{"id":49764,"date":"2015-10-28T12:14:30","date_gmt":"2015-10-28T14:14:30","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=49764"},"modified":"2023-03-30T23:56:29","modified_gmt":"2023-03-31T02:56:29","slug":"probleminha-dividindo-um-retangulo","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/probleminha-dividindo-um-retangulo\/","title":{"rendered":".Probleminha: Dividindo um ret\u00e2ngulo"},"content":{"rendered":"<p style=\"margin:20px\">\n<div class=\"g1n\">\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\"\/>\nDividimos um ret\u00e2ngulo em quatro outros ret\u00e2ngulos com os lados paralelos aos lados do ret\u00e2ngulo maior, conforme indica a figura.<br \/>\nSe as \u00e1reas de tr\u00eas desses quatro ret\u00e2ngulos s\u00e3o como as indicadas pela figura, qual a \u00e1rea do quarto\u00a0ret\u00e2ngulo?<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"size-medium wp-image-34448 aligncenter\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2015\/04\/ret\u00e2ngulo-300x220.jpg\" style=\"border:0px;\" alt=\"ret\u00e2ngulo\" width=\"300\" height=\"220\" srcset=\"https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2015\/04\/ret\u00e2ngulo-300x220.jpg 300w, https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2015\/04\/ret\u00e2ngulo.jpg 377w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/p>\n<p style=\"margin:10px\">\n<\/div>\n<p style=\"margin:30px\">\n<div class=\"g1n\">\n<p class=\"tit\" style=\"line-height:30px;margin: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<img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2015\/10\/f001.png\" style=\"border:0px;\"  alt=\"f001\" width=\"209\" height=\"209\" class=\"aligncenter size-full wp-image-49766\" srcset=\"https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2015\/10\/f001.png 209w, https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2015\/10\/f001-150x150.png 150w\" sizes=\"auto, (max-width: 209px) 100vw, 209px\" \/><\/p>\n<p>Pela figura, temos que:<\/p>\n<ul>\n<li>[tex]a \\cdot b=6[\/tex]; (I)<\/li>\n<li>[tex]b \\cdot c=14[\/tex]; (II)<\/li>\n<li>[tex]d \\cdot c=35[\/tex]. (III)<\/li>\n<\/ul>\n<p>Multiplicando as equa\u00e7\u00f5es (I) e (III) obtemos que:<br \/>\n[tex]\\qquad \\qquad  a \\cdot b \\cdot c \\cdot d= 210[\/tex].<br \/>\nPor (II):<br \/>\n[tex]\\qquad \\qquad b \\cdot c=14[\/tex].<br \/>\nLogo<br \/>\n[tex]\\qquad \\qquad a \\cdot d \\cdot 14=210[\/tex]<br \/>\n[tex]\\qquad \\qquad a \\cdot d= 15[\/tex].<br \/>\nPortanto a \u00e1rea procurada \u00e9 15.<\/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 pelo COM <strong> Os N\u00f3bregas<\/strong>.<\/p>\n<\/div>\n<p style=\"margin:30px\">\n<div class=\"g1n\" style=\"line-height:30px;\">\n<p class=\"tit\" style=\"line-height:30px;margin: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\"\/>\nSejam [tex]x[\/tex], [tex]y[\/tex], [tex]a[\/tex] e [tex]b[\/tex] como indicados abaixo. <\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2015\/10\/f002.jpg\" alt=\"f002\" width=\"450\" height=\"343\" style=\"border:0px;\" class=\"aligncenter size-full wp-image-49772\" srcset=\"https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2015\/10\/f002.jpg 450w, https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2015\/10\/f002-300x229.jpg 300w\" sizes=\"auto, (max-width: 450px) 100vw, 450px\" \/><\/p>\n<p style=\"margin:10px\">\n<p>Note que, nesse caso, temos:<\/p>\n<ul>\n<li>[tex]y \\cdot b = 6[\/tex];<\/li>\n<li>[tex]a \\cdot b = 14[\/tex];<\/li>\n<li>[tex]a \\cdot x = 35[\/tex].<\/li>\n<\/ul>\n<p>Multiplicando todas express\u00f5es do lado esquerdo e todos valores do lado direito, obtemos:<br \/>\n[tex]\\qquad \\qquad a^2 \\cdot b^2 \\cdot x \\cdot y = (a \\cdot b)^2 \\cdot x \\cdot y = 6 \\cdot 14 \\cdot 35[\/tex]<br \/>\nComo [tex]a \\cdot b = 14[\/tex], ent\u00e3o<br \/>\n[tex]\\qquad \\qquad x \\cdot y = \\dfrac{6 \\cdot 14 \\cdot 35}{14 \\cdot 14}=15[\/tex],<br \/>\nque \u00e9 justamente a \u00e1rea  procurada.<\/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:30px\">\n<div class=\"g3\">\nParticiparam da discuss\u00e3o os Clubes: <strong>Felipe Alencar Barros; MIRIM APRENDIZ; Os N\u00f3bregas<\/strong>.\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Problema (Indicado a partir do 9\u00ba ano do E. F.) Dividimos um ret\u00e2ngulo em quatro outros ret\u00e2ngulos com os lados paralelos aos lados do ret\u00e2ngulo maior, conforme indica a figura. Se as \u00e1reas de tr\u00eas desses quatro ret\u00e2ngulos s\u00e3o como as indicadas pela figura, qual a \u00e1rea do quarto\u00a0ret\u00e2ngulo? Solu\u00e7\u00e3o 1 Pela figura, temos que: &hellip; <\/p>\n<p><a class=\"more-link btn\" href=\"https:\/\/clubes.obmep.org.br\/blog\/probleminha-dividindo-um-retangulo\/\">Continue lendo<\/a><\/p>\n","protected":false},"author":165,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"template-onecolumn.php","meta":{"footnotes":""},"class_list":["post-49764","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/49764","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\/165"}],"replies":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/comments?post=49764"}],"version-history":[{"count":20,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/49764\/revisions"}],"predecessor-version":[{"id":185885,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/49764\/revisions\/185885"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=49764"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}