We have 
Radius r = 1m
θ = 90º
length of arc LE = arc FG = arc HI = arc JK 
$=2\mathrm{\pi r}\times \frac{\mathrm{\theta }}{{360}^{\circ }}$
$=2\mathrm{\pi r}\times \frac{{90}^{\circ }}{{360}^{\circ }}$
$=\frac{\mathrm{\pi }}{2}\mathrm{m}$
Now  KL = GH = 4 – (1 + 1) = 2 m
  JI = EF = 17 – (1 + 1) = 15 m
The perimeter of the shaded portion 
= arc LE + EF + arc FG + GH + arc HI + IJ + arc JK + KL
$=\frac{\mathrm{\pi }}{2}+15+\frac{\mathrm{\pi }}{2}+2+\frac{\mathrm{\pi }}{2}+15+\frac{\mathrm{\pi }}{2}+2$
            = 2p + 34
            = 34 + 6.28 = 40.28 m