Question

giúp em mấy bài hình này với

Answer 1

\(2,\)

\(a,\left\{{}\begin{matrix}\widehat{CAB}=\widehat{AHB}\left(=90^0\right)\\\widehat{ABC}.chung\end{matrix}\right.\Rightarrow\Delta ABC\sim\Delta HBA\left(g.g\right)\)

\(b,BC=\sqrt{AB^2+AC^2}=\sqrt{6^2+8^2}=10\left(cm\right)\left(pytago\right)\\ \Delta ABC\sim\Delta HBA\left(g.g\right)\\ \Rightarrow\dfrac{AC}{AH}=\dfrac{BC}{AB}\Rightarrow AH=\dfrac{AB\cdot AC}{BC}=\dfrac{48}{10}=4,8\left(cm\right)\)

 

Answer 2

\(3,\)

\(a,\left\{{}\begin{matrix}\widehat{A}=\widehat{H}\left(=90^0\right)\\\widehat{I_1}=\widehat{I_2}\left(đối.đỉnh\right)\end{matrix}\right.\Rightarrow\Delta ABI\sim\Delta HCI\left(g.g\right)\)

\(b,\Delta ABI\sim\Delta HCI\left(g.g\right)\\ \Rightarrow\widehat{IBA}=\widehat{ICH}\)

Mà \(\widehat{IBA}=\widehat{IBC}\left(BI.là.phân.giác.\widehat{ABC}\right)\)

\(\Rightarrow\widehat{IBC}=\widehat{ICH}\)

\(c,BC=\sqrt{AB^2+AC^2}=\sqrt{100}=10\left(cm\right)\)

Vì BI là phân giác góc ABC nên \(\dfrac{AB}{BC}=\dfrac{AI}{CI}=\dfrac{6}{10}=\dfrac{3}{5}\Rightarrow AI=\dfrac{3}{5}CI\)

Mà \(AI+CI=AC=8\)

\(\Rightarrow\dfrac{3}{5}CI+CI=8\\ \Rightarrow\dfrac{8}{5}CI=8\Rightarrow CI=5\left(cm\right)\\ \Rightarrow AI=\dfrac{3}{5}\cdot5=3\left(cm\right)\)

 

 

Answer 3

\(1,\)

\(a,\left\{{}\begin{matrix}\widehat{BDC}=\widehat{BHC}\left(=90^0\right)\\\widehat{BCD}.chung\end{matrix}\right.\Rightarrow\Delta BDC\sim\Delta HBC\left(g.g\right)\)

\(b,\Delta BDC\sim\Delta HBC\left(cm.trên\right)\\ \Rightarrow\dfrac{BC}{HC}=\dfrac{DC}{BC}\Rightarrow HC=\dfrac{BC^2}{DC}=\dfrac{15^2}{25}=\dfrac{225}{25}=9\left(cm\right)\\ \Rightarrow HD=DC-HC=25-9=16\left(cm\right)\)