{"id":219584,"date":"2024-06-02T02:36:07","date_gmt":"2024-06-02T05:36:07","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=219584"},"modified":"2024-06-02T02:36:07","modified_gmt":"2024-06-02T05:36:07","slug":"a-problema-para-ajudar-na-escola-uma-relacao-entre-areas","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/a-problema-para-ajudar-na-escola-uma-relacao-entre-areas\/","title":{"rendered":"(A) Problema para ajudar na escola: Uma rela\u00e7\u00e3o entre \u00e1reas"},"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:25px;\">\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\"\/>\n(Adaptado da <strong>Junior Mathematical Olympiad &#8211; 2014<\/strong>) Um ponto [tex]P[\/tex] tomado  no interior de um paralelogramo [tex]ABCD [\/tex] define os tri\u00e2ngulos [tex]APB [\/tex], [tex]BPC [\/tex], [tex]CPD [\/tex] e [tex]DPA[\/tex].<br \/>\nQue rela\u00e7\u00e3o existe entre as \u00e1reas desses quatro tri\u00e2ngulos? Justifique sua resposta.\n<\/div>\n<p style=\"margin:30px\">\n<div class=\"cb36 g1n\" style=\"margin:30px;line-height:30px; width:80%; 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=\"80\" height=\"91\" class=\"alignleft size-full wp-image-1001\" \/> <\/p>\n<p class=\"c3 ti3\" style=\"line-height:25px;margin:10px\"> Uma ajuda<\/p>\n<p style=\"margin:10px\">\n<p>O <em>applet <\/em>abaixo pode ajudar&#8230;<br \/>\nPara utiliz\u00e1-lo, basta movimentar o ponto [tex]P[\/tex].<br \/>\nObserve que [tex]V[\/tex] denota a soma das \u00e1reas dos tri\u00e2ngulos coloridos de verde e [tex]L[\/tex] denota a soma das \u00e1reas dos tri\u00e2ngulos coloridos com a cor alaranjada.<\/p>\n<p style=\"text-align:center\">\n<iframe loading=\"lazy\" src=\"https:\/\/www.geogebra.org\/material\/iframe\/id\/C8HF642K\/width\/518\/height\/435\/border\/ffffff\/smb\/false\/stb\/false\/stbh\/false\/ai\/false\/asb\/false\/sri\/false\/rc\/false\/ld\/false\/sdz\/false\/ctl\/false\" width=\"518\" height=\"435\" style=\"border: 2px solid #800000;\"> <\/iframe><br \/>\n<span style=\"font-size:small\">OBMEP_srg, criado com o <a href=\"http:\/\/www.geogebra.org\/\" target=\"_blank\" >GeoGebra<\/a><\/span><\/p>\n<p style=\"margin:10px\">\n<\/div>\n<p style=\"margin:30px\">\n<div class=\"cb36 g1n\" style=\"margin:30px;line-height:30px; width:60%; margin-left:auto; margin-right:auto;\">\n<p><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=\"80\" height=\"91\" class=\"alignleft size-full wp-image-1001\" \/> <\/p>\n<p class=\"c3 ti3\" style=\"line-height:25px;margin:10px\"> Um lembrete<\/p>\n<p style=\"margin:10px\">\n<p>Cada diagonal de um paralelogramo divide-o em dois tri\u00e2ngulos congruentes, ou seja, em dois tri\u00e2ngulos com a mesma \u00e1rea.<\/p>\n<p style=\"margin:10px\">\n<\/div>\n<p style=\"margin:30px\">\n<div class=\"g1n\" style=\"line-height:27px;\">\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: 20px;border: 0 none\"\/>\nTome um ponto [tex]P[\/tex] no interior de um paralelogramo [tex]ABCD [\/tex], conforme ilustra a figura abaixo. Denominaremos as \u00e1reas dos tri\u00e2ngulos  [tex]APB [\/tex], [tex]BPC [\/tex], [tex]CPD [\/tex] e [tex]DPA[\/tex] por [tex]\\textcolor{#009900}T[\/tex], [tex]\\textcolor{#ff6600}M[\/tex], [tex]\\textcolor{#FFD700}H[\/tex] e [tex]\\textcolor{#CC66FF}E[\/tex], respectivamente.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2018\/06\/prob_214.png\" style=\"border: none;\"  alt=\"\" width=\"440\" height=\"348\" class=\"aligncenter size-full wp-image-105990\" srcset=\"https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2018\/06\/prob_214.png 440w, https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2018\/06\/prob_214-300x237.png 300w\" sizes=\"auto, (max-width: 440px) 100vw, 440px\" \/><\/p>\n<p>Utilizando o <em>applet<\/em>, podemos perceber que [tex]\\boxed{\\textcolor{#009900}{T}+\\textcolor{#FFD700}{H}=\\textcolor{#CC66FF}{E}+\\textcolor{#ff6600}{M}}[\/tex], e \u00e9 essa rela\u00e7\u00e3o que vamos provar.<br \/>\nCom efeito, ao tra\u00e7armos dois segmentos passando por [tex]P[\/tex] e paralelos, respectivamente, aos lados [tex]\\overline{AB} \\, [\/tex] e [tex] \\, \\overline{AD}[\/tex] do paralelogramo [tex]ABCD [\/tex], obtemos quatro paralelogramos que decomp\u00f5em cada um dos tri\u00e2ngulos  [tex]APB [\/tex], [tex]BPC [\/tex], [tex]CPD [\/tex] e [tex]DPA[\/tex] em dois tri\u00e2ngulos, cujas \u00e1reas denotaremos por [tex]\\textcolor{#009900}{T_1}[\/tex], [tex]\\textcolor{#009900}{T_2}[\/tex], [tex]\\textcolor{#ff6600}{M_1}[\/tex], [tex]\\textcolor{#ff6600}{M_2}[\/tex], [tex]\\textcolor{#FFD700}{H_1}[\/tex], [tex]\\textcolor{#FFD700}{H_2}[\/tex], [tex]\\textcolor{#CC66FF}{E_1}[\/tex] e [tex]\\textcolor{#CC66FF}{E_2}[\/tex], conforme ilustra a pr\u00f3xima figura.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2018\/06\/prob_214_1.png\" style=\"border: none;\" alt=\"\" width=\"439\" height=\"352\" class=\"aligncenter size-full wp-image-105999\" srcset=\"https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2018\/06\/prob_214_1.png 439w, https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2018\/06\/prob_214_1-300x241.png 300w\" sizes=\"auto, (max-width: 439px) 100vw, 439px\" \/><\/p>\n<p>Perceba que os segmentos [tex]\\overline{AP} \\, [\/tex], [tex]\\overline{BP} \\, [\/tex], [tex]\\overline{CP} \\, [\/tex] e [tex]\\overline{DP} \\, [\/tex] s\u00e3o diagonais dos quatro paralelogramos obtidos; assim, pelo <strong>Lembrete<\/strong>, temos que:<\/p>\n<ul>\n<li>[tex]\\textcolor{#CC66FF}{E_1}=\\textcolor{#009900}{T_1}[\/tex];<\/li>\n<li>[tex]\\textcolor{#009900}{T_2}=\\textcolor{#ff6600}{M_1}[\/tex];<\/li>\n<li>[tex]\\textcolor{#ff6600}{M_2}=\\textcolor{#FFD700}{H_2}[\/tex];<\/li>\n<li>[tex]\\textcolor{#FFD700}{H_1}=\\textcolor{#CC66FF}{E_2}.[\/tex]<\/li>\n<\/ul>\n<p>Dessa forma:<br \/>\n[tex]\\qquad \\qquad \\begin{align*}\\textcolor{#009900}{T}+\\textcolor{#FFD700}{H}&#038;=\\left(\\textcolor{#009900}{T_1}+\\textcolor{#009900}{T_2}\\right)+\\left(\\textcolor{#FFD700}{H_1}+\\textcolor{#FFD700}{H_2}\\right)\\\\<br \/>\n&#038;=\\left(\\textcolor{#CC66FF}{E_1}+\\textcolor{#ff6600}{M_1}\\right)+\\left(\\textcolor{#CC66FF}{E_2}+\\textcolor{#ff6600}{M_2}\\right)\\\\<br \/>\n&#038;=\\left(\\textcolor{#CC66FF}{E_1}+\\textcolor{#CC66FF}{E_2}\\right)+\\left(\\textcolor{#ff6600}{M_1}+\\textcolor{#ff6600}{M_2}\\right)\\\\<br \/>\n&#038;=\\textcolor{#CC66FF}{E}+\\textcolor{#ff6600}{M}<br \/>\n\\end{align*}[\/tex]<br \/>\nPortanto, de fato, [tex] \\, \\fcolorbox{black}{#eee0e5}{$T+H=E+M$} \\, .[\/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\/Uma-rela\u00e7\u00e3o-entre-\u00e1reas.pdf\" title=\" Arquivo PDF\" download=\"Clubes_ Uma rela\u00e7\u00e3o entre \u00e1reas.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) (Adaptado da Junior Mathematical Olympiad &#8211; 2014) Um ponto [tex]P[\/tex] tomado no interior de um paralelogramo [tex]ABCD [\/tex] define os tri\u00e2ngulos [tex]APB [\/tex], [tex]BPC [\/tex], [tex]CPD [\/tex] e [tex]DPA[\/tex]. Que rela\u00e7\u00e3o existe entre &hellip; <\/p>\n<p><a class=\"more-link btn\" href=\"https:\/\/clubes.obmep.org.br\/blog\/a-problema-para-ajudar-na-escola-uma-relacao-entre-areas\/\">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-219584","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/219584","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=219584"}],"version-history":[{"count":1,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/219584\/revisions"}],"predecessor-version":[{"id":219585,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/219584\/revisions\/219585"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=219584"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}