Experiment with hyperbolic geometry (solutions):

Plotting and measuring: applet at

http://www.math.psu.edu/dlittle/java/geometry/spherical/toolbox.html

You can set the background to white under View>Background. This may work better for copying/printing.

Q1: Start off by drawing and measuring a few triangles.

·  Select Constructions > Plot Point, and plot 3 points by clicking.

·  Select Constructions > Draw Segment, and click endpoints to connect (A to B, B to C, C to A).

·  Select Measurements > Measure Triangle, and it will give all three side lengths and angle measures.

Select Edit > Move Point, drag the vertices of your triangle around, and watch the measurements change. The angle sum appears to always be less than 180˚, as expected.

Q2: Draw a pair of parallel lines by constructing alternate interior angles equal. You can select File > New to clear the previous construction

·  Constructions > Draw Infinite Line. Click on two spots on the disk, and you'll get points and a line.
·  Constructions>Draw Ray at Specific Angle.
·  Enter a value for the angle (default is 45)
·  Click B, then A, then somewhere on the disk.
·  You'll get ray AC and ray AB forming an angle.
·  Constructions>Draw Ray at Specific Angle.
·  Keep the same value as above.
·  Click A, then C, then somewhere on the disk, to the side of ray AC which is opposite B. (You want alternate angles).
·  You should get a ray CD.
·  You may need to extend ray CD into an infinite line by choosing Constructions > Draw Infinite Line, and clicking points C and D.
·  The lines should be non-intersecting; i.e. parallel.
·  Measurements > Measure Angle and measure angles BAC and DCA to verify that alternate interior angles are equal.
·  Now, Edit > Move point
·  Slide point C around. The angles should stay fixed by construction. Can you ever get CD to intersect AB, or are they always parallel? (Being the same line doesn't count!)
They will always be parallel – the only time you'll intersect is at the instant they're the same line.


Q3: A parallel to a parallel may not be parallel!

·  Draw infinite line AB.
·  Constructions > Draw Perpendicular to line AB through point B by first clicking point B, then the line.
·  Same thing through AB through point B.
·  You may need to Edit>Move Point a bit to bring points C and D closer. Get something that looks basically like the image to the right.
·  Draw a perpendicular to line AD through point D (which will give you a point E). Because alternate interior angles are equal (both 90), line AB and line DE are parallel.
·  Also draw a perpendicular to line BC through point C (which will give you a point F). Because alternate interior angles are equal (both 90), line AB and line CF are parallel.
·  Go to Edit > Move Point, and slide points C and D around. You should be able to get things so that line DE and line CF intersect. This means that (1) you have two distinct lines through the same point (the point of intersection) that are parallel to a given line, and therefore (2) lines which are parallel to a third line are not necessarily parallel to each other. (These are intersecting at point G.)

Q4: Triangle area: (answers will vary, of course)

·  Plot 3 points (A, B, and C).
·  Draw segments A to B, B to C, C to A.
·  Measure triangle ABC (gives all sides and all angles).
·  Record segment lengths:
AB = 1.836
BC = 1.893
CA = 2.699
·  Draw perpendicular to segment AB through C.
·  Measure distance CD
CD = 1.831
·  Calculate value of .5(base)(height), using AB as the base, and CD as the height:
.5(AB)(CD) = .5(1.836)(1.831) = 1.689
·  Draw perpendicular to segment BC through A.
·  Measure distance AE
AE = 1.775
·  Calculate value of .5(base)(height), using BC as the base, and AE as the height:
.5(BC)(AE) = .5(1.893)(1.775) = 1.680

You should recognize .5(base)(height) as the formula for the area of a triangle in Euclidean geometry. Does it work as the area formula in hyperbolic geometry? The value changes depending on what side you use for the base (this does not happen in Euclidean geometry).


Q5: SAS congruence

·  Start by constructing two segments of the same length: Constructions> Draw Segment of Specific Length.
·  Then, with vertices at points A (on one segment) and C (on the other), Constructions> Draw Ray at a Specific Angle (pick an angle). [ Click B to A to somewhere, then D to C to somewhere.]
·  By default, it should put point on those rays that are the same length as the original segments. (You can measure distance to check, and move point if needed.)
·  This means SAS = SAS by construction.
·  Form triangles by drawing segment EB and segment FD.
·  Measure triangle ABE and triangle CDF.
·  Are the triangles congruent? YEP.
AB = DE = 2 constructed, then rays at angle A and angle E. Length AC = EF = 2 as well. Triangle ABC congruent to triangle FDE.
·  Since ASA congruence can be proven from SAS congruence, we should have that as well. Construct a pair of triangles with ASA = ASA (construct a matching side for each, then construct matching pairs of angles at each endpoint). Measure. Congruent? (Hope so!)
For this one, first segments AB = CD = 3. Then, angles at A and C (45), then angles at B and D (5). Find intersection point was used to get vertices I and J. Triangle ABI congruent to triangle CDJ