{"id":171710,"date":"2020-12-07T01:12:39","date_gmt":"2020-12-07T04:12:39","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=171710"},"modified":"2024-06-26T21:34:36","modified_gmt":"2024-06-27T00:34:36","slug":"problema-para-ajudar-na-escola-area-de-um-losango","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/problema-para-ajudar-na-escola-area-de-um-losango\/","title":{"rendered":".Problema para ajudar na escola: \u00c1rea de um losango"},"content":{"rendered":"<p class=\"g2\" ><strong><a href=\"http:\/\/clubes.obmep.org.br\/blog\/?p=220992\" 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; margin-bottom:40px;\">\n<p class=\"tit\" style=\"line-height:20px;margin:10px\">Problema<br \/>\n<span style=\"font-size:12px;color:#0000FF\"> (A partir da 3\u00aa s\u00e9rie do E. M.- <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\"\/>\nDeterminar a \u00e1rea do  losango <em><strong>ABCD<\/strong><\/em> sabendo que as coordenadas dos v\u00e9rtices <strong>A<\/strong>, <strong>B<\/strong> e <strong>C<\/strong> em um mesmo plano cartesiano xOy s\u00e3o dadas por:<\/p>\n<ul>\n<li><strong>A=(10,3)<\/strong> ;  <strong>B=(5,3)<\/strong>  e  <strong>C=(2,-1)<\/strong>.<\/li>\n<\/ul>\n<\/div>\n<p style=\"margin:40px\">\n<div class=\"cb36 g1\" style=\"margin:30px;line-height:27px; width:85%; margin-left:auto; margin-right:auto; \">\n<p class=\"c3 ti3\" style=\"line-height:20px; margin-bottom: 10px;\"> AJUDA<\/p>\n<p style=\"margin:10px\">\n<table style=\" height:auto; margin-left: auto; margin-right: auto;width:540px;\">\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=\"95\" height=\"115\" \/><\/td>\n<td class=\"bf\">\n<div class=\"fala\" style=\"width:400px; background-color: #FFF5EE;\">\nPara resolver este problema vamos utilizar no\u00e7\u00f5es b\u00e1sicas de plano cartesiano.<br \/>\nTalvez o v\u00eddeo abaixo possa ajudar!\n<\/div>\n<\/td>\n<\/tr>\n<\/table>\n<hr style=\"height:2px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 8px;border: 0 none\"\/>\n<hr style=\"height:2px; width:100%;background-color:#800000; margin-top: 8px; margin-bottom: 30px;border: 0 none\"\/>\n<p style=\"margin:10px; text-align:center\">\n<iframe loading=\"lazy\" width=\"640\" height=\"360\" src=\"https:\/\/www.youtube.com\/embed\/gIAod8Ns2Uw?rel=0\" allowfullscreen><\/iframe><br \/>\n<strong>Referencial Cartesiano<\/strong> (Abcissas e ordenadas)<\/p>\n<hr style=\"height:2px; width:100%;background-color:#800000; margin-top: 10px; margin-bottom: 8px;border: 0 none\"\/>\n<hr style=\"height:2px; width:100%;background-color:#800000; margin-top: 8px; margin-bottom: 20px;border: 0 none\"\/>\n<p class=\"c3 ti3\" style=\"line-height:20px; margin-bottom: 10px;\"> Lembretes e nota\u00e7\u00f5es<\/p>\n<p>Em um plano cartesiano [tex]xOy[\/tex], considere os pontos [tex]A=\\left(x_A,y_A\\right)\\,[\/tex] e [tex]\\,B=\\left(x_B,y_B\\right)\\,.[\/tex]<br \/>\n<span class=\"c3\"><strong>(1)<\/strong><\/span> A dist\u00e2ncia entre os pontos [tex]A\\,[\/tex] e [tex]\\,B[\/tex], denotada por [tex]d_{AB}[\/tex], \u00e9 definida por:<br \/>\n[tex]\\qquad d_{AB}=\\sqrt{\\left(x_A-x_B\\right)^2+\\left(y_A-y_B\\right)^2\\,}[\/tex].<br \/>\n<span class=\"c3\"><strong>(2)<\/strong><\/span> O ponto m\u00e9dio dos pontos [tex]A\\,[\/tex] e [tex]\\,B[\/tex], denotado por [tex]M_{AB}[\/tex], \u00e9 o ponto definido por:<br \/>\n[tex]\\qquad M_{AB}=\\left(\\dfrac{x_A+x_B}{2}\\,,\\, \\dfrac{y_A+y_B}{2}\\right)[\/tex].<br \/>\n<span class=\"c3\"><strong>(3)<\/strong><\/span> Um losango possui os quatro lados congruentes.<br \/>\n<span class=\"c3\"><strong>(4)<\/strong><\/span> Em todo losango, as diagonais intersectam-se perpendicularmente nos respectivos pontos m\u00e9dios.<br \/>\n<strong><span class=\"c3\" style=\"font-size: 25px;\"> &#10000; <\/span><\/strong> Denotaremos o segmento de reta definido por dois pontos, digamos [tex]X[\/tex] e [tex]Y[\/tex], por [tex]\\overline{XY}[\/tex] e seu respectivo comprimento por [tex]XY[\/tex].<\/p>\n<p style=\"margin:10px\">\n<\/div>\n<p style=\"margin:40px\">\n<div class=\"g1n\" style=\"line-height:25px;\">\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: 1px; margin-bottom: 20px;border: 0 none\"\/>\nA partir dos <strong>Lembretes (3) e (4)<\/strong>, podemos decompor um losango em quatro tri\u00e2ngulos ret\u00e2ngulos congruentes ou em dois tri\u00e2ngulos, n\u00e3o necessariamente ret\u00e2ngulos, congruentes, conforme ilustram as duas imagens a seguir.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2020\/11\/prob493_3.png\" style=\"border:none\" alt=\"\" width=\"432\" height=\"304\" class=\"aligncenter size-full wp-image-171738\" srcset=\"https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2020\/11\/prob493_3.png 432w, https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2020\/11\/prob493_3-300x211.png 300w\" sizes=\"auto, (max-width: 432px) 100vw, 432px\" \/><\/p>\n<p>Dessa forma, sequer precisamos do quarto v\u00e9rtice do paralelogramo do problema para calcular a sua \u00e1rea. Assim, vamos inicialmente fixar um plano cartesiano [tex]xOy[\/tex] e representar os pontos [tex]A,B\\,[\/tex] e [tex]\\, C\\,[\/tex] para melhor visualizarmos a solu\u00e7\u00e3o.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2020\/11\/prob493_1.png\" style=\"border:none\" alt=\"\" width=\"500\" height=\"254\" class=\"aligncenter size-full wp-image-171733\" \/><\/p>\n<p>I &#8211; Vamos determinar o ponto m\u00e9dio comum das duas diagonais do losango utilizando os v\u00e9rtices [tex]A\\,[\/tex] e [tex]\\,C[\/tex], para podermos calcular a altura do tri\u00e2ngulo [tex]ABC[\/tex]. A \u00e1rea do losango ser\u00e1, ent\u00e3o, duas vezes a \u00e1rea desse tri\u00e2ngulo.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2020\/11\/prob493_4.png\"  style=\"border:none\"  alt=\"\" width=\"500\" height=\"254\" class=\"aligncenter size-full wp-image-171740\" \/><\/p>\n<ul>\n<li>Ponto m\u00e9dio da diagonal [tex]\\overline{AC}[\/tex]: <\/li>\n<p>[tex]\\qquad M_{AC}=\\left(\\dfrac{x_A+x_C}{2}\\,,\\, \\dfrac{y_A+y_C}{2}\\right)\\\\<br \/>\n\\qquad M_{AC}=\\left(\\dfrac{10+2}{2}\\,,\\, \\dfrac{3-1}{2}\\right)\\\\<br \/>\n\\qquad M_{AC}=\\left(6\\,,\\,1\\right).<br \/>\n[\/tex]<\/ul>\n<p>Vamos agora determinar os comprimentos dos segmentos [tex]\\overline{AC}\\,[\/tex] e [tex]\\,\\overline{BM}[\/tex], respectivamente, base e altura do tri\u00e2ngulo [tex]ABC\\,.[\/tex]<\/p>\n<ul>\n<li> Comprimento do segmento [tex]\\overline{AC}\\,[\/tex]: <\/li>\n<p>[tex]\\qquad d_{AC}=\\sqrt{\\left(x_A-x_C\\right)^2+\\left(y_A-y_C\\right)^2\\,}\\\\<br \/>\n\\qquad d_{AC}=\\sqrt{\\left(10-2\\right)^2+\\left(3-(-1)\\right)^2\\,}\\\\<br \/>\n\\qquad d_{AC}=\\sqrt{8^2+4^2\\,}=\\sqrt{64+16\\,}\\\\<br \/>\n\\qquad d_{AC}=\\sqrt{80\\,}.[\/tex]<\/p>\n<li> Comprimento do segmento [tex]\\overline{BM}\\,[\/tex]: <\/li>\n<p>[tex]\\qquad d_{BM}=\\sqrt{\\left(x_B-x_M\\right)^2+\\left(y_B-y_M\\right)^2\\,}\\\\<br \/>\n\\qquad d_{BM}=\\sqrt{\\left(5-6\\right)^2+\\left(3-1\\right)^2\\,}\\\\<br \/>\n\\qquad d_{BM}=\\sqrt{(-1)^2+2^2\\,}=\\sqrt{1+4\\,}\\\\<br \/>\n\\qquad d_{BM}=\\sqrt{5\\,}.[\/tex]<\/p>\n<li> \u00c1rea do tri\u00e2ngulo [tex]ABC[\/tex]: <\/li>\n<p>[tex]\\qquad \\textcolor{#0099CC}{S_{ABC}=\\dfrac{base \\times altura}{2}}\\\\<br \/>\n\\qquad \\textcolor{#0099CC}{S_{ABC}=\\dfrac{\\sqrt{80\\,}\\times \\sqrt{5\\,}}{2}}\\\\<br \/>\n\\qquad \\textcolor{#0099CC}{S_{ABC}=\\dfrac{\\sqrt{400\\,}}{2}}\\\\<br \/>\n\\qquad \\textcolor{#0099CC}{S_{ABC}=10}.\\\\<br \/>\n[\/tex]<\/ul>\n<p>Portanto, a \u00e1rea do losango [tex]ABCD[\/tex] \u00e9 [tex]\\fcolorbox{black}{#d5e4ef}{$2 \\times 10=20$}[\/tex] unidades de \u00e1rea.<\/p>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 30px; margin-bottom: 8px;border: 0 none\"\/>\n<hr style=\"height:1px; width:100%;background-color:#800000; margin-top: 8px; margin-bottom: 20px;border: 0 none\"\/>\n<p>II &#8211; Vamos utilizar o ponto m\u00e9dio comum das duas diagonais do losango j\u00e1 calculado, [tex]M_{AC}=\\left(6\\,,\\,1\\right)[\/tex], para calcularmos a medida do cateto [tex]\\overline{MA}\\,[\/tex] do tri\u00e2ngulo ret\u00e2ngulo [tex]BMA\\, .[\/tex] A medida de [tex]\\, \\overline{MB}[\/tex] j\u00e1 est\u00e1 calculada: [tex] d_{BM}=\\sqrt{5\\,}\\,.[\/tex] A \u00e1rea do losango ser\u00e1, dessa vez, quatro vezes a \u00e1rea desse tri\u00e2ngulo.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2020\/11\/prob493_5.png\" style=\"border:none\" alt=\"\" width=\"500\" height=\"256\" class=\"aligncenter size-full wp-image-171779\" srcset=\"https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2020\/11\/prob493_5.png 619w, https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2020\/11\/prob493_5-300x154.png 300w\" sizes=\"auto, (max-width: 500px) 100vw, 500px\" \/><\/p>\n<ul>\n<li> Comprimento do segmento [tex]\\overline{MA}\\,[\/tex]: <\/li>\n<p>[tex]\\qquad d_{MA}=\\sqrt{\\left(x_M-x_A\\right)^2+\\left(y_M-y_A\\right)^2\\,}\\\\<br \/>\n\\qquad d_{MA}=\\sqrt{\\left(6-10\\right)^2+\\left(1-3\\right)^2\\,}\\\\<br \/>\n\\qquad d_{MA}=\\sqrt{(-4)^2+(-2)^2\\,}=\\sqrt{16+4\\,}\\\\<br \/>\n\\qquad d_{MA}=\\sqrt{20\\,}.[\/tex]<\/p>\n<li> \u00c1rea do tri\u00e2ngulo [tex]BMA[\/tex]: <\/li>\n<p>[tex]\\qquad \\textcolor{#009900}{S_{BMA}=\\dfrac{base \\times altura}{2}}\\\\<br \/>\n\\qquad \\textcolor{#009900}{S_{BMA}=\\dfrac{\\sqrt{20\\,}\\times \\sqrt{5\\,}}{2}}\\\\<br \/>\n\\qquad \\textcolor{#009900}{S_{BMA}=\\dfrac{\\sqrt{100\\,}}{2}}\\\\<br \/>\n\\qquad \\textcolor{#009900}{S_{BMA}=5}.\\\\<br \/>\n[\/tex]<\/ul>\n<p>Portanto, a \u00e1rea do losango [tex]ABCD[\/tex] \u00e9 [tex]\\fcolorbox{black}{#bfb}{$4 \\times 5=20$}[\/tex] unidades de \u00e1rea.<\/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\/2020\/12\/\u00c1rea-de-um-losango_c.pdf\" title=\" Arquivo PDF\" download=\"Clubes_ \u00c1rea de um losango.pdf\"><span style=\"color: #FFFAFA; font-family: georgia; font-size: 16px;\"><strong>Arquivo <\/strong><\/span><\/a>\n <\/p>\n<\/div>\n<div style=\"clear: both;\"><\/div>\n","protected":false},"excerpt":{"rendered":"<p>&#9999; Link do problema para dispositivos da Apple. Problema (A partir da 3\u00aa s\u00e9rie do E. M.- N\u00edvel de dificuldade: M\u00e9dio) Determinar a \u00e1rea do losango ABCD sabendo que as coordenadas dos v\u00e9rtices A, B e C em um mesmo plano cartesiano xOy s\u00e3o dadas por: A=(10,3) ; B=(5,3) e C=(2,-1). AJUDA Para resolver este &hellip; <\/p>\n<p><a class=\"more-link btn\" href=\"https:\/\/clubes.obmep.org.br\/blog\/problema-para-ajudar-na-escola-area-de-um-losango\/\">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-171710","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/171710","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=171710"}],"version-history":[{"count":83,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/171710\/revisions"}],"predecessor-version":[{"id":220994,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/171710\/revisions\/220994"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=171710"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}