{"id":3659,"date":"2025-07-19T06:40:47","date_gmt":"2025-07-19T06:40:47","guid":{"rendered":"https:\/\/blogs.ua.es\/dimates\/?p=3659"},"modified":"2025-07-27T09:45:21","modified_gmt":"2025-07-27T09:45:21","slug":"rectangulo-dividido","status":"publish","type":"post","link":"https:\/\/blogs.ua.es\/dimates\/2025\/07\/19\/rectangulo-dividido\/","title":{"rendered":"Rect\u00e1ngulo dividido"},"content":{"rendered":"<pre>Problema 11 del concurso Marat\u00f3 de problemes 2025\nSe dirige a una edad de: 14-15 a\u00f1os\n<\/pre>\n\n\n<p>El rect\u00e1ngulo ABCD de la figura est\u00e1 dividido en 6 rect\u00e1ngulos iguales por 5 segmentos paralelos al lado BC.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"300\" height=\"300\" src=\"https:\/\/blogs.ua.es\/dimates\/files\/2025\/07\/391.rectangulo.png\" alt=\"\" class=\"wp-image-3660\" \/><\/figure>\n\n\n\n<p>El punto E del segmento BC tiene la propiedad de que el segmento AE divide al \u00faltimo de los rect\u00e1ngulos peque\u00f1os, el que contiene al segmento BC, en dos partes exactamente iguales.<\/p>\n\n\n\n<p>Si suponemos que el tri\u00e1ngulo que forma el segmento AE en el primer rect\u00e1ngulo peque\u00f1o, el que contiene al segmento AD, tiene un \u00e1rea de 1 cm\u00b2, calcula el \u00e1rea del rect\u00e1ngulo ABCD.<\/p>\n\n\n\n<p>Soluci\u00f3n: <a href=\"https:\/\/blogs.ua.es\/dimates\/2025\/07\/27\/solucion-a-rectangulo-dividido\/\">Aqu\u00ed<\/a>.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Problema 11 del concurso Marat\u00f3 de problemes 2025 Se dirige a una edad de: 14-15 a\u00f1os El rect\u00e1ngulo ABCD de la figura est\u00e1 dividido en 6 rect\u00e1ngulos iguales por 5 segmentos paralelos al lado BC. El punto E del segmento BC tiene la propiedad de que el segmento AE divide al \u00faltimo de los rect\u00e1ngulos [&hellip;]<\/p>\n","protected":false},"author":4267,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[2242030,2242026,1738,2849],"tags":[],"class_list":["post-3659","post","type-post","status-publish","format-standard","hentry","category-marato","category-marato-de-problemes","category-olimpiadas","category-problemas"],"_links":{"self":[{"href":"https:\/\/blogs.ua.es\/dimates\/wp-json\/wp\/v2\/posts\/3659","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/blogs.ua.es\/dimates\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/blogs.ua.es\/dimates\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/blogs.ua.es\/dimates\/wp-json\/wp\/v2\/users\/4267"}],"replies":[{"embeddable":true,"href":"https:\/\/blogs.ua.es\/dimates\/wp-json\/wp\/v2\/comments?post=3659"}],"version-history":[{"count":2,"href":"https:\/\/blogs.ua.es\/dimates\/wp-json\/wp\/v2\/posts\/3659\/revisions"}],"predecessor-version":[{"id":3666,"href":"https:\/\/blogs.ua.es\/dimates\/wp-json\/wp\/v2\/posts\/3659\/revisions\/3666"}],"wp:attachment":[{"href":"https:\/\/blogs.ua.es\/dimates\/wp-json\/wp\/v2\/media?parent=3659"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blogs.ua.es\/dimates\/wp-json\/wp\/v2\/categories?post=3659"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blogs.ua.es\/dimates\/wp-json\/wp\/v2\/tags?post=3659"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}