Octagonal mat

Octagonal mat formed from a square plate with a side of 40 cm so that every corner cut the isosceles triangle with leg 3.6 cm. What is the content area of one mat?

Correct result:

S =  1574.08 cm²

Solution:

$$\begin{gathered} A=3.6^2\mathrm{/}2={\displaystyle \frac{162}{25}}=6.48 \\ S=40^2-4\cdot A=40^2-4\cdot 6.48={\displaystyle \frac{39352}{25}}={\displaystyle \frac{39352}{25}}{\text{cm}}^2=1574.08{\text{cm}}^2 \end{gathered}$$

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