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.
\(S_{AHB}=\dfrac{1}{2}.AH.BH=\dfrac{1}{2}.7,2.9,6=34,56\left(cm^2\right)\)
\(S_{BCD}=\dfrac{1}{2}.BC.CD=\dfrac{1}{2}.9.12=54\left(cm^2\right)\)
\(\dfrac{S_{AHB}}{S_{BCD}}=\dfrac{34,56}{54}=0,64\)
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