{"id":128111,"date":"2019-06-05T23:25:04","date_gmt":"2019-06-06T02:25:04","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=128111"},"modified":"2024-06-12T20:05:23","modified_gmt":"2024-06-12T23:05:23","slug":"problema-para-ajudar-na-escola-uma-area-colorida-de-um-circulo","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/problema-para-ajudar-na-escola-uma-area-colorida-de-um-circulo\/","title":{"rendered":".Problema para ajudar na escola: Uma \u00e1rea colorida de um c\u00edrculo"},"content":{"rendered":"<p class=\"g2\" ><strong><a href=\"http:\/\/clubes.obmep.org.br\/blog\/?p=220318\" 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-bottom: 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: F\u00e1cil<\/span>)<\/span><\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 20px;border: 0 none\"\/>\n<p>(<strong>XVIII OPM<\/strong>, 1999 &#8211; Adaptado)  A \u00e1rea do c\u00edrculo da figura abaixo mede [tex]40 \\, cm^2.[\/tex]<br \/>\nSe as medidas dos \u00e2ngulos [tex]A\\hat{O}B[\/tex] e [tex]C\\hat{O}D[\/tex] s\u00e3o respectivamente [tex]60^\\circ [\/tex] e  [tex]30^\\circ[\/tex],  determine a medida da \u00e1rea colorida do c\u00edrculo.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2019\/04\/prob309.png\" style=\"border: none;\" alt=\"\" width=\"280\" height=\"268\" class=\"aligncenter size-full wp-image-125996\" \/><\/p>\n<p style=\"margin:10px\">\n<\/div>\n<p style=\"margin:40px\">\n<div class=\"g1n\" style=\"line-height:27px;\">\n<p class=\"tit\" style=\"line-height:30px; margin-bottom: 10px;\">Solu\u00e7\u00e3o 1 <\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 20px;border: 0 none\"\/>\n<p>Podemos resolver este problema estabelecendo propor\u00e7\u00f5es entre \u201ca \u00e1rea total do c\u00edrculo\u201d (o todo) e \u201ccada regi\u00e3o colorida\u201d (as partes). O par\u00e2metro que utilizaremos \u00e9 a medida dos \u00e2ngulos que definem as tr\u00eas regi\u00f5es. As partes s\u00e3o definidas por \u00e2ngulos de [tex]30^\\circ [\/tex] e  [tex]60^\\circ[\/tex], de acordo com os dados do problema, e o todo \u00e9 definido por um \u00e2ngulo de [tex]360^\\circ [\/tex], uma volta completa.<br \/>\nCom base nessa propor\u00e7\u00e3o, determinaremos as medidas das respectivas \u00e1reas das duas regi\u00f5es coloridas do c\u00edrculo resolvendo duas regrinhas de tr\u00eas simples.<\/p>\n<table style=\"border: 1px solid #ffffff; width:85%; height:auto; text-align:center;  margin-left: auto; margin-right:auto;\">\n<tr class=\"bf\">\n<td style=\"border: 1px solid #ffffff; border-right: 2px solid #800000;\"><strong>\u00c1rea verde: <\/strong>[tex]\\textcolor{#329865}{A_{v}}[\/tex]<\/p>\n<p style=\"margin:20px\">\n<p> [tex]\\begin{array}{c c c}<br \/>\n \\qquad 40 \\, cm^2  &#038; \\text{&#8212;&#8212;&#8212;&#8212;&#8211;} &#038; 360^\\circ \\qquad \\\\<br \/>\n \\qquad \\textcolor{#329865}{A_{v}} &#038; \\text{&#8212;&#8212;&#8212;&#8212;&#8211;} &#038; 30^\\circ \\qquad  \\end{array}[\/tex]<\/p>\n<p style=\"text-align:left; padding-left:20px;\">Dessa forma, obtemos que<br \/>\n[tex]\\quad \\textcolor{#329865}{A_{v}} \\times 360^\\circ= 40 \\times 30^\\circ[\/tex],<br \/>\ndonde:<br \/>\n [tex]\\quad \\textcolor{#329865}{A_{v}} =\\dfrac{ 40 \\times 30}{360}=\\dfrac{10}{3} \\, cm^2[\/tex].<\/p>\n<\/td>\n<td style=\"border: 1px solid #ffffff; \"><strong>\u00c1rea lil\u00e1s: <\/strong>[tex]\\textcolor{#9A67C3}{A_l}[\/tex]<\/p>\n<p style=\"margin:20px\">\n<p> [tex]\\begin{array}{c c c}<br \/>\n \\quad 40 \\, cm^2  &#038; \\text{&#8212;&#8212;&#8212;&#8212;&#8211;} &#038; 360^\\circ \\qquad \\\\<br \/>\n \\quad \\textcolor{#9A67C3}{A_{l}} &#038; \\text{&#8212;&#8212;&#8212;&#8212;&#8211;} &#038; 60^\\circ \\qquad  \\end{array}[\/tex]<\/p>\n<p style=\"text-align:left; padding-left:20px;\">Dessa forma, obtemos que<br \/>\n[tex]\\quad \\textcolor{#9A67C3}{A_l} \\times 360^\\circ= 40 \\times 60^\\circ[\/tex],<br \/>\ndonde:<br \/>\n [tex]\\quad \\textcolor{#9A67C3}{A_{l}} =\\dfrac{ 40 \\times 60}{360}=\\dfrac{40}{6}=\\dfrac{20}{3} \\, cm^2[\/tex].<\/p>\n<\/td>\n<\/tr>\n<td style=\"border: 1px solid #ffffff; border-right: 2px solid #800000;\"> <\/td>\n<\/tr>\n<\/table>\n<p>Portanto, a medida da \u00e1rea [tex]S[\/tex] da regi\u00e3o colorida do c\u00edrculo pode ser assim calculada:<br \/>\n[tex] \\qquad \\qquad S=\\textcolor{#329865}{A_{v}} + \\textcolor{#9A67C3}{A_l}=\\dfrac{10}{3}+\\dfrac{20}{3}=\\dfrac{30}{3}= \\, \\fcolorbox{black}{#eee0e5}{$10 \\, cm^2$} \\, .[\/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:40px\">\n<div class=\"g1n\" style=\"line-height:27px;\">\n<p class=\"tit\" style=\"line-height:30px; margin-bottom: 10px;\">Solu\u00e7\u00e3o 2 <\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 20px;border: 0 none\"\/>\nPodemos rotacionar as duas regi\u00f5es coloridas em torno do centro [tex]O[\/tex] do c\u00edrculo, sem sobrep\u00f4-las, de modo a obter uma regi\u00e3o que corresponde a um quarto do c\u00edrculo, j\u00e1 que ter\u00edamos um setor circular de [tex]90^\\circ.[\/tex]<br \/>\nAssim, a medida da \u00e1rea [tex]S[\/tex] da regi\u00e3o colorida \u00e9 dada por:<br \/>\n[tex] \\qquad \\qquad S=\\dfrac{40}{4}= \\, \\fcolorbox{black}{#eee0e5}{$10 \\, cm^2$} \\, .[\/tex]<\/p>\n<hr style=\"height:2px; width:100%;background-color:#800000; margin-top: 30px; margin-bottom: 4px;border: 0 none\"\/>\n<hr style=\"height:2px; width:100%;background-color:#800000; margin-top: 4px; margin-bottom: 50px;border: 0 none\"\/>\n<div class=\"g1n\" style=\"border: 2px dashed #800000; text-align:center; padding: 10px; width:85%; margin-left: auto; margin-right: auto;\" >\n<p class=\"c3 ti3\"> Um applet para ajudar <\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 1px; margin-bottom: 20px;border: 0 none\"\/>\n<div style=\"text-align:justify;\">\nVoc\u00ea pode rotacionar as duas regi\u00f5es coloridas para obter uma regi\u00e3o que corresponde a um quarto do c\u00edrculo utilizando o applet abaixo. <\/p>\n<p style=\"margin:10px\">\n<p><strong>Instru\u00e7\u00f5es<\/strong>:<br \/>\n<strong>(1)<\/strong> Espere o aplicativo carregar completamente.<br \/>\n<strong>(2)<\/strong> Para fazer a rota\u00e7\u00e3o da regi\u00e3o colorida de verde, clique no ponto <span style=\"color:#329865\"><strong>C<\/strong><\/span>, mantenha o mouse pressionado e fa\u00e7a o movimento.<br \/>\n<strong>(3)<\/strong> Para fazer a rota\u00e7\u00e3o da regi\u00e3o colorida de lil\u00e1s, clique no ponto <span style=\"color:#9A67C3\"><strong>A<\/strong><\/span>, mantenha o mouse pressionado e fa\u00e7a o movimento.<br \/>\n<strong>(4)<\/strong> Para retornar \u00e0 posi\u00e7\u00e3o inicial, clique no centro das setinhas circulares que aparecem no canto superior direito  do aplicativo.<\/div>\n<p style=\"margin:30px\">\n<p><iframe loading=\"lazy\" scrolling=\"no\" title=\"\u00c1rea colorida de um c\u00edrculo\" src=\"https:\/\/www.geogebra.org\/material\/iframe\/id\/ywc2gynn\/width\/556\/height\/555\/border\/ffffff\/sfsb\/false\/smb\/false\/stb\/false\/stbh\/false\/ai\/false\/asb\/false\/sri\/true\/rc\/false\/ld\/false\/sdz\/false\/ctl\/false\" width=\"556px\" height=\"555px\" style=\"border:2px solid #800000\"> <\/iframe><\/p>\n<p><span style=\"font-size: 14px;\">OBMEP_ srdg, criado com o <a href=\"http:\/\/www.geogebra.org\/\" target=\"_blank\">GeoGebra<\/a><\/span><\/p>\n<p style=\"margin:10px\">\n<\/div>\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<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\/Uma-\u00e1rea-colorida-de-um-c\u00edrculo.pdf\" title=\" Arquivo PDF\" download=\"Clubes_ Uma \u00e1rea colorida de um c\u00edrculo.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: F\u00e1cil) (XVIII OPM, 1999 &#8211; Adaptado) A \u00e1rea do c\u00edrculo da figura abaixo mede [tex]40 \\, cm^2.[\/tex] Se as medidas dos \u00e2ngulos [tex]A\\hat{O}B[\/tex] e [tex]C\\hat{O}D[\/tex] s\u00e3o respectivamente [tex]60^\\circ [\/tex] e [tex]30^\\circ[\/tex], determine a medida &hellip; <\/p>\n<p><a class=\"more-link btn\" href=\"https:\/\/clubes.obmep.org.br\/blog\/problema-para-ajudar-na-escola-uma-area-colorida-de-um-circulo\/\">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-128111","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/128111","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=128111"}],"version-history":[{"count":22,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/128111\/revisions"}],"predecessor-version":[{"id":220322,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/128111\/revisions\/220322"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=128111"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}