jmc

To make the $8$ triangles have the same area, the base must be divided into $8$ segments of length $1$ inch each. Define points $A$, $B_0$, $B_1$, $B_2$, $B_3$, and $B_4$ as in the figure. For $0\le k\le 3$, the perimeter $P(k)$ of triangle $A{B}_k{B}_{k+1}$ in inches is $$P(k)=1+\sqrt{10^2+k^2}+\sqrt{10^2+(k+1{)}^2},$$where each distance $A{B}_k$ is calculated by applying the Pythagorean theorem to right triangle $A{B}_0{B}_k$. Since $P(k)$ increases as $k$ increases, its largest value is $P(3)=1+\sqrt{100+3^2}+\sqrt{100+4^2}$, which to the nearest hundredth is $\boxed{22.21}$ inches.