当CQ= 1 2 时,即Q为CC1中点,此时可得PQ∥AD1,AP=QD1=√5/2,
故可得截面APQD1为等腰梯形,故②正确;
由上图当点Q向C移动时,满足0<CQ< 1/ 2 ,只需在DD1上取点M满足AM∥PQ,
即可得截面为四边形APQM,故①正确;
③当CQ= 3 /4 时,如图,
延长DD1至N,使D1N= 1/2 ,连接AN交A1D1于S,连接NQ交C1D1于R,连接SR,
可证AN∥PQ,由△NRD1∽△QRC1,可得C1R:D1R=C1Q:D1N=1:2,故可得C1R= 1 /3 ,故正确;故可知当 3/ 4 <CQ<1时,只需点Q上移即可,此时的截面形状仍然上图所示的APQRS,显然为五边形,故错误;
④当CQ=1时,Q与C1重合,取A1D1的中点F,连接AF,可证PC1∥AF,且PC1=AF,
可知截面为APC1F为菱形,故其面积为1/2 AC1•PF=√6/2,故正确.
故答案为:①②④
追答![](https://video.ask-data.xyz/img.php?b=https://iknow-pic.cdn.bcebos.com/5bafa40f4bfbfbed7a31140973f0f736afc31f5f?x-bce-process=image%2Fresize%2Cm_lfit%2Cw_600%2Ch_800%2Climit_1%2Fquality%2Cq_85%2Fformat%2Cf_auto)
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