10. As shown in the figure, in square $\mathrm{ABCD}$, the area of isosceles right triangle $\mathrm{AEF}$ is $1 \mathrm{~cm}^2$, and the area of rectangle $\mathrm{EFGH}$ is $10 \mathrm{~cm}^2$. What is the area of square $\mathrm{ABCD}$ in $\mathrm{cm}^2$?
【Analysis】It is known that the height on the hypotenuse of an isosceles right triangle is half of the hypotenuse.  
Thus, since the area of the isosceles right triangle $\mathrm{AEF}$ is 1, we know its hypotenuse $\mathrm{EF}$ is 2. Since the area of rectangle $\mathrm{EFGH}$ is 10, then $\mathrm{FG}$ is 5.  
Therefore, the area of the isosceles right triangle $\mathrm{BFG}$ is $\frac{1}{2} \times 5 \times 2.5=6.25$  
Thus, the area of square $\mathrm{ABCD}$ is $1 \times 2+6.25 \times 2+10=24.5$