{"id":219664,"date":"2024-06-02T08:39:16","date_gmt":"2024-06-02T11:39:16","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=219664"},"modified":"2024-06-02T08:39:16","modified_gmt":"2024-06-02T11:39:16","slug":"a-problema-para-ajudar-na-escola-um-triangulo-e-um-hexagono","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/a-problema-para-ajudar-na-escola-um-triangulo-e-um-hexagono\/","title":{"rendered":"(A) Problema para ajudar na escola: Um tri\u00e2ngulo e um hex\u00e1gono"},"content":{"rendered":"<p class=\"ti3\"  style=\"text-align:center;\">Clique no bot\u00e3o abaixo para visualizar o problema.<\/p>\n<div style=\"text-align:center;margin-top:10px;\">\n<input class=\"bo1 c1 cb13\" style=\"width:150px;\" value=\"Problema\" title=\"Clique aqui\" type=\"button\" onclick=    \"if(this.parentNode.getElementsByTagName('div')[0].style.display != ''){this.parentNode.getElementsByTagName('div')[0].style.display = '';this.value = 'Ocultar';} else{this.parentNode.getElementsByTagName('div')[0].style.display = 'none'; this.value = 'Problema';}\" \/>  <\/p>\n<div style=\"display: none;\">\n<p style=\"margin:10px\">\n<div class=\"g1n\" style=\"line-height:24px;\">\n<p class=\"tit\" style=\"line-height:25px;margin:10px\">Problema<br \/>\n<span style=\"font-size:12px;color:#00BF00\"> (A partir do 9\u00ba ano do E. F. &#8211; <span style=\"font-weight: normal;\"> N\u00edvel de dificuldade: M\u00e9dio<\/span>)<\/span><\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 20px;border: 0 none\"\/>\nNa figura, vemos um hex\u00e1gono regular cuja \u00e1rea \u00e9  [tex] 36 \\, \\text{m}^2[\/tex].<br \/>\nQual \u00e9 a \u00e1rea do tri\u00e2ngulo colorido?<br \/>\nExpresse essa \u00e1rea em  [tex] \\text{cm}^2[\/tex] e em [tex]\\text{m}^2[\/tex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2018\/06\/prob208.png\" style=\"border: none;\" alt=\"\" width=\"300\" height=\"259\" class=\"aligncenter size-full wp-image-104198\" \/><\/p>\n<p style=\"margin:10px\">\n<\/div>\n<p style=\"margin:30px\">\n<div class=\"g1n\" style=\"line-height:25px;\">\n<p class=\"tit\" style=\"line-height:30px;margin:10px\">Solu\u00e7\u00e3o <\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 30px;border: 0 none\"\/>\n<ul>\n<li>A partir do seu centro, um hex\u00e1gono regular pode ser decomposto em seis tri\u00e2ngulos equil\u00e1teros congruentes cujo comprimento dos lados \u00e9 o comprimento dos lados do pr\u00f3prio hex\u00e1gono.<\/li>\n<\/ul>\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_srdg, criado com o <a href=\"http:\/\/www.geogebra.org\/\" target=\"_blank\" >GeoGebra<\/a><\/p>\n<ul>\n<li>Observe, agora, que o tri\u00e2ngulo colorido da figura dada no problema corresponde \u00e0 metade do paralelogramo definido por dois dos seis tri\u00e2ngulos equil\u00e1teros que decomp\u00f5em o hex\u00e1gono original.<\/li>\n<\/ul>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2018\/08\/prob_208.png\" style=\"border: none;\"  alt=\"\" width=\"300\" height=\"264\" class=\"aligncenter size-full wp-image-108467\" \/><\/p>\n<p>Dessa forma, a \u00e1rea [tex]A[\/tex] do tri\u00e2ngulo colorido pode ser assim calculada:<\/p>\n<p style=\"margin:10px\">[tex]\\qquad A=\\dfrac{\\text{\u00c1rea do paralelogramo}}{2}[\/tex]<\/p>\n<p style=\"margin:10px\">[tex]\\qquad A=\\dfrac{2 \\times \\text{\u00c1rea do tri\u00e2ngulo equil\u00e1tero}}{2}[\/tex]<\/p>\n<p style=\"margin:10px\">[tex]\\qquad A=\\dfrac{\\cancel{2} \\times \\text{\u00c1rea do tri\u00e2ngulo equil\u00e1tero}}{\\cancel{2}}[\/tex]<\/p>\n<p style=\"margin:10px\">[tex]\\qquad A= \\text{\u00c1rea do tri\u00e2ngulo equil\u00e1tero}[\/tex]<\/p>\n<p style=\"margin:10px\">[tex]\\qquad A=\\dfrac{\\text{\u00c1rea do hex\u00e1gono}}{6}[\/tex]<\/p>\n<p style=\"margin:10px\">[tex]\\qquad A=\\dfrac{36}{6}[\/tex]<\/p>\n<p>e, portanto, a \u00e1rea do tri\u00e2ngulo colorido \u00e9 [tex]\\, \\fcolorbox{black}{#edd2ff}{$6\\, \\text{m}^2$}\\, .[\/tex]<br \/>\nPara finalizarmos a solu\u00e7\u00e3o do problema, precisamos converter [tex]6\\, \\text{m}^2\\, [\/tex] em [tex]\\text{cm}^2\\, .[\/tex]<br \/>\nO esqueminha abaixo pode ajudar!<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2018\/08\/prob_208_1.png\" alt=\"\" width=\"288\" height=\"263\" class=\"aligncenter size-full wp-image-108492\" \/><\/p>\n<ul>\n<li>\u00c1rea do tri\u00e2ngulo colorido em [tex]\\text{cm}^2:[\/tex]<br \/>\n[tex]\\qquad 6\\, \\text{m}^2 \\times 10000 \\longmapsto\\fcolorbox{black}{#edd2ff}{$60000\\, \\text{cm}^2$}\\, .[\/tex]<\/li>\n<\/ul>\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:30px\">\n<div class=\"g3\">\nParticipou da discuss\u00e3o o Clube <strong>OCTETO MATEM\u00c1TICO<\/strong>.<\/div>\n<p style=\"margin:20px\">\n<table style=\" height:auto; margin-left: auto; margin-right: auto;width:640px;\">\n<tr>\n<td class=\"bf\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1001 alignright\" style=\"border: 0px;\" alt=\"\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2013\/01\/menino.jpg\" width=\"80\" height=\"115\" \/><\/td>\n<td class=\"bf\">\n<p class=\"fala\" style=\"width:520px;background-color: #FFF5EE; padding:10px; font-family: calibri; font-size: 13px;color: #1C1C1C; line-height: 20px; \"> Se for conveniente, voc\u00ea pode obter um arquivo PDF desta p\u00e1gina, com o problema e a solu\u00e7\u00e3o, clicando no bot\u00e3o abaixo.<br \/>\nVoc\u00ea pode abrir o arquivo diretamente no seu navegador (<em>Chrome<\/em>, <em>Edge<\/em>, <em>Firefox<\/em>, <em>Safari<\/em>, entre outros), mas tamb\u00e9m pode utilizar o software gratuito <em><strong>Adobe Acrobat Reader<\/strong><\/em>.<br \/>\nCaso o dispositivo que voc\u00ea est\u00e1 utilizando n\u00e3o tenha o <strong><em>Acrobat Reader<\/em><\/strong> instalado, \u00e9 s\u00f3 clicar <a href=\"http:\/\/get.adobe.com\/br\/reader\/\" rel=\"noopener\" target=\"_blank\"><strong>AQUI<\/strong><\/a> para fazer o download adequado ao seu dispositivo.<\/p>\n<\/td>\n<\/tr>\n<\/table>\n<div style=\" float:right;  width: 100px;\">\n<p style=\"text-align: center; border-style: dotted; border-color :#FFFAFA; border-width :thin; width:90px;  height:40px;  padding: 5px; background-color: #800000;\">\n<a href=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2019\/12\/Um-tri\u00e2ngulo-e-um-hex\u00e1gono.pdf\" title=\" Arquivo PDF\" download=\"Clubes_ Um tri\u00e2ngulo e um hex\u00e1gono.pdf\"><span style=\"color: #FFFAFA; font-family: georgia; font-size: 16px;\"><strong>Arquivo <\/strong><\/span><\/a>\n <\/p>\n<\/div>\n<\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Clique no bot\u00e3o abaixo para visualizar o problema. Problema (A partir do 9\u00ba ano do E. F. &#8211; N\u00edvel de dificuldade: M\u00e9dio) Na figura, vemos um hex\u00e1gono regular cuja \u00e1rea \u00e9 [tex] 36 \\, \\text{m}^2[\/tex]. Qual \u00e9 a \u00e1rea do tri\u00e2ngulo colorido? Expresse essa \u00e1rea em [tex] \\text{cm}^2[\/tex] e em [tex]\\text{m}^2[\/tex]. Solu\u00e7\u00e3o A partir do &hellip; <\/p>\n<p><a class=\"more-link btn\" href=\"https:\/\/clubes.obmep.org.br\/blog\/a-problema-para-ajudar-na-escola-um-triangulo-e-um-hexagono\/\">Continue lendo<\/a><\/p>\n","protected":false},"author":2,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"template-onecolumn.php","meta":{"footnotes":""},"class_list":["post-219664","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/219664","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\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/comments?post=219664"}],"version-history":[{"count":1,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/219664\/revisions"}],"predecessor-version":[{"id":219665,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/219664\/revisions\/219665"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=219664"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}