Problem:
A regular octagon ABCDEFGH has an area of one square unit. What is the area of the rectangle ABEF?
Answer Choices:
A. 1−22​​
B. 42​​
C. 2​−1
D. 21​
E. 41+2​​
Solution:
Let O be the intersection of the diagonals of ABEF. Since the octagon is regular, â–³AOB has area 1/8. Since O is the midpoint of AE,â–³OAB and â–³BOE have the same area. Thus â–³ABE has area 1/4, so ABEF has area 1/2.
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 â–³AIH and â–³JOI are congruent. By symmetry, the area in the center square IJKL is the sum of the areas of â–³AIH,â–³CJB, â–³EKD, and â–³GLF. Thus the area of rectangle ABEF is half the area of the octagon.
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