{"id":171592,"date":"2020-11-19T04:02:17","date_gmt":"2020-11-19T06:02:17","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=171592"},"modified":"2023-07-24T21:51:17","modified_gmt":"2023-07-25T00:51:17","slug":"problema-intersecao-de-hexagonos","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/problema-intersecao-de-hexagonos\/","title":{"rendered":".Problema: Interse\u00e7\u00e3o de hex\u00e1gonos"},"content":{"rendered":"<p style=\"margin:10px\">\n<div class=\"g1n\" style=\"line-height:28px;padding-bottom:15px\">\n<p class=\"tit\" style=\"line-height:25px;margin:10px\">Problema<br \/>\n<span style=\"font-size:12px;color:#800000\"> (Indicado a partir do 9\u00ba ano do E. F.)<\/span><\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 20px;border: 0 none\"\/>\nA figura a seguir mostra dois hex\u00e1gonos regulares [tex]ABCDEF[\/tex] e [tex]BGHICJ[\/tex] com dois v\u00e9rtices coincidentes. <\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2020\/09\/hexagonos.png\" style=\"border:none\" alt=\"\" width=\"265\" height=\"220\" class=\"aligncenter size-full wp-image-165764\" \/><\/p>\n<p>O hex\u00e1gono menor tem \u00e1rea igual a [tex]\\dfrac{3\\sqrt{3}}{2}\\;\\text{cm}^2.[\/tex]<br \/>\n<strong>Qual a \u00e1rea do hex\u00e1gono maior?<\/strong><\/p>\n<p style=\"margin:10px;\">\n<p style=\"font-size: 12px;text-align:right;line-height:10px;\">Adaptado de <strong> PSS para professor substituto SEMEC<\/strong> &#8211; Teresina.<\/p>\n<\/div>\n<p style=\"margin:40px\">\n<div class=\"cb36 g1n\" style=\"margin:30px;line-height:25px; width:60%; margin-left:auto; margin-right:auto; \">\n<p class=\"c3 ti3\" style=\"line-height:25px;margin:10px\"> Ajudas<\/p>\n<p><span class=\"c3\"><strong> (1)<\/strong> <\/span> Um tri\u00e2ngulo equil\u00e1tero de lado medindo [tex]l\\text{ cm}[\/tex] tem altura dada por [tex]h = \\dfrac{l\\sqrt{3}}{2}\\text{ cm}[\/tex] e \u00e1rea dada por [tex]A = \\dfrac{l^2\\sqrt{3}}{4} \\text{ cm}^2.[\/tex]<\/p>\n<p style=\"margin:10px\">\n<p><span class=\"c3\"><strong> (2)<\/strong> <\/span> Dado um hex\u00e1gono regular de lado medindo [tex]l \\text{ cm}[\/tex], este pode ser dividido em seis tri\u00e2ngulos equil\u00e1teros de lado medindo [tex]l \\text{ cm}.[\/tex]<\/p>\n<p style=\"text-align:center; line-height:20px;font-size:14px; margin:30px\">\n<iframe loading=\"lazy\" title=\"Decomposi\u00e7\u00e3o do hex\u00e1gono\" src=\"https:\/\/www.geogebra.org\/material\/iframe\/id\/ebw7wevg\/width\/466\/height\/492\/border\/ffffff\/smb\/false\/stb\/false\/stbh\/false\/ai\/false\/asb\/false\/sri\/false\/rc\/false\/ld\/false\/sdz\/false\/ctl\/false\" width=\"500px\" height=\"540px\" style=\"border: 2px solid #0099ff;\"> <\/iframe><br \/>\nOBMEP_srg, criado com o <a href=\"http:\/\/www.geogebra.org\/\" target=\"_blank\" >GeoGebra<\/a><\/p>\n<p style=\"margin:10px\">\n<\/div>\n<p style=\"margin:40px\">\n<div class=\"g1n\" style=\"line-height:28px;\">\n<p class=\"tit\" style=\"line-height:28px;margin:10px\">Solu\u00e7\u00e3o<\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 20px;border: 0 none\"\/>\n<p>Seja [tex]l[\/tex] a medida em cent\u00edmetros do lado do menor hex\u00e1gono. Pela <strong>Ajuda 2<\/strong>, esse hex\u00e1gono pode ser dividido em seis tri\u00e2ngulos equil\u00e1teros de lado [tex]l[\/tex]. <\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2020\/11\/fig008.png\" alt=\"\" style=\"border:none\" width=\"190\" height=\"200\" class=\"aligncenter size-full wp-image-171600\" \/><\/p>\n<p>Como a \u00e1rea do menor hex\u00e1gono \u00e9 [tex]\\dfrac{3\\sqrt{3}}{2}\\text{ cm}^2[\/tex], ent\u00e3o a \u00e1rea [tex]A_t[\/tex] de cada um desses tri\u00e2ngulos \u00e9 dada por:<br \/>\n [tex]\\quad A_t=\\dfrac{\\frac{3\\sqrt{3}}{2}}{6} \\\\<br \/>\n\\quad A_t= \\dfrac{\\sqrt{3}}{4} \\text{ cm}^2.[\/tex]<br \/>\nMas, pela <strong>Ajuda 1<\/strong>, a \u00e1rea de um tri\u00e2ngulo equil\u00e1tero de lado medindo [tex]l\\text{ cm}[\/tex] \u00e9 [tex]\\dfrac{l^2\\sqrt{3}}{4} \\;\\text{cm}^2[\/tex]; assim, em particular, [tex]A_t=\\dfrac{l^2\\sqrt{3}}{4}[\/tex] e, portanto, segue que:<br \/>\n[tex]\\qquad \\dfrac{l^2\\sqrt{3}}{4} = \\dfrac{\\sqrt{3}}{4}\\\\<br \/>\n\\qquad l^2 = 1\\\\<br \/>\n\\qquad l = 1. [\/tex]<br \/>\nAssim, a medida [tex]l[\/tex] do lado do menor hex\u00e1gono \u00e9 [tex]l=1 \\text{ cm}[\/tex].<br \/>\nNovamente pela <strong>Ajuda 1<\/strong>, a altura [tex]h[\/tex] de cada um desses tri\u00e2ngulos equil\u00e1teros \u00e9 dada por:<br \/>\n [tex]\\qquad h = \\dfrac{l\\sqrt{3}}{2} = \\dfrac{\\sqrt{3}}{2} \\text{ cm}[\/tex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2020\/11\/fig009.png\" style=\"border:none\" alt=\"\" width=\"209\" height=\"204\" class=\"aligncenter size-full wp-image-171601\" \/><\/p>\n<p>Agora precisamos encontrar a medida do lado do maior hex\u00e1gono. Para isso, observe a figura abaixo, onde [tex]O[\/tex] \u00e9 o centro do hex\u00e1gono [tex]BGHICJ[\/tex] e perceba que a <strong>Ajuda 2<\/strong> garante que esse hex\u00e1gono tamb\u00e9m pode ser dividido em seis tri\u00e2ngulos equil\u00e1teros. <\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2020\/11\/fig012.png\"style=\"border:none\"  alt=\"\" width=\"493\" height=\"395\" class=\"aligncenter size-full wp-image-171604\" srcset=\"https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2020\/11\/fig012.png 493w, https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2020\/11\/fig012-300x240.png 300w\" sizes=\"auto, (max-width: 493px) 100vw, 493px\" \/><\/p>\n<p>Note que o lado [tex]BC[\/tex] do maior hex\u00e1gono \u00e9 igual \u00e0 soma das alturas dos tri\u00e2ngulos equil\u00e1teros [tex]BOJ[\/tex] e [tex]COJ[\/tex] e como estes s\u00e3o iguais, sendo [tex]L[\/tex] a medida do lado do maior hex\u00e1gono, temos<br \/>\n[tex]\\qquad L = 2h\\\\<br \/>\n\\qquad L =  2\\dfrac{\\sqrt{3}}{2}\\\\<br \/>\n\\qquad L = \\sqrt{3}\\text{ cm}. [\/tex]<\/p>\n<p>Com isso, utilizando uma vez mais a<strong> Ajuda 1<\/strong>, temos que a \u00e1rea [tex]A_T[\/tex] de cada um dos tri\u00e2ngulos equil\u00e1teros que comp\u00f5em o hex\u00e1gono maior \u00e9  dada por:<br \/>\n [tex]\\qquad A_T=\\dfrac{(\\sqrt{3})^2\\sqrt{3}}{4}=\\dfrac{3\\,\\sqrt{3}}{4} \\text{ cm}^2.[\/tex]<br \/>\nLogo, sendo [tex]A[\/tex] a \u00e1rea do maior hex\u00e1gono, temos<br \/>\n[tex]\\qquad A = 6\\cdot \\left(\\dfrac{3\\sqrt{3}}{4}\\right)\\\\<br \/>\n\\qquad \\fcolorbox{black}{#eee0e5}{$A = \\dfrac{9\\sqrt{3}}{2}\\text{ cm}^2$}\\,.[\/tex]<\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 15px;border: 0 none\"\/>\n<p class=\"g2\" style=\"line-height:20px;\">Solu\u00e7\u00e3o elaborada pelos <strong>Moderadores do Blog<\/strong>.<\/p>\n<\/div>\n<p style=\"margin:10px\">\n","protected":false},"excerpt":{"rendered":"<p>Problema (Indicado a partir do 9\u00ba ano do E. F.) A figura a seguir mostra dois hex\u00e1gonos regulares [tex]ABCDEF[\/tex] e [tex]BGHICJ[\/tex] com dois v\u00e9rtices coincidentes. O hex\u00e1gono menor tem \u00e1rea igual a [tex]\\dfrac{3\\sqrt{3}}{2}\\;\\text{cm}^2.[\/tex] Qual a \u00e1rea do hex\u00e1gono maior? Adaptado de PSS para professor substituto SEMEC &#8211; Teresina. Ajudas (1) Um tri\u00e2ngulo equil\u00e1tero de lado &hellip; <\/p>\n<p><a class=\"more-link btn\" href=\"https:\/\/clubes.obmep.org.br\/blog\/problema-intersecao-de-hexagonos\/\">Continue lendo<\/a><\/p>\n","protected":false},"author":254,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"template-onecolumn.php","meta":{"footnotes":""},"class_list":["post-171592","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/171592","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/users\/254"}],"replies":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/comments?post=171592"}],"version-history":[{"count":15,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/171592\/revisions"}],"predecessor-version":[{"id":171599,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/171592\/revisions\/171599"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=171592"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}