Question

Cho hình thang ABCD có đáy =3/5 CD,hai đường chéo AC và BD cắt nhau tại O.Biết diện tích tam giác   ODC=45cm ². Tính diện tích ABCD

Answer 1

AB/DC=3/5 hay AB=3/5 DC

S ABC =3/5 S ADC vì   - AB =3/5 CD

                                  - chiều cao hạ từ C & A bằng nhau vì

                                    cùng bằng chiều cao hình thang 

mà S ABC 3/5 S ADC chung đáy AC 

suy ra chiều cao hạ từ B xuống AC= chiều cao hạ từ D xuống AC

mặt khác tam giác BOC cũng có chiều cao hạ từ B xuống OC

tam giác DOC có chiều cao hạ từ D xuống OC

mà S ODC & S BOC có chung đáy OC mà chiều cao BO=3/5  OD

suy ra S BOC =3/5 S DOC

S BOC= 45x 3/5=27 (cm2) hay cũng chính là S của AOD

S ADC= 45+ 27= 72 (cm2)

S ABC= 72 x 3/5=43,2

S ABCD= 43,2 +72=115,2 

Answer 2

Xet 2 tam giac ABC va ABC co chieu cao la chieu cao cua hinh thang va co day AB =3/5 DC suy ra dien h ABC=3/5dien h ADC.XetABC va ADC co chung dayAC va co dien h ABC/ADC=3/5 suy ra BK=3/5 DH.Xet 2tam giac OBC va ODC co chung day OC va co chieu cao BK =3/5 DH suy ra dien h OBC =3/5 ODC =45x3/5=27[cm2].dien h BDC la:45+27=72[cm2].Dien h ABCD=3/5.Vay dien h ABCD la:72+[72x3/5]=115.2[cm2]

Answer 3

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