(19)(本小题满分12分)解:(1)连接BD,设AC与BD的交点为F,连接EF因为E为B1D中点,F为BD中点,所以EF∥BB1,因为BB1⊥平面ABCD,所以EF⊥平面ABCD,又因为EF在平面ACE内,所以平面ACE⊥平面ABCD6分(Ⅱ)由于四边形ABCD是菱形,所以以F为坐标原点,分别以FC,FD,FE为x轴,By轴,z轴,建立空间直角坐标系.设FA=a,FD=b,则有a2+b2=1,于是可求出平面ADE的一个法向量n=(b,-a,-2ab)平面ACE的一个法向量n:=(0,1,O)由题意知cos60°=cos |=解得即菱形ABCD为正方形B所以三棱锥C=1DE的体积×EFx2×AD×CD=……12分12

解题导语】这是一篇说明文。文幸讲述了书香环境对于孩子成长的好处32B推理判断题。根据第二段中“ Radiation effect is a situationwhere children grow up around books, but they don't readbooks. But somehow books benefit then, even thoughthey don't read them as much as their parents wish themo”可知,辐射效应在本文中指在书香环境中长大的孩子虽然他们没有像父母期待的那样读那么多的书,但他们仍然从书中受益良多。由此推断出,书香环境对青少年的良好影响,故选B这A理判断题。根据第三段中“ If we grow up in a house,ina home where parents enjoy books, where books are givenas birthday presents and valued, this is something thatbecomes a part of our identity and gives us this lifelongurge to always come close to books and read more than we I则ld呵可知,如果我们生长在父母热爱书、珍惜书的家庭中,这将成为我们认知中的一部分,让我们终身都有接近书,多读书的渴望、故选A现点态度题,根据最后一段中“ So keep shelves piled with30y所以书柜里装满书吧,由此可知,作者对于书香环境是支持的,故选DsC主旨大意题,根据全文的大意,尤其是第一段中“weknow that reading is good for children. Now, a new studysuggests that just being around books has its benefits. "pJ,本文主要讲述了书香环境对孩子成长的好处。故选

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