4. In the figure, there is an octagon $A_{1} A_{2} A_{3} A_{4} A_{5} A_{6} A_{7} A_{8}$. Calculate the area of this octagon if the following conditions are met: $\left|A_{1} A_{2}\right|=\left|A_{6} A_{7}\right|=0.3 \mathrm{dm}, \quad\left|A_{2} A_{3}\right|=\left|A_{5} A_{6}\right|=150 \mathrm{~mm}, \quad\left|A_{3} A_{4}\right|=\left|A_{4} A_{5}\right|=5 \sqrt{2} \mathrm{~cm}$, $\left|\nless A_{1} A_{2} A_{3}\right|=\left|\Varangle A_{3} A_{4} A_{5}\right|=\left|\Varangle A_{5} A_{6} A_{7}\right|=\left|\Varangle A_{7} A_{8} A_{1}\right|=90^{\circ} \mathrm{and}\left|A_{7} A_{8}\right|=\left|A_{8} A_{1}\right|$.

![](https://cdn.mathpix.com/cropped/2024_06_03_a289088c89d3b43dc20dg-5.jpg?height=602&width=1079&top_left_y=1658&top_left_x=366)
4. Divide the octagon into parts:

- Two isosceles right triangles ($A_{3} A_{4} A_{5}$ and $A_{1} A_{7} A_{8}$)
- Rectangle $A_{2} A_{3} A_{5} A_{6}$

$\Delta A_{3} A_{4} A_{5}$ is an isosceles right triangle, so $P_{\Delta A_{3} A_{4} A_{5}}=\frac{1}{2} \cdot(5 \sqrt{2})^{2}=25 \text{ cm}^{2}$

and $\left|A_{3} A_{5}\right|=\sqrt{\left|A_{3} A_{4}\right|^{2}+\left|A_{4} A_{5}\right|^{2}}=\sqrt{100}=10 \text{ cm}$

$\overline{A_{3} A_{5}}$ is a side of the rectangle $A_{2} A_{3} A_{5} A_{6}$ and part of the hypotenuse of the isosceles right triangle

$A_{1} A_{7} A_{8}$

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so $P_{A_{2} A_{3} A_{5} A_{6}}=10 \cdot 15=150 \text{ cm}^{2}$

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and $\left|A_{1} A_{7}\right|=3+10+3=16 \text{ cm}$

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It follows that $2\left|A_{1} A_{8}\right|^{2}=\left|A_{1} A_{2}\right|^{2}$, so $\left|A_{1} A_{8}\right|=8 \sqrt{2} \text{ cm}$

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and $P_{\Delta A_{1} A_{7} A_{8}}=\frac{1}{2} \cdot(8 \sqrt{2})^{2}=64 \text{ cm}^{2}$.

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Finally, $P=25+150+64=239 \text{ cm}^{2}$.

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