6. Rectangle $A B C D$ has a perimeter of 16 meters. On each of its sides, a square is drawn with that side as its edge. It is known that the sum of the areas of these four squares is 68 square meters. Find the area of rectangle $A B C D$
6.【Solution】As shown in the figure, extend $A_{1} D_{1}$ to the right and $C_{1} E_{1}$ upwards, intersecting at point $\mathrm{E}$. The area of the square $A_{1} B C_{1} E$ is equal to the square of half the perimeter of the rectangle $\mathrm{ABCD}$, which is 64 square centimeters. The rectangle $\mathrm{ABCD}$ is congruent to $D_{1} D E_{1} E D C C_{1} E_{1}$. The sum of the areas of the squares $A_{1} A D D_{1}$ and $D C C_{1} E_{1}$ is equal to half the sum of the areas of the four given squares, which is $\frac{1}{2} \times 68=34$ square centimeters. $64-34=30$ square centimeters should be equal to twice the area of the rectangle $\mathrm{ABCD}$. Therefore, the area of $\mathrm{ABCD}$ is $\frac{1}{2} \times 30=15$ square centimeters.