Exercise 2. In the figure below, the triangles ABO, BCO, CDO, DEO, and FEO are right isosceles triangles. We assume that $\mathrm{OA}=8 \mathrm{~cm}$. Determine the area of $A O F$ in $\mathrm{cm}^{2}$.

![](https://cdn.mathpix.com/cropped/2024_06_04_ff4f5d9cabdcb92d54e8g-2.jpg?height=626&width=1095&top_left_y=1178&top_left_x=496)
Solution to Exercise 2 This triangle has base $[\mathrm{OF}]$ and height $[\mathrm{AB}]$. Using the Pythagorean theorem in the isosceles triangle $OAB$ at $B$, we find that $OB=AB=OA / \sqrt{2}=8 / \sqrt{2}$. Similarly, $\mathrm{OC}=4, \mathrm{OD}=4 / \sqrt{2}, \mathrm{OE}=2, \mathrm{OF}=\sqrt{2}$, so the area of $\mathrm{AOF}$ is $\frac{1}{2} \times \sqrt{2} \times \frac{8}{\sqrt{2}}=4 \mathrm{~cm}^{2}$.