5) A circular porthole with a radius of $20 \mathrm{~cm}$ is gridded with bars so that the four squares at the center have a side length of $10 \mathrm{~cm}$. The total length of the bars is
(A) $80 \sqrt{3} \mathrm{~cm}$
(B)
(D) $80(\sqrt{3}+1) \mathrm{~cm} \quad$ (E) $210 \mathrm{~cm}$.
(C) $200 \mathrm{~cm}$

![](https://cdn.mathpix.com/cropped/2024_04_17_c3d7afe478310a3de07bg-1.jpg?height=216&width=216&top_left_y=627&top_left_x=2555)
5) The answer is $\mathbf{( D )}$.

Applying the Pythagorean theorem to the right triangle $O A B$ we obtain

$$
A B=\sqrt{20^{2}-10^{2}} \mathrm{~cm}=10 \sqrt{3} \mathrm{~cm}
$$

Four of the bars have a length of $2 \cdot A B$ and two bars are as long as the diameter of the circle. Therefore, the total length is

$$
L=(40+40+4 \cdot 2 \cdot 10 \sqrt{3}) \mathrm{cm}=80(\sqrt{3}+1) \mathrm{cm}
$$