{"id":7390,"date":"2012-02-12T19:57:12","date_gmt":"2012-02-12T19:57:12","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=7390"},"modified":"2022-01-16T21:52:05","modified_gmt":"2022-01-16T21:52:05","slug":"determina-a-area-de-um-octogono-regular","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=7390","title":{"rendered":"Determina a \u00e1rea de um oct\u00f3gono regular"},"content":{"rendered":"<p><ul id='GTTabs_ul_7390' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_7390' class='GTTabs_curr'><a  id=\"7390_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_7390' ><a  id=\"7390_1\" onMouseOver=\"GTTabsShowLinks('Resolu\u00e7\u00e3o'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Resolu\u00e7\u00e3o<\/a><\/li>\n<\/ul>\n\n<div class='GTTabs_divs GTTabs_curr_div' id='GTTabs_0_7390'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p>Determina a \u00e1rea de um oct\u00f3gono regular, sabendo que o lado do pol\u00edgono \u00e9 4 cm e o ap\u00f3tema \u00e9 $2\\left( {1 + \\sqrt 2 } \\right)$ cm.<\/p>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_7390' onClick='GTTabs_show(1,7390)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_7390'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><!--more--><\/p>\n<p>\u00a0<a href=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/02\/OctogonoRegular2.png\"><img loading=\"lazy\" decoding=\"async\" data-attachment-id=\"7391\" data-permalink=\"https:\/\/www.acasinhadamatematica.pt\/?attachment_id=7391\" data-orig-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/02\/OctogonoRegular2.png\" data-orig-size=\"335,341\" data-comments-opened=\"1\" data-image-meta=\"{&quot;aperture&quot;:&quot;0&quot;,&quot;credit&quot;:&quot;&quot;,&quot;camera&quot;:&quot;&quot;,&quot;caption&quot;:&quot;&quot;,&quot;created_timestamp&quot;:&quot;0&quot;,&quot;copyright&quot;:&quot;&quot;,&quot;focal_length&quot;:&quot;0&quot;,&quot;iso&quot;:&quot;0&quot;,&quot;shutter_speed&quot;:&quot;0&quot;,&quot;title&quot;:&quot;&quot;}\" data-image-title=\"Oct\u00f3gono\" data-image-description=\"\" data-image-caption=\"\" data-medium-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/02\/OctogonoRegular2-294x300.png\" data-large-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/02\/OctogonoRegular2.png\" class=\"alignright wp-image-7391 size-full\" title=\"Oct\u00f3gono\" src=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/02\/OctogonoRegular2.png\" alt=\"\" width=\"335\" height=\"341\" srcset=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/02\/OctogonoRegular2.png 335w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/02\/OctogonoRegular2-294x300.png 294w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/02\/OctogonoRegular2-147x150.png 147w\" sizes=\"auto, (max-width: 335px) 100vw, 335px\" \/><\/a>Comecemos por determinar a \u00e1rea do tri\u00e2ngulo [ABO]:<\/p>\n<p>$$\\begin{array}{*{20}{l}}<br \/>\n{{A_{[ABO]}}}&amp; = &amp;{\\frac{{\\overline {AB}\u00a0 \\times \\overline {OM} }}{2}} \\\\<br \/>\n{}&amp; = &amp;{\\frac{{4 \\times 2\\left( {1 + \\sqrt 2 } \\right)}}{2}} \\\\<br \/>\n{}&amp; = &amp;{4\\left( {1 + \\sqrt 2 } \\right)\\,\\,c{m^2}}<br \/>\n\\end{array}$$<\/p>\n<p>Logo a \u00e1rea do oct\u00f3gono \u00e9:<\/p>\n<p>$$\\begin{array}{*{20}{l}}<br \/>\nA&amp; = &amp;{8 \\times {A_{[ABO]}}} \\\\<br \/>\n{}&amp; = &amp;{8 \\times 4\\left( {1 + \\sqrt 2 } \\right)} \\\\<br \/>\n{}&amp; = &amp;{32\\left( {1 + \\sqrt 2 } \\right)\\,\\,c{m^2}}<br \/>\n\\end{array}$$<\/p>\n<\/p>\n<hr \/>\n<h5>Sobre a corre\u00e7\u00e3o dos dados do enunciado <sup>(1)<\/sup><\/h5>\n<p><sup>(1)<\/sup> Este aditamento vem no seguimento do coment\u00e1rio submetido no dia 26-07-2015.<\/p>\n<\/p>\n<p><a href=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/02\/Octogono-validacao.png\"><img loading=\"lazy\" decoding=\"async\" data-attachment-id=\"12448\" data-permalink=\"https:\/\/www.acasinhadamatematica.pt\/?attachment_id=12448\" data-orig-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/02\/Octogono-validacao.png\" data-orig-size=\"1128,862\" data-comments-opened=\"1\" data-image-meta=\"{&quot;aperture&quot;:&quot;0&quot;,&quot;credit&quot;:&quot;&quot;,&quot;camera&quot;:&quot;&quot;,&quot;caption&quot;:&quot;&quot;,&quot;created_timestamp&quot;:&quot;0&quot;,&quot;copyright&quot;:&quot;&quot;,&quot;focal_length&quot;:&quot;0&quot;,&quot;iso&quot;:&quot;0&quot;,&quot;shutter_speed&quot;:&quot;0&quot;,&quot;title&quot;:&quot;&quot;,&quot;orientation&quot;:&quot;0&quot;}\" data-image-title=\"Oct\u00f3gono regular\" data-image-description=\"\" data-image-caption=\"\" data-medium-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/02\/Octogono-validacao-300x229.png\" data-large-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/02\/Octogono-validacao-1024x783.png\" class=\"alignright wp-image-12448\" src=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/02\/Octogono-validacao-300x229.png\" alt=\"Oct\u00f3gono regular\" width=\"500\" height=\"382\" srcset=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/02\/Octogono-validacao-300x229.png 300w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/02\/Octogono-validacao-1024x783.png 1024w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/02\/Octogono-validacao.png 1128w\" sizes=\"auto, (max-width: 500px) 100vw, 500px\" \/><\/a>Seja \\(\\left[ {A,B,C,D,E,F,G,H} \\right]\\) um oct\u00f3gono regular, inscrito numa circunfer\u00eancia de raio \\(r\\), cujo comprimento do lado \u00e9 \\(\\overline {AB} = \\overline {BC} = \\cdots = \\overline {HA} = 4\\;cm\\).<\/p>\n<p>Vamos mostrar que o comprimento do ap\u00f3tema \u00e9, de facto, \\(a = \\overline {OM} = \\overline {ON} = \\cdots = 2\\left( {1 + \\sqrt 2 } \\right)\\;cm\\).<\/p>\n<p>Conv\u00e9m reparar que:<\/p>\n<ul>\n<li>o oct\u00f3gono foi dividido em oito tri\u00e2ngulos is\u00f3sceles, congruentes entre si;<\/li>\n<li>cada um destes tri\u00e2ngulos is\u00f3sceles foi dividido, pela altura relativa ao lado do oct\u00f3gono, em dois tri\u00e2ngulos ret\u00e2ngulos congruentes;<\/li>\n<li>os \u00e2ngulos de amplitude \\(\\alpha \\) e \\(\\beta \\) s\u00e3o complementares, isto \u00e9, \\(\\alpha + \\beta = 90^\\circ \\);<\/li>\n<li>\\(\\alpha = \\frac{{A\\widehat OC}}{4} = \\frac{{90^\\circ }}{4} = \\frac{{45^\\circ }}{2} = 22,5^\\circ \\) e \u00a0\\(\\beta = 90^\\circ &#8211; \\alpha = 90^\\circ &#8211; \\frac{{90^\\circ }}{4} = \\frac{{270^\\circ }}{4} = \\frac{{135^\\circ }}{2} = 67,5^\\circ \\).<\/li>\n<\/ul>\n<p>O tri\u00e2ngulo ret\u00e2ngulo (em O) [AOC] \u00e9 is\u00f3sceles, sendo o comprimento da sua hipotenusa:<\/p>\n<p>\\[\\overline {AC} \u00a0= \\sqrt {{r^2} + {r^2}} \u00a0= \\sqrt {2{r^2}} \u00a0= \\sqrt 2 \u00a0\\times r\\]<\/p>\n<p>Logo, \\(\\overline {OP} = \\overline {PC} = \\frac{{\\overline {AC} }}{2} = \\frac{{\\sqrt 2 }}{2} \\times r\\).<\/p>\n<p>Neste momento \u00e9 conveniente reparar que os tri\u00e2ngulos [OPQ] e [CPB] s\u00e3o congruentes (ALA), pois, como vimos imediatamente acima, \\(\\overline {OP} = \\overline {PC} = \\frac{{\\overline {AC} }}{2} = \\frac{{\\sqrt 2 }}{2} \\times r\\), pelo que ser\u00e1 \\(\\overline {OQ} = \\overline {BC} = 4\\).<\/p>\n<\/p>\n<p>Os tri\u00e2ngulos ret\u00e2ngulos [OPQ] e [OBN] s\u00e3o semelhantes (AA), pelo que os comprimentos dos lados correspondentes s\u00e3o diretamente proporcionais:<\/p>\n<p>\\[\\frac{{\\overline {OP} }}{{\\overline {ON} }} = \\frac{{\\overline {PQ} }}{{\\overline {BN} }} = \\frac{{\\overline {OQ} }}{{\\overline {OB} }}\\]<\/p>\n<p>Usando a primeira e terceira raz\u00f5es da igualdade acima, vem:<\/p>\n<p>\\[\\begin{array}{*{20}{l}}{\\frac{{\\overline {OP} }}{{\\overline {ON} }} = \\frac{{\\overline {OQ} }}{{\\overline {OB} }}}&amp; \\Leftrightarrow &amp;{\\frac{{\\frac{{\\sqrt 2 \\times r}}{2}}}{a} = \\frac{4}{r}}\\\\{}&amp; \\Leftrightarrow &amp;{a = \\frac{{\\sqrt 2 \\times {r^2}}}{8}}\\end{array}\\]<\/p>\n<p>Aplicando o Teorema de Pit\u00e1goras no tri\u00e2ngulo ret\u00e2ngulo [AOM], temos: \\({r^2} = {2^2} + {a^2}\\).<br \/>\nSubstituindo \\({r^2}\\) na igualdade anterior, vem:<\/p>\n<p>\\[\\begin{array}{*{20}{l}}{a = \\frac{{\\sqrt 2 \\times \\left( {4 + {a^2}} \\right)}}{8}}&amp; \\Leftrightarrow &amp;{\\sqrt 2 \\times {a^2} &#8211; 8a + 4\\sqrt 2 = 0}\\\\{}&amp; \\Leftrightarrow &amp;{a = \\frac{{8 \\mp \\sqrt {64 &#8211; 4 \\times \\sqrt 2 \\times 4\\sqrt 2 } }}{{2\\sqrt 2 }}}\\\\{}&amp; \\Leftrightarrow &amp;{a = \\frac{{8 \\mp \\sqrt {32} }}{{2\\sqrt 2 }}}\\\\{}&amp; \\Leftrightarrow &amp;{a = \\frac{{8 \\mp 4\\sqrt 2 }}{{2\\sqrt 2 }}}\\\\{}&amp; \\Leftrightarrow &amp;{a = \\frac{8}{{2\\sqrt 2 }} \\mp \\frac{{4\\sqrt 2 }}{{2\\sqrt 2 }}}\\\\{}&amp; \\Leftrightarrow &amp;{a = \\frac{4}{{\\sqrt 2 }} \\times \\frac{{\\sqrt 2 }}{{\\sqrt 2 }} \\mp 2}\\\\{}&amp; \\Leftrightarrow &amp;{a = 2\\sqrt 2 \\mp 2}\\\\{}&amp; \\Leftrightarrow &amp;{a = 2 \\times \\left( {\\sqrt 2 \\mp 1} \\right)}\\end{array}\\]<\/p>\n<\/p>\n<p>No tri\u00e2ngulo [OAM], como \\(\\beta &gt; \\alpha \\), ent\u00e3o \\(\\overline {OM} &gt; \\overline {AM} \\Leftrightarrow a &gt; 2\\).<\/p>\n<p>Portanto, a solu\u00e7\u00e3o da equa\u00e7\u00e3o do 2.\u00ba grau que nos interessa \u00e9 \\(a = 2 \\times \\left( {\\sqrt 2 + 1} \\right)\\).<\/p>\n<\/p>\n<p>Assim, conclui-se que \\(a = 2 \\times \\left( {1 + \\sqrt 2 } \\right)\\;cm\\) e comprova-se a corre\u00e7\u00e3o dos dados do enunciado.<\/p><\/p>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_7390' onClick='GTTabs_show(0,7390)'>&lt;&lt; Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Enunciado Resolu\u00e7\u00e3o Enunciado Determina a \u00e1rea de um oct\u00f3gono regular, sabendo que o lado do pol\u00edgono \u00e9 4 cm e o ap\u00f3tema \u00e9 $2\\left( {1 + \\sqrt 2 } \\right)$ cm. Resolu\u00e7\u00e3o &gt;&gt; Resolu\u00e7\u00e3o&#46;&#46;&#46;<\/p>\n","protected":false},"author":1,"featured_media":20422,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[213,97,278],"tags":[426,108,283],"series":[],"class_list":["post-7390","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-9--ano","category-aplicando","category-circunferencia-e-poligonos","tag-9-o-ano","tag-area","tag-poligono-regular"],"views":22256,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/02\/9V1Pag037-1_520x245.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/7390","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=7390"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/7390\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/20422"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=7390"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=7390"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=7390"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=7390"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}