallel.

You could also have used the side-splitter theorem, if you remembered that one. It states that if a line is parallel to one side of a triangle and intersects the other two sides, then it divides those sides proportionally. Same idea, but you don't need to mention angle A or SAS.

You didn't need to write a complete proof. If you did write a complete proof, any mistake could cost 1 of the 2 points.

End of Part II

How did you do?

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