![\"\"](\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/10\/lunulas-b.jpg\" "\\"L\u00fanulas\\"")
<\/a>L\u00fanulas de Hip\u00f3crates<\/p><\/div>\n
\n
\u00a0De acordo com as \u00e1reas assinaladas na figura, vem:<\/p>\n
$$\\begin{array}{*{20}{l}} {{A_{L\u00fanulas}}}& = &{{A_1} + {A_2}}\\\\ {}& = &{\\left[ {\\left( {{A_1} + {A_3}} \\right) + \\left( {{A_2} + {A_4}} \\right)} \\right] – \\left[ {\\left( {{A_3} + {A_4} + {A_5}} \\right) – {A_{\\left[ {ABC} \\right]}}} \\right]}\\\\ {}& = &{\\left( {\\frac{{{A_{C\u00edrculo – M\u00e9dio}}}}{2} + \\frac{{{A_{C\u00edrculo – Pequeno}}}}{2}} \\right) – \\left( {\\frac{{{A_{C\u00edrculo – Grande}}}}{2} – {A_{\\left[ {ABC} \\right]}}} \\right)}\\\\ {}& = &{\\underbrace {\\left( {\\frac{{{A_{C\u00edrculo – M\u00e9dio}}}}{2} + \\frac{{{A_{C\u00edrculo – Pequeno}}}}{2} – \\frac{{{A_{C\u00edrculo – Grande}}}}{2}} \\right)}_{0{\\rm{ }}(*)} + {A_{\\left[ {ABC} \\right]}}}\\\\ {}& = &{{A_{\\left[ {ABC} \\right]}}} \\end{array}$$<\/p>\n<\/p>\n
(*)
\nDe acordo com o Teorema de Pit\u00e1goras, temos:<\/p>\n
$$\\begin{array}{*{20}{l}} {\\frac{{{A_{C\u00edrculo – M\u00e9dio}}}}{2} + \\frac{{{A_{C\u00edrculo – Pequeno}}}}{2} – \\frac{{{A_{C\u00edrculo – Grande}}}}{2}}& = &{\\frac{{\\pi\u00a0 \\times {{\\left( {\\frac{4}{2}} \\right)}^2}}}{2} + \\frac{{\\pi\u00a0 \\times {{\\left( {\\frac{3}{2}} \\right)}^2}}}{2} – \\frac{{\\pi\u00a0 \\times {{\\left( {\\frac{5}{2}} \\right)}^2}}}{2}}\\\\ {}& = &{\\frac{{\\pi\u00a0 \\times {4^2}}}{8} + \\frac{{\\pi\u00a0 \\times {3^2}}}{8} – \\frac{{\\pi\u00a0 \\times {5^2}}}{8}}\\\\ {}& = &{\\frac{\\pi }{8}\\left( {{4^2} + {3^2} – {5^2}} \\right)}\\\\ {}& = &0 \\end{array}$$<\/p>\n
Note que este resultado n\u00e3o depende das medidas particulares dos catetos do tri\u00e2ngulo ret\u00e2ngulo, mas sim do facto de o tri\u00e2ngulo [ABC] ser ret\u00e2ngulo.<\/p>\n<\/p>\n
![\"\"](\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/10\/lunulas.jpg\" "\\"L\u00fanulas\\"")
<\/a>