{"id":151159,"date":"2020-04-11T16:56:00","date_gmt":"2020-04-11T19:56:00","guid":{"rendered":"http:\/\/clubes.obmep.org.br\/blog\/?page_id=151159"},"modified":"2023-06-11T00:53:03","modified_gmt":"2023-06-11T03:53:03","slug":"problema-a-area-do-paralelogramo","status":"publish","type":"page","link":"https:\/\/clubes.obmep.org.br\/blog\/problema-a-area-do-paralelogramo\/","title":{"rendered":".Problema: A \u00e1rea do paralelogramo"},"content":{"rendered":"<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:#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: 1px; margin-bottom: 20px;border: 0 none\"\/>\n<p style=\"margin:10px\">\n<p><span style=\"font-size:12px\">(GERONIMO, J.R; FRANCO, V.S, <strong>Geometria Plana e Espacial: um estudo axiom\u00e1tico<\/strong>, 2ed, EDUEM, Maring\u00e1, PR, 2010)<\/span> Um paralelogramo tem dois lados medindo [tex]18~ cm[\/tex] e [tex]8~ cm[\/tex], e um \u00e2ngulo medindo [tex]30^{\\circ}[\/tex].<br \/>\n<strong>Determine a \u00e1rea do paralelogramo.<\/strong><\/p>\n<p style=\"margin:10px\">\n<\/div>\n<p style=\"margin:30px\">\n<table style=\" height:auto; text-align:justify ;margin-left: auto; margin-right: auto; width: 580px;\">\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<p class=\"fala\" style=\"width:450px;background-color: #eee0e5;\" ><span class=\"c3\"> <strong>Nota\u00e7\u00e3o:<\/strong><\/span> Denotaremos o segmento definido por dois pontos, digamos [tex]X[\/tex] e [tex] Y[\/tex],  por [tex]\\overline {XY} [\/tex] e o seu comprimento por [tex]XY[\/tex].<\/p>\n<\/td>\n<\/tr>\n<\/table>\n<p style=\"margin:30px\">\n<div class=\"g1n\" style=\"line-height:28px;\">\n<p class=\"tit\" style=\"line-height:25px;margin:10px\">Solu\u00e7\u00e3o<\/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>Considere o paralelogramo [tex]ABCD[\/tex] com [tex]\\overline {AD}[\/tex] paralelo a [tex]\\overline {BC}[\/tex], [tex]D\\widehat{A}B=30^\\circ[\/tex], [tex]AB= 8~cm[\/tex]  e [tex]BC=AD=18 ~cm\\,.[\/tex] <\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2020\/04\/paralelogramoinicio.png\" style=\"border:none\" alt=\"\" width=\"610\" height=\"231\" class=\"aligncenter size-full wp-image-151153\" \/><\/p>\n<p>Tra\u00e7ando a altura por [tex]D[\/tex] relativa ao lado [tex]\\overline {AB}[\/tex], obtemos o tri\u00e2ngulo ret\u00e2ngulo [tex]ADE[\/tex].<br \/>\nNote que, como os \u00e2ngulos [tex]D\\widehat{A}E[\/tex] e [tex]A\\widehat{E}D[\/tex] medem [tex]30^\\circ[\/tex] e [tex]90^\\circ[\/tex], respectivamente, conclu\u00edmos que a medida do \u00e2ngulo [tex] E\\widehat{D}A[\/tex]  \u00e9 [tex] 60^\\circ[\/tex]. <span style=\"font-size:12px;\">(Lembre-se de que a soma das medidas dos \u00e2ngulos internos de um tri\u00e2ngulo \u00e9 [tex]180^\\circ[\/tex].)<\/span><\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2020\/04\/paralelogramored1.png\" style=\"border:none\"  alt=\"\" width=\"610\" height=\"231\" class=\"aligncenter size-full wp-image-151155\" srcset=\"https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2020\/04\/paralelogramored1.png 610w, https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2020\/04\/paralelogramored1-300x114.png 300w\" sizes=\"auto, (max-width: 610px) 100vw, 610px\" \/><\/p>\n<p>Agora, veja que<br \/>\n [tex]\\qquad \\dfrac{1}{2} = sen(30^\\circ) = \\dfrac{DE}{AD}=  \\dfrac{DE}{18} [\/tex],<br \/>\nassim,<br \/>\n [tex]\\qquad DE = \\dfrac{18}{2}= 9~cm [\/tex].<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" src=\"http:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2020\/04\/paralelogramored2.png\" style=\"border:none\" alt=\"\" width=\"610\" height=\"266\" class=\"aligncenter size-full wp-image-151156\" srcset=\"https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2020\/04\/paralelogramored2.png 610w, https:\/\/clubes.obmep.org.br\/blog\/wp-content\/uploads\/2020\/04\/paralelogramored2-300x131.png 300w\" sizes=\"auto, (max-width: 610px) 100vw, 610px\" \/><\/p>\n<p>Logo, a \u00e1rea [tex]S[\/tex] do paralelogramo [tex]ABCD[\/tex] \u00e9 dada por<br \/>\n[tex]\\qquad S = AB \\cdot DE = 8~cm \\cdot 9~cm= \\fcolorbox{black}{#eee0e5}{$72 \\,cm^{2}$}\\,.  [\/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:10px\">\n","protected":false},"excerpt":{"rendered":"<p>Problema (Indicado a partir do 9\u00ba ano do E. F.) (GERONIMO, J.R; FRANCO, V.S, Geometria Plana e Espacial: um estudo axiom\u00e1tico, 2ed, EDUEM, Maring\u00e1, PR, 2010) Um paralelogramo tem dois lados medindo [tex]18~ cm[\/tex] e [tex]8~ cm[\/tex], e um \u00e2ngulo medindo [tex]30^{\\circ}[\/tex]. Determine a \u00e1rea do paralelogramo. Nota\u00e7\u00e3o: Denotaremos o segmento definido por dois pontos, &hellip; <\/p>\n<p><a class=\"more-link btn\" href=\"https:\/\/clubes.obmep.org.br\/blog\/problema-a-area-do-paralelogramo\/\">Continue lendo<\/a><\/p>\n","protected":false},"author":732,"featured_media":0,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"template-onecolumn.php","meta":{"footnotes":""},"class_list":["post-151159","page","type-page","status-publish","hentry","nodate","item-wrap"],"_links":{"self":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/151159","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\/732"}],"replies":[{"embeddable":true,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/comments?post=151159"}],"version-history":[{"count":4,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/151159\/revisions"}],"predecessor-version":[{"id":194270,"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/pages\/151159\/revisions\/194270"}],"wp:attachment":[{"href":"https:\/\/clubes.obmep.org.br\/blog\/wp-json\/wp\/v2\/media?parent=151159"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}