Problem:
A regular octagon ABCDEFGH has an area of one square unit. What is the area of the rectangle ABEF ?
Answer Choices:
A. 1β22ββ
B. 42ββ
C. 2ββ1
D. 21β
E. 41+2ββ Solution:
Let O be the intersection of the diagonals of ABEF. Since the octagon is regular, β³AOB has area 1/8. Since O is the midpoint of AE,β³OAB and β³BOE have the same area. Thus β³ABE has area 1/4, so ABEF has area 1/2β.
OR
Let O be the intersection of the diagonals of the square IJKL. Rectangles ABJI,JCDK,KEFL, and LGHI are congruent. Also IJ=AB=AH, so the right isosceles triangles β³AIH and β³JOI are congruent. By symmetry, the area in the center square IJKL is the sum of the areas of β³AIH,β³CJB, β³EKD, and β³GLF. Thus the area of rectangle ABEF is half the area of the octagon.