Problem:
Regular octagon ABCDEFGH has area n. Let m be the area of quadrilateral ACEG. What is nm ?
Answer Choices:
A. 42
B. 22
C. 43
D. 532
E. 322 Solution:
ACEG is a square. WLOG AB=1, then using Law of Cosines, AC2=[ACEG]=12+12−2cos135=2+2. The area of the octagon is just [ACEG] plus the area of the four congruent (by symmetry) isosceles triangles, all an angle of 135 in between two sides of length 1. Now,
nm=2+2+4⋅21sin1352+2=2+222+2=22
The answer is (B)22.
OR
Refer to the diagram. Call one of the side lengths of the square s. Since quadrilateral ACEG is a square, the area of the square would just be s2, which we can find by applying Law of Cosines on one of the four triangles. Assume each of the sides of the octagon has length 1 . Since each angle measures 135∘ in an octagon, then s2=12+12−2∗cos(135∘)=2+2
There are many ways to find the area of the octagon, but one way is to split the octagon into two trapezoids and one rectangle. We can easily compute AF to be 1+2 from splitting one of the sides into two 45−45−90 triangles. So the area of the octagon is 2∗21+2+1+2⇒2+22.