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【语篇类型】记叙文[文]【主题语境】人与自[章]我—一顿晚餐【文章[来]大意】作者是一位抚[自]养四个孩子的单身母[知]亲,生活过得很艰辛[嘛],温饱都成了难题,[答]但是当女儿请求让邻[案]居家的五个孩子留下[网]吃晚饭时,作者还是[文]同意了。这顿晚饭会[章]带给作者什么样的感[来]受和发现呢?【写作[自]指导】根据 Paragraph[知] I的开头"晚餐快要结束时,我[嘛]感觉好多了"可知,第一段可以描[答]述作者在晚餐快要结[案]束时的心情以及晚餐[网]结束后的场景,再结[文]合均 aragraph2[章]开头中的" we left two for vou"可知,第一段还应写[来]到作者发现居然还有[自]肉丸剩下的内容。P[知]aragraph2[嘛]的开头是"‘那是我们吃过的最[答]好吃的肉丸,所以我[案]们留下了两颗给你,[网]邻居家的一个男孩子[文]说",由此可推断出,本[章]段续写主要描述作者[来]听到孩予的话后的深[自]思、对孩子们的赞叹[知]以及对两颗肉丸的处[嘛]理,如让邻居家的孩[答]子带回家等One passible versionPa[案]ragraph IBy the end of the meal I felt so much better. The troubles burdening[网]me when I first walked through the door had gone. The children had lickedthe[文]ir plates clean. We were all full and happy, and granother[章]'s company. I moved to clear the table with my oldest daughter.[来] Mymouth dropped open. Two meatballs[自] sat uneaten on the plate. How was thapossib[知]le? My daughter put her hand on my shoulder,[嘛] "What,s wrong, MomDid everybody[答] eat? "Yes. It was delicious[案]! "the children repliedPa[网]ragraph 2The best meatballs[文] ice ever had, so we left two for you, "one neighborb[章]oy said. I was astonishe[来]d. How had I managed to feed everyone Had thekids cut the meatballs[自] in half? Maybe some of them had only eatenspag[知]hetti. What kind and considera[嘛]te children![答] That night left me with anoverflo[案]wing heart and great hope for the future. i gave the leftovers[网] to theneighb[文]or children for lunch the next day. Things were hard now, but theywould[章] get better. For us. For the neighbor children

19.(1)证明：[来]因为在等腰梯形AB[自]CD中，AB∥DC[知],点E为AB的中点[嘛]，△BCE是等边三[答]角形，所以AD=A[案]E=CE=2,AB[网]=4,∠A=∠CE[文]B=60°，∠BC[章]D=120°，所以[来]四边形AECD与四[自]边形BCDE是菱形[知]，所以EC⊥BD.[嘛]即BF⊥CE,DF[答]⊥CE.(3分)又[案]BF,DFC平面B[网]DF,BF∩DF=[文]F,所以CE⊥平面[章]BDF.又CEC平[来]面AECD,所以平[自]面BDF⊥平面AE[知]CD.（5分)(2[嘛])解：因为平面BC[答]E⊥平面AECD,[案]平面BCE∩平面A[网]ECD=CE,且B[文]F⊥CE,所以BF[章]⊥平面AECD,所[来]以BF⊥DF,以B[自]F,CE,DF两两[知]垂直以点F为坐标原[嘛]点，F元，F,F为[答]x轴、y轴、z轴的[案]正方向，建立如图所[网]示的空间直角坐标系[文].以点F为坐标原点[章]，FC,FD,FB[来]为x轴、y轴、z轴[自]的正方向，建立如图[知]所示的空间直角坐标[嘛]系个211/40B[答]DHE则B(0,0[案],3),C(1,0[网],0),D(0,3[文],0),E(-1,[章]0,0),所以ED[来]=(1,3,0),[自]因为点c是BC的中[知]点，由中点公式，可[嘛]得G320,分》，[答]得G=(30(8分[案])设平面DEG的一[网]个法向量为n=(x[文],y,z),n·E[章]元=0，nrx+3[来]y=0,由令y=1[自],得x=-√3,2[知]=3，得n=(-3[嘛],1,3).(10[答]分)又由(1)知m[案]=(1,0,0)是[网]平面BDF的法向量[文]，所以cos(m,[章]nl=m·n339[来]n1×/1313即[自]平面DEG与平面B[知]DF所成锐二面角的[嘛]余弦值为3913([答]12分)