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Data publicarii: 07 Noiembrie, 2014

PROBLEMA 1.2

Suport teoretic:

Teorema bisectoarei,lungimea bisectoarei,cerc circumscris.

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:                            

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

Raspuns:

  1. 20cm.;
  2. 30cm.;
  3. {30}(\sqrt{3}+1)cm.;{30}(\sqrt{3}+1)cm.;   
  4. {300}\pi{cm}^{2};{300}\pi{cm}^{2};  
  5. {10}(3-\sqrt{3})\;cm.{10}(3-\sqrt{3})\;cm.

Rezolvare:

  1. {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};
  2. {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};  
  3. {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};                                                                                    
  4. BC=2RBC=2R \Leftrightarrow\Leftrightarrow 20\sqrt{3}=2R20\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};          
  5. \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}.

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