Cap. 22 - Áreas de Regiones Circulares ⇒ Solucionário:Racso - Cap XXII - Problemas de Geometria y Como Resolverlos - I Edição - Ex:16 Tópico resolvido

[petras]

Problema Proposto
16 - Seja P = ( [tex3]\frac{2\sqrt{10}}{5}, \frac{\sqrt{{10}}}{5}[/tex3]) o ponto onde
o círculo e ol segmento são tangentes. A área
da região sombreada é?

Resposta

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E) [tex3]\frac{5}{2}-\frac{\pi }{2}[/tex3]

:

[petras]

[tex3]\mathsf{
\triangle APH: \frac{2\sqrt10}{5}^2+\frac{\sqrt10}{5}^2 = AP^2 \implies \boxed{AP = \sqrt2}\\
\triangle APB: \frac{\sqrt10}{5}^2 =\frac{2\sqrt10}{5}. HB \implies \boxed{HB=\frac{\sqrt{10}}{10}} \\
Utilizando~Pitágoras: \\
\triangle APB: \boxed{PB = \frac{\sqrt2}{2}}\\
\triangle ACP \sim \triangle BAP \implies \boxed{AC = \sqrt10} ~e~\boxed{CP = 2\sqrt2}\\
CB = CP + PB \implies\boxed{CB = \frac{5\sqrt2}{2}}\\
\boxed{AB = \frac{2\sqrt10} {5}+\frac{\sqrt10}{10} = \frac{\sqrt{10 }} {2}}\\
S= S_{\triangle ABC}-S_{setorADE} = \frac{\frac{\sqrt{10}}{2}.\sqrt{10}}{2}-\frac{\pi .(\sqrt2)^2}{4}=\frac{10}{4}-\frac{2\pi}{4}\\
\therefore \boxed{\color{red}S = \frac{5}{2}-\frac{\pi}{2}}










}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\mathsf{
\triangle APH: \frac{2\sqrt10}{5}^2+\frac{\sqrt10}{5}^2 = AP^2 \implies \boxed{AP = \sqrt2}\\
\triangle APB: \frac{\sqrt10}{5}^2 =\frac{2\sqrt10}{5}. HB \implies \boxed{HB=\frac{\sqrt{10}}{10}} \\
Utilizando~Pitágoras: \\
\triangle APB: \boxed{PB = \frac{\sqrt2}{2}}\\
\triangle ACP \sim \triangle BAP \implies \boxed{AC = \sqrt10} ~e~\boxed{CP = 2\sqrt2}\\
CB = CP + PB \implies\boxed{CB = \frac{5\sqrt2}{2}}\\
\boxed{AB = \frac{2\sqrt10} {5}+\frac{\sqrt10}{10} = \frac{\sqrt{10 }} {2}}\\
S= S_{\triangle ABC}-S_{setorADE} = \frac{\frac{\sqrt{10}}{2}.\sqrt{10}}{2}-\frac{\pi .(\sqrt2)^2}{4}=\frac{10}{4}-\frac{2\pi}{4}\\
\therefore \boxed{\color{red}S = \frac{5}{2}-\frac{\pi}{2}}










}[/tex3]