June 2023 Geometry Regents Part III

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.

June 2023 Geometry, Part III

Each correct answer is worth up to 4 credits. Partial credit can be given. Work must be shown or explained.

31. Cape Canaveral, Florida is where NASA launches rockets into space. As modeled in the diagram below, a person views the launch of a rocket from observation area A, 3280 feet away from launch pad B. After launch, the rocket was sighted at C with an angle of elevation of 15^o. The rocket was later sighted at D with an angle of elevation of 31^o.

Determine and state, to the nearest foot, the distance the rocket traveled between the two sightings, C and D.

Answer:

To find the length of CD, you need to find the length of BD and BC and then subtract BD - BC. In both cases, you have the angle and the adjacent side and you are looking for the opposite side, so you want to use tangent ratios.

tan 15 = x/3280, so x = 3280 * tan 15 = 878.873..

tan 31 = y/3280, so y = 3280 * tan 31 = 1970.822..

CD = 1970.822 - 878.873 = 1091.949 = 1092 feet.

33. TA small can of soup is a right circular cylinder with a base diameter of 7 cm and a height of 9 cm. A large container is also a right circular cylinder with a base diameter of 9 cm and a height of 13 cm. Determine and state the volume of the small can and the volume of the large container to the nearest cubic centimeter.

What is the minimum number of small cans that must be opened to fill the large container? Justify your answer.

Answer:

I can't tell you (because even if I remembered, I couldn't disclose it) the number of students who missed up the second part of the question.

The question gives diameters, but the formula uses radius. Halve each number.

V(sm) = π (3.5)²(9) = 346.36... = 346 cm³

V(lg) = π (4.5)²(13) = 827.024... = 827 cm³

For the second part, divide the volume of the larger can by the volume of the smaller can. And then round UP. If you don't round up, you won't fill the larger can.

827/346 = 2.39..., so you would need to open 3 cans.

34. Parallelogram MATH has vertices M(-7,-2), A(0,4), T(9,2), and H(2,-4).
Prove that parallelogram MATH is a rhombus.
[The use of the set of axes below is optional.]

Determine and state the area of MATH.

Answer:

You probably want to plot the points just to make it easier to visualize, but it isn't necessary.

You must either show that the four sides have the same length, or you must show that the diagonals are perpendicular and bisect each other. It is not enough to show that the diagonals are perpendicular because the diagonals of a kite are perpendicular also.

MA: √( (-7 - 0)² + (-2 - 4)² ) = √ (85)

AT: √( (0 - 9)² + (4 - 2)² ) = √ (85)

TH: √( (9 - 2)² + (2 - -4)² ) = √ (85)

MA: √( (2 - -7)² + (-4 - -2)² ) = √ (85)

All four sides are congruent, so MATH is a rhombus. This concluding statment is required.

The easiest way to find the area of the rhombus is to draw a rectangle around it that has points M, A, T, and H on each of its sides. Four triangles will be formed. Find the area of the rectangle and subtract the areas of the four triangles.

You have the numbers that you need -- you found the rise and run when you did the distance formula!

The rectangle has an area of 16 * 8 = 128. The four triangles have areas of 1/2(6)(7) = 21, 1/2(9)(2) = 9, 1/2(6)(7) = 21, and 1/2(9)(2) = 9.

So 128 - (21 + 9 + 21 + 9) = 68

You could also have found the lengths of the two diagonals and multiplied 1/2 d₁d₂.

End of Part III

How did you do?

Questions, comments and corrections welcome.

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