Problema (nível 3): O João tem um terreno triangular, representado na figura ao lado, cujos lados medem $5$, $5$ e $6$ metros. O terreno está dividido em dois triângulos retângulos: uma piscina azul e uma área relvada verde. Qual é a área da piscina? (image1)


Conteúdo referenciado:
[image1]
\begin{pspicture*}(-1.464365580943121,-1.044628099173556)(11.370928536703936,5.567136606708796) \pspolygon[linewidth=0.pt](0.,0.)(3.,4.)(6.,0.) \pspolygon[linewidth=0.pt,fillcolor=black,fillstyle=solid,opacity=0.25](2.0102597404546136,2.680346320606151)(2.2099134198484625,2.5306060610607646)(2.359653679393849,2.7302597404546134)(2.16,2.88) \pspolygon[linestyle=none,fillcolor=azulclarologotipo,fillstyle=solid](2.16,2.88)(3.,4.)(6.,0.) \pspolygon[linestyle=none,fillcolor=verdeescurologotipo,fillstyle=solid](2.16,2.88)(6.,0.)(0.,0.) \pspolygon[linewidth=0.pt,fillcolor=black,fillstyle=solid,opacity=0.25](2.359653679393849,2.7302597404546134)(2.5093939389392355,2.929913419848462)(2.3097402595453866,3.0796536793938487)(2.16,2.88) \psline[linewidth=0.4pt](6.,0.)(3.839847819050873,2.8798858580056135) \psline[linewidth=0.4pt](2.16,2.88)(6.,0.) \psline[linewidth=0.4pt](2.16,2.88)(3.,4.) \psline[linewidth=0.4pt](3.,4.)(6.,0.) \psline[linewidth=0.4pt](6.,0.)(2.16,2.88) \rput[tl](4.45,2.4){$5$} \rput[tl](1.3,2.4){$5$} \rput[tl](2.9238697131745255,-0.2210986874088502){$6$} \psline[linewidth=0.4pt](2.16,2.88)(6.,0.) \psline[linewidth=0.4pt](6.,0.)(0.,0.) \psline[linewidth=0.4pt](0.,0.)(2.16,2.88) \psdots[dotsize=1pt 0,dotstyle=*](6.,0.) \psdots[dotsize=1pt 0,dotstyle=*](0.,0.) \psdots[dotsize=1pt 0,dotstyle=*](3.,4.) \psdots[dotsize=1pt 0,dotstyle=*](2.16,2.88) \end{pspicture*}

Resolução: Seja $I$ o pé da perpendicular relativa ao lado $[AB]$. (image2) Como o triângulo $[ABC]$ é isósceles, pois $\overline{CB}=\overline{CA}=5$, tem-se $\overline{AI}=\overline{BI}=3$. O triângulo $[CIA]$ é retângulo e o Teorema de Pitágoras garante que $$ \overline{CI}^2=\overline{CA}^2-\overline{IA}^2=4^2, $$ ou seja, $\ds \overline{CI}=4$. Pelo critério $AA$ observa-se que os triângulos $[CIB]$ e $[AHB]$ são semelhantes e $$ \frac{\overline{AH}}{\overline{CI}}=\frac{\overline{HB}}{\overline{IB}}=\frac{\overline{AB}}{\overline{CB}}, $$ ou seja, $$ \frac{\overline{AH}}{4}=\frac{\overline{HB}}{3}=\frac{6}{5}. $$ Assim conclui-se que $\ds \overline{AH}=\frac{24}{5}$ e $\overline{HB}=\frac{18}{5}$. Portanto $\ds \overline{CH}=5-\frac{18}{5}=\frac{7}{5}$ e $$\mbox{área de } [CHA]=\frac{1}{2}\overline{CH}\times \overline{HA}=\frac{1}{2}.\frac{7}{5}\times \frac{24}{5}=\frac{84}{25}\,\mbox{m}^2.$$


Conteúdo referenciado:
[image2]
\begin{pspicture*}(-1.464365580943121,-1.044628099173556)(11.370928536703936,5.567136606708796) \pspolygon[linewidth=0.pt](0.,0.)(3.,4.)(6.,0.) \pspolygon[linewidth=0.pt,fillcolor=black,fillstyle=solid,opacity=0.25](2.0102597404546136,2.680346320606151)(2.2099134198484625,2.5306060610607646)(2.359653679393849,2.7302597404546134)(2.16,2.88) \pspolygon[linewidth=0.4pt,linecolor=qqzzff,fillcolor=qqzzff,fillstyle=solid,opacity=0.1](2.16,2.88)(3.,4.)(6.,0.) \pspolygon[linewidth=0.4pt,linecolor=zzffqq,fillcolor=zzffqq,fillstyle=solid,opacity=0.5](2.16,2.88)(6.,0.)(0.,0.) \pspolygon[linewidth=0.pt,fillcolor=black,fillstyle=solid,opacity=0.25](2.359653679393849,2.7302597404546134)(2.5093939389392355,2.929913419848462)(2.3097402595453866,3.0796536793938487)(2.16,2.88) \psline[linewidth=0.4pt](6.,0.)(3.,0.) \psline[linewidth=0.4pt](2.16,2.88)(6.,0.) \psline[linewidth=0.4pt](2.16,2.88)(3.,4.) \psline[linewidth=0.4pt](3.,4.)(6.,0.) \psline[linewidth=0.4pt](6.,0.)(2.16,2.88) \rput[tl](4.406222654350996,2.8259601361205613){5} \rput[tl](1.2415167719980549,2.320077783179385){5} \rput[tl](2.9238697131745255,-0.2210986874088502){6} \psline[linewidth=0.4pt](2.16,2.88)(6.,0.) \psline[linewidth=0.4pt](6.,0.)(0.,0.) \psline[linewidth=0.4pt](0.,0.)(2.16,2.88) \psline[linewidth=0.4pt,linestyle=dashed,dash=3pt 3pt](3.,4.)(3.,0.) \psdots[dotsize=1pt 0,dotstyle=*](6.,0.) \rput(6.2,0){$A$} \psdots[dotsize=1pt 0,dotstyle=*](0.,0.) \rput(-0.3,0){$B$} \psdots[dotsize=1pt 0,dotstyle=*](3.,4.) \rput[bl](2.8415167719980547,4.155371900826443){$C$} \psdots[dotsize=1pt 0,dotstyle=*](3.,0.) \rput[bl](3.0885755955274665,0.1){$I$} \psdots[dotsize=1pt 0,dotstyle=*](2.16,2.88) \rput[bl](1.8,2.9200777831793845){$H$} \end{pspicture*}