Mensagem não lida por joaopcarv » Qui 28 Nov, 2019 17:57
Mensagem não lida
por joaopcarv » Qui 28 Nov, 2019 17:57
Sejam [tex3]\mathsf{a}[/tex3]
e [tex3]\mathsf{b}[/tex3]
os lados de [tex3]\mathsf{\triangle ABE}[/tex3]
e [tex3]\mathsf{\triangle BCD}[/tex3]
, respectivamente. [tex3]\mathsf{\measuredangle{ABD} \ = \ \underbrace{\mathsf{\measuredangle{ABC}}}_{= \ 60^\circ} \ + \ \underbrace{\mathsf{\measuredangle{CBD}}}_{= \ 60^\circ} \ = \ 120^\circ.}[/tex3]
Pela lei do cosseno em [tex3]\mathsf{\triangle ABD}[/tex3]
, tem-se: [tex3]\mathsf{(\overline{\underbrace{AD}_{(l)}})^2 \ = \ (\overline{\underbrace{AB}_{(a)}})^2 \ + \ (\overline{\underbrace{BD}_{(b)}})^2 \ - \ 2 \cdot (\overline{\underbrace{AB}_{(a)}}) \cdot (\overline{\underbrace{BD}_{(b)}}) \cdot \cos(\underbrace{\measuredangle{CBD}}_{= 120^\circ}):}[/tex3]
[tex3]\boxed{\mathsf{l^2 \ = \ a^2 \ + \ b^2 \ + \ a\cdot b}}[/tex3]
Agora, pelas áreas:
[tex3]\mathsf{A_{ACDE} \ = \ A_{\triangle ABE} \ + \ A_{\triangle ABC} \ + \ A _{\mathsf{\triangle BCD}}.}[/tex3]
Calculando as áreas:
[tex3]\mathsf{A_{\triangle ABE} \ = \ \dfrac{a^2 \cdot \sqrt{3}}{4}, A_{\triangle BCD} \ = \ \dfrac{b^2 \cdot \sqrt{3}}{4}, A_{\triangle ABC}\ = \ \dfrac{a \cdot b \cdot \sen(60^\circ)}{2} \ = \ \dfrac{a\cdot b \cdot \sqrt{3}}{4}}[/tex3]
.
Logo, a área total é:
[tex3]\mathsf{A_{ACDE} \ = \ \dfrac{\overbrace{(a^2 \ + \ b^2 \ + \ a\cdot b)}^{l^2} \ \cdot \sqrt{3}}{4} \ = \ \boxed{\mathsf{\dfrac{l^2 \cdot \sqrt{3}}{4}}}}[/tex3]
Vou resumir um pouco...
Pelas áreas:
[tex3]\mathsf{A_{ECMN} \ = \ \dfrac{l^2 \cdot \sqrt{3}}{4} \ - \ A_{\triangle EDN} \ - \ A_{\triangle ACM}.}[/tex3]
[tex3]\mathsf{A_{\triangle ACM} \ = \ \dfrac{(a\ + \ b) \cdot \ \frac{b}{2} \ \cdot \sen(60^\circ)}{2} \ = \ \dfrac{(a \ + \ b) \cdot b \cdot \sqrt{3}}{8}}[/tex3]
.
Pela lei do cosseno em [tex3]\mathsf{\triangle ABD}[/tex3]
, tem-se: [tex3]\mathsf{\overline{AC} \ = \ \sqrt{a^2 \ + \ b^2 \ - \ a\cdot b}}[/tex3]
, e, além disso, sendo [tex3]\mathsf{\alpha \ = \ \measuredangle{BAC}, \cos(\alpha) \ = \ \dfrac{2\cdot a \ - \ b}{2 \cdot \ (\sqrt{a^2 \ + \ b^2 \ - \ a\cdot b})}}[/tex3]
.
Portanto, [tex3]\mathsf{\sen(\alpha) \ = \ \dfrac{\sqrt3 \cdot b}{2 \cdot \ (\sqrt{a^2 \ + \ b^2 \ - \ a\cdot b})}}[/tex3]
.
Temos então que [tex3]\mathsf{\sen(\measuredangle(\underbrace{EAB}_{= \ 60^\circ}) \ + \ \alpha) \ = \ \sen(60^\circ) \cdot \cos(\alpha) \ + \ \cos(60^\circ) \cdot \sen(\alpha) \ = \ \dfrac{\ a \cdot \sqrt{3}}{2 \cdot (\sqrt{a^2 \ + \ b^2 \ - \ a\cdot b})}.}[/tex3]
Logo:
[tex3]\mathsf{A_{\triangle EDN} \ = \ \dfrac{(\sqrt{a^2 \ + \ b^2 \ - \ a\cdot b}) \cdot \ \frac{a}{2} \ \cdot \sen(60^\circ \ + \ \alpha)}{2} \ = \ \dfrac{\cancel{(\sqrt{a^2 \ + \ b^2 \ - \ a\cdot b})} \cdot \ a \ \cdot \frac{\ a \cdot \sqrt{3}}{2 \cdot \cancel{(\sqrt{a^2 \ + \ b^2 \ - \ a\cdot b})}}}{4} \ = \ \dfrac{a^2 \cdot \sqrt{3}}{8}}[/tex3]
Por fim, temos:
[tex3]\mathsf{A_{ECMN} \ = \ \dfrac{l^2 \cdot \sqrt{3}}{4} \ - \ \cancelto{\frac{a^2 \cdot \sqrt{3}}{8}}{A_{\triangle EDN}} \ - \ \cancelto{\frac{(a \ + \ b) \cdot b \cdot \sqrt{3}}{8}}{A_{\triangle ACM}}}[/tex3]
[tex3]\mathsf{A_{ECMN} \ = \ \dfrac{l^2 \cdot \sqrt{3}}{4} \ - \ \dfrac{\sqrt{3}}{8} \cdot \ \cancelto{l^2}{(a^2 \ + \ b^2 \ + \ a\cdot b)}}[/tex3]
[tex3]\boxed{\boxed{\mathsf{A_{ECMN} \ = \ \dfrac{l^2 \cdot \sqrt{3}}{8}}}}[/tex3]
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