jmc

Let the width be $w$ and the length be $l$. Then, the original perimeter is $2(w+l)$. After the increase, the new width and new length are $1.1w$ and $1.1l$, so the new perimeter is $2(1.1w+1.1l)=2.2(w+l)$. Therefore, the percent change is $$\begin{array}{rl}\frac{2.2(w+l)-2(w+l)}{2(w+l)}& =\frac{.2(w+l)}{2(w+l)}\\ & =\frac{.2}{2}\\ & =10\%\end{array}$$ $\boxed{10\%}$