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2022-04-17 20:17:06
Kẻ đường cao AH và BK (H,K∈DC)
Ta có AB=BC=AD⇒hình thang ABCD là hình thang cân
Ta lại có chuviABCD=AB+BC+CD+AD=3AB+22⇒3AB=52-22=30⇒AB=BC=AD=10(cm)
Xét △AHD và △BKC có:
∠D=∠C
BC=AD
∠AHD=∠BKC=90
Suy ra △AHD = △BKC( cạnh huyền góc nhọn)
⇒KC=DH
Ta có ∠AHD=∠BKC=∠HAB=90(vì AB//HK)⇒ABKH là hình chữ nhật⇒AB=HK=10(cm)
Ta có DC=DH+HK+KC⇒22=2DC+10⇒2DC=12⇒DC=6(cm)
Ta có △AHD vuông tại H⇒AD2=AH2+HD2⇒100=AH2+36⇒AH2=100-36=64⇒AH=8(cm)
Vậy chiều cao hình thang là 8cm