The line A B in the following diagram represents the side of a square
having the same area as the cross. I have shown elsewhere, as stated,
how to make a square and equilateral triangle of equal area. I need not
go, therefore, into the preliminary question of finding the dimensions
of the triangle that is to equal our cross. We will assume that we have
already found this, and the question then becomes, How are we to cut up
one of these into pieces that will form the other?
First draw the line A B where A and B are midway between the extremities
of the two side arms. Next make the lines D C and E F equal in length to
half the side of the triangle. Now from E and F describe with the same
radius the intersecting arcs at G and draw F G. Finally make I K equal
to H C and L B equal to A D. If we now draw I L, it should be parallel
to F G, and all the six pieces are marked out. These fit together and
form a perfect equilateral triangle, as shown in the second diagram. Or
we might have first found the direction of the line M N in our triangle,
then placed the point O over the point E in the cross and turned round
the triangle over the cross until the line M N was parallel to A B. The
piece 5 can then be marked off and the other pieces in succession.
[Illustration]
I have seen many attempts at a solution involving the assumption that
the height of the triangle is exactly the same as the height of the
cross. This is a fallacy: the cross will always be higher than the
triangle of equal area.