{"id":1803,"date":"2020-10-18T16:27:24","date_gmt":"2020-10-18T16:27:24","guid":{"rendered":"https:\/\/blogs.ua.es\/dimates\/?p=1803"},"modified":"2020-10-24T07:48:46","modified_gmt":"2020-10-24T07:48:46","slug":"cociente-de-areas","status":"publish","type":"post","link":"https:\/\/blogs.ua.es\/dimates\/2020\/10\/18\/cociente-de-areas\/","title":{"rendered":"Cociente de \u00e1reas"},"content":{"rendered":"<pre>Problema 6 del concurso marat\u00f3 de problemes 2019\r\nSe dirige a una edad de: 14-15 a\u00f1os<\/pre>\n<p>En un tri\u00e1ngulo rect\u00e1ngulo ABC, la hipotenusa AC se divide en 12 partes iguales.<\/p>\n<p>De cada una de ellas se traza una paralela al cateto BC, hasta el cateto AB, dividiendo el tri\u00e1ngulo en 12 pol\u00edgonos (un tri\u00e1ngulo y 11 trapecios).<\/p>\n<p>Hemos marcado dos de ellos, el cuarto a partir del v\u00e9rtice A, y el cuarto a partir del cateto BC.<\/p>\n<p>\u00bfCu\u00e1nto vale el resultado de dividir el \u00e1rea del mayor de los dos entre el otro?<br \/>\n<img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/blogs.ua.es\/dimates\/files\/2020\/10\/163.Cocientedeareas.png\" alt=\"\" width=\"300\" height=\"300\" class=\"aligncenter size-full wp-image-1804\" srcset=\"https:\/\/blogs.ua.es\/dimates\/files\/2020\/10\/163.Cocientedeareas.png 300w, https:\/\/blogs.ua.es\/dimates\/files\/2020\/10\/163.Cocientedeareas-150x150.png 150w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><br \/>\nSoluci\u00f3n: <a href=\"https:\/\/blogs.ua.es\/dimates\/2020\/10\/24\/solucion-a-cociente-de-areas\/\">Aqu\u00ed<\/a>.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Problema 6 del concurso marat\u00f3 de problemes 2019 Se dirige a una edad de: 14-15 a\u00f1os En un tri\u00e1ngulo rect\u00e1ngulo ABC, la hipotenusa AC se divide en 12 partes iguales. De cada una de ellas se traza una paralela al cateto BC, hasta el cateto AB, dividiendo el tri\u00e1ngulo en 12 pol\u00edgonos (un tri\u00e1ngulo y [&hellip;]<\/p>\n","protected":false},"author":4267,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[2242026,1738,2849],"tags":[],"class_list":["post-1803","post","type-post","status-publish","format-standard","hentry","category-marato-de-problemes","category-olimpiadas","category-problemas"],"_links":{"self":[{"href":"https:\/\/blogs.ua.es\/dimates\/wp-json\/wp\/v2\/posts\/1803","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=1803"}],"version-history":[{"count":2,"href":"https:\/\/blogs.ua.es\/dimates\/wp-json\/wp\/v2\/posts\/1803\/revisions"}],"predecessor-version":[{"id":1810,"href":"https:\/\/blogs.ua.es\/dimates\/wp-json\/wp\/v2\/posts\/1803\/revisions\/1810"}],"wp:attachment":[{"href":"https:\/\/blogs.ua.es\/dimates\/wp-json\/wp\/v2\/media?parent=1803"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/blogs.ua.es\/dimates\/wp-json\/wp\/v2\/categories?post=1803"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/blogs.ua.es\/dimates\/wp-json\/wp\/v2\/tags?post=1803"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}