$ \left(1\right)\because $四边形$EFHG$是正方形,且$AD\bot BC$,
$\therefore EF$∥$BC,EG=EF=MD($设为$\lambda )$,
$\therefore \triangle AEF$∽$\triangle ABC$,$AM=80-\lambda $;
$\therefore EF:BC=AM:AD$,
即$\lambda :120=\left(80-\lambda \right):80$,
解得:$\lambda =48\left(cm\right)$,
即这个正方形的边长为$48cm$.
(2)设矩形$EFHG$的面积为$\lambda $,
由(1)知:$EF:BC=AM:AD$,
即$EF:120=\left(80-x\right):80$,
解得:$EF=120-1.5x$,
$\therefore \lambda =x\left(120-1.5x\right)=-1.5x^{2}+120x$,
$\therefore $当$x=-\dfrac{120}{2\times \left(-1.5\right)}=40$时,$\lambda $取得最大值,
$\lambda $的最大值$=-1.5\times 1600+120\times 40=2400\left(cm^{2}\right)$.
标签:EFHG