The area of the square is 82 D 64, whereas the rectangle, which seems to have the same
constituent parts, has an area 5 Ð 13 D 65, and so the area has apparently been increased
by one square unit. The puzzle is easy to explain. The points a, b, c, and d do not all lie
on the diagonal of the rectangle, but instead are the vertices of a parallelogram whose
area is exactly equal to the extra unit of area.
The construction can be carried out with any square whose sides are equal to the
Fibonacci number F2k. When the square is partitioned as in the diagram, the pieces can be
re-formed to produce a rectangle having a slot in the shape of a slim parallelogram (ourgure is exaggerated). The identity F2k1F2kC1 1 D F2
2k can be interpreted as asserting
that the area of the rectangle minus the area of the parallelogram is precisely equal to the
area of the original square. It can be shown that the height of the parallelogram—that is, the width of the slot at its widest point—is
The area of the square is 82 D 64, whereas the rectangle, which seems to have the same
constituent parts, has an area 5 Ð 13 D 65, and so the area has apparently been increased
by one square unit. The puzzle is easy to explain. The points a, b, c, and d do not all lie
on the diagonal of the rectangle, but instead are the vertices of a parallelogram whose
area is exactly equal to the extra unit of area.
The construction can be carried out with any square whose sides are equal to the
Fibonacci number F2k. When the square is partitioned as in the diagram, the pieces can be
re-formed to produce a rectangle having a slot in the shape of a slim parallelogram (ourgure is exaggerated). The identity F2k1F2kC1 1 D F2
2k can be interpreted as asserting
that the area of the rectangle minus the area of the parallelogram is precisely equal to the
area of the original square. It can be shown that the height of the parallelogram—that is, the width of the slot at its widest point—is
การแปล กรุณารอสักครู่..

The area of the square is 82 D 64, whereas the rectangle, which seems to have the same
constituent parts, has an area 5 Ð 13 D 65, and so the area has apparently been increased
by one square unit. The puzzle is easy to explain. The points a, b, c, and d do not all lie
on the diagonal of the rectangle, but instead are the vertices of a parallelogram whose
area is exactly equal to the extra unit of area.
The construction can be carried out with any square whose sides are equal to the
Fibonacci number F2k. When the square is partitioned as in the diagram, the pieces can be
re-formed to produce a rectangle having a slot in the shape of a slim parallelogram (ourgure is exaggerated). The identity F2k1F2kC1 1 D F2
2k can be interpreted as asserting
that the area of the rectangle minus the area of the parallelogram is precisely equal to the
area of the original square. It can be shown that the height of the parallelogram—that is, the width of the slot at its widest point—is
การแปล กรุณารอสักครู่..
