*The following are the questions and answers (and commentary) for part of the New York State Algebra Regents exam. If you have any questions or comments (or corrections), please add them in the*

**Comments**section.My apologies for typos, particularly if they are in the questions, because then the answers are subject to change.

### June 2017, Geometry (Common Core), Part II

**25.*** Given: Trapezoid JKLM with JK || ML
Using a compass and straightedge, construct the altitude from vertex J to ML. {Leave all construction marks.]
*

There are several methods that work. One of them is to start at point J. Make an arc that cuts across LM in two places. You are allowed to use the straightedge to extend LM, if necessary. (If your arc goes outside the trapezoid, you MUST extend the line.)

From each of the points where the arc intersects LM, make an arc. At the two points of intersection, use the straightedge to draw the altitude, which will intersect point J.

Note: you only need the arcs to intersect once because you already have J.

**26.***Determine and state, in terms of pi, the area of a sector that intercepts a 40 ^{o} arc of a circle with a radius of 4.5.
*

The area of a sector of a circle is the area of the circle times the measure of the central angle divided by 360^{o}. (In other words, you multiply the area by the fraction of the circle represented.)

A = (40/360) * pi * (4.5)^{2}

A = 2.25 pi or A = 9/4 pi.

You could also have found the area by formula A = ½ r^{2} ~~O~~, where ~~O~~ is the central angle, measured in **radians**.

40^{o} = 40 (pi ) / (180) = 2 pi / 9

A = ½ (4.5)^{2} (2 pi / 9) = (1/2) (20.25) (2 / 9) pi = 2.25 pi, which is the same answer.

**27. ***The diagram below shows two figures. Figure A is a right triangular prism and figure B is an oblique triangular prism. The base of figure A has a height of 5 and a length of 8 and the height of prism A is 14. The base of figure B has a height of 8 and a length of 5 and the height of prism B is 14.
*

*Use Cavalieri’s Principle to explain why the volumes of these two triangular prisms are equal.
[Image omitted.]
*

First, yes, you needed to refer or appeal to the principle, even if you didn’t use the name.

Second, you had to be very specific about the language, particularly using words like base, length, and height.

Third, the triangle bases are NOT congruent, even if the area of the Bases is the same.

Fourth, the oblique prism has a height of 14. This is not the slant height, or the length of the sides. In other words, the surface areas are different, the prisms are not congruent, etc.

Those were some misunderstandings I came across speaking to students.

What was necessary to state was this: the areas of the base were equal and the heights of the prisms are the same, therefore the Volumes must be equal.

You could have mentioned that the area of the cross sections will be equal at every level, but it would not have been necessary.

You could have found the area of the triangles and the volumes of the prisms, but that was not necessary. And be careful if you calculate them incorrectly. (For instance, leaving out the ½, or using 1/3, which is for pyramids. A triangular prism is not a pyramid!)

**28. ***When volleyballs are purchased, they are not fully inflated. A partially inflated volleyball can be modeled by a sphere whose volume is approximately 180 in ^{3}. After being fully inflated, its volume is approximately 294 in^{3}. To the nearest tenth of an inch, how much does the radius increase when the volleyball is fully inflated?
*

You need to use the formula V = (4 / 3) (pi) (r^{3}) for each volume, and solve for r.

Then subtract the two radii to find the increase. Do NOT find a ratio.

This was a lot of work for only 2 points, with plenty of places for an error to sneak in, but there was nothing “tricky” about it. It was just a lot of steps.

Look at the image below. The increase is 0.6 inches.

**29. ***In right triangle ABC shown below (image omitted), altitude CD is drawn to hypotenuse AB.
Explain why triangle ABC ~ triangle ACD.
*

This is an explanation, not a proof. You need to have reasons, back up what you write, but you don’t need to be so formal.

Simplest answer: If you said it was true because of the **Right Triangle Altitude Theorem** (and stated what that says), that was sufficient. You didn’t have to prove it. You already know it’s a theorem. (You don’t have to prove Pythagorean Theorem every time you use it, right?)

Otherwise, you can prove it using AA, or AAA, but you didn’t need the third angle. If you do this, you need three things for two points: two pairs of congruent angles and a statement that the are similar because of AA.

Angle CDA is a right angle because of the altitude. Angle ACB is a right angle of the given triangle. Angle A is in both triangles. The two triangles have two pairs of angles that have the same measure so they are similar by AA.

**30. ***Triangle ABC and triangle DEF are drawn below. (image omitted)
If AB = DE, AC = DF, and <A = <D, write a sequence of transformations that maps triangle ABC onto triangle DEF.
*

You can see that there needs to be a rotation and a translation. That answer isn’t good enough.

Consider this: if there was a coordinate plane, you would have been expected to give amounts, directions, etc. This is true here as well. You can do it using rigid motions.

Translate triangle ABC along vector CF, mapping C to point F. Then rotate ABC around point C until point A is mapped onto point D.

**31. ***Line n is represented by the equation 3x + 4y = 20. Determine and state the equation of line p, the image of line n, after a dilation of scale factor 1/3, centered at the point (4, 2).
[The use of the set of axes below is optional.]
*

*Explain your answer.
*

If you substitute (4, 2) into 3x + 4y = 20, you get 3(4) + 4(2) = 20, 12 + 8 = 20, 20 = 20.

Therefore, (4, 2) is a point on line n. (You also would have noticed this if you graphed the line.)

The dilation of a line centered at a point on the line will not affect the line at all. (One third of the infinite length is infinity. One third of its 0 width is zero. One third of its 0 distance from the center is still zero.)

So the equation for p is 3x + 4y = 20.

If you rewrote it in slope-intercept form for some reason, you would have y = -3/4 x + 5

End of **Part II**

How did you do?

Comments, questions, corrections and concerns are all welcome.

Typos happen.

## 5 comments:

I could always re-take the regents, no worries. Those questions really stumped me. I'm disappointed

Other than explaining Cavalieri's Principle,, you'll see more question of similar types.

There's a lot of similarities among the tests.

My daughter got an 84 is that good?

My daughter got an 84 is that good?

That's not for me to say. I consider anything off 80 to be very good, but sometimes students have higher expectations for themselves.

It's nothing to be unhappy about, and she can always take it again if she wants to try for a higher score. Only the highest score will count.

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