Regular octagon pad

You need to make a pad in the shape of a regular octagon with a side length of 4 cm. What is the minimum diameter of the circle-shaped semi-finished product from which we make the pad, and what will be the percentage of waste? (Round the results to 1 decimal place)

Correct answer:

D =  10.5 cm
p =  10 %

Step-by-step explanation:

$$\begin{gathered} a=4\text{cm} \\ n=8 \\ \alpha =\frac{360}{2\cdot n}=\frac{360}{2\cdot 8}=\frac{45}{2}=22.5{}^{\circ } \\ \tan \alpha =a/2:h \\ h=\frac{a}{2\cdot \tan \alpha }=\frac{a}{2\cdot \tan 22.5°}=\frac{4}{2\cdot \tan 22.5°}=\frac{4}{2\cdot 0.414214}=4.82843\text{cm} \\ \sin \alpha =a/2:r \\ r=\frac{a}{2\cdot \sin \alpha }=\frac{a}{2\cdot \sin 22.5°}=\frac{4}{2\cdot \sin 22.5°}=\frac{4}{2\cdot 0.382683}=5.22625\text{cm} \\ D=2\cdot r=2\cdot 5.2263=10.5\text{cm} \end{gathered}$$

$$\begin{gathered} S_1=\pi \cdot r^2=3.1416\cdot 5.2263^2\doteq 85.8085{\text{cm}}^2 \\ {S}_0=\frac{a\cdot h}{2}=\frac{4\cdot 4.8284}{2}\doteq 9.6569{\text{cm}}^2 \\ {S}_2=n\cdot S_0=8\cdot 9.6569\doteq 77.2548{\text{cm}}^2 \\ p=100\cdot \frac{S_1-S_2}{S_1}=100\cdot \frac{85.8085-77.2548}{85.8085}=10\% \end{gathered}$$

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