Problema (nível 4): Um rectângulo foi dividido como se indica na figura. Sabendo que a circunferência é tangente aos três lados, quanto mede a área da região sombreada? (image1)


Conteúdo referenciado:
[image1]
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Resolução: A soma das áreas dos triângulos $[AED]$ e $[BCE]$ é ${ \frac{\overline{DE}\cdot \overline{AD}}{2}+\frac{\overline{EC}\cdot\overline{BC}}{2}=1}$. Resta determinar a área do círculo. Sejam $O$ o centro e $r$ o raio do círculo. Note-se que, pelo Teorema de Pitágoras, $\overline{BE}=\sqrt{2}$. Além disso, o qua­dri­lá­te­ro $[OHEG]$ é um quadrado de lado $r$, pois os dois triângulos sombreados são isósceles rectângulos e por isso $A\widehat{E}B=90^o$. (image2) Solução 1: Por outro lado, ${\displaystyle O\widehat{F}B=O\widehat{H}B=90^0}$ e $\overline{OF}=\overline{OH}=r$ logo $\overline{HB}=\overline{FB}=1$ e assim $1=\overline{HB}=\overline{EB}-\overline{EH}=\sqrt{2}-r$, pelo que $r=\sqrt{2}-1$ e a área do círculo é $\pi r^2=\pi(3-2\sqrt{2})$. Solução 2: Por outro lado, aplicando o Teorema de Pitágoras ao triângulo rectângulo $[OHE]$, vem $\overline{OE}^{2}=\overline{EH}^{2}+\overline{HO}^{2}$. Como $\overline{OE}=1-r$ e $\overline{EH}=\overline{HO}=r$, então $r^{2}+2r-1=0$, pelo que $r=\sqrt{2}-1$ e a área do círculo é $\pi r^2=\pi(3-2\sqrt{2})$. Solução 3: Por outro lado, os triângulos $[OHE]$ e $[BCE]$ são isósceles rectângulos, logo $\frac{\overline{EH}}{\overline{EC}}=\frac{\overline{EO}}{\overline{EB}}$ e assim $\sqrt{2}r=1-r$, pelo que $r=\sqrt{2}-1$ e a área do círculo é $\pi r^2=\pi(3-2\sqrt{2})$. Portanto, a área da região sombreada é $\pi(3-2\sqrt{2})+1$.


Conteúdo referenciado:
[image2]
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