4. On a horizontal line that divides the plane into upper and lower half-planes, a segment $AB$ with a length of $72 \, \text{cm}$ is drawn. Using the endpoints of the segment, in the upper half-plane, a regular (equilateral) triangle $A M_{1} M_{2}$ and a regular pentagon $M_{5} M_{6} M_{7} M_{8} M_{9}$ are drawn, and in the lower half-plane, a regular quadrilateral (square) $M_{2} M_{3} M_{4} M_{5}$ and a regular hexagon $M_{9} M_{10} M_{11} M_{12} M_{13} B$ are drawn. Here, $M_{2}, M_{5}$, and $M_{9}$ are on the segment $AB$, point $M_{2}$ is between $A$ and $M_{5}$, and point $M_{9}$ is between $M_{5}$ and $B$. The lengths of the sides of the regular polygons are in the same ratio as the corresponding numbers of their sides. Calculate the length of the broken line

$$
L \equiv A M_{1} M_{2} M_{3} M_{4} M_{5} M_{6} M_{7} M_{8} M_{9} M_{10} M_{11} M_{12} M_{13} B
$$
Solution. Let the side of the equilateral triangle $A M_{1} M_{2}$ be denoted by $a$, the side of the square $M_{2} M_{3} M_{4} M_{5}$ by $b$, the side of the regular pentagon $M_{5} M_{6} M_{7} M_{8} M_{9}$ by $c$, and the side of the regular hexagon $M_{9} M_{10} M_{11} M_{12} M_{13} B$ by $d$.

From the condition of the problem, it follows that $a: b: c: d=3: 4: 5: 6$, that is,

$$
a=3 k, b=4 k, c=5 k, d=6 k
$$

We have $72=a+b+c+d$, so $72=3 k+4 k+5 k+6 k$, i.e., $k=4$. The length of the broken line $L$ is:

$$
2 a+3 b+4 c+5 d=6 k+12 k+20 k+30 k=68 k=68 \cdot 4=272 \text{ cm}
$$

## VII section