{"id":8057,"date":"2012-04-08T22:36:41","date_gmt":"2012-04-08T21:36:41","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=8057"},"modified":"2022-01-30T19:39:53","modified_gmt":"2022-01-30T19:39:53","slug":"um-trapezio-isosceles","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=8057","title":{"rendered":"Um trap\u00e9zio is\u00f3sceles"},"content":{"rendered":"<p><ul id='GTTabs_ul_8057' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_8057' class='GTTabs_curr'><a  id=\"8057_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_8057' ><a  id=\"8057_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_8057'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p><a href=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/04\/12pag224-79.jpg\"><img loading=\"lazy\" decoding=\"async\" data-attachment-id=\"8060\" data-permalink=\"https:\/\/www.acasinhadamatematica.pt\/?attachment_id=8060\" data-orig-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/04\/12pag224-79.jpg\" data-orig-size=\"459,165\" 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=\"Trap\u00e9zio\" data-image-description=\"\" data-image-caption=\"\" data-large-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/04\/12pag224-79.jpg\" class=\"alignright wp-image-8060\" title=\"Trap\u00e9zio\" src=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/04\/12pag224-79.jpg\" alt=\"\" width=\"400\" height=\"144\" srcset=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/04\/12pag224-79.jpg 459w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/04\/12pag224-79-300x107.jpg 300w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/04\/12pag224-79-150x53.jpg 150w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/04\/12pag224-79-400x143.jpg 400w\" sizes=\"auto, (max-width: 400px) 100vw, 400px\" \/><\/a>[ABCD] \u00e9 um trap\u00e9zio is\u00f3sceles de \u00e1rea $5\\sqrt 2 \\,\\,c{m^2}$.<\/p>\n<p>Os \u00e2ngulos agudos medem 45\u00ba.<\/p>\n<p>Seja $x$ (em cm) a altura do trap\u00e9zio e $P(x)$ o seu per\u00edmetro (em cm).<\/p>\n<ol>\n<li>Exprima $\\overline {DH} $ e $\\overline {CK} $ em fun\u00e7\u00e3o de $x$.<\/li>\n<li>Exprima $\\overline {AD} $ e $\\overline {BC} $ em fun\u00e7\u00e3o de $x$.<\/li>\n<li>Utilize a \u00e1rea do trap\u00e9zio para exprimir $\\overline {AB} $ em fun\u00e7\u00e3o de $x$.<\/li>\n<li>Mostre que $$P(x) = 2\\sqrt 2 x + \\frac{{10\\sqrt 2 }}{x}$$<\/li>\n<li>Encontre o valor de $x$ para o qual o per\u00edmetro do trap\u00e9zio \u00e9 m\u00ednimo.<br \/>\nQual \u00e9 ent\u00e3o a medida dos lados do trap\u00e9zio? Interprete graficamente.<\/li>\n<\/ol>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_8057' onClick='GTTabs_show(1,8057)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_8057'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><!--more--><\/p>\n<ol>\n<li><a href=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/04\/12pag224-79.jpg\"><img loading=\"lazy\" decoding=\"async\" data-attachment-id=\"8060\" data-permalink=\"https:\/\/www.acasinhadamatematica.pt\/?attachment_id=8060\" data-orig-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/04\/12pag224-79.jpg\" data-orig-size=\"459,165\" 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=\"Trap\u00e9zio\" data-image-description=\"\" data-image-caption=\"\" data-large-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/04\/12pag224-79.jpg\" class=\"alignright wp-image-8060\" title=\"Trap\u00e9zio\" src=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/04\/12pag224-79.jpg\" alt=\"\" width=\"400\" height=\"144\" srcset=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/04\/12pag224-79.jpg 459w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/04\/12pag224-79-300x107.jpg 300w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/04\/12pag224-79-150x53.jpg 150w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/04\/12pag224-79-400x143.jpg 400w\" sizes=\"auto, (max-width: 400px) 100vw, 400px\" \/><\/a>Como os tri\u00e2ngulos [ADH] e [BCK] s\u00e3o ret\u00e2ngulos is\u00f3sceles, temos: $\\overline {DH}\u00a0 = \\overline {AK}\u00a0 = x$ e $\\overline {CK}\u00a0 = \\overline {AK}\u00a0 = x$, com $x &gt; 0$.<br \/>\n\u00ad<\/li>\n<li>Aplicando o teorema de Pit\u00e1goras nos tri\u00e2ngulos ret\u00e2ngulos referidos anteriormente, vem: $\\overline {AD}\u00a0 = \\overline {BC}\u00a0 = x\\sqrt 2 $.<br \/>\n\u00ad<\/li>\n<li>Ora, $$\\begin{array}{*{20}{l}}<br \/>\n{{A_{[ABCD]}} = 5\\sqrt 2 }&amp; \\Leftrightarrow &amp;{\\frac{{\\overline {CD}\u00a0 + \\overline {AB} }}{2} \\times \\overline {AH}\u00a0 = 5\\sqrt 2 } \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{\\frac{{\\left( {\\overline {AB}\u00a0 + 2x} \\right) + \\overline {AB} }}{2} \\times x = 5\\sqrt 2 } \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{\\left( {\\overline {AB}\u00a0 + x} \\right)x = 5\\sqrt 2 } \\\\<br \/>\n{}&amp; \\Leftrightarrow &amp;{\\overline {AB}\u00a0 = \\frac{{5\\sqrt 2\u00a0 &#8211; {x^2}}}{x}}<br \/>\n\\end{array}$$<br \/>\nComo $\\overline {AB}\u00a0 &gt; 0$, ent\u00e3o $x &gt; 0 \\wedge 5\\sqrt 2\u00a0 &#8211; {x^2} &gt; 0 \\Leftrightarrow x &gt; 0 \\wedge x &lt; \\sqrt[4]{{50}}$.<br \/>\n\u00ad<\/li>\n<li>O per\u00edmetro do trap\u00e9zio pode ser expresso por: $$\\begin{array}{*{20}{l}}<br \/>\n{P(x)}&amp; = &amp;{2\\left( {\\overline {AB}\u00a0 + \\overline {DH}\u00a0 + \\overline {AD} } \\right)} \\\\<br \/>\n{}&amp; = &amp;{2\\left( {\\frac{{5\\sqrt 2\u00a0 &#8211; {x^2}}}{x} + x + x\\sqrt 2 } \\right)} \\\\<br \/>\n{}&amp; = &amp;{2\\left( {\\frac{{5\\sqrt 2 }}{x} &#8211; x + x + x\\sqrt 2 } \\right)} \\\\<br \/>\n{}&amp; = &amp;{2\\sqrt 2 x + \\frac{{10\\sqrt 2 }}{x}}<br \/>\n\\end{array}$$ com $0 &lt; x &lt; \\sqrt[4]{{50}}$.<br \/>\n\u00ad<\/li>\n<li>Como\u00a0\\[\\begin{array}{*{20}{l}}<br \/>\n{P&#8217;\\left( x \\right)}&amp; = &amp;{{{\\left( {2\\sqrt 2 x + \\frac{{10\\sqrt 2 }}{x}} \\right)}^\\prime }} \\\\<br \/>\n{}&amp; = &amp;{2\\sqrt 2\u00a0 &#8211; \\frac{{10\\sqrt 2 }}{{{x^2}}}} \\\\<br \/>\n{}&amp; = &amp;{2\\sqrt 2\u00a0 \\times \\frac{{{x^2} &#8211; 5}}{{{x^2}}}}<br \/>\n\\end{array}\\] temos:<\/p>\n<table class=\" aligncenter\" style=\"width: 80%;\" border=\"0\" align=\"center\">\n<tbody>\n<tr>\n<td style=\"text-align: center; border: #00008b 1px solid;\">$x$<\/td>\n<td style=\"text-align: center; border: #00008b 1px solid;\">$0$<\/td>\n<td style=\"text-align: center; border: #00008b 1px solid;\"><\/td>\n<td style=\"text-align: center; border: #00008b 1px solid;\">$\\sqrt 5 $<\/td>\n<td style=\"text-align: center; border: #00008b 1px solid;\"><\/td>\n<td style=\"text-align: center; border: #00008b 1px solid;\">$\\sqrt[4]{{50}}$<\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center; border: #00008b 1px solid;\">Sinal de $P'(x) = 2\\sqrt 2\u00a0 \\times \\frac{{{x^2} &#8211; 5}}{{{x^2}}}$<\/td>\n<td style=\"text-align: center; background-color: #a9a9a9; border: #00008b 1px solid;\"><\/td>\n<td style=\"text-align: center; border: #00008b 1px solid;\">$-$<\/td>\n<td style=\"text-align: center; border: #00008b 1px solid;\">$0$<\/td>\n<td style=\"text-align: center; border: #00008b 1px solid;\">$+$<\/td>\n<td style=\"text-align: center; background-color: #a9a9a9; border: #00008b 1px solid;\"><\/td>\n<\/tr>\n<tr>\n<td style=\"text-align: center; border: #00008b 1px solid;\">Varia\u00e7\u00e3o de $P$<\/td>\n<td style=\"text-align: center; background-color: #a9a9a9; border: #00008b 1px solid;\"><\/td>\n<td style=\"text-align: center; border: #00008b 1px solid;\">$ \\searrow $<\/td>\n<td style=\"text-align: center; border: #00008b 1px solid;\">$4\\sqrt {10} $<\/td>\n<td style=\"text-align: center; border: #00008b 1px solid;\">$ \\nearrow $<\/td>\n<td style=\"text-align: center; background-color: #a9a9a9; border: #00008b 1px solid;\"><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>$$P(\\sqrt 5 ) = 2\\sqrt 2\u00a0 \\times \\sqrt 5\u00a0 + \\frac{{10\\sqrt 2 }}{{\\sqrt 5 }} = 2\\sqrt {10}\u00a0 + 2\\sqrt {10}\u00a0 = 4\\sqrt {10} $$<\/p>\n<p>O per\u00edmetro do trap\u00e9zio \u00e9 m\u00ednimo para $x = \\sqrt 5 $ cent\u00edmetros.<br \/>\nNessa situa\u00e7\u00e3o, temos:<br \/>\n$$\\overline {AD}\u00a0 = \\overline {BC}\u00a0 = \\sqrt 5\u00a0 \\times \\sqrt 2\u00a0 = \\sqrt {10} $$<br \/>\n$$\\overline {AB}\u00a0 = \\frac{{5\\sqrt 2\u00a0 &#8211; {{\\left( {\\sqrt 5 } \\right)}^2}}}{{\\sqrt 5 }} = \\frac{{5\\sqrt {10}\u00a0 &#8211; 5\\sqrt 5 }}{5} = \\sqrt {10}\u00a0 &#8211; \\sqrt 5 $$<br \/>\n$$\\overline {CD}\u00a0 = \\overline {AB}\u00a0 + 2\\sqrt 5\u00a0 = \\sqrt {10}\u00a0 &#8211; \\sqrt 5\u00a0 + 2\\sqrt 5\u00a0 = \\sqrt {10}\u00a0 + \\sqrt 5 $$<br \/>\nem cent\u00edmetros.<br \/>\n\u00ad<\/li>\n<\/ol>\n<p style=\"text-align: center;\"><script 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Os \u00e2ngulos agudos medem 45\u00ba. Seja $x$ (em cm) a altura do trap\u00e9zio e $P(x)$ o seu per\u00edmetro (em cm). Exprima&#46;&#46;&#46;<\/p>\n","protected":false},"author":1,"featured_media":21144,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[226,97,292],"tags":[427,145,144],"series":[],"class_list":["post-8057","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-12--ano","category-aplicando","category-calculo-diferencial","tag-12-o-ano","tag-derivadas-2","tag-extremos-relativos"],"views":3246,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/04\/12V2Pag224-79_520x245.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/8057","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=8057"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/8057\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/21144"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=8057"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=8057"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=8057"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=8057"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}