$解过点B作BH⊥CC_{1}于点H$
$∵∠BCC_{1}=45°,BC=5cm,∴易得BH=\frac{\sqrt{2}}{2}BC=\frac{5\sqrt{2}}{2} cm$
$∵正方形纸板ABCD在投影面上的正投影为四边形A_{1}B_{1}C_{1}D_{1}$
$∴B_{1}C_{1}=BH=\frac{5\sqrt{2}}{2}cm,C_{1}D_{1}=CD=5cm$
$易得四边形A_{1}B_{1}C_{1}D_{1}为矩形$
$∴四边形A_{1}B_{1}C_{1}D_{1}的面积为\frac{5\sqrt{2}}{2}×5=\frac{25\sqrt{2}}{2}(cm^{2})$
