The given figure shows nine squares. The area of square $\mathrm{A}$ is $1 \mathrm{~cm}^{2}$ and that of square B is $81 \mathrm{~cm}^{2}$. What is the area, in $\mathrm{cm}^{2}$, of square I?
(a) 196
(d) 324
(b) 256
(e) 361
(c) 289

![](https://cdn.mathpix.com/cropped/2024_05_01_280b19a87fa67143351eg-016.jpg?height=406&width=462&top_left_y=291&top_left_x=1368)
The correct option is (d).

The side of $A$ measures $\sqrt{1}=1 \mathrm{~cm}$ and the side of $B$ measures $\sqrt{81}=9 \mathrm{~cm}$. Now we have:

- Side of $G=$ side of $B$ - side of $A=9-1=8 \mathrm{~cm}$.
- Side of $C=$ side of $B+$ side of $A=1+9=10 \mathrm{~cm}$.
- Side of $F=$ side of $G$ - side of $A=8-1=7 \mathrm{~cm}$.
- Side of $H=$ side of $G+$ side of $F=8+7=15 \mathrm{~cm}$.
- Side of $B+$ side of $C=$ side of $G+$ side of $F+$ side of $E$, therefore, $9+10=8+7+$ side of $E$, which means side of $\mathrm{E}=4 \mathrm{~cm}$.
- Side of $D=$ side of $C+$ side of $E=10+4=14 \mathrm{~cm}$.
- Side of $I=$ side of $E+$ side of $D=18 \mathrm{~cm}$.

Finally, the area of $I$ measures $18^{2}=324 \mathrm{~cm}^{2}$.