Ta có \(S\)ADFE = DF.FE \(\left(1\right)\)
Xét \(\Delta DBF\) và \(\Delta E\)\(FC\) có :
\(\hat{BDF}\) \(=\) \(\hat{FEC}\) \(\left(=90°\right)\)
\(\hat{DFB}\) \(=\) \(\hat{ECF}\) ( cùng phụ với \(\hat{EFC}\))
\(\Rightarrow\Delta DBF\approx\Delta EFC\)(\(G-G\) )
\(\Rightarrow\dfrac{DF}{EC}=\dfrac{BD}{FE}\)
\(\Leftrightarrow\dfrac{DF}{25}=\dfrac{20}{FE}\)
\(\Rightarrow DF.FE=20.25=500\) \(\left(2\right)\)
\(\underrightarrow{\left(1\right)\left(2\right)}\) \(S\)ADFE =500 (\(cm^2\))
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