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{"id":6691,"date":"2011-04-04T23:23:38","date_gmt":"2011-04-04T22:23:38","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=6691"},"modified":"2022-01-17T18:19:54","modified_gmt":"2022-01-17T18:19:54","slug":"a-largura-de-um-rio","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=6691","title":{"rendered":"A largura de um rio"},"content":{"rendered":"<p><ul id='GTTabs_ul_6691' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_6691' class='GTTabs_curr'><a  id=\"6691_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_6691' ><a  id=\"6691_1\" onMouseOver=\"GTTabsShowLinks('Resolu\u00e7\u00e3o'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Resolu\u00e7\u00e3o<\/a><\/li>\n<\/ul>\n\n<div class='GTTabs_divs GTTabs_curr_div' id='GTTabs_0_6691'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p><a href=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/04\/8Pag133-8.jpg\"><img loading=\"lazy\" decoding=\"async\" data-attachment-id=\"6692\" data-permalink=\"https:\/\/www.acasinhadamatematica.pt\/?attachment_id=6692\" data-orig-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/04\/8Pag133-8.jpg\" data-orig-size=\"590,449\" data-comments-opened=\"1\" data-image-meta=\"{&quot;aperture&quot;:&quot;0&quot;,&quot;credit&quot;:&quot;&quot;,&quot;camera&quot;:&quot;HP pstc4380&quot;,&quot;caption&quot;:&quot;&quot;,&quot;created_timestamp&quot;:&quot;0&quot;,&quot;copyright&quot;:&quot;&quot;,&quot;focal_length&quot;:&quot;0&quot;,&quot;iso&quot;:&quot;0&quot;,&quot;shutter_speed&quot;:&quot;0&quot;,&quot;title&quot;:&quot;&quot;}\" data-image-title=\"Rio\" data-image-description=\"\" data-image-caption=\"\" data-large-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/04\/8Pag133-8.jpg\" class=\"alignright wp-image-6692\" title=\"Rio\" src=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/04\/8Pag133-8.jpg\" alt=\"\" width=\"520\" height=\"396\" srcset=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/04\/8Pag133-8.jpg 590w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/04\/8Pag133-8-300x228.jpg 300w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/04\/8Pag133-8-150x114.jpg 150w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/04\/8Pag133-8-400x304.jpg 400w\" sizes=\"auto, (max-width: 520px) 100vw, 520px\" \/><\/a>Para determinarmos a largura de um rio sem o atravessarmos, seguimos o m\u00e9todo esquematizado na figura. Considera que os \u00e2ngulos ABC e CED s\u00e3o geometricamente iguais.<\/p>\n<p>Qual \u00e9 a largura do rio?<\/p>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_6691' onClick='GTTabs_show(1,6691)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_6691'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><!--more--><\/p>\n<p><a href=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/04\/8Pag133-8.jpg\"><img loading=\"lazy\" decoding=\"async\" data-attachment-id=\"6692\" data-permalink=\"https:\/\/www.acasinhadamatematica.pt\/?attachment_id=6692\" data-orig-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/04\/8Pag133-8.jpg\" data-orig-size=\"590,449\" data-comments-opened=\"1\" data-image-meta=\"{&quot;aperture&quot;:&quot;0&quot;,&quot;credit&quot;:&quot;&quot;,&quot;camera&quot;:&quot;HP pstc4380&quot;,&quot;caption&quot;:&quot;&quot;,&quot;created_timestamp&quot;:&quot;0&quot;,&quot;copyright&quot;:&quot;&quot;,&quot;focal_length&quot;:&quot;0&quot;,&quot;iso&quot;:&quot;0&quot;,&quot;shutter_speed&quot;:&quot;0&quot;,&quot;title&quot;:&quot;&quot;}\" data-image-title=\"Rio\" data-image-description=\"\" data-image-caption=\"\" data-large-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/04\/8Pag133-8.jpg\" class=\"alignright wp-image-6692\" title=\"Rio\" src=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/04\/8Pag133-8.jpg\" alt=\"\" width=\"520\" height=\"396\" srcset=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/04\/8Pag133-8.jpg 590w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/04\/8Pag133-8-300x228.jpg 300w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/04\/8Pag133-8-150x114.jpg 150w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/04\/8Pag133-8-400x304.jpg 400w\" sizes=\"auto, (max-width: 520px) 100vw, 520px\" \/><\/a>Como os \u00e2ngulos ACB e DCE s\u00e3o verticalmente opostos, ent\u00e3o s\u00e3o geometricamente iguais.<\/p>\n<p>Logo, os tri\u00e2ngulos [ABC] e [CDE] s\u00e3o semelhantes, pois possuem dois \u00e2ngulos geometricamente iguais, cada um a cada um, de um para o outro dos tri\u00e2ngulos.<\/p>\n<p>Sendo semelhantes, os lados correspondentes desses tri\u00e2ngulos t\u00eam comprimentos diretamente proporcionais.<\/p>\n<p>Assim, temos:<\/p>\n<p>\\[\\frac{\\overline{AB}}{4}=\\frac{40}{5}\\Leftrightarrow \\overline{AB}=\\frac{4\\times 40}{5}\\Leftrightarrow \\overline{AB}=32\\]<\/p>\n<p>Logo, o rio tem 32 m de largura.<\/p><\/p>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_6691' onClick='GTTabs_show(0,6691)'>&lt;&lt; Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Enunciado Resolu\u00e7\u00e3o Enunciado Para determinarmos a largura de um rio sem o atravessarmos, seguimos o m\u00e9todo esquematizado na figura. Considera que os \u00e2ngulos ABC e CED s\u00e3o geometricamente iguais. Qual \u00e9 a largura do&#46;&#46;&#46;<\/p>\n","protected":false},"author":1,"featured_media":20515,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[100,97,151],"tags":[149],"series":[],"class_list":["post-6691","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-8--ano","category-aplicando","category-semelhanca-de-triangulos-8--ano","tag-semelhanca-de-triangulos"],"views":3821,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2011\/04\/8V1Pag133-8_520x245.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/6691","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=6691"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/6691\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/20515"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=6691"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=6691"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=6691"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=6691"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}