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 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, use a compass and straightedge to construct the altitude from C to AB.

[Leave all construction marks.]

Answer:

An altitude is a perpendicular line, and the way to create one is to make a perpendicular bisector. But first we need a segment with two endpoints to bisect. Do NOT use A and B.

The first step is to put the compass on point C. Open it wide enough to swing past line AB in two places. These two points will be used for your perpendicular bisector.

Since we already have point C, you only have to make arcs below the line to find another point. If you make them above the line, that is okay, because they will line up with point C.

Draw a line from point C to the point below the line you just found. That line will contain the altitude of the triangle.

Look at this image. I didn’t use my own because “faking” construction in MS Paint is not an easy task, and it sometimes doesn’t come out just right.

26. Triangles ABC and DEF are graphed on the set of axes below.

Describe a sequence of transformations that maps △ABC onto △DEF.

Describe a sequence of transformations that maps △ABC onto △DEF.

Answer:

This was an odd question because a “sequence” is not needed. It is one simple rotation. No translations, reflections or dilations needed afterward.

A rotation of 90 degrees clockwise about the origin maps △ABC onto △DEF.

You need to say rotation, how much, and about which center to get both credits.

You also could’ve said, for instance, a rotation of 90 degrees clockwise (or 270 CCW) around point A or point C, and then listed the required translation to move onto DEF.

27. Line segment PQ has endpoints P(–5,1) and Q(5,6), and point R is on PQ. Determine and state the coordinates of R, such that PR:RQ = 2:3.

[The use of the set of axes below is optional.]

Answer:

If the ratio is 2:3 that PR is 2/5 of the length of line segment and RQ is 3/5.

Find the difference between the two x-coordinates. Multiply that number by 2/5. Then add the result to -5. Do the same for the two y-coordinates. Find the differnce. Multiply by 2/5, but add the result to 1, because P is at (-5,1).

5 – -5 = 10; 2/5(10) = 4. -5 + 4 = -1. The x-coordinate of R is -1. (If you plot the line, you can see what the y value will be.)

6 – 1 = 5. 2/5(5) = 2. 1 + 2 = 3. The y-coordinate of R is 3. The location of R is (-1.3)

If you plot P and Q and draw the line with a straightedge, you will see that the line goes through the point (-1,3).

28. A circle has a radius of 6.4 inches. Determine and state, to the nearest square inch, the area of a sector whose arc measures 80°.

Answer:

The area of a circle is πr2. The area of a sector is the area of the full circle times the measure of the central angle divided by 360 degrees.

A = (80/360)π(6.4)2 = 28.5955…

The area is approximately 29 square inches.

29. A large snowman is made of three spherical snowballs with radii of 1 foot, 2 feet, and 3 feet, respectively. Determine and state the amount of snow, in cubic feet, that is used to make the snowman.

[Leave your answer in terms of p.]

Answer:

The formula for the Volume of a sphere is 4/3 π r3. You will need to use the formula 3 times and then add the results together.

Do NOT use the value of pi or press the pi key on your calculator when calculating.

V = 4/3 π (1)3 + 4/3 π (1)3 + 4/3 π (1)3

V = 4/3 π ( (1)3 + (2)3 + (3)3)

V = 4/3 π (1 + 8 + 27)

V = 4/3 π (36)

V = 48 π

Interesting note: 1 + 2 + 3 = 6, and (1)3 + (2)3 + (3)3 = 62

30.

In the diagram below of right triangle ACB, altitude CD is drawn to hypotenuse AB, AD = 2 and AC = 6.

Determine and state the length of AB.

Answer:

Some students look at this and see the Right Triangle Altitude Theorem, except we don’t immediately know the altitude. You can find it with Pythagorean Theorem, but remember to leave the answer as a radical. Do NOT round it, because an error will creep into your final answer.

The large right triangle is divided into two smaller right triangles. All three triangles are similar, and their corresponding angles are congruent. This means that their corresponding sides are proportional.

In triangle ACD, we know the hypotenuse and the short leg.

In triangle ABC, we know the short leg and we are looking for the hypotenuse.

Set up a proportion:

Short/hyp = short/hyp>

2 / 6 = 6 / AB

2 AB = 36

AB = 18

AB has a length of 18.

If you used the Pythagorean Theorem, you would’ve found that CD has a length of √(32). Then DB would equal 16. Finally, AB = AD + DB = 2 + 16 = 18. Same result.

31. Triangle RST has vertices with coordinates R(-3,-2), S(3,2) and T(4,-4). Determine and state an equation of the line parallel to RT that passes through point S. [The use of the set of axes below is optional.]

Answer:

You can graph it but it isn’t likely to help, except to tell you that between what two numbers the y-intercept is, but you can’t read it precisely because it’s a fraction.

First, find the slope of RT. Then use that slope and point S to come up with the equation of the line. If you remember Point-Slope Form, it’s not to difficult. If you use Slope-Intercept Form, it’s a little trickier because of the fractions. Use fractions. Don’t round decimals!

The slope of RT = (-4 – -2) / (4 – -3) = -2/7, which isn’t a very nice fraction.

Point-Slope Form, using S: y – 2 = -2/7(x – 3)

Slope-Intercept Form:

y = mx + b

2 = (-2/7)(3) + b

2 = -6/7 + b

20/7 = b

y = -2/7 x + 20/7

20/7 could be written as a mixed number, 2 6/7, but not as a decimal because it’s infinite.

End of Part II

How did you do?

Questions, comments and corrections welcome.

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