![\"\"](\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2010\/11\/pag-32-7.jpg\")
<\/a>Comecemos por determinar o comprimento da hipotenusa:\n$$\\begin{array}{*{35}{l}}
\n{{h}^{2}}={{6}^{2}}+{{8}^{2}} & \\Leftrightarrow\u00a0 & {{h}^{2}}=36+64\u00a0 \\\\
\n{} & \\Leftrightarrow\u00a0 & {{h}^{2}}=100\u00a0 \\\\
\n{} & Logo, & h=10\u00a0 \\\\
\n\\end{array}$$<\/p>\n
Tendo em conta que a f\u00f3rmula para calcular a \u00e1rea de um c\u00edrculo \u00e9 ${{A}_{C\\acute{i}rculo}}=\\pi {{r}^{2}}$, temos:<\/p>\n
\u00c1rea do semic\u00edrculo pequeno<\/strong>: \\[{{A}_{SCp}}=\\frac{\\pi \\times {{3}^{2}}}{2}=\\frac{9\\pi }{2}\\,\\,c{{m}^{2}}\\]<\/p>\n\u00c1rea do semic\u00edrculo m\u00e9dio<\/strong>: \\[{{A}_{SCm}}=\\frac{\\pi \\times {{4}^{2}}}{2}=\\frac{16\\pi }{2}\\,\\,c{{m}^{2}}\\]<\/p>\n\u00c1rea do semic\u00edrculo grande<\/strong>: \\[{{A}_{SCg}}=\\frac{\\pi \\times {{5}^{2}}}{2}=\\frac{25\\pi }{2}\\,\\,c{{m}^{2}}\\]
\n\u00ad<\/p>\n<\/li>\n\nComparando as \u00e1reas determinadas, conclui-se: \\[{{A}_{SCp}}+{{A}_{SCm}}=\\frac{9\\pi }{2}+\\frac{16\\pi }{2}=\\frac{25\\pi }{2}={{A}_{SCg}}\\]<\/p>\n
\nA soma das \u00e1reas dos semic\u00edrculos sobre os catetos \u00e9 igual \u00e0 \u00e1rea do semic\u00edrculo sobre a hipotenusa.<\/strong><\/p>\n<\/blockquote>\n<\/li>\n<\/ol>\n\u00ad
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