{"id":4009,"date":"2010-10-10T03:38:19","date_gmt":"2010-10-10T02:38:19","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=4009"},"modified":"2022-01-21T02:11:57","modified_gmt":"2022-01-21T02:11:57","slug":"rascunho-10","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=4009","title":{"rendered":"Um trap\u00e9zio ret\u00e2ngulo"},"content":{"rendered":"<p><ul id='GTTabs_ul_4009' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_4009' class='GTTabs_curr'><a  id=\"4009_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_4009' ><a  id=\"4009_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_4009'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p>Represente o trap\u00e9zio [ABCD], ret\u00e2ngulo em A e D, tal que: \\[\\begin{matrix}<br \/>\n\\overline{AB}=5,4\\,cm; &amp; \\overline{BC}=3\\,cm &amp; e &amp; \\overline{CD}=5\\,cm\u00a0 \\\\<br \/>\n\\end{matrix}\\]<\/p>\n<p>Determine um valor aproximado:<\/p>\n<ol>\n<li>do comprimento do lado [AD];<\/li>\n<li>da amplitude do \u00e2ngulo agudo adjacente ao lado [CB];<\/li>\n<li>da amplitude do \u00e2ngulo ACB.<\/li>\n<\/ol>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_4009' onClick='GTTabs_show(1,4009)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_4009'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><!--more--><\/p>\n<p><a href=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2010\/10\/49.jpg\"><img loading=\"lazy\" decoding=\"async\" data-attachment-id=\"4014\" data-permalink=\"https:\/\/www.acasinhadamatematica.pt\/?attachment_id=4014\" data-orig-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2010\/10\/49.jpg\" data-orig-size=\"361,169\" data-comments-opened=\"1\" data-image-meta=\"{&quot;aperture&quot;:&quot;0&quot;,&quot;credit&quot;:&quot;&quot;,&quot;camera&quot;:&quot;&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=\"Trap\u00e9zio rect\u00e2ngulo\" data-image-description=\"\" data-image-caption=\"\" data-large-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2010\/10\/49.jpg\" class=\"aligncenter wp-image-4014 size-full\" title=\"Trap\u00e9zio rect\u00e2ngulo\" src=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2010\/10\/49.jpg\" alt=\"\" width=\"361\" height=\"169\" srcset=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2010\/10\/49.jpg 361w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2010\/10\/49-300x140.jpg 300w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2010\/10\/49-150x70.jpg 150w\" sizes=\"auto, (max-width: 361px) 100vw, 361px\" \/><\/a><\/p>\n<ol>\n<li>Seja E a proje\u00e7\u00e3o ortogonal do ponto C sobre [AB].<br \/>\nAssim, $\\overline{EB}=\\overline{AB}-\\overline{DC}=0,4\\,cm$.<br \/>\nAplicando o Teorema de Pit\u00e1goras no tri\u00e2ngulo ret\u00e2ngulo [EBC], vem: \\[\\overline{EC}=\\sqrt{{{3}^{2}}-{{(0,4)}^{2}}}=\\sqrt{9-{{\\left( \\frac{2}{5} \\right)}^{2}}}=\\sqrt{\\frac{225-4}{25}}=\\frac{\\sqrt{221}}{5}\\]<br \/>\nLogo, $\\overline{AD}=\\overline{EC}\\simeq 2,97\\,cm$.<br \/>\n\u00ad<\/li>\n<li>Esse \u00e2ngulo \u00e9 um \u00e2ngulo agudo do tri\u00e2ngulo ret\u00e2ngulo [EBC].<br \/>\nComo\u00a0\\[sen\\,E\\hat{B}C=\\frac{\\overline{EC}}{\\overline{BC}}=\\frac{\\frac{\\sqrt{221}}{5}}{3}=\\frac{\\sqrt{221}}{15}\\] ent\u00e3o \\[E\\hat{B}C=se{{n}^{-1}}(\\frac{\\sqrt{221}}{15})\\simeq 82,3{}^\\text{o}\\]<br \/>\n\u00ad<\/li>\n<li>Relativamente aos tri\u00e2ngulos ret\u00e2ngulos [ACE] e [ECB], temos:<br \/>\n\\[tg\\,A\\hat{C}E=\\frac{\\overline{AE}}{\\overline{EC}}=\\frac{5}{\\frac{\\sqrt{221}}{5}}=\\frac{25\\sqrt{221}}{221}\\]<br \/>\n\\[sen\\,E\\hat{C}B=\\frac{\\overline{EB}}{\\overline{BC}}=\\frac{0,4}{3}=\\frac{2}{15}\\]<br \/>\nDado que $A\\hat{C}B=A\\hat{C}E+E\\hat{C}B$, ent\u00e3o $A\\hat{C}B=t{{g}^{-1}}(\\frac{25\\sqrt{221}}{221})+se{{n}^{-1}}(\\frac{2}{15})\\simeq 66,9{}^\\text{o}$.<\/li>\n<\/ol>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_4009' onClick='GTTabs_show(0,4009)'>&lt;&lt; Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Enunciado Resolu\u00e7\u00e3o Enunciado Represente o trap\u00e9zio [ABCD], ret\u00e2ngulo em A e D, tal que: \\[\\begin{matrix} \\overline{AB}=5,4\\,cm; &amp; \\overline{BC}=3\\,cm &amp; e &amp; \\overline{CD}=5\\,cm\u00a0 \\\\ \\end{matrix}\\] Determine um valor aproximado: do comprimento do lado [AD]; da&#46;&#46;&#46;<\/p>\n","protected":false},"author":1,"featured_media":20803,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[98,97,99],"tags":[422,423],"series":[],"class_list":["post-4009","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-11--ano","category-aplicando","category-trigonometria","tag-11-o-ano","tag-trigonometria"],"views":1911,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2010\/10\/11V1Pag095-49_520x245.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/4009","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=4009"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/4009\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/20803"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=4009"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=4009"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=4009"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=4009"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}