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?
The goal of this investigation
is to select the correct point on the vertical line which is formed by
folding the paper in half so that it divides the paper into three equal
sections-two trapezoids and one triangle.
Solve/Investigate the Problem
First I will divide the paper
into three equal sections and then use my knowledge of equal fractional
parts to begin exploring the idea of equal sections.
of the Problem
I begin by folding the paper
into two equal sections. In other words, I formed a perpendicular bisector
of the side of the rectangle.
I then bisected each of the
rectangles’ sides to divide the side of the paper into four equal
sections. At this point I know
that I have to fold the paper in order to connect the bisectors to the
opposite corners. In this way, the point where these folds meet is
two-thirds down the original fold.
The height of the triangle
formed is now 2/3 and the length is 1. Using the formula A=bh/2, the area
of the triangle is now 1/3 the area of the paper.
Now the problem turns to determining
if the areas of the two trapezoids are equal to two thirds of the paper. You
can determine this by looking at the fact that you have divided the paper
into two equal sections. Since the right triangle formed by each side of
the equilateral triangle is equal to 1/6 of the total area, the rest of the
rectangular section must add up to 1/2 the total area. 1/6 +x = 1/2, so
There are some other, really interesting
relationships amongst the sections that are created by the folds. By folding the paper again (by the
same process of bisecting the vertical sides and then folding the paper to
these points) you can fold the paper into a total of six sections. By doing
this, now you can look at both the triangles and the quadrilaterals that
are formed and their relationship to the total area of the paper.
Here are some relationships
that I found:
Now that I have cut the figure
into six equal sections, the square sections are now 1/6 of the total area.
The equilateral triangle in the center has a side length that is 1/2 of the
length of the larger equilateral triangle. As a result, the area of the
smaller triangle is 1/3 the area of the smaller triangle, or 1/9 the total
area of the rectangle.
Another relationship is between
the trapezoid and the right triangle that is formed. Since the smaller
equilateral triangle is 1/9 the total area, the right triangle that is
formed by the bisection of the figure yields 1/18 of the total area or 1/6
of each of the three sections.
Extensions of the Problem
The extensions above can be
taken even further by dividing the three sections again into 12 and then 24
sections and look at the figures that are created by these divisions.