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Data publicarii: 20 Noiembrie, 2009

GEOMETRIE-16

Suport teoretic:

Teorema bisectoarei, lungimea bisectoarei, perimetrul unui triunghi, cerc circumscris unui triunghi.

Enunt:

Se da triunghiul dreptunghic ABC, in care:

$mas(\hat{A})={90}^{\circ},\;{mas(\hat{C})}={30}^{\circ}\;si\;{M}\in{(AC)},$ mas(\hat{A})={90}^{\circ},\;{mas(\hat{C})}={30}^{\circ}\;si\;{M}\in{(AC)},

astfel incat AM = 10cm., iar [BM] este bisectoarea unghiului B.

Sa se afle:

lungimea bisectoarei [BM];
lungimea catetei [AC];
perimetrul triunghiului ABC;
aria cercului circumscris triunghiului ABC;
lungimea segmentului [BN], unde N apartine segmentului (BC) si [AN] este bisectoarea unghiului A.

Raspuns:

20cm.;
30cm.;
${30}(\sqrt{3}+1)cm.;$ {30}(\sqrt{3}+1)cm.;
${300}\pi{cm}^{2};$ {300}\pi{cm}^{2};
${10}(3-\sqrt{3})\;cm.$ {10}(3-\sqrt{3})\;cm.

Rezolvare:

${mas}(\widehat{ABM})={30}^{\circ}\Rightarrow$ {mas}(\widehat{ABM})={30}^{\circ}\Rightarrow ${BM}=2{AM}=2\cdot{10cm}={20cm};$ {BM}=2{AM}=2\cdot{10cm}={20cm};
${mas}(\widehat{MBC})={mas}(\widehat{MCB})={30}^{\circ}\Rightarrow$ {mas}(\widehat{MBC})={mas}(\widehat{MCB})={30}^{\circ}\Rightarrow ${MC}={BM}={20cm}$ {MC}={BM}={20cm} $\Rightarrow$ \Rightarrow ${AC}={AM}+{MC}={10cm}+{20cm}={30cm};$ {AC}={AM}+{MC}={10cm}+{20cm}={30cm};
${AB}^{2}={BM}^{2}-{AM}^{2}=400-100=300$ {AB}^{2}={BM}^{2}-{AM}^{2}=400-100=300 $\Rightarrow{AB}={10}{\sqrt{3}}{cm};$ \Rightarrow{AB}={10}{\sqrt{3}}{cm}; ${BC}^{2}={AB}^{2}+{AC}^{2}=300+900=1200$ {BC}^{2}={AB}^{2}+{AC}^{2}=300+900=1200 $\Rightarrow{BC}={20}{\sqrt{3}{cm};}$ \Rightarrow{BC}={20}{\sqrt{3}{cm};} ${AB+BC+AC }=\cdots ={30}(\sqrt{3}+{cm};$ {AB+BC+AC }=\cdots ={30}(\sqrt{3}+{cm};
$BC=2R$ BC=2R $\Leftrightarrow$ \Leftrightarrow $20\sqrt{3}=2R$ 20\sqrt{3}=2R $\Leftrightarrow{R}=10{\sqrt{3}}{cm}$ \Leftrightarrow{R}=10{\sqrt{3}}{cm} $\Rightarrow$ \Rightarrow ${A}{[ABC]}={\pi}{R^2}={\pi}{({10}{\sqrt{3})}^{2}}={300}{\pi}{cm}{2};$ {A}{[ABC]}={\pi}{R^2}={\pi}{({10}{\sqrt{3})}^{2}}={300}{\pi}{cm}{2};
$\frac{AB}{AC}=\frac{BN}{NC}$ \frac{AB}{AC}=\frac{BN}{NC} $(teorema\;bisectoarei)$ (teorema\;bisectoarei) $\Leftrightarrow\frac{AB}{AC+AB}=\frac{BN}{NC+BN}\Leftrightarrow{\frac{10\sqrt{3}}{30+10\sqrt{3}}}={\frac{BN}{{20}{\sqrt{3}}}\Leftrightarrow\cdots\Leftrightarrow{BN} ={10}(3 -\sqrt{3})}{cm}.$ \Leftrightarrow\frac{AB}{AC+AB}=\frac{BN}{NC+BN}\Leftrightarrow{\frac{10\sqrt{3}}{30+10\sqrt{3}}}={\frac{BN}{{20}{\sqrt{3}}}\Leftrightarrow\cdots\Leftrightarrow{BN} ={10}(3 -\sqrt{3})}{cm}.