Four diagonals of a regular octagon with side length 2 intersect as shown. Find the area of the shaded region.
The triangle just to the left of the blue area is a 45-45-90 triangle....and the side opposite the 90° angle has a length of 2√2 units = √8 units
And the blue area is a parallelogram with equal sides of this length
And the acute angle in the parallelogram = 45°
So....its area =
2 * (1/2) (√8)^2 sin (45°) =
8 / √2 units^2 = 8 √2 / 2 units^2 = 4 √2 units^2