(1)人教版八年级数学下册92页第14题是这样叙述的:如图1,?ABCD中,过对角线BD上一点P作EF∥BC,HG∥AB

2025-12-17 08:56:41
推荐回答(1个)
回答1:

(1)∵?ABCD中,EF∥BC,HG∥AB,
∴S△ABD=S△BCD,S△PBE=S△PBG,S△PDH=S△PDF
∴S?AEPH=S?PGCF,S?ABGH=S?EBCF,S?AEFD=S?HGCD
故答案为:?AEPH 和?PGCF 或?ABGH 和?EBCF 或?AEFD 和?HGCD;

(2)根据(1)可得:S△ABC=S△ADC,S△PAE=S△PAG,S△PCH=S△PCF
∵S?BHPE=3,S?PFDG=5,
∴S△PAC=S△PAG+S△PCF+S?PFDG-S△ACD=S△PAG+S△PCF+S?PFDG-

1
2
S?ABCD=S△PAG+S△PCF+S?PFDG-
1
2
(2S△PAG+2S△PCF+S?BHPE+S?PFDG)=S?PFDG-
1
2
(S?BHPE+S?PFDG)=1;
故答案为:1;

(3)∵①②③④四个平行四边形面积的和为14,
∴S1+S2+S3+S4=14,
∵四边形ABCD的面积为11,
∴S5=11-14×
1
2
=4,
∴S菱形EFGH=S1+S2+S3+S4+S5=18,
∵菱形EFGH的一个内角为30°,
∴设边长为x,
则x?xsin30°=18,
解得:x=6,
∴菱形EFGH的周长为24.
故答案为:24.