(1)∵正方形ABCD与正方形DEFG,点A、D、E三点共线,
∴AD=CD,DG=DE,
∵S
△ADG=
AD×DG,S
△DCE=
DE×CD,
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/d439b6003af33a8744f49899c55c10385243b59c?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
∴S
△ADG=S
△DCE,
故答案为:=;
(2)把△DCE绕点D顺时针旋转90°,使CD与AD重合,E旋转到E'的位置,
∵四边形GDEF为正方形,∠GDE=90°,DG=DE=DE′,
∴G、D、E'在一直线上,且AD为△AGE'的中线,
∴S
△ADG=S
△ADE'=S
△CDE,
∴S
△ADG=S
△DCE,
故答案为:=;
(3)把△DCF绕点C顺时针旋转90°,使CD与AC重合,F旋转到F'的位置,
∵四边形BCFG为正方形,∠BCF=90°,BC=CF=CF′,
∴B、C、F'在一直线上,且AC为△ABF'的中线,
∴S
△CDF=S
△ACF'=S
△ABC,
![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/8c1001e93901213f75aad65f57e736d12e2e959c?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
同理:S
△AEK=S
△HBG=S
△ABC,
所以△AKE,△CDF,△BGH的面积和为S
△ABC的3倍,
又AC长为5,边AB长为4,
∴S
阴影部分面积=3S
△ABC=3×
AB×AC×sin∠ABC,
当∠ABC最大时△AKE,△CDF,△BGH的面积和最大,
即当AB⊥BC时,S
△ABC最大值为:
×5×4=10
∴△AKE,△CDF,△BGH的面积和的最大值为10×3=30.
故答案为:30.