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C

cuncon2395

cho ABC vuong tai A co phan giac BD , co AD=4can10 , CD=5can10 Tinh BD

vì BD là phân giác
[TEX]\Rightarrow \frac{AB}{BC}=\frac{AD}{DC} \Rightarrow \frac{AB}{BC}=\frac{4\sqrt{10}}{5\sqrt{10}}=\frac{4}{5} \Rightarrow AB=4k, BC=5k[/TEX]
lại có [TEX]AB^2+AC^2=BC^2 \Rightarrow 16k^2+(4\sqrt{10}+5\sqrt{10})^2=25k^2 \Rightarrow 810=9k^2 \Rightarrow k^2=90 \Rightarrow k=3\sqrt{10}[/TEX]
[TEX]\Rightarrow AB=4k=4.3\sqrt{10}=12\sqrt{10} (cm) [/TEX]

[TEX]BD^2=AD^2+AB^2 \Rightarrow BD^2=160+1440=1600 \Rightarrow BD=40 (cm)[/TEX]
 
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