{"id":23284,"date":"2022-11-22T10:07:54","date_gmt":"2022-11-22T10:07:54","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=23284"},"modified":"2022-11-24T19:07:34","modified_gmt":"2022-11-24T19:07:34","slug":"area-e-perimetro-de-um-losango","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=23284","title":{"rendered":"\u00c1rea e per\u00edmetro de um losango"},"content":{"rendered":"\n<p><ul id='GTTabs_ul_23284' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_23284' class='GTTabs_curr'><a  id=\"23284_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_23284' ><a  id=\"23284_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_23284'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p>Determina a \u00e1rea e o per\u00edmetro de um losango, sabendo que as diagonais t\u00eam 2 cm e 5 cm de comprimento.<\/p>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_23284' onClick='GTTabs_show(1,23284)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_23284'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><!--more--><\/p>\n<p><a href=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2022\/11\/8_Pag058-11.png\"><img loading=\"lazy\" decoding=\"async\" data-attachment-id=\"23286\" data-permalink=\"https:\/\/www.acasinhadamatematica.pt\/?attachment_id=23286\" data-orig-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2022\/11\/8_Pag058-11.png\" data-orig-size=\"690,330\" 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;,&quot;orientation&quot;:&quot;0&quot;}\" data-image-title=\"8_Pag058-11\" data-image-description=\"\" data-image-caption=\"\" data-large-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2022\/11\/8_Pag058-11.png\" class=\"alignright wp-image-23286\" src=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2022\/11\/8_Pag058-11-300x143.png\" alt=\"\" width=\"380\" height=\"182\" srcset=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2022\/11\/8_Pag058-11-300x143.png 300w, https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2022\/11\/8_Pag058-11.png 690w\" sizes=\"auto, (max-width: 380px) 100vw, 380px\" \/><\/a>Comecemos por determinar o comprimento do lado do losango, em cent\u00edmetros, por aplica\u00e7\u00e3o do Teorema se Pit\u00e1goras no tri\u00e2ngulo ret\u00e2ngulo [BCE]:<\/p>\n<p>\\[\\begin{array}{*{20}{l}}{\\overline {BC} }&amp; = &amp;{\\sqrt {{{\\overline {BE} }^2} + {{\\overline {CE} }^2}} }\\\\{}&amp; = &amp;{\\sqrt {{1^2} + {{2,5}^2}} }\\\\{}&amp; = &amp;{\\sqrt {1 + 6,25} }\\\\{}&amp; = &amp;{\\sqrt {7,25} }\\end{array}\\]<\/p>\n<p>Assim, o losango tem \\(P = 4 \\times \\overline {BC} = 4 \\times \\sqrt {7,25} \\)\u00a0cm de per\u00edmetro.<br \/>A \u00e1rea do losango \u00e9 \\(A = \\frac{{\\overline {AC} \\times \\overline {BD} }}{2} = \\frac{{5 \\times 2}}{2} = 5\\) cm<sup>2<\/sup>.<\/p>\n<p>\u00a0<\/p>\n<h6>Outra forma de apresentar o per\u00edmetro do losango<\/h6>\n<p>O comprimento do lado do losango, em cent\u00edmetros, por aplica\u00e7\u00e3o do Teorema se Pit\u00e1goras no tri\u00e2ngulo ret\u00e2ngulo [BCE], \u00e9:<\/p>\n<p>\\[\\begin{array}{*{20}{l}}{\\overline {BC} }&amp; = &amp;{\\sqrt {{{\\overline {BE} }^2} + {{\\overline {CE} }^2}} }\\\\{}&amp; = &amp;{\\sqrt {{1^2} + {{\\left( {\\frac{5}{2}} \\right)}^2}} }\\\\{}&amp; = &amp;{\\sqrt {1 + \\frac{{25}}{4}} }\\\\{}&amp; = &amp;{\\sqrt {\\frac{{29}}{4}} }\\\\{}&amp; = &amp;{\\frac{{\\sqrt {29} }}{2}}\\end{array}\\]<\/p>\n<p>Assim, o losango tem \\(P = 4 \\times \\overline {BC} = 4 \\times \\frac{{\\sqrt {29} }}{2} = 2\\sqrt {29} \\) cm de per\u00edmetro.<\/p>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_23284' onClick='GTTabs_show(0,23284)'>&lt;&lt; Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Enunciado Resolu\u00e7\u00e3o Enunciado Determina a \u00e1rea e o per\u00edmetro de um losango, sabendo que as diagonais t\u00eam 2 cm e 5 cm de comprimento. Resolu\u00e7\u00e3o &gt;&gt; Resolu\u00e7\u00e3o &lt;&lt; Enunciado<\/p>\n","protected":false},"author":1,"featured_media":23287,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_memberships_contains_paid_content":false,"footnotes":""},"categories":[100,97,682],"tags":[424,108,67,335,118],"series":[],"class_list":["post-23284","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-8--ano","category-aplicando","category-teorema-de-pitagoras","tag-8-o-ano","tag-area","tag-geometria","tag-perimetro","tag-teorema-de-pitagoras"],"views":262,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2022\/11\/8_Pag058-11_520x245.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/23284","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=23284"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/23284\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/23287"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=23284"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=23284"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=23284"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=23284"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}