$易知tan∠ABC=\frac {AC}{BC}=\sqrt {3}$
$∴∠ABC=60°$
$易知AB=\sqrt {AC^{2}+BC^{2}}=2$
$∴∠A'BC'=∠ABC=60°,A'C'=AC=\sqrt {3}$
$∠A''C''B''=∠ACB=90°$
$∴∠ABA'=180°-∠A'BC'=120°$
$∴A经过路线长\frac {120π×2}{180}+\frac {90×π×\sqrt {3}}{180}=\frac {4π}{3}+\frac {\sqrt {3}}{2}π$
$面积为\frac {120π×2^{2}}{360}+\frac {1}{2}×1×\sqrt {3}+\frac {90×π×(\sqrt {3})^{2}}{360}=\frac {25}{12}π+\frac {\sqrt {3}}{2}$