6. In pentagon $A B C D E \quad A B=A E=C D=1, \quad B C+D E=1$, $\angle A B C=\angle A E D=90^{\circ}$. Find the area of the pentagon.
6. Answer: 1. Solution. Extend the segment $C B$ beyond point $B$ to segment $B F=D E$, and then draw segments $C A, C E$ and $A F$. It is easy to see that triangles $C D E$ and $A B F$ are equal. Therefore, $C E=A F$ and the areas of the pentagon and quadrilateral $A E C F$ coincide. But triangles $A E C$ and $A F C$ are equal by three sides, and the area of triangle $A F C$ is $1 / 2$. Therefore, the area of quadrilateral $A E C F$ is 1. Grading criteria. Correct solution - 7 points. In all other cases - 0 points.