**New York State Geometry Regents exam**. Since I typed this up as quickly as I could, there are no images, graphs or diagrams included. I'll edit them as I am able to.

A common theme: there were several questions involving the **equation of a circle**, as there always are, as well as several questions which not only involved the **Pythagorean Theorem**, but very specifically a **3-4-5** triangle or a multiple of it.

Link to Part I.

I hope everyone did well. Here is are the (most of the) open-ended questions. Again, sorry, no images.

Part II

**29. ** The Proof. Image to be uploaded later.

Very simply, you are given two pairs of congruent sides. The included angles,
BCA and DCE are congruent because they are vertical angles. Therefore ABC is
congruent to DEC by SAS.

**30. **
Using a compass and straightedge, construct the perpendicular bisector of side
AR in triangle ART show below. [Leave all construction marks.] (image omitted)

I can’t really show this, but keep in mind that point T is irrelevant. It is
NOT on the bisector. Ignore the triangle and focus on segment AR. Create the
perpendicular bisector as you would with any other segment.

**31. **
Determine and state the measure, in degrees, of an interior angle of a regular
decagon.

First,
even if you have this memorized, you still need to show your work.

Second,
there are two methods, both of which are acceptable.

Interior: (10 – 2) * 180 = 1440. 1440
/ 10 = 144 degrees for each interior angle.

Or
exterior: 360/10 = 36 degrees per exterior angle. Interior angle = 180 – 36 =
144 degrees.

**32. **
Write an equation of a line that is parallel to the line whose equation I s 3y
= x + 6 and that passes through the point (-3, 4).

Note
that it says “an equation” and doesn’t say that it has to be in any specific
form.

Easiest
method, plug in (-3, 4) into the equation and find the new constant. (I won’t
call it “b”.)

3(4) =
-3 + c

12 = -3
+ c

15 = c

3y = x
+ 15 is an acceptable answer

In
slope-intercept form, it is y = 1/3x + 5.

**33.** In
the diagram below, secants PQR and PST are drawn to a circle from point P.
(image omitted)

If PR =
24, PQ = 6, and PS = 8, determine and state the length of PT.

Not a
trick question. No extra adding or subtracting.

(PQ)(PR)
= (PS)(PT)

(6)(24)
= (8)(x)

144 =
8x

18 = x

PT is
18.

**34. **The
slope of QR is (x -1)/4 and the slope of ST is 8/3. If QR is perpendicular to
ST, determine and state the value of x.

If QR
is perpendicular to ST, then the slope of QR is -3/8. So

(x-1)/4 = -3/8

8(x-1)
= -12

8x – 8
= -12

8x = -4

X =
-1/2

**Part
III**

35.
Quadrilateral HYPE have vertices H(2,3), Y(1, 7),
P(-2, 7) and E(-2, 4). State and label the coordinates of the vertices of
H’’Y’’P’’E’’ after the composition of transformations r _{x-axis} o T _{5,
-3}.

Remember:
the translation comes before the reflection. If you do it the other way around,
you will lose half-credit (2 points). You MUST label the coordinates of the final
points. Keep this in mind if you are using the graph!

H(2, 3)
> H’(7, 0) > H’’(7, 0)

Y(1, 7)
> Y’(6, 4) > Y’’(6, -4)

P(-2, 7)
> P’(3, 4) > P’’(3, -4)

E(-2, 4)
> E’(3, 1) > E’’(3, -1)

**36. ** On
the set of axes below, graph two horizontal lines whose y-intercepts are (0,
-2) and (0, 6), respectively.

Graph
the locus of points equidistant from these horizontal lines.

Graph
the locus of points 3 units from the y-axis.

State
the coordinates of the points that satisfy both loci.

The
first locus is the horizontal line y = 2.

The
second locus is comprised of the vertical lines x = -3 and x = 3.

The
coordinates of the points that satisfy both conditions are (-3, 2) and (3, 2).

**37.** In
the diagram below, a right circular cone with a radius of 3 inches has a slant
height of 5 inches and a right cylinder with a radius of 4 inches has a height
of 6 inches.

(image omitted)

Determine
and state the number of full cones of water needed to completely fill the
cylinder with water.

You
need to know the Volume of each solid. You can leave them in terms of pi.

First,
you need to know the height of the cone – you were given the *slant height*.

Notice
that *one more time* the problem
involves a 3-4-5 right triangle. Satisfy yourself however you have to that the
height of the cone is 4 inches. (State
that it’s a right triangle with dimensions 3-4-5, using Pythagorean Theorem.)

The
cone has V = 1/3 (pi) r^{2 }h = 1/3 (pi) (9) (4) = 12 pi

The
cylinder has V = (pi) r^{2} h = (pi) (16)(6) =
96 pi.

Divide
96pi/12pi = 8. (Pi will cancel out.)

8 cones
will fill up the cylinder.

Part IV

**38.** In
the diagram below, right triangle RSU is inscribed in circle O, and UT is the
altitude drawn to hypotenuse RS. The length of RT is 16 more than the length of
TS and TU = 15.

Find
the length of TS.

Find,
in simplest radical form, the length of RU.

(Image
omitted)

You
need to use the Right Triangle Altitude Theorem, and Pythagorean Theorem. And
then simplify the radicals.

First
part: x / 15 = 15 / x + 16

Cross-multiply:
x^{2} + 16x = 225

So x^{2}
+ 16x – 225 = 0

(x +
25)(x – 9) = 0

X = -25
or x = 9.

Discard
the negative answer and TS = 9.

This
makes TR = 9 + 16 = 25.

RU =
the square root of (25^{2} + 15^{2}) = the square root of (625
+ 225) = the square root of (850).

The
largest perfect square that is a factor of 850 is 25, which goes into 850 34
times.

RU =
(25)^{.5}(34)^{.5} = 5(34)^{.5}. (That is, 5 times the square root of 34.)

Yes,
that was a crazy problem, but gettable if you worked your way through it.

I
double-checked that answer twice.

For
instance, you can find the length of US, which is the square root of 306. And
the square root of (306 + 850) = square root of 1156 = 34, which is the length
of the diameter of circle O. We know this because ST is 9 and TR is 25, and 9 +
25 = 34.

## 2 comments:

Interesting - when I took the geometry Regents in (about) 1953 there was no coordinate geometry. It was all Euclid.

It's changed a bit. It's been like this (in some form or other) since I started teaching, but it's not like when I was in school, either.

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