1. **Problem Statement:** Find the area of a regular octagon where the length of the diagonal connecting two opposite vertices is 2 ft. 2. **Formula and Important Rules:** For a regular octagon, the longest diagonal (connecting opposite vertices) is equal to $2a(1+\sqrt{2})$, where $a$ is the side length. 3. **Find the side length $a$:** Given the diagonal $d = 2$ ft, $$ d = 2a(1+\sqrt{2}) = 2 $$ Divide both sides by 2: $$ \cancel{2}a(1+\sqrt{2}) = \cancel{2} $$ $$ a(1+\sqrt{2}) = 1 $$ Solve for $a$: $$ a = \frac{1}{1+\sqrt{2}} $$ Rationalize the denominator: $$ a = \frac{1}{1+\sqrt{2}} \times \frac{1-\sqrt{2}}{1-\sqrt{2}} = \frac{1-\sqrt{2}}{1 - 2} = \frac{1-\sqrt{2}}{-1} = \sqrt{2} - 1 $$ 4. **Calculate the area $A$ of the regular octagon:** The area formula is $$ A = 2(1+\sqrt{2})a^2 $$ Substitute $a = \sqrt{2} - 1$: $$ a^2 = (\sqrt{2} - 1)^2 = 2 - 2\sqrt{2} + 1 = 3 - 2\sqrt{2} $$ So, $$ A = 2(1+\sqrt{2})(3 - 2\sqrt{2}) $$ Expand: $$ A = 2[(1)(3) + (1)(-2\sqrt{2}) + (\sqrt{2})(3) + (\sqrt{2})(-2\sqrt{2})] $$ $$ = 2[3 - 2\sqrt{2} + 3\sqrt{2} - 4] $$ $$ = 2[-1 + \sqrt{2}] $$ $$ = 2\sqrt{2} - 2 $$ 5. **Final answer:** The area of the regular octagon is $$ \boxed{2\sqrt{2} - 2 \text{ square feet}} $$