Mensagem não lida por joaopcarv » 15 Jan 2022, 13:56
Mensagem não lida
por joaopcarv » 15 Jan 2022, 13:56
Sejam [tex3]\mathsf{r}[/tex3]
o raio do setor [tex3]\mathsf{COQ}[/tex3]
e [tex3]\mathsf{R}[/tex3]
o raio da semicircunferência [tex3]\mathsf{AB.}[/tex3]
No [tex3]\mathsf{\triangle COQ}[/tex3]
, sendo [tex3]\mathsf{\widehat{CQO} \ = \ 90^\circ, \ \overline{CQ} \ = \ \overline{OQ} \ = \ r}[/tex3]
e [tex3]\mathsf{\overline{OC} \ = \ R:}[/tex3]
[tex3]\mathsf{r^2 \ + \ r^2 \ = \ R^2 \ \rightarrow \ R \ = \ r \cdot \sqrt{2}.}[/tex3]
Temos que [tex3]\mathsf{\widehat{ACB} \ = \ 90^\circ}[/tex3]
porque [tex3]\mathsf{\overline{AB}}[/tex3]
é diâmetro. Então, sejam [tex3]\mathsf{\widehat{ACQ} \ = \ \theta}[/tex3]
e [tex3]\mathsf{\widehat{QCE} \ = \ \psi}[/tex3]
, tais que [tex3]\mathsf{\theta \ + \ \psi \ = \ 90^\circ.}[/tex3]
Em [tex3]\mathsf{\triangle ACQ, \ \widehat{CAQ} \ = \ 90^\circ \ - \ \underbrace{\widehat{ACQ}}_{= \ \theta} \ \therefore \ \widehat{CAQ} \ = \ \psi.}[/tex3]
Tendo que [tex3]\mathsf{\overline{CQ} \ = \ r}[/tex3]
e [tex3]\mathsf{\overline{AQ} \ = \ \underbrace{\overline{AO}}_{= \ r \cdot \sqrt{2}} \ - \ \underbrace{\overline{OQ}}_{= \ r} \ \therefore \ \overline{AQ} \ = \ r \cdot \big(\sqrt{2} \ - \ 1\big):}[/tex3]
[tex3]\mathsf{\overline{AC}^2 \ = \ r^2 \ + \ \big(r \cdot \big(\sqrt{2} \ - \ 1\big)\big)^2 \ \therefore \ \overline{AC} \ = \ r \cdot \sqrt{2} \cdot \bigg(\sqrt{2 \ - \ \sqrt{2}}\bigg)}[/tex3]
Então:
[tex3]\mathsf{\sin(\psi) \ = \ \dfrac{\overbrace{\overline{CQ}}^{= \cancel{r}}}{\underbrace{\overline{AC}}_{= \cancel{r} \cdot \sqrt{2} \cdot \Big(\sqrt{2 \ - \ \sqrt{2}}\Big)}}}[/tex3]
[tex3]\mathsf{\sin(\psi) \ = \ \dfrac{1}{\sqrt{2} \cdot \Big(\sqrt{2 \ - \ \sqrt{2}}\Big)}}[/tex3]
[tex3]\mathsf{\cos(\psi) \ = \ \dfrac{\overbrace{\overline{QQ}}^{= \cancel{r} \cdot \big(\sqrt{2} \ - \ 1\big)}}{\underbrace{\overline{AC}}_{= \cancel{r} \cdot \sqrt{2} \cdot \Big(\sqrt{2 \ - \ \sqrt{2}}\Big)}}}[/tex3]
[tex3]\mathsf{\cos(\psi) \ = \ \dfrac{\big(\sqrt{2} \ - \ 1\big)}{\sqrt{2} \cdot \Big(\sqrt{2 \ - \ \sqrt{2}}\Big)}}[/tex3]
O triângulo [tex3]\mathsf{\triangle CEQ}[/tex3]
é isósceles [tex3]\mathsf{\Big(\overline{QC} \ = \ \overline{QE} \ = \ r\Big)}[/tex3]
, portanto [tex3]\mathsf{\widehat{QCE} \ = \ \widehat{CEQ} \ = \ \psi}[/tex3]
. Disso, temos que [tex3]\mathsf{\widehat{CQE} \ = \ \big(180^\circ \ - \ 2\cdot \psi\big) \ \therefore \ \sin\Big(\widehat{CQE}\Big) \ = \ \sin(2\cdot \psi):}[/tex3]
[tex3]\mathsf{\sin\Big(\widehat{CQE}\Big) \ = \ 2 \cdot \sin(\psi) \cdot \cos(\psi)}[/tex3]
[tex3]\mathsf{\sin\Big(\widehat{CQE}\Big) \ = \ \cancel{2} \cdot \dfrac{1}{\cancel{\sqrt{2}} \cdot \Big(\sqrt{2 \ - \ \sqrt{2}}\Big)} \cdot \dfrac{\big(\sqrt{2} \ - \ 1\big)}{\cancel{\sqrt{2}} \cdot \Big(\sqrt{2 \ - \ \sqrt{2}}\Big)}}[/tex3]
[tex3]\mathsf{\sin\Big(\widehat{CQE}\Big) \ = \ \dfrac{\big(\sqrt{2} \ - \ 1\big)}{\big(2 \ - \ \sqrt{2}\big)}}[/tex3]
[tex3]\mathsf{\sin\Big(\widehat{CQE}\Big) \ = \ \dfrac{\sqrt{2}}{2} \ \therefore \ \widehat{CQE} \ = \ \dfrac{\pi}{4} \ \ \Bigg(\widehat{CQE} \ < \ \dfrac{\pi}{2}\Bigg).}[/tex3]
A área [tex3]\mathsf{S_1}[/tex3]
, delimitada por [tex3]\mathsf{CE}[/tex3]
, é a área do setor [tex3]\mathsf{CQE}[/tex3]
menos a área do [tex3]\mathsf{\triangle CQE:}[/tex3]
[tex3]\mathsf{S_1 \ = \ \dfrac{\widehat{CQE} \cdot r^2}{2} \ - \ \dfrac{\sin\Big(\widehat{CQE}\Big) \cdot r^2}{2}}[/tex3]
[tex3]\mathsf{S_1 \ = \ \dfrac{\frac{\pi}{4} \cdot r^2}{2} \ - \ \dfrac{\sin\Big(\frac{\pi}{4} \Big) \cdot r^2}{2}}[/tex3]
[tex3]\boxed{\mathsf{S_1 \ = \ r^2 \cdot \Bigg(\dfrac{\pi}{8} \ - \ \dfrac{\sqrt{2}}{4}\Bigg)}}[/tex3]
Já a área [tex3]\mathsf{S_2}[/tex3]
, delimitada por [tex3]\mathsf{AC}[/tex3]
, é a área do setor [tex3]\mathsf{AOC}[/tex3]
menos a área do [tex3]\mathsf{\triangle AOC.}[/tex3]
O ângulo [tex3]\mathsf{\widehat{AOC} \ = \ \dfrac{\pi}{4}}[/tex3]
, determinado pelo [tex3]\mathsf{\triangle COQ}[/tex3]
(isósceles e retângulo):
[tex3]\mathsf{S_2 \ = \ \dfrac{\widehat{AOC} \cdot R^2}{2} \ - \ \dfrac{\sin\Big(\widehat{AOC}\Big) \cdot R^2}{2}}[/tex3]
[tex3]\mathsf{S_2 \ = \ \dfrac{\frac{\pi}{4} \cdot 2 \cdot r^2}{2} \ - \ \dfrac{\sin\Big(\frac{\pi}{4}\Big) \cdot 2 \cdot r^2}{2}}[/tex3]
[tex3]\boxed{\mathsf{S_2 \ = \ 2 \cdot r^2 \cdot \Bigg(\dfrac{\pi}{8} \ - \ \dfrac{\sqrt{2}}{4}\Bigg)}}[/tex3]
Fazendo a relação entre as áreas pedidas:
[tex3]\mathsf{\dfrac{S_1}{S_2} \ = \ \dfrac{\cancel{r^2 \cdot \Bigg(\dfrac{\pi}{8} \ - \ \dfrac{\sqrt{2}}{4}\Bigg)}}{2 \cdot \cancel{r^2 \cdot \Bigg(\dfrac{\pi}{8} \ - \ \dfrac{\sqrt{2}}{4}\Bigg)}}}[/tex3]
[tex3]\boxed{\boxed{\mathsf{\dfrac{S_1}{S_2} \ = \ \dfrac{1}{2}}}}[/tex3]
Editado pela última vez por
joaopcarv em 15 Jan 2022, 14:04, em um total de 1 vez.
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