Investigations | Geometry | Quadrilaterals Additional Investigations
| Paper Folding
Take a rectangular sheet of paper. Fold it in half to make a crease down the center
of the sheet from top to bottom. Then, select a point on the sheet and make a crease
from the upper right corner to the point; now make a crease from the upper left
corner to the point.
How would the point be selected so that the triangle formed by the top of the sheet
and the two slant creases has the same area as each of the lateral trapezoids?
First, get some sheets of paper and do some folding to get a feel for the problem.
It's possible to make a fold across any two points, and a point is indicated where two
creases cross or where a crease intersects an edge of the paper.
Folds can be used to bisect a line segment. For example, the bottom of the page is a
line segment. We can match the corners of the page together to form a crease that
is the perpendicular bisector. Proof?
Second, folding to trisect a line segment (e.g. folding the paper into thirds) is
probably a guessing game. If you claim it is a "folding" construction you should have
a proof that the fold trisects the segment (exactly, not approximately.)
Third, of course you will want to switch to a line drawing representation for analysis
and proof at some point. Use similarity concepts to show an exact folding
construction for the desired configuration.
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To begin this investigation, I constructed a rectangle using GSP. Using
the midpoint key, I constructed the midpoint segment (which happens
to be the vertical line of symmetry of the rectangle). The question
posed is where should I place a point on the line (center line) to create
a triangle whose area is equal to either trapezoid formed.
m AB = 4.00 cm
m CA = 5.50 cm
C
G
D
m CD = 4.00 cm
m BD = 5.50 cm
Area
CED = 7.35 cm2
Area FBDE = 7.32 cm2
EF = 1.83 cm
GE = 3.68 cm
E
EF
GE
= 0.50
A
F
B
The figure above, shows the original rectangle (ABDC) whose opposite
sides are congruent. Segment GF is the perpendicular bisector of AB as
well as CD. Points F and G are the midpoints of AB and CD. I then
placed a point on segment FG and constructed segment CE and DE.
This formed a triangle. I then was able to move point E up and down
segment FG to find where the area of triangle CED = trapezoid EFBD.
I then measured the areas of the triangle and trapezoid and compared
them as I moved point E up and down segment GF. Once I got equal
areas, I compared the lengths of FE and GE. I found that FE was
exactly ½ of GE. Thus, point E must be located 1/3 of the way down
on segment GF.
To do this activity using paper folding, I would have the students fold
their papers into thirds (like folding a letter on 8 ½ by 11 paper into an
envelope). They then would have to fold their paper in half vertically
(forming the vertical line of symmetry). This is not an exact activity,
yet would give them a tangible activity regarding the areas formed.