{"id":106406,"date":"2018-07-30T16:43:10","date_gmt":"2018-07-30T19:43:10","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=106406"},"modified":"2024-06-02T03:03:17","modified_gmt":"2024-06-02T06:03:17","slug":"problema-para-ajudar-na-escola-porcentagem-da-area","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/problema-para-ajudar-na-escola-porcentagem-da-area\/","title":{"rendered":".Problema para ajudar na escola: Porcentagem da \u00e1rea"},"content":{"rendered":"<p class=\"g2\" ><strong><a href=\"http:\/\/clubes.obmep.org.br\/blog\/?p=219614\" 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: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(<strong>ONEM &#8211; 2011<\/strong>) O tri\u00e2ngulo [tex]ABC[\/tex] mostrado na figura \u00e9 equil\u00e1tero e os pontos [tex]D, E, F[\/tex] s\u00e3o pontos m\u00e9dios dos segmentos [tex]\\overline{CA}[\/tex], [tex]\\overline{AB}[\/tex], [tex]\\overline{BC}[\/tex], respectivamente.<br \/>\nPor outro lado, [tex]G, H, I[\/tex] s\u00e3o pontos m\u00e9dios dos segmentos [tex]\\overline{FD}[\/tex], [tex]\\overline{DE}[\/tex], [tex]\\overline{EF}[\/tex], respectivamente.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2018\/04\/prob188.png\" style=\"border: none;\" alt=\"\" width=\"243\" height=\"218\" class=\"aligncenter size-full wp-image-101984\" \/><\/p>\n<p>Nessas condi\u00e7\u00f5es, a \u00e1rea do tri\u00e2ngulo [tex]GHI[\/tex] corresponde a que porcentagem da \u00e1rea do tri\u00e2ngulo [tex]ABC[\/tex]?<\/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:90%; 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=\"96\" class=\"alignleft size-full wp-image-1001\" \/> <\/p>\n<p class=\"c3 ti3\" style=\"line-height:25px;margin:10px\"> Lembretes <\/p>\n<p style=\"margin:10px\">\n<p><span class=\"c3\"> <strong>(1)<\/strong> <\/span> <strong>Caso de congru\u00eancia L.A.L. (lado &#8211; \u00e2ngulo &#8211; lado):<\/strong> Se dois tri\u00e2ngulos t\u00eam ordenadamente congruentes dois lados e o \u00e2ngulo por eles definido, ent\u00e3o estes tri\u00e2ngulos s\u00e3o congruentes.<br \/>\n<span class=\"c3\"> <strong>(2)<\/strong> <\/span> Os \u00e2ngulos da base de um tri\u00e2ngulo is\u00f3sceles s\u00e3o congruentes.<br \/>\n<span class=\"c3\"> <strong>(3)<\/strong> <\/span> <strong>Caso de congru\u00eancia L.L.L. (lado &#8211; lado- lado):<\/strong> Se dois tri\u00e2ngulos t\u00eam ordenadamente congruentes os tr\u00eas lados, ent\u00e3o estes tri\u00e2ngulos s\u00e3o congruentes.<br \/>\n<span style=\"font-size:11px;\"> (Se voc\u00ea n\u00e3o se lembra desses resultados, clique <a href=\"http:\/\/clubes.obmep.org.br\/blog\/?p=92799\" target=\"_blank\" title=\"Sala para leitura: Um pouco sobre congru\u00eancia de tri\u00e2ngulos\"> <strong>AQUI<\/strong><\/a><\/span>.)\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\"\/>\nObserve, inicialmente, que <\/p>\n<ul>\n<li>como o tri\u00e2ngulo [tex]ABC[\/tex] \u00e9 equil\u00e1tero, os segmentos [tex]\\overline{AB}~[\/tex], [tex]\\overline{BC}~[\/tex] e [tex]~\\overline{CA}[\/tex] t\u00eam o mesmo comprimento, digamos [tex]l[\/tex];<\/li>\n<li>como [tex]D, E, F[\/tex] s\u00e3o pontos m\u00e9dios dos segmentos [tex]\\overline{CA}[\/tex], [tex]\\overline{AB}[\/tex] e [tex]\\overline{BC}[\/tex], os segmentos [tex]\\overline{AE}~[\/tex], [tex]\\overline{EB}~[\/tex], [tex]\\overline{BF}~[\/tex], [tex]\\overline{FC}~[\/tex], [tex]\\overline{CD}~[\/tex] e [tex]\\overline{DA}~[\/tex] t\u00eam comprimentos iguais a  [tex]\\frac{l}{2}~.[\/tex]<\/li>\n<li>como [tex]\\Delta ABC[\/tex] \u00e9 um tri\u00e2ngulo equil\u00e1tero, cada um de seus \u00e2ngulos internos mede [tex]60^{\\circ}~.[\/tex] <\/li>\n<\/ul>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2018\/07\/prob188_1.png\" style=\"border: none;\" alt=\"\" width=\"241\" height=\"239\" class=\"aligncenter size-full wp-image-106417\" srcset=\"https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2018\/07\/prob188_1.png 241w, https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2018\/07\/prob188_1-150x150.png 150w\" sizes=\"auto, (max-width: 241px) 100vw, 241px\" \/><\/p>\n<p style=\"margin:10px\">\n<p>Com isso, pelo Lembrete <span class=\"c3\"> <strong>(1)<\/strong><\/span>, os tri\u00e2ngulos [tex]DAE~[\/tex], [tex]EBF~[\/tex] e [tex]FCD[\/tex] s\u00e3o congruentes.<br \/>\nMais do que isso, como esses tri\u00e2ngulos s\u00e3o is\u00f3sceles, pelo  Lembrete <span class=\"c3\"> <strong>(2)<\/strong><\/span>, seus \u00e2ngulos de base t\u00eam a mesma medida. Assim, como a medida do terceiro \u00e2ngulo de cada tri\u00e2ngulo \u00e9 [tex]60^{\\circ}~[\/tex], a medida de cada \u00e2ngulo de base ser\u00e1 [tex]\\dfrac{180^{\\circ}-60^{\\circ}}{2}=60^{\\circ}[\/tex] e, ent\u00e3o, os tr\u00eas tri\u00e2ngulos s\u00e3o equil\u00e1teros.<br \/>\nPortanto, na realidade, temos quatro tri\u00e2ngulos congruentes: [tex]DAE~[\/tex], [tex]EBF~[\/tex], [tex]FCD~[\/tex] e [tex]DEF~.[\/tex] <\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2018\/07\/prob188_3.png\" style=\"border: none;\" alt=\"\" width=\"241\" height=\"241\" class=\"aligncenter size-full wp-image-106429\" srcset=\"https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2018\/07\/prob188_3.png 241w, https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2018\/07\/prob188_3-150x150.png 150w\" sizes=\"auto, (max-width: 241px) 100vw, 241px\" \/><\/p>\n<p>Como os quatro tri\u00e2ngulos [tex]DAE~[\/tex], [tex]EBF~[\/tex], [tex]FCD~[\/tex] e [tex]DEF~[\/tex] t\u00eam a mesma \u00e1rea, se denotarmos as \u00e1reas dos tri\u00e2ngulos [tex]DEF~[\/tex] e [tex]~ABC[\/tex] por, respectivamente,  [tex][DEF]~[\/tex] e [tex]~[ABC][\/tex], segue que:<br \/>\n [tex]\\qquad [DEF]=\\dfrac{[ABC]}{4}\\qquad \\textcolor{#800000}{(i)}~.[\/tex]<br \/>\nDe maneira an\u00e1loga, mantendo a nota\u00e7\u00e3o [tex][ ~~][\/tex] para indicar \u00e1reas,  podemos concluir que:<\/p>\n<p style=\"margin:10px\"> [tex]\\qquad [GHI]=\\dfrac{[DEF]}{4}\\qquad \\textcolor{#800000}{(ii)}~[\/tex],<\/p>\n<p>assim, de [tex]\\textcolor{#800000}{(i)}~[\/tex] e de [tex]\\textcolor{#800000}{(ii)}~[\/tex], segue que:<\/p>\n<p style=\"margin:10px\"> [tex]\\qquad [GHI]\\stackrel{\\textcolor{#800000}{(ii)}}{=}\\dfrac{[DEF]}{4}\\stackrel{\\textcolor{#800000}{(i)}}{=}\\dfrac{\\dfrac{[ABC]}{4}}{4}=\\dfrac{[ABC]}{16}~,[\/tex]<\/p>\n<p>ou seja,  [tex][GHI]=\\dfrac{[ABC]}{16}~.[\/tex]<br \/>\nPara expressarmos essa rela\u00e7\u00e3o entre \u00e1reas percentualmente, basta fazermos uma regrinha de tr\u00eas simples. Observe:<\/p>\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][ABC][\/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]100\\%[\/tex]<\/td>\n<\/tr>\n<tr class=\"bf\">\n<td style=\"border: 1px solid #ffffff\"> [tex][GHI][\/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]?[\/tex]<\/td>\n<\/tr>\n<\/table>\n<p>Substituindo a rela\u00e7\u00e3o [tex][GHI]=\\dfrac{[ABC]}{16}~[\/tex], segue que<\/p>\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][ABC][\/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]100\\%[\/tex]<\/td>\n<\/tr>\n<tr class=\"bf\">\n<td style=\"border: 1px solid #ffffff\"> [tex]\\dfrac{[ABC]}{16}[\/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]?[\/tex]<\/td>\n<\/tr>\n<\/table>\n<p>e, assim,<br \/>\n[tex]\\qquad ?=\\dfrac{100\\% \\times \\dfrac{[ABC]}{16}}{[ABC]}=\\dfrac{100\\% \\times\\cancel{[ABC]}}{16 \\times\\cancel{[ABC]}}=\\dfrac{100}{16}\\%~.[\/tex]<\/p>\n<p style=\"margin:20px\">Logo, a \u00e1rea do tri\u00e2ngulo [tex]GHI[\/tex] corresponde a [tex]\\fcolorbox{black}{#eee0e5}{$6,25\\%$}~[\/tex] da \u00e1rea do tri\u00e2ngulo [tex]ABC.[\/tex]<\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000;  margin-top: 30px; margin-bottom: 8px;  border: 0 none\"\/>\n<hr style=\"height:3px; width:100%;background-color:#800000;  margin-top: 8px; margin-bottom: 8px;  border: 0 none\"\/>\n<hr style=\"height:1px; width:100%;background-color:#800000;  margin-top: 8px; margin-bottom: 30px;  border: 0 none\"\/>\n<p>Se voc\u00ea n\u00e3o conseguiu obter a equa\u00e7\u00e3o [tex]\\textcolor{#800000}{(ii)},~[GHI]=\\dfrac{[DEF]}{4}~[\/tex], clique no bot\u00e3o abaixo e acompanhe a dedu\u00e7\u00e3o.<\/p>\n<div style=\"text-align:center;\">\n<input class=\"buts2 bbg3\" value=\"Mostrar justificativas\" 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 = 'Mostrar justificativas';}\" \/>    <\/p>\n<div  style=\"display: none; text-align:justify;\">\nComo o tri\u00e2ngulo [tex]DFE[\/tex] \u00e9 equil\u00e1tero, seus lados t\u00eam o mesmo comprimento: [tex]\\dfrac{l}{2}[\/tex]. Dessa forma, como  [tex]G, H, I[\/tex] s\u00e3o pontos m\u00e9dios dos segmentos [tex]\\overline{FD}[\/tex], [tex]\\overline{DE}[\/tex], [tex]\\overline{EF}[\/tex], respectivamente, ent\u00e3o  os segmentos [tex]\\overline{DH}~[\/tex], [tex]\\overline{HE}~[\/tex], [tex]\\overline{EI}~[\/tex], [tex]\\overline{IF}~[\/tex], [tex]\\overline{FG}~[\/tex] e [tex]\\overline{GD}~[\/tex] t\u00eam comprimentos iguais a  [tex]\\frac{l}{4}~.[\/tex]<br \/>\nComo j\u00e1 sabemos que os \u00e2ngulos com v\u00e9rtices em [tex]D, E, F[\/tex] medem  [tex]60^{\\circ}~[\/tex], ent\u00e3o pelo Lembrete <span class=\"c3\"> <strong>(1)<\/strong><\/span> podemos concluir que os tri\u00e2ngulos [tex]DHG~[\/tex], [tex]EIH~[\/tex] e [tex]FGI[\/tex] s\u00e3o congruentes.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2018\/07\/prob188_4.png\"  style=\"border: none;\" alt=\"\" width=\"353\" height=\"312\" class=\"aligncenter size-full wp-image-106476\" srcset=\"https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2018\/07\/prob188_4.png 353w, https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2018\/07\/prob188_4-300x265.png 300w\" sizes=\"auto, (max-width: 353px) 100vw, 353px\" \/><\/p>\n<p>Por outro lado, veja que esses tr\u00eas tri\u00e2ngulos s\u00e3o is\u00f3sceles; logo, pelo Lembrete <span class=\"c3\"> <strong>(2)<\/strong><\/span>, seus \u00e2ngulos de base t\u00eam a mesma medida. Com isso, como a medida do terceiro \u00e2ngulo de cada tri\u00e2ngulo \u00e9 [tex]60^{\\circ}~[\/tex], a medida de cada \u00e2ngulo de base ser\u00e1 [tex]\\dfrac{180^{\\circ}-60^{\\circ}}{2}=60^{\\circ}.[\/tex] Assim, os tr\u00eas tri\u00e2ngulos  [tex]DHG~[\/tex], [tex]EIH~[\/tex] e [tex]FGI[\/tex] s\u00e3o equil\u00e1teros, e, pelo Lembrete <span class=\"c3\"> <strong>(3)<\/strong><\/span>, os tri\u00e2ngulos [tex]DHG~[\/tex], [tex]EIH~[\/tex], [tex]FGI~[\/tex] e [tex]GHI[\/tex] s\u00e3o congruentes.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2018\/07\/prob188_5.png\" style=\"border: none;\" alt=\"\" width=\"342\" height=\"303\" class=\"aligncenter size-full wp-image-106480\" srcset=\"https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2018\/07\/prob188_5.png 342w, https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2018\/07\/prob188_5-300x266.png 300w\" sizes=\"auto, (max-width: 342px) 100vw, 342px\" \/><\/p>\n<p>Como os tri\u00e2ngulos [tex]DHG~[\/tex], [tex]EIH~[\/tex], [tex]FGI~[\/tex] e [tex]GHI[\/tex] t\u00eam a mesma \u00e1rea, se denotarmos as \u00e1reas dos tri\u00e2ngulos [tex]DEF~[\/tex] e [tex]~GHI[\/tex] por, respectivamente,  [tex][DEF]~[\/tex] e [tex]~[GHI][\/tex], segue que  [tex]\\boxed{[GHI]=\\dfrac{[DEF]}{4}}~.[\/tex] <\/p>\n<p style=\"margin:30px\">\n<\/div>\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: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\/Porcentagem-da-\u00e1rea.pdf\" title=\" Arquivo PDF\" download=\"Clubes_ Porcentagem da \u00e1rea.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) (ONEM &#8211; 2011) O tri\u00e2ngulo [tex]ABC[\/tex] mostrado na figura \u00e9 equil\u00e1tero e os pontos [tex]D, E, F[\/tex] s\u00e3o pontos m\u00e9dios dos segmentos [tex]\\overline{CA}[\/tex], [tex]\\overline{AB}[\/tex], [tex]\\overline{BC}[\/tex], respectivamente. Por outro lado, [tex]G, H, I[\/tex] s\u00e3o &hellip; <\/p>\n<p><a class=\"more-link btn\" href=\"https:\/\/clubes.obmep.org.br\/blog\/problema-para-ajudar-na-escola-porcentagem-da-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-106406","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/106406","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=106406"}],"version-history":[{"count":94,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/106406\/revisions"}],"predecessor-version":[{"id":219616,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/106406\/revisions\/219616"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=106406"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}