2022英语周报34期答案

12.C【解析】如图所示,取AC的中点O,连接PO,OB,由OP⊥AC,OB⊥AC知∠POB为二面角P-AC-B的平面角,故cos∠POB=y6在平面PBO内过点P作BO的垂线交BO的延长线于点D,cos∠POD3OP⊥AC,OB⊥AC得AC⊥平面PBO,所以AC⊥PD,得PD⊥平面ABC,OP=√AP2-OA2=√5-2=√3,所以OD=PCos∠POD=3×6=2,DP=√OP-OD=1,所以OA=OC=OD=OB=√,则四边形ABCD为正方形,从而易联想到将此三棱锥补形成长方体ABCD-EFGP.所以三棱锥P-ABC的外接球即为长方体ABCD-PEFG的外接球,且PB为该外接球的一条直径,又PB=√AD2 CD2 PD2=3,现该外接球的半径R=2,所以该三棱锥的外接球的表面积为S=4R2=4x×=9故选CFP5“---r----------==BD

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