Olhe o anexo :
- JOAOLIMADOTUTORBRASIL.jpg (61.68 KiB) Exibido 1518 vezes
Na face [tex3]\mathsf{\triangle ABD} \ \rightarrow[/tex3]
Como [tex3]\mathsf{\overline{AB}\perp \overline{AD}}[/tex3]
, podemos usar Pitágoras para achar [tex3]\mathsf{\overline{BD}}[/tex3]
:
[tex3]\mathsf{\overline{BD}^² \ = \ \overline{AB}^² \ +\ \overline{AD}^² \ \rightarrow}[/tex3]
[tex3]\mathsf{\overline{BD}^² \ = \ 4^2 \ + \ 3^2 \ \rightarrow \ \boxed{\mathsf{\overline{BD} \ = \ 5 \ |u|}}}[/tex3]
Na face [tex3]\mathsf{\triangle ACD} \ \rightarrow[/tex3]
Da mesma forma, como [tex3]\mathsf{ \overline{AC}\perp \overline{AD}}[/tex3]
, para acharmos [tex3]\mathsf{\overline{CD}}[/tex3]
:
[tex3]\mathsf{\overline{CD}^² \ = \ \overline{AC}^² \ +\ \overline{AD}^² \ \rightarrow}[/tex3]
[tex3]\mathsf{\overline{CD}^² \ = \ 4^2 \ + \ 3^2 \ \rightarrow \ \boxed{\mathsf{\overline{CD} \ = \ 5 \ |u|}}}[/tex3]
Agora, observe que eu destaquei o [tex3]\mathsf{\triangle BCD}[/tex3]
, que isósceles ([tex3]\mathsf{\overline{BD} \ = \ \overline{CD}.}[/tex3]
) Sendo isósceles, a altura [tex3]\mathsf{\overline{DM}}[/tex3]
relativa à base [tex3]\mathsf{\overline{BC}}[/tex3]
é fincada no ponto médio [tex3]\mathsf{M}[/tex3]
da mesma.
Aplicando Pitágoras em [tex3]\mathsf{\triangle BDM}[/tex3]
, por exemplo, para acharmos a altura [tex3]\mathsf{\overline{DM}}[/tex3]
:
[tex3]\mathsf{\overline{BD}^2 \ = \ \overline{DM}^2 \ + \bigg(\frac{\overline{BD}}{2}\bigg)^2 \ \rightarrow}[/tex3]
[tex3]\mathsf{5^2 \ = \ \overline{DM}^2 + \bigg(\dfrac{4}{2}\bigg)^2 \ \rightarrow}[/tex3]
[tex3]\mathsf{\overline{DM}^2 \ = \ 25 \ - \ 4 \ \rightarrow}[/tex3]
[tex3]\mathsf{\overline{DM}^2 \ = \ 21 \ \rightarrow \ \boxed{\mathsf{\overline{DM} \ = \ \sqrt{21} \ |u|}}} \ \rightarrow[/tex3]
Altura relativa ao lado [tex3]\mathsf{\overline{BC}.}[/tex3]
Agora, vamos calcular o segmento verde ([tex3]\mathsf{\overline{AM}}[/tex3]
) do desenho [tex3]\rightarrow[/tex3]
Ele é a altura da base triangular equilátera [tex3]\mathsf{\triangle ABC}[/tex3]
. Logo :
[tex3]\mathsf{\overline{AM} \ = \ \dfrac{l \ \cdot \ \sqrt{3}}{2} \ \rightarrow}[/tex3]
Sendo [tex3]\mathsf{l \ = \ 4 \ |u|}[/tex3]
:
[tex3]\mathsf{\overline{AM} \ = \ \dfrac{4 \ \cdot \ \sqrt{3}}{2} \ \rightarrow} \ \boxed{\mathsf{\overline{AM} \ = \ 2 \ \cdot \ \sqrt{3} \ |u|}}[/tex3]
[tex3]\dots[/tex3]
A distância de [tex3]\mathsf{A}[/tex3]
à face [tex3]\mathsf{ \triangle BCD}[/tex3]
, [tex3]\mathsf{\overline{AN}}[/tex3]
é perpendicular à mesma e é fincada em um ponto (ponto [tex3]\mathsf{N}[/tex3]
) da altura [tex3]\mathsf{\overline{DM}}[/tex3]
dessa face (que já calculamos anteriormente).
Veja dois triângulos retângulos formados com essa distância, um com cateto vermelho [tex3]\mathsf{\overline{DN}}[/tex3]
e outro com cateto amarelo [tex3]\mathsf{\overline{MN} \dots}[/tex3]
[tex3]\mathsf{\overline{DN} \ + \ \overline{NM} \ = \ \cancelto{\sqrt{21} \ |u|}{\overline{DM}} \dots}[/tex3]
Pitágoras em [tex3]\mathsf{\triangle AMN}[/tex3]
:
[tex3]\mathsf{\overline{AM}^2 \ = \ \overline{AN}^2 + \overline{MN}^2 \rightarrow}[/tex3]
[tex3]\mathsf{(2\cdot \sqrt{3})^2 \ = \ \overline{AN}^2 + \overline{MN}^2 \rightarrow}[/tex3]
[tex3]\mathsf{12 \ = \ \overline{AN}^2 + \overline{MN}^2 \ (I)}[/tex3]
Pitágoras em [tex3]\mathsf{\triangle ADN}[/tex3]
:
[tex3]\mathsf{\overline{AD}^2 \ = \ \overline{AN}^2 + \overline{DN}^2 \rightarrow}[/tex3]
[tex3]\mathsf{3^2 \ = \ \overline{AN}^2 + \overline{DN}^2 \rightarrow}[/tex3]
[tex3]\mathsf{9 \ = \ \overline{AN}^2 + \overline{DN}^2 \ (II)}[/tex3]
[tex3]\dots[/tex3]
[tex3]\begin{cases} \mathsf{12 \ = \ \overline{AN}^2 + \overline{MN}^2 \\
9 \ \ = \ \ \overline{AN}^2 + \overline{DN}^2} \end{cases}[/tex3]
[tex3]\mathsf{3 \ = \ \overline{MN}^2 \ - \ \overline{DN}^2} \rightarrow[/tex3]
Dos produtos notáveis :
[tex3]\mathsf{3 \ = \ \cancelto{\sqrt{21} \ |u|}{(\overline{MN} \ + \ \overline{ND})} \ \cdot \ (\overline{MN} \ - \ \overline{ND}) \rightarrow}[/tex3]
[tex3]\mathsf{\dfrac{3}{\sqrt{21}} \ = \ \overline{MN} \ - \ \overline{ND}}[/tex3]
[tex3]\begin{cases} \mathsf{\sqrt{21} \ = \ \overline{MN} \ + \ \overline{ND} \\
\dfrac{3}{\sqrt{21}} \ \ = \ \ \overline{MN} \ - \ \overline{ND}} \end{cases}[/tex3]
[tex3]\boxed{\mathsf{\overline{MN} \ = \ \dfrac{12}{\sqrt{21}} \ |u|, \ \overline{DN} \ = \ \dfrac{9}{\sqrt{21}} \ |u|}}[/tex3]
Por fim, em qualquer um dos Pitágoras:
[tex3]\mathsf{12 \ = \ \overline{AN}^2 \ + \ \Bigg(\cancelto{\frac{12}{\sqrt{21}}}{\overline{MN}}\Bigg)^2 \ \rightarrow}[/tex3]
[tex3]\boxed{\boxed{\mathsf{\overline{AN} \ = \ \dfrac{6 \ \cdot \ \sqrt{7}}{7} \ |u|}}}[/tex3]
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