{"id":10237,"date":"2012-10-12T14:33:41","date_gmt":"2012-10-12T13:33:41","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=10237"},"modified":"2022-01-20T22:57:19","modified_gmt":"2022-01-20T22:57:19","slug":"uma-caixa-com-latas-de-refrigerante","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=10237","title":{"rendered":"Uma caixa com latas de refrigerante"},"content":{"rendered":"<p><ul id='GTTabs_ul_10237' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_10237' class='GTTabs_curr'><a  id=\"10237_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_10237' ><a  id=\"10237_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_10237'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<div id=\"attachment_10243\" style=\"width: 310px\" class=\"wp-caption alignright\"><a href=\"https:\/\/www.acasinhadamatematica.pt\/?attachment_id=10243\" rel=\"attachment wp-att-10243\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-10243\" data-attachment-id=\"10243\" data-permalink=\"https:\/\/www.acasinhadamatematica.pt\/?attachment_id=10243\" data-orig-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/10\/caixalatas.jpg\" data-orig-size=\"310,156\" 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=\"Base da caixa\" data-image-description=\"\" data-image-caption=\"\" data-large-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/10\/caixalatas.jpg\" class=\"size-medium wp-image-10243\" title=\"Base da caixa\" src=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/10\/caixalatas-300x150.jpg\" alt=\"\" width=\"300\" height=\"150\" srcset=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/10\/caixalatas-300x150.jpg 300w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/10\/caixalatas-150x75.jpg 150w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/10\/caixalatas.jpg 310w\" sizes=\"auto, (max-width: 300px) 100vw, 300px\" \/><\/a><p id=\"caption-attachment-10243\" class=\"wp-caption-text\">Base da caixa<\/p><\/div>\n<p>Imagine que algu\u00e9m pensou acondicionar latas de $75$ cl de refrigerante numa caixa prism\u00e1tica cuja base \u00e9 um paralelogramo obliqu\u00e2ngulo, como mostra a figura.<\/p>\n<ol>\n<li>Se o raio da base de cada lata medir $4$ cm, qual \u00e9 a \u00e1rea da base da caixa?<br \/>\n<strong>Sugest\u00e3o<\/strong>: No esquema, marcaram-se v\u00e1rios raios de circunfer\u00eancias. Recorrendo aos seus\u00a0conhecimentos sobre tri\u00e2ngulos (acut\u00e2ngulos ou ret\u00e2ngulos) ou ao Teorema de Pit\u00e1goras, determine o comprimento e a altura do paralelogramo.<\/li>\n<li>As latas t\u00eam $16$ cm de altura. Qual ser\u00e1 o volume da caixa se tiver a altura das latas?<\/li>\n<li>Determine uma medida para a efic\u00e1cia desta caixa e compare-a com a efic\u00e1cia da caixa paralelepip\u00e9dica que, em alternativa, poderia ser usada para guardar as seis latas.<\/li>\n<\/ol>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_10237' onClick='GTTabs_show(1,10237)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_10237'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><!--more--><\/p>\n<p style=\"text-align: center;\"><a href=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/10\/caixalatas-a.png\"><img loading=\"lazy\" decoding=\"async\" data-attachment-id=\"10245\" data-permalink=\"https:\/\/www.acasinhadamatematica.pt\/?attachment_id=10245\" data-orig-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/10\/caixalatas-a.png\" data-orig-size=\"877,470\" 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=\"Base da caixa\" data-image-description=\"\" data-image-caption=\"\" data-large-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/10\/caixalatas-a.png\" class=\"aligncenter  wp-image-10245\" title=\"Base da caixa\" src=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/10\/caixalatas-a.png\" alt=\"\" width=\"702\" height=\"376\" srcset=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/10\/caixalatas-a.png 877w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/10\/caixalatas-a-300x160.png 300w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/10\/caixalatas-a-150x80.png 150w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/10\/caixalatas-a-400x214.png 400w\" sizes=\"auto, (max-width: 702px) 100vw, 702px\" \/><\/a><\/p>\n<\/p>\n<ol>\n<li>A altura e o comprimento da base do pol\u00edgono base da caixa s\u00e3o dados, respetivamente, por: $h = \\overline {LC}\u00a0 + \\overline {CM}\u00a0 + \\overline {AH} $ e $b = \\overline {FG}\u00a0 + \\overline {GI}\u00a0 + \\overline {IJ} $.\n<p>Comecemos por determinar a altura (em cm) do tri\u00e2ngulo equil\u00e1tero [ABC]:<br \/>\n$$\\overline {CM}\u00a0 = \\sqrt {{{\\overline {CB} }^2} &#8211; {{\\overline {MB} }^2}}\u00a0 = \\sqrt {{8^2} &#8211; {4^2}}\u00a0 = \\sqrt {64 &#8211; 16}\u00a0 = \\sqrt {48}\u00a0 = \\sqrt {16 \\times 3}\u00a0 = 4\\sqrt 3 $$<\/p>\n<p>Portanto, a altura (em cm) do paralelogramo \u00e9: $h = 4 + 4\\sqrt 3\u00a0 + 4 = 8 + 4\\sqrt 3 $.<\/p>\n<p>Ora, uma leitura atenta da figura permite concluir que os tri\u00e2ngulos ret\u00e2ngulos [BCM] e [BIJ] s\u00e3o geometricamente iguais (<strong>ALA<\/strong> &#8211; note que os catetos [BM] e [BI] s\u00e3o geometricamente iguais, bem como os \u00e2ngulos adjacentes a esses lados, cada um a cada um).<br \/>\nConsequentemente, temos: $\\overline {IJ}\u00a0 = \\overline {CM}\u00a0 = 4\\sqrt 3 $.<\/p>\n<p>Por outro lado, os tri\u00e2ngulos [DFG] e [CMB] s\u00e3o semelhantes (<strong>AA<\/strong>), pelo que os comprimentos dos lados correspondentes s\u00e3o diretamente proporcionais: $$\\frac{{\\overline {FG} }}{{\\overline {BM} }} = \\frac{{\\overline {DG} }}{{\\overline {CM} }}$$<br \/>\nDonde, substituindo os valores conhecidos, resulta:<br \/>\n$$\\begin{array}{*{20}{l}} \u00a0 {\\frac{{\\overline {FG} }}{4} = \\frac{4}{{4\\sqrt 3 }}}&amp; \\Leftrightarrow &amp;{\\overline {FG}\u00a0 = \\frac{4}{{\\sqrt 3 }}} \\\\ \u00a0 {}&amp; \\Leftrightarrow &amp;{\\overline {FG}\u00a0 = \\frac{4}{{\\sqrt 3 }} \\times \\frac{{\\sqrt 3 }}{{\\sqrt 3 }}} \\\\ \u00a0 {}&amp; \\Leftrightarrow &amp;{\\overline {FG}\u00a0 = \\frac{{4\\sqrt 3 }}{3}} \\end{array}$$<br \/>\nAssim, o comprimento (em cm) da base do paralelogramo \u00e9:<br \/>\n$$b = \\frac{{4\\sqrt 3 }}{3} + 16 + 4\\sqrt 3\u00a0 = 16 + \\frac{{16\\sqrt 3 }}{3}$$<br \/>\nPortanto, a \u00e1rea (em cm<sup>2<\/sup>) da base da caixa \u00e9:<br \/>\n$${A_b} = \\left( {16 + \\frac{{16\\sqrt 3 }}{3}} \\right) \\times \\left( {8 + 4\\sqrt 3 } \\right) = 128 + 64\\sqrt 3\u00a0 + \\frac{{128\\sqrt 3 }}{3} + \\frac{{64 \\times 3}}{3} = 192 + \\frac{{320\\sqrt 3 }}{3}$$<\/p>\n<\/li>\n<li>Como a caixa tem a altura das latas, o seu volume (em cm<sup>3<\/sup>) \u00e9:<br \/>\n$${V_C} = \\left( {192 + \\frac{{320\\sqrt 3 }}{3}} \\right) \\times 16 = 3072 + \\frac{{5120\\sqrt 3 }}{3}$$<\/li>\n<li>O volume das seis latas (em cm<sup>3<\/sup>) \u00e9 ${V_{Latas}} = 6 \\times \\left( {\\pi\u00a0 \\times {4^2} \\times 16} \\right) = 1536\\pi $, pelo que a efic\u00e1cia desta caixa \u00e9:<br \/>\n$$E{f_1} = \\frac{{1536\\pi }}{{3072 + \\frac{{5120\\sqrt 3 }}{3}}} \\approx 0,80$$<br \/>\nQuanto \u00e0 caixa paralelepip\u00e9dica, a sua base ser\u00e1 um ret\u00e2ngulo com $24$ cm de comprimento e $16$ cm de largura (Porqu\u00ea?).<br \/>\nAssim, o seu volume (em cm<sup>3<\/sup>) \u00e9 ${V_{C &#8211; P}} = \\left( {24 \\times 16} \\right) \\times 16 = 6144$, pelo que a sua efic\u00e1cia \u00e9:<br \/>\n$$E{f_2} = \\frac{{1536\\pi }}{{6144}} \\approx 0,79$$<\/p>\n<p>A caixa paralelepip\u00e9dica \u00e9 ligeiramente menos eficaz.<\/p>\n<\/li>\n<\/ol>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_10237' onClick='GTTabs_show(0,10237)'>&lt;&lt; Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Enunciado Resolu\u00e7\u00e3o Enunciado Imagine que algu\u00e9m pensou acondicionar latas de $75$ cl de refrigerante numa caixa prism\u00e1tica cuja base \u00e9 um paralelogramo obliqu\u00e2ngulo, como mostra a figura. Se o raio da base de cada&#46;&#46;&#46;<\/p>\n","protected":false},"author":1,"featured_media":20791,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[321,97,322],"tags":[429,67,430,326],"series":[],"class_list":["post-10237","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-10-o-ano","category-aplicando","category-modulo-inicial","tag-10-o-ano","tag-geometria","tag-modulo-inicial","tag-triangulos-semelhantes"],"views":3945,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2012\/10\/10V1Pag039-31_520x245.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/10237","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=10237"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/10237\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/20791"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=10237"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=10237"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=10237"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=10237"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}