Câu hỏi của Khủng Long Xanh

a.Xét tam giác AHB và tam giác BCD, có:^AHB = ^BCD = 90 độ^ABH = ^BDC ( slt )Vậy tam giác AHB đồng dạng tam giác BCD ( g.g ) (1)b.\(BD=\sqrt{9^2+12^2}=15\left(cm\right)\)\(\left(1\right)\Rightarrow\dfrac{AH}{BC}=\dfrac{AB}{BD}=\dfrac{HB}{CD}\)\(\Leftrightarrow\dfrac{AH}{9}=\dfrac{12}{15}=\dfrac{HB}{12}\)\(\Rightarrow AH=\dfrac{12.9}{15}=7,2\left(cm\right)\)\(\Rightarrow HB=\dfrac{12.12}{15}=9,6\left(cm\right)\)c. \(S_{AHB}=\dfrac{1}{2}.AH.BH=\dfrac{1}{2}.7,2.9,6=34,56\left(cm^2\right)\)\(S_{BCD}=\dfrac{1}{2}.BC.CD=\dfrac{1}{2}.9.12=54\left(cm^2\right)\)\(\dfrac{S_{AHB}}{S_{BCD}}=\dfrac{34,56}{54}=0,64\)Câu trả lời: a.Xét tam giác AHB và tam giác BCD, có:^AHB = ^BCD = 90 độ^ABH = ^BDC ( slt )Vậy tam giác AHB đồng dạng tam giác BCD ( g.g ) (1)b.\(BD=\sqrt{9^2+12^2}=15\left(cm\right)\)\(\left(1\right)\Rightarrow\dfrac{AH}{BC}=\dfrac{AB}{BD}=\dfrac{HB}{CD}\)\(\Leftrightarrow\dfrac{AH}{9}=\dfrac{12}{15}=\dfrac{HB}{12}\)\(\Rightarrow AH=\dfrac{12.9}{15}=7,2\left(cm\right)\)\(\Rightarrow HB=\dfrac{12.12}{15}=9,6\left(cm\right)\)c. \(S_{AHB}=\dfrac{1}{2}.AH.BH=\dfrac{1}{2}.7,2.9,6=34,56\left(cm^2\right)\)\(S_{BCD}=\dfrac{1}{2}.BC.CD=\dfrac{1}{2}.9.12=54\left(cm^2\right)\)\(\dfrac{S_{AHB}}{S_{BCD}}=\dfrac{34,56}{54}=0,64\)