{"id":94230,"date":"2018-01-16T15:29:39","date_gmt":"2018-01-16T17:29:39","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=94230"},"modified":"2023-04-15T04:43:26","modified_gmt":"2023-04-15T07:43:26","slug":"problema-os-arcos-das-circunferencias","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/problema-os-arcos-das-circunferencias\/","title":{"rendered":".Problema: Os arcos das circunfer\u00eancias"},"content":{"rendered":"<p style=\"margin:30px\">\n<div class=\"g1n\" style=\"line-height:30px;\">\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\"\/>\nUma circunfer\u00eancia [tex]\\Gamma_1[\/tex] possui um arco de [tex]45^\\circ[\/tex] com o mesmo comprimento de um arco de [tex]30^\\circ[\/tex] de uma outra circunfer\u00eancia [tex]\\Gamma_2[\/tex].<br \/>\nQual \u00e9 a raz\u00e3o entre as \u00e1reas dos c\u00edrculos determinados pelas circunfer\u00eancias  [tex]\\Gamma_1[\/tex] e [tex]\\Gamma_2[\/tex]?<\/p>\n<p style=\"margin:20px\">\n<\/div>\n<p style=\"margin:30px\">\n<div class=\"cb31 g1n\" style=\"margin:30px;line-height:30px; width:65%; margin-left:auto; margin-right:auto; \">\n<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2013\/01\/menino.jpg\" style=\"border:none\"  alt=\"explicador_p\" width=\"99\" height=\"120\" class=\"alignleft size-full wp-image-1001\" \/><\/p>\n<p class=\"c3 ti3\" style=\"line-height:25px;margin:10px\"> Ajuda<\/p>\n<p>A um arco de circunfer\u00eancia podemos associar duas medidas distintas:<br \/>\n<span class=\"c3\" style=\"font-size: 25px;\"> &#10000; <\/span> a sua medida angular, [tex]\\alpha[\/tex];<br \/>\n<span class=\"c3\" style=\"font-size: 25px;\"> &#10000; <\/span> a sua  medida linear, [tex]c[\/tex].<br \/>\nAmbas podem ser obtidas a partir da medida angular e do comprimento da circunfer\u00eancia que define o arco por meio de uma regra de tr\u00eas simples:<\/p>\n<p style=\"margin:10px\">\n<table style=\"border: 1px solid #ffffff; width:40%; height:auto;text-align:center;  margin-left: auto; margin-right:auto;\">\n<tr class=\"bf\">\n<td style=\"border: 1px solid #ffffff; \"> [tex]2\\pi \\, r[\/tex] <\/td>\n<td style=\"border: 1px solid #ffffff\"> &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8211;<\/td>\n<td style=\"border: 1px solid #ffffff\"> [tex]360^\\circ[\/tex] <\/td>\n<\/tr>\n<tr class=\"bf\">\n<td style=\"border: 1px solid #ffffff\"> [tex]c[\/tex] <\/td>\n<td style=\"border: 1px solid #ffffff\"> &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8211;<\/td>\n<td style=\"border: 1px solid #ffffff\"> [tex]\\alpha[\/tex] <\/td>\n<\/tr>\n<\/table>\n<p>Se voc\u00ea n\u00e3o se lembra disso, n\u00e3o faz mal: clique <a  title=\"Uma ajudinha\" href=\"http:\/\/clubes.obmep.org.br\/blog\/?p=86521_#O1\" onclick=\"window.open(this.href, '_blank', 'width=900,height=580,left=0,top=0'); return false\"><span class=\"c3\"> <strong>AQUI<\/strong><\/span><\/a>.<\/span> <\/p>\n<p style=\"margin:10px\">\n<\/div>\n<p style=\"margin:30px\">\n<div class=\"g1n\" style=\"line-height:40px;\">\n<p class=\"tit\" style=\"line-height:25px;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\"\/>\nSejam [tex]r_1[\/tex] e [tex]r_2[\/tex] os raios dos c\u00edrculos [tex]\\Gamma_1[\/tex] e [tex]\\Gamma_2[\/tex], respectivamente.  <\/p>\n<ul>\n<li>Vamos, inicialmente, fazer duas regras de tr\u00eas para obtermos os comprimentos dos arcos definidos em [tex]\\Gamma_1[\/tex] e [tex]\\Gamma_2[\/tex].<\/li>\n<table style=\"border: 1px solid #ffffff; width:75%; height:auto;text-align:center;\">\n<tr class=\"bf\">\n<td style=\"border: 1px solid #ffffff; \"> [tex]2\\pi \\, r_1[\/tex] <\/td>\n<td style=\"border: 1px solid #ffffff\"> &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;-<\/td>\n<td style=\"border: 1px solid #ffffff\"> [tex]360^\\circ[\/tex] <\/td>\n<td class=\"bf\" style=\"width:30%\"><\/td>\n<td style=\"border: 1px solid #ffffff; \"> [tex]2\\pi \\, r_2[\/tex] <\/td>\n<td style=\"border: 1px solid #ffffff\"> &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;-<\/td>\n<td style=\"border: 1px solid #ffffff\"> [tex]360^\\circ[\/tex] <\/td>\n<\/tr>\n<tr class=\"bf\">\n<td style=\"border: 1px solid #ffffff\"> [tex]c_1[\/tex]<\/td>\n<td style=\"border: 1px solid #ffffff\"> &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;-<\/td>\n<td style=\"border: 1px solid #ffffff\"> [tex]45^\\circ[\/tex] <\/td>\n<td class=\"bf\"><\/td>\n<td style=\"border: 1px solid #ffffff\"> [tex]c_2[\/tex]<\/td>\n<td style=\"border: 1px solid #ffffff\"> &#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;&#8212;-<\/td>\n<td style=\"border: 1px solid #ffffff\"> [tex]30^\\circ[\/tex] <\/td>\n<\/tr>\n<tr class=\"bf\">\n<td class=\"bf\" colspan=\"3\" style=\"text-align:justify; line-height:40px;\">[tex]\\textcolor{#800000}{(i)}[\/tex] Assim, temos que<br \/>\n[tex]\\qquad 2\\pi \\, r_1 \\cdot 45^\\circ=c_1 \\cdot 360^\\circ[\/tex]<br \/>\n[tex]\\qquad c_1=\\dfrac{45}{360}\\cdot2\\pi{r_1}[\/tex]<br \/>\n[tex]\\qquad \\boxed{c_1=\\dfrac{r_1\\pi}{4}} \\, .[\/tex]<\/td>\n<td class=\"bf\"><\/td>\n<td colspan=\"3\" class=\"bf\" style=\"text-align:justify; line-height:40px;\">[tex]\\textcolor{#800000}{(ii)}[\/tex] Agora, temos que<br \/>\n[tex]\\qquad2\\pi \\, r_2 \\cdot 30^\\circ=c_2 \\cdot 360^\\circ[\/tex]<br \/>\n[tex]\\qquad c_2=\\dfrac{30}{360}\\cdot 2\\pi{r_2}[\/tex]<br \/>\n[tex]\\qquad\\boxed{c_2=\\dfrac{r_2\\pi}{6}} \\, .[\/tex]<\/td>\n<\/tr>\n<\/table>\n<li>Como sabemos que os dois arcos possuem o mesmo comprimento, segue que:<br \/>\n[tex]\\qquad c_1=c_2[\/tex]<\/p>\n<p style=\"margin:10px\">\n<p>[tex]\\qquad\\dfrac{r_1\\pi}{4}=\\dfrac{r_2\\pi}{6}[\/tex]<\/p>\n<p style=\"margin:10px\">\n<p>[tex]\\qquad \\dfrac{r_1}{r_2}=\\dfrac{2}{3}[\/tex]. <\/li>\n<li>Vamos, finalmente, calcular a raz\u00e3o entre as \u00e1reas dos c\u00edrculos determinados por [tex]\\Gamma_1[\/tex] e [tex]\\Gamma_2[\/tex]:<\/li>\n<p>[tex]\\qquad \\dfrac{\\text{\u00c1rea de }{\\Gamma_1}}{\\text{\u00c1rea de }{\\Gamma_2}}=\\dfrac{\\pi \\left(r_1\\right)^2}{\\pi \\left(r_2\\right)^2}=\\dfrac{ \\left(r_1\\right)^2}{\\left(r_2\\right)^2}=<br \/>\n\\left(\\dfrac{ r_1}{r_2}\\right)^2=\\left(\\dfrac{2}{3}\\right)^2=\\fcolorbox{black}{#eee0e5}{$\\dfrac{4}{9}$} \\, .[\/tex]\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","protected":false},"excerpt":{"rendered":"<p>Problema (Indicado a partir do 9\u00ba ano do E. F.) Uma circunfer\u00eancia [tex]\\Gamma_1[\/tex] possui um arco de [tex]45^\\circ[\/tex] com o mesmo comprimento de um arco de [tex]30^\\circ[\/tex] de uma outra circunfer\u00eancia [tex]\\Gamma_2[\/tex]. Qual \u00e9 a raz\u00e3o entre as \u00e1reas dos c\u00edrculos determinados pelas circunfer\u00eancias [tex]\\Gamma_1[\/tex] e [tex]\\Gamma_2[\/tex]? Ajuda A um arco de circunfer\u00eancia podemos associar &hellip; <\/p>\n<p><a class=\"more-link btn\" href=\"https:\/\/clubes.obmep.org.br\/blog\/problema-os-arcos-das-circunferencias\/\">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-94230","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/94230","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=94230"}],"version-history":[{"count":8,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/94230\/revisions"}],"predecessor-version":[{"id":187130,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/94230\/revisions\/187130"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=94230"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}