{"id":120726,"date":"2019-01-23T14:10:22","date_gmt":"2019-01-23T16:10:22","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=120726"},"modified":"2024-06-08T23:43:03","modified_gmt":"2024-06-09T02:43:03","slug":"problema-para-ajudar-na-escola-coracao-tem-perimetro-e-area","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/problema-para-ajudar-na-escola-coracao-tem-perimetro-e-area\/","title":{"rendered":".Problema para ajudar na escola: Cora\u00e7\u00e3o tem per\u00edmetro? E \u00e1rea?"},"content":{"rendered":"<p class=\"g2\" ><strong><a href=\"http:\/\/clubes.obmep.org.br\/blog\/?p=220070\" rel=\"noopener\" target=\"_blank\">&#9999; <span style=\"color:#A0522D;font-size:14px;\">Link do problema para dispositivos da <em>Apple<\/em>.<\/span><\/a><\/strong><\/p>\n<div class=\"g1n\" style=\"line-height:25px;\">\n<p class=\"tit\" style=\"line-height:20px;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\"\/>\n<p>O cora\u00e7\u00e3o mostrado nas figuras foi desenhado em uma malha quadriculada, utilizando-se quatro arcos de circunfer\u00eancias: [tex]\\stackrel{\\textstyle\\frown}{\\mathrm{AB}}[\/tex] , [tex]\\stackrel{\\textstyle\\frown}{\\mathrm{BC}}[\/tex] , [tex]\\stackrel{\\textstyle\\frown}{\\mathrm{CD}}[\/tex] , [tex]\\stackrel{\\textstyle\\frown}{\\mathrm{DA}}[\/tex], cujos centros est\u00e3o indicados por pontos azuis na forma de cruz.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2018\/10\/prob260.png\" alt=\"\" width=\"481\" height=\"363\" class=\"aligncenter size-full wp-image-113878\" srcset=\"https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2018\/10\/prob260.png 481w, https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2018\/10\/prob260-300x226.png 300w\" sizes=\"auto, (max-width: 481px) 100vw, 481px\" \/><\/p>\n<p>Sabendo-se que os lados dos quadradinhos da malha medem [tex]1 \\, \\text{cm}[\/tex], determine:<\/p>\n<p><strong>(a)<\/strong> O per\u00edmetro do cora\u00e7\u00e3o.<\/p>\n<p><strong>(b)<\/strong> A \u00e1rea do cora\u00e7\u00e3o.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2018\/10\/prob260_1.png\" alt=\"\" width=\"557\" height=\"433\" class=\"aligncenter size-full wp-image-113879\" srcset=\"https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2018\/10\/prob260_1.png 557w, https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2018\/10\/prob260_1-300x233.png 300w\" sizes=\"auto, (max-width: 557px) 100vw, 557px\" \/><\/p>\n<p style=\"margin:20px;\">\n<\/div>\n<p style=\"margin:30px\">\n<div class=\"g1n\" style=\"line-height:35px;\">\n<p class=\"tit\" style=\"line-height:20px;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><strong>(a)<\/strong> Observe as circunfer\u00eancias de centros nos pontos [tex]C_1, \\, C_2, \\, C_3, \\, C_4[\/tex], a partir das quais foram tra\u00e7ados os arcos [tex]\\stackrel{\\textstyle\\frown}{\\mathrm{AB}}[\/tex] , [tex]\\stackrel{\\textstyle\\frown}{\\mathrm{BC}}[\/tex] , [tex]\\stackrel{\\textstyle\\frown}{\\mathrm{CD}}[\/tex] , [tex]\\stackrel{\\textstyle\\frown}{\\mathrm{DA}}[\/tex] que definem o cora\u00e7\u00e3o.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2019\/01\/cora\u00e7\u00e3o2.png\" alt=\"\" width=\"461\" height=\"449\" class=\"aligncenter size-full wp-image-120468\" srcset=\"https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2019\/01\/cora\u00e7\u00e3o2.png 461w, https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2019\/01\/cora\u00e7\u00e3o2-300x292.png 300w\" sizes=\"auto, (max-width: 461px) 100vw, 461px\" \/><\/p>\n<p>Note que cada uma dessas circunfer\u00eancias tem raio [tex]r=2 \\, \\text{cm}[\/tex]; logo, cada uma tem per\u00edmetro [tex]\\boxed{p=2\\pi \\, r=4\\pi \\, \\text{cm}} \\, .[\/tex]<br \/>\nSe denotarmos o per\u00edmetro do cora\u00e7\u00e3o por [tex]\\textcolor{red}{p_{cor}}[\/tex], segue que:<br \/>\n[tex]\\qquad \\textcolor{red}{p_{cor}}=\\dfrac{3}{4}p+\\dfrac{1}{4}p+\\dfrac{1}{4}p+\\dfrac{3}{4}p[\/tex]<\/p>\n<p style=\"margin:8px;\">\n<p>[tex]\\qquad \\textcolor{red}{p_{cor}}=\\dfrac{8}{4}p[\/tex]<br \/>\n[tex]\\qquad \\textcolor{red}{p_{cor}}=2p[\/tex]<br \/>\n[tex]\\qquad \\textcolor{red}{p_{cor}}=2\\times(4\\pi)[\/tex]<br \/>\n[tex]\\qquad \\textcolor{red}{p_{cor}}=8\\pi.[\/tex]<br \/>\nAssim, o per\u00edmetro do cora\u00e7\u00e3o \u00e9 [tex] \\, \\fcolorbox{black}{#eee0e5}{$8\\pi \\, \\text{cm}$} \\, .[\/tex] Se  utilizarmos [tex]3,14[\/tex] como aproxima\u00e7\u00e3o para [tex]\\pi[\/tex], o per\u00edmetro do cora\u00e7\u00e3o ser\u00e1 aproximadamente [tex]25,12 \\, \\text{cm}.[\/tex] <\/p>\n<p style=\"margin:20px\">\n<p><strong>(b)<\/strong> A \u00e1rea do cora\u00e7\u00e3o corresponde \u00e0 soma das \u00e1reas [tex]A_1, \\, A_2, \\, A_3[\/tex] indicadas na figura abaixo: [tex]\\boxed{\\textcolor{red}{A_{cor}}=A_1+A_2+A_3}.[\/tex]<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2019\/01\/cora\u00e7\u00e3o3.png\" alt=\"\" width=\"441\" height=\"358\" class=\"aligncenter size-full wp-image-120488\" srcset=\"https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2019\/01\/cora\u00e7\u00e3o3.png 441w, https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2019\/01\/cora\u00e7\u00e3o3-300x244.png 300w\" sizes=\"auto, (max-width: 441px) 100vw, 441px\" \/><\/p>\n<p>Observe que:<\/p>\n<ul>\n<li>Cada uma das \u00e1reas [tex]A_1[\/tex] e [tex]A_2[\/tex] corresponde \u00e0 \u00e1rea de [tex]\\frac{3}{4}[\/tex] de um c\u00edrculo de raio [tex]r=2 \\, \\text{cm}[\/tex]. Assim:<\/li>\n<p style=\"margin:10px\">\n<p> [tex]\\qquad A_1=A_2=\\dfrac{3}{4}\\pi \\, r^2[\/tex]<\/p>\n<p style=\"margin:10px\">\n<p>[tex]\\qquad A_1=A_2=\\dfrac{3}{4}\\pi \\, 2^2[\/tex]<\/p>\n<p style=\"margin:10px\">\n<p>[tex]\\qquad A_1=A_2=\\dfrac{3}{\\cancel{4}}\\pi \\, \\cancel{4}[\/tex]<\/p>\n<p style=\"margin:10px\">\n<p>[tex]\\qquad \\boxed{A_1=A_2=3\\pi \\, \\text{cm}^2}.[\/tex]<\/p>\n<li>A \u00e1rea [tex]A_3[\/tex] corresponde \u00e0 diferen\u00e7a entre a \u00e1rea de um quadrado de lado [tex]l=4 \\, \\text{cm}[\/tex] e duas \u00e1reas de [tex]\\frac{1}{4}[\/tex] de um c\u00edrculo de raio [tex]r=2 \\, \\text{cm}[\/tex]. Logo:<\/li>\n<p style=\"margin:10px\">\n<p> [tex]\\qquad A_3=l^2-2\\times \\dfrac{1}{4}\\pi \\, r^2[\/tex]<\/p>\n<p style=\"margin:10px\">\n<p> [tex]\\qquad A_3=4^2-2\\times \\dfrac{1}{4}\\pi \\, 2^2[\/tex]<\/p>\n<p style=\"margin:10px\">\n<p> [tex]\\qquad A_3=16-2\\times \\dfrac{1}{\\cancel{4}}\\pi \\, \\cancel{4}[\/tex]<\/p>\n<p style=\"margin:10px\">\n<p> [tex]\\qquad \\boxed{A_3=(16-2\\pi )\\,  \\, \\text{cm}^2}.[\/tex]<\/ul>\n<p>Dessa forma,<br \/>\n[tex]\\qquad \\textcolor{red}{A_{cor}}=A_1+A_2+A_3[\/tex]<br \/>\n[tex]\\qquad \\textcolor{red}{A_{cor}}=3\\pi+3\\pi+16-2\\pi[\/tex]<br \/>\n[tex]\\qquad \\textcolor{red}{A_{cor}}=4\\pi+16 \\, .[\/tex]<br \/>\nPortanto, a \u00e1rea do cora\u00e7\u00e3o \u00e9 [tex] \\, \\fcolorbox{black}{#eee0e5}{$(4\\pi+16) \\, \\text{cm}^2$} \\, [\/tex]. Se  utilizarmos [tex]3,14[\/tex] como aproxima\u00e7\u00e3o para [tex]\\pi[\/tex], a \u00e1rea do cora\u00e7\u00e3o \u00e9 aproximadamente[tex]28,56 \\, \\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: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\/Cora\u00e7\u00e3o-tem-per\u00edmetro.pdf\" title=\" Arquivo PDF\" download=\"Clubes_ Cora\u00e7\u00e3o tem per\u00edmetro.pdf\"><span style=\"color: #FFFAFA; font-family: georgia; font-size: 16px;\"><strong>Arquivo <\/strong><\/span><\/a>\n <\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>&#9999; Link do problema para dispositivos da Apple. Problema (A partir do 9\u00ba ano do E. F. &#8211; N\u00edvel de dificuldade: M\u00e9dio) O cora\u00e7\u00e3o mostrado nas figuras foi desenhado em uma malha quadriculada, utilizando-se quatro arcos de circunfer\u00eancias: [tex]\\stackrel{\\textstyle\\frown}{\\mathrm{AB}}[\/tex] , [tex]\\stackrel{\\textstyle\\frown}{\\mathrm{BC}}[\/tex] , [tex]\\stackrel{\\textstyle\\frown}{\\mathrm{CD}}[\/tex] , [tex]\\stackrel{\\textstyle\\frown}{\\mathrm{DA}}[\/tex], cujos centros est\u00e3o indicados por pontos azuis na forma de &hellip; <\/p>\n<p><a class=\"more-link btn\" href=\"https:\/\/clubes.obmep.org.br\/blog\/problema-para-ajudar-na-escola-coracao-tem-perimetro-e-area\/\">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-120726","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/120726","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=120726"}],"version-history":[{"count":17,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/120726\/revisions"}],"predecessor-version":[{"id":220072,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/120726\/revisions\/220072"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=120726"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}