Hình:
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a/ Có: \(BC=\sqrt{AB^2+AC^2}=\sqrt{4,5^2+6^2}=7,5\left(cm\right)\)
Xét tg ABC và tg DEC có:
\(\widehat{BAC}=\widehat{CDE}=90^o\)
\(\widehat{C}:chung\)
=> tg ABC ~ tg DEC (g.g)
=> \(\dfrac{AC}{DC}=\dfrac{BC}{EC}\)=> EC = \(\dfrac{BC\cdot DC}{AC}=\dfrac{7,5\cdot2}{6}=2,5\left(cm\right)\)
b/ Có: \(DE=\sqrt{EC^2-DC^2}=\sqrt{2,5^2-2^2}=1,5\left(cm\right)\)
=> \(S_{\Delta DEC}=\dfrac{1}{2}\cdot DE\cdot DC=\dfrac{1}{2}\cdot1,5\cdot2=1,5\left(cm^2\right)\)
c/ đề đúng ?