{"id":220644,"date":"2024-06-23T06:37:31","date_gmt":"2024-06-23T09:37:31","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=220644"},"modified":"2024-06-23T06:37:31","modified_gmt":"2024-06-23T09:37:31","slug":"a-problema-para-ajudar-na-escola-area-de-um-quadrilatero-com-pontas","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/a-problema-para-ajudar-na-escola-area-de-um-quadrilatero-com-pontas\/","title":{"rendered":"(A) Problema para ajudar na escola: \u00c1rea de um quadril\u00e1tero com pontas"},"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:20px;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: 1px; margin-bottom: 20px;border: 0 none\"\/>\n(<strong>ONEM<\/strong>, 2004 &#8211; Adaptado) Na figura, vemos uma circunfer\u00eancia de centro em [tex]O[\/tex] que tangencia os lados [tex]\\overline{AD}[\/tex] e [tex]\\overline{BC}[\/tex] do ret\u00e2ngulo [tex]ABCD[\/tex] nos pontos [tex]E[\/tex] e [tex]F[\/tex], respectivamente.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2019\/09\/prob372_1.png\" style=\"border: none;\"  alt=\"\" width=\"488\" height=\"251\" class=\"aligncenter size-full wp-image-135629\" srcset=\"https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2019\/09\/prob372_1.png 488w, https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2019\/09\/prob372_1-300x154.png 300w\" sizes=\"auto, (max-width: 488px) 100vw, 488px\" \/><\/p>\n<p>Se o ret\u00e2ngulo [tex]ABCD[\/tex] tem uma \u00e1rea de [tex]40\\, cm^2[\/tex], qual \u00e9 a medida da \u00e1rea do quadril\u00e1tero colorido  [tex]OCEB[\/tex] ?<\/p>\n<p style=\"margin:10px;\">\n<p style=\"margin:10px\">\n<\/div>\n<p style=\"margin:40px\">\n<div class=\"cb36 g1\" style=\"margin:30px;line-height:25px; width:75%; margin-left:auto; margin-right:auto; \"><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=\"65\" height=\"77\" class=\"alignleft size-full wp-image-1001\" \/> <\/p>\n<p class=\"c3 ti3\" style=\"line-height:20px;margin:10px\"> Lembretes e nota\u00e7\u00f5es <\/p>\n<p style=\"margin:10px\">\n<p><span class=\"c3\"> <strong>(1)<\/strong> <\/span>  Uma reta tangente a uma circunfer\u00eancia \u00e9 perpendicular ao raio no seu ponto de tang\u00eancia.<br \/>\n<span class=\"c3\"> <strong>(2)<\/strong> <\/span>  Dado um ponto [tex]P[\/tex] fora de uma dada reta [tex]r[\/tex],  existe uma \u00fanica reta perpendicular a [tex]r[\/tex]  passando por [tex]P[\/tex] .<br \/>\n<span class=\"c3\"> <strong>(3)<\/strong> <\/span>  Para facilitar a nota\u00e7\u00e3o, denotaremos o comprimento em cent\u00edmetros do raio da circunfer\u00eancia da figura por [tex]r[\/tex] e os comprimentos em cent\u00edmetros dos lados [tex]\\overline{AE}[\/tex] e [tex]\\overline{ED}[\/tex]  por [tex]a[\/tex] e [tex]b[\/tex] respectivamente.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2019\/11\/prob372_2.png\" style=\"border: none;\"  alt=\"\" width=\"503\" height=\"255\" class=\"aligncenter size-full wp-image-140602\" srcset=\"https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2019\/11\/prob372_2.png 503w, https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2019\/11\/prob372_2-300x152.png 300w\" sizes=\"auto, (max-width: 503px) 100vw, 503px\" \/><\/p>\n<p style=\"margin:10px\">\n<\/div>\n<p style=\"margin:40px\">\n<div class=\"g1n\" style=\"line-height:29px;\">\n<p class=\"tit\" style=\"line-height:28px;margin:10px\">Solu\u00e7\u00e3o 1<\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 1px; margin-bottom: 30px;border: 0 none\"\/>\n<p style=\"margin:10px\">\n<p>Observe, inicialmente, que  os segmentos [tex]\\overline{AD}[\/tex] e [tex]\\overline{BC}[\/tex] s\u00e3o paralelos, j\u00e1 que eles s\u00e3o lados opostos de um ret\u00e2ngulo. Dessa forma, utilizando os <span class=\"c3\"> <strong>Lembretes (1)<\/strong><\/span> e <span class=\"c3\"> <strong>(2)<\/strong><\/span>, podemos concluir que:<\/p>\n<ul>\n<li>os pontos [tex]O[\/tex], [tex]E[\/tex] e [tex]F[\/tex] s\u00e3o colineares; <\/li>\n<li>a reta determinada por [tex]O[\/tex], [tex]E[\/tex] e [tex]F[\/tex] \u00e9 perpendicular a [tex]\\overline{AD}[\/tex] e a [tex]\\overline{BC}[\/tex];<\/li>\n<li>os segmentos [tex]\\overline{AB}[\/tex] e [tex]\\overline{DC}[\/tex] t\u00eam comprimento [tex]2r[\/tex].<\/li>\n<\/ul>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2019\/11\/prob372_4.png\"  style=\"border: none;\"   alt=\"\" width=\"504\" height=\"251\" class=\"aligncenter size-full wp-image-140612\" srcset=\"https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2019\/11\/prob372_4.png 504w, https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2019\/11\/prob372_4-300x149.png 300w\" sizes=\"auto, (max-width: 504px) 100vw, 504px\" \/><\/p>\n<p>Por outro lado, observe que a \u00e1rea do ret\u00e2ngulo [tex]ABCD[\/tex]  pode ser decomposta como a soma das \u00e1reas dos tri\u00e2ngulos [tex]ABE[\/tex], [tex]BCO[\/tex]  e [tex]CDE[\/tex]  e do quadril\u00e1tero [tex]OCEB[\/tex]. Portanto, calcularemos as medidas [tex]A_1[\/tex], [tex]A_2[\/tex]  e [tex]A_3[\/tex] das respectivas \u00e1reas desses tr\u00eas tri\u00e2ngulos para obtermos a \u00e1rea [tex]\\textcolor{#FF3366}{A}[\/tex] solicitada no problema.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2019\/11\/prob372_6.png\"  style=\"border: none;\"   alt=\"\" width=\"517\" height=\"260\" class=\"aligncenter size-full wp-image-140620\" srcset=\"https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2019\/11\/prob372_6.png 517w, https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2019\/11\/prob372_6-300x151.png 300w\" sizes=\"auto, (max-width: 517px) 100vw, 517px\" \/><\/p>\n<p>Vamos aos c\u00e1lculos!<\/p>\n<table class=\"bf\" style=\"margin-left:auto; margin-right:auto;width:55%\">\n<tr>\n<td class=\"bf\" style=\"border: 1px solid #ffffff; border-right: 2px solid #800000;\">\n<ul>\n<li>Tri\u00e2ngulo [tex]ABE[\/tex]:<\/li>\n<p>[tex]\\;A_1=\\dfrac{a \\cdot 2r}{2}=a \\cdot r[\/tex].<\/ul>\n<\/td>\n<td class=\"bf\" style=\"border: 1px solid #ffffff; border-right: 2px solid #800000;\">\n<ul>\n<li>Tri\u00e2ngulo [tex]BCO[\/tex]:<\/li>\n<p>[tex]\\;A_2=\\dfrac{(a+b) \\cdot r}{2}[\/tex].<\/ul>\n<\/td>\n<td class=\"bf\">\n<ul>\n<li>Tri\u00e2ngulo [tex]CDE[\/tex]:<\/li>\n<p>[tex]\\;A_3=\\dfrac{b \\cdot 2r}{2}=b \\cdot r[\/tex].<\/ul>\n<\/td>\n<\/tr>\n<\/table>\n<p>Como [tex]\\boxed{A_1+A_2+A_3+\\textcolor{#FF3366}{A}=(a+b) \\cdot 2r}[\/tex], segue que:<br \/>\n[tex]\\qquad a \\cdot r+\\dfrac{(a+b) \\cdot r}{2}+b \\cdot r+\\textcolor{#FF3366}{A}=(a+b) \\cdot 2r[\/tex]<br \/>\n[tex]\\qquad (a+b) \\cdot r+\\dfrac{(a+b) \\cdot r}{2}+\\textcolor{#FF3366}{A}=(a+b) \\cdot 2r[\/tex]<br \/>\n[tex]\\qquad 2 \\cdot (a+b) \\cdot r+(a+b) \\cdot r+2 \\cdot \\textcolor{#FF3366}{A}=4 \\cdot (a+b) \\cdot r[\/tex]<br \/>\n[tex]\\qquad 3 \\cdot  (a+b) \\cdot r+2 \\cdot \\textcolor{#FF3366}{A}=4 \\cdot (a+b) \\cdot r[\/tex]<br \/>\n[tex]\\qquad 2 \\cdot \\textcolor{#FF3366}{A}=4 \\cdot (a+b) \\cdot r-3 \\cdot (a+b) \\cdot r[\/tex]<br \/>\n[tex]\\qquad 2 \\cdot \\textcolor{#FF3366}{A}= (a+b) \\cdot r[\/tex]<br \/>\n[tex]\\qquad \\textcolor{#FF3366}{A}= \\dfrac{(a+b) \\cdot r}{2}. \\qquad \\qquad \\textcolor{#800000}{(i)}[\/tex]<br \/>\nMas sabemos que o ret\u00e2ngulo [tex]ABCD[\/tex] tem uma \u00e1rea de [tex]40\\, cm^2[\/tex]; assim:<br \/>\n[tex]\\qquad 40=(a+b)\\cdot 2r[\/tex]<br \/>\n[tex]\\qquad (a+b)\\cdot r=20. \\qquad \\qquad \\textcolor{#800000}{(ii)}[\/tex]<br \/>\nSubstituindo [tex] \\textcolor{#800000}{(ii)}[\/tex] em [tex] \\textcolor{#800000}{(i)}[\/tex], obtemos:<br \/>\n[tex]\\qquad \\textcolor{#FF3366}{A}= \\dfrac{20}{2}=10[\/tex].<br \/>\nLogo, a medida da \u00e1rea do quadril\u00e1tero colorido  [tex]\\textcolor{#FF3366}{OCEB}[\/tex] \u00e9 [tex]\\, \\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:29px;\">\n<p class=\"tit\" style=\"line-height:28px;margin:10px\">Solu\u00e7\u00e3o 2<\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 1px; margin-bottom: 30px;border: 0 none\"\/>\n<p style=\"margin:10px\">\n<p>Observe, inicialmente, que  os segmentos [tex]\\overline{AD}[\/tex] e [tex]\\overline{BC}[\/tex] s\u00e3o paralelos, j\u00e1 que eles s\u00e3o lados opostos de um ret\u00e2ngulo. Dessa forma, utilizando os <span class=\"c3\"> <strong>Lembretes (1)<\/strong><\/span> e <span class=\"c3\"> <strong>(2)<\/strong><\/span>, podemos concluir que:<\/p>\n<ul>\n<li>os pontos [tex]O[\/tex], [tex]E[\/tex] e [tex]F[\/tex] s\u00e3o colineares; <\/li>\n<li>a reta determinada por [tex]O[\/tex], [tex]E[\/tex] e [tex]F[\/tex] \u00e9 perpendicular a [tex]\\overline{AD}[\/tex] e a [tex]\\overline{BC}[\/tex];<\/li>\n<li>os segmentos [tex]\\overline{AB}[\/tex] e [tex]\\overline{DC}[\/tex] t\u00eam comprimento [tex]2r[\/tex].<\/li>\n<\/ul>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2019\/11\/prob372_4.png\"  style=\"border: none;\"   alt=\"\" width=\"504\" height=\"251\" class=\"aligncenter size-full wp-image-140612\" srcset=\"https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2019\/11\/prob372_4.png 504w, https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2019\/11\/prob372_4-300x149.png 300w\" sizes=\"auto, (max-width: 504px) 100vw, 504px\" \/><\/p>\n<p>Para esta solu\u00e7\u00e3o, observe que a \u00e1rea do quadril\u00e1tero [tex]OCEB[\/tex]  pode ser decomposta como a soma das \u00e1reas dos tri\u00e2ngulos [tex]EBO[\/tex] e [tex]EOC[\/tex]. Portanto, calcularemos as medidas [tex]A_1[\/tex]  e [tex]A_2[\/tex] das \u00e1reas desses dois tri\u00e2ngulos para obtermos a \u00e1rea [tex]\\textcolor{#FF3366}{A}[\/tex] solicitada no problema.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2019\/11\/prob372_7.png\" style=\"border: none;\"   alt=\"\" width=\"493\" height=\"244\" class=\"aligncenter size-full wp-image-140684\" srcset=\"https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2019\/11\/prob372_7.png 493w, https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2019\/11\/prob372_7-300x148.png 300w\" sizes=\"auto, (max-width: 493px) 100vw, 493px\" \/><\/p>\n<p>Em ambos os casos, utilizaremos o segmento [tex]OE[\/tex] como base.<\/p>\n<table class=\"bf\"  style=\"margin-left:auto; margin-right:auto; width:60%;\">\n<tr>\n<td class=\"bf\" style=\"border: 1px solid #ffffff; border-right: 2px solid #800000;\">\n<ul>\n<li>Tri\u00e2ngulo [tex]EBO[\/tex]:<\/li>\n<p>[tex]\\quad A_1=\\dfrac{a \\cdot r}{2}[\/tex].<\/ul>\n<\/td>\n<td class=\"bf\" style=\"border: 1px solid #ffffff;\">\n<ul>\n<li>Tri\u00e2ngulo [tex]EOC[\/tex]:<\/li>\n<p>[tex]\\quad A_2=\\dfrac{b \\cdot r}{2}[\/tex].<\/ul>\n<\/td>\n<\/tr>\n<\/table>\n<p>Como [tex]\\boxed{\\textcolor{#FF3366}{A}=A_1+A_2}[\/tex], segue que:<br \/>\n[tex]\\qquad\\textcolor{#FF3366}{A}=\\dfrac{a \\cdot r}{2}+\\dfrac{b \\cdot r}{2}\\\\<br \/>\n\\qquad \\textcolor{#FF3366}{A}= \\dfrac{(a+b) \\cdot r}{2}. \\qquad \\qquad \\textcolor{#800000}{(iii)}[\/tex]<br \/>\nMas sabemos que o ret\u00e2ngulo [tex]ABCD[\/tex] tem uma \u00e1rea de [tex]40\\, cm^2[\/tex]; assim:<br \/>\n[tex]\\qquad 40=(a+b)\\cdot 2r[\/tex]<br \/>\n[tex]\\qquad (a+b)\\cdot r=20. \\qquad \\qquad \\textcolor{#800000}{(iv)}[\/tex]<br \/>\nSubstituindo [tex] \\textcolor{#800000}{(iv)}[\/tex] em [tex] \\textcolor{#800000}{(iii)}[\/tex], obtemos:<br \/>\n[tex]\\qquad \\textcolor{#FF3366}{A}= \\dfrac{20}{2}=10[\/tex].<br \/>\nLogo, a medida da \u00e1rea do quadril\u00e1tero colorido  [tex]\\textcolor{#FF3366}{OCEB}[\/tex] \u00e9 [tex]\\, \\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: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\/\u00c1rea-de-um-quadril\u00e1tero-com-pontas.pdf\" title=\" Arquivo PDF\" download=\"Clubes_ \u00c1rea de um quadril\u00e1tero com pontas.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) (ONEM, 2004 &#8211; Adaptado) Na figura, vemos uma circunfer\u00eancia de centro em [tex]O[\/tex] que tangencia os lados [tex]\\overline{AD}[\/tex] e [tex]\\overline{BC}[\/tex] do ret\u00e2ngulo [tex]ABCD[\/tex] nos pontos [tex]E[\/tex] e [tex]F[\/tex], respectivamente. Se o ret\u00e2ngulo [tex]ABCD[\/tex] &hellip; <\/p>\n<p><a class=\"more-link btn\" href=\"https:\/\/clubes.obmep.org.br\/blog\/a-problema-para-ajudar-na-escola-area-de-um-quadrilatero-com-pontas\/\">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-220644","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/220644","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=220644"}],"version-history":[{"count":1,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/220644\/revisions"}],"predecessor-version":[{"id":220645,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/220644\/revisions\/220645"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=220644"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}