10. (3 points) The area of square $A B C D$ is 9 square centimeters, and the area of square $E F G H$ is 64 square centimeters. As shown in the figure, side $B C$ lies on $E H$. Given that the area of triangle $A C G$ is 6.75 square centimeters, the area of triangle $A B E$ is $\qquad$ square centimeters.
【Analysis】Extend $A B$ to intersect with $F G$ at $M$, as shown in the figure. Let the area of square $A B C D$ be $a^2$, $E B=b$, and $C H=c$. Express $B H$ as $C H+B C$, which is $M G$. The area of triangle $A B C$ + the area of trapezoid $B C G M$ - the area of triangle $A M G$ = the area of triangle $A C G$. Use the area formulas for the trapezoid and triangles, along with the known area of triangle $A C G$, to set up a relationship. Given that the area of square $A B C D$ is 9, find the value of $a^2$ to be 9. Substitute the value of $a^2$ into the equation to get the value of $\frac{1}{2} a b$, which is the area of triangle $A B E$.
【Solution】Solution: Extend $A B$ to intersect with $F G$ at point $M$, as shown in the figure:
Let the side length of square $A B C D$ be $a$ cm, $E B=b$ cm, and $C H=c$ cm,
then $A B=B C=a$ cm, $B M=E H=E B+B C+C H=(a+b+c)$ cm, $M G=B H=(a+c)$ cm,
Since $S_{\triangle A C G}=S_{\triangle A B C}+S_{\text {trapezoid } B C G M}-S_{\triangle A M G}=6.75$,
we have $\frac{1}{2} a^{2}+\frac{1}{2}(a+b+c)(2 a+c)-\frac{1}{2}(2 a+b+c)(a+c)=6.75$,
simplifying gives: $\frac{1}{2} a^{2}+\frac{1}{2} a b=6.75$,
Given that the area of square $A B C D$ is 9 square cm, i.e., $a^{2}=9$,
thus $S_{\triangle A B E}=\frac{1}{2} A B \cdot E B=\frac{1}{2} a b=6.75-\frac{1}{2} \times 9=6.75-4.5=2.25$ (square cm).
Answer: The area of triangle $A B E$ is 2.25 square cm.
The answer is: 2.25.