Seems that I never got around to posting Parts 2, 3, and 4 of the non-Common Core Geometry test. Sorry for the delay.
Here are the questions, with answers and explanations, for the **New York State Geometry Regents** (not Common Core) exam, Part 2. There were 6 questions, each worth 2 credits. Partial credit may be earned for correct work on a problem without a solution, or for a problem with a solution that contains **one** computational or conceptual error. All work must be shown. In general, a correct answer without any work is worth 1 credit, unless that answer is given as a choice and an explanation is required.

Link to Part 1

### Part 2

**29. ** *The image of after a reflection through the origin is R'S'. If the coordinates of the endpoints of are R(2,-3) and S(5,1), state and label the coordinates of R' and S '.*[The use of the set of axes below is optional.]

In a reflection about the origin, P(x, y) -> P'(-x, -y)

R(2, -3) -> R'(-2, 3)

S(5, 1) - > S'(-5, -1)

**30. ** *A paper container in the shape of a right circular cone has a radius of 3 inches and a height of 8 inches. Determine and state the number of cubic inches in the volume of the cone, in terms
of π.*

The Volume of a Right Circular Cone is found using, V = (1/3)Bh, where B is the area of the base. The area of the Base is πr^{2}, so V = (1/3)πr^{2}h = (1/3)π(3)^{2}(8) = 24π.

**31. ** *In isosceles triangle RST shown below, RS = RT, M and N are midpoints of RS and RT, respectively, and MN is drawn. If MN = 3.5 and the perimeter of triangle RST is 25, determine and
state the length of NT.*

MN is the midsegment, so ST is twice as long. Since MN = 3.5, ST = 7. RST is isosceles and the perimeter is 25, so x + x + 7 = 25. 2x + 7 = 25, 2x = 18, x = 9.

RS and RT have lengths of 9. N is the midpoint of RT, so NT is 1/2 of RT. NT = 4.5

**32. ***In the diagram below, ABC is equilateral.
Using a compass and straightedge, construct a new equilateral triangle congruent to ABC in
the space below.*
[Leave all construction marks.]

Make a point on the lower portion of the page. Call it P. Use the compass to measure the length of AB. Make a little arc. Go back to the point you made and, without changing the compass, make an arc. Make a point on the arc. Call it Q. Use the straightedge to make PQ. Still not changing the compass, make an arc from point P above the middle of the line segment. Make a similar arc from point Q. Where the two arcs intersect, make a point. Call it R. Use the straightedge to make PR and QR. PQR is an equilateral triangle.

For a visual reference, look at *Method 1* on this MathBits page.

**33. ***Write an equation of the line that is perpendicular to the line whose equation is 2y = 3x + 12 and that passes through the origin.*

In **slope-intercept form**, the given line is y = (3/2)x + 6. A line perpendicular to it would have to have slope = -2/3. If it goes through the origin, the y-intercept is 0.

Therefore, y = (-2/3)x.

**34. ***Rectangle KLMN has vertices K(0,4), L(4,2), M(1,-4), and N(-3,-2). Determine and state the coordinates of the point of intersection of the diagonals.*

The diagonals of a rectangle **bisect** each other, so you just need to find the midpoint of either of the diagonals.

( (0 + 1)/2, (4 + (-4))/2 ) = (1/2, 0) or

( (4 + (-3))/2, (2 + -2)/2 ) = (1/2, 0).