{"id":23313,"date":"2022-11-22T23:54:51","date_gmt":"2022-11-22T23:54:51","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=23313"},"modified":"2022-12-13T01:24:29","modified_gmt":"2022-12-13T01:24:29","slug":"um-hexagono-regular-3","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=23313","title":{"rendered":"Um hex\u00e1gono regular"},"content":{"rendered":"\n<p><ul id='GTTabs_ul_23313' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_23313' class='GTTabs_curr'><a  id=\"23313_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_23313' ><a  id=\"23313_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_23313'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p><img decoding=\"async\" class=\"alignright\" src=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2022\/11\/8_Pag058-15-300x245.png\" \/>Na figura est\u00e1 representado um hex\u00e1gono regular, com 5 cm de lado, inscrito numa circunfer\u00eancia de centro O e um quadrado circunscrito \u00e0 mesma circunfer\u00eancia.<\/p>\n<ol>\n<li>Decomp\u00f5e o hex\u00e1gono em tri\u00e2ngulos, em que um dos v\u00e9rtices \u00e9 O e o lado oposto \u00e9 um dos lados do pol\u00edgono. Como classificas cada um desses tri\u00e2ngulos quanto aos lados? Justifica.<\/li>\n<li>Determina:<\/li>\n<\/ol>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>o comprimento do ap\u00f3tema do hex\u00e1gono;<\/li>\n<li>a \u00e1rea do hex\u00e1gono;<\/li>\n<li>o per\u00edmetro do quadrado.<\/li>\n<\/ol>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_23313' onClick='GTTabs_show(1,23313)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_23313'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><!--more--><\/p>\n<ol>\n<li><a href=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2022\/11\/8_Pag058-15_R.png\"><img loading=\"lazy\" decoding=\"async\" data-attachment-id=\"23319\" data-permalink=\"https:\/\/www.acasinhadamatematica.pt\/?attachment_id=23319\" data-orig-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2022\/11\/8_Pag058-15_R.png\" data-orig-size=\"1249,1254\" 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=\"8_Pag058-15_R\" data-image-description=\"\" data-image-caption=\"\" data-large-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2022\/11\/8_Pag058-15_R-1020x1024.png\" class=\"alignright wp-image-23319\" src=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2022\/11\/8_Pag058-15_R-300x300.png\" alt=\"\" width=\"320\" height=\"321\" srcset=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2022\/11\/8_Pag058-15_R-300x300.png 300w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2022\/11\/8_Pag058-15_R-1020x1024.png 1020w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2022\/11\/8_Pag058-15_R-150x150.png 150w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2022\/11\/8_Pag058-15_R-768x771.png 768w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2022\/11\/8_Pag058-15_R-80x80.png 80w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2022\/11\/8_Pag058-15_R-320x320.png 320w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2022\/11\/8_Pag058-15_R.png 1249w\" sizes=\"auto, (max-width: 320px) 100vw, 320px\" \/><\/a>O hex\u00e1gono est\u00e1 decomposto em seis tri\u00e2ngulos equil\u00e1teros e geometricamente iguais, conforme se explicar\u00e1 de seguida.<br \/>Ora, o \u00e2ngulo AOB \u00e9 um sexto de um \u00e2ngulo giro e, por isso, a sua amplitude \u00e9 60\u00b0.<br \/>Podemos tamb\u00e9m reparar que os segmentos de reta [AO] e [BO] s\u00e3o raios da mesma circunfer\u00eancia. Consequentemente, os \u00e2ngulos internos do tri\u00e2ngulo [ABO] opostos a esses lados ser\u00e3o geometricamente iguais, pois, num tri\u00e2ngulo, a lados iguais op\u00f4em-se \u00e2ngulos iguais.<br \/>Assim, podemos determinar a amplitude dos \u00e2ngulos OAB e OBA:<br \/>\\[O\\widehat AB = O\\widehat BA = \\frac{{180^\\circ &#8211; A\\widehat OB}}{2} = \\frac{{180^\\circ &#8211; 60^\\circ }}{2} = 60^\\circ \\]<br \/>Desta forma, como o tri\u00e2ngulo [AOB] \u00e9 equi\u00e2ngulo, ent\u00e3o ser\u00e1 tamb\u00e9m equil\u00e1tero.<br \/>Por isso, o hex\u00e1gono est\u00e1 decomposto em seis tri\u00e2ngulos equil\u00e1teros e geometricamente iguais.<br \/><br \/><\/li>\n<li>\u00a0<\/li>\n<\/ol>\n<ol style=\"list-style-type: lower-alpha;\">\n<li>O ap\u00f3tema do hex\u00e1gono \u00e9 o segmento de reta [OM].<br \/>Conv\u00e9m recordar que \\(\\overline {OA} = \\overline {OB} = \\overline {AB} = r = 5\\) cm.<br \/>Aplicando o Teorema de Pit\u00e1goras no tri\u00e2ngulo ret\u00e2ngulo [ABO], temos:<br \/>\\[\\begin{array}{*{20}{l}}{\\overline {OM} }&amp; = &amp;{\\sqrt {{{\\overline {OB} }^2} &#8211; {{\\overline {BM} }^2}} }\\\\{}&amp; = &amp;{\\sqrt {{5^2} &#8211; {{\\left( {\\frac{5}{2}} \\right)}^2}} }\\\\{}&amp; = &amp;{\\sqrt {25 &#8211; \\frac{{25}}{4}} }\\\\{}&amp; = &amp;{\\sqrt {\\frac{{4 \\times 25}}{4} &#8211; \\frac{{25}}{4}} }\\\\{}&amp; = &amp;{\\sqrt {3 \\times \\frac{{25}}{4}} }\\\\{}&amp; = &amp;{\\frac{5}{2}\\sqrt 3 }\\end{array}\\]<br \/>Portanto, o comprimento do ap\u00f3tema do hex\u00e1gono \u00e9\u00a0\\({a_p} = \\overline {OM} = \\frac{5}{2}\\sqrt 3 = \\sqrt {18,75} \\) cm.<br \/><br \/><\/li>\n<li>A \u00e1rea do hex\u00e1gono \u00e9 \\({A_{{\\rm{Hex\u00e1gono}}}} = \\frac{{75\\sqrt 3 }}{2} = 15\\sqrt {18,75} \\) cm<sup>2<\/sup>:<br \/>\\[{A_{{\\rm{Hex\u00e1gono}}}} = 6 \\times {A_{\\left[ {ABO} \\right]}} = 6 \\times \\frac{{\\overline {AB} \\times \\overline {OM} }}{2} = 6 \\times \\frac{{5 \\times \\frac{5}{2}\\sqrt 3 }}{2} = 3 \\times 5 \\times \\frac{5}{2}\\sqrt 3 = \\frac{{75\\sqrt 3 }}{2}\\]<\/li>\n<li>O per\u00edmetro do quadrado \u00e9 \\({P_{{\\rm{Quadrado}}}} = 4 \\times \\left( {2 \\times \\overline {AB} } \\right) = 4 \\times 10 = 40\\) cm.<\/li>\n<\/ol>\n<p>\u00a0<\/p>\n<p>\u00a0<\/p>\n<p>\u00a0<\/p>\n<p>\u00a0<\/p>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_23313' onClick='GTTabs_show(0,23313)'>&lt;&lt; Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Enunciado Resolu\u00e7\u00e3o Enunciado Na figura est\u00e1 representado um hex\u00e1gono regular, com 5 cm de lado, inscrito numa circunfer\u00eancia de centro O e um quadrado circunscrito \u00e0 mesma circunfer\u00eancia. Decomp\u00f5e o hex\u00e1gono em tri\u00e2ngulos, em&#46;&#46;&#46;<\/p>\n","protected":false},"author":1,"featured_media":23315,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[100,97,682],"tags":[424,108,67,335,118],"series":[],"class_list":["post-23313","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-8--ano","category-aplicando","category-teorema-de-pitagoras","tag-8-o-ano","tag-area","tag-geometria","tag-perimetro","tag-teorema-de-pitagoras"],"views":220,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2022\/11\/8_Pag058-15_520x245.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/23313","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=23313"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/23313\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/23315"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=23313"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=23313"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=23313"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=23313"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}