{"id":13420,"date":"2018-02-01T22:05:43","date_gmt":"2018-02-01T22:05:43","guid":{"rendered":"https:\/\/www.acasinhadamatematica.pt\/?p=13420"},"modified":"2022-01-17T15:07:47","modified_gmt":"2022-01-17T15:07:47","slug":"area-de-um-segmento-de-circulo","status":"publish","type":"post","link":"https:\/\/www.acasinhadamatematica.pt\/?p=13420","title":{"rendered":"\u00c1rea de um segmento de c\u00edrculo"},"content":{"rendered":"<p><ul id='GTTabs_ul_13420' class='GTTabs' style='display:none'>\n<li id='GTTabs_li_0_13420' class='GTTabs_curr'><a  id=\"13420_0\" onMouseOver=\"GTTabsShowLinks('Enunciado'); return true;\"  onMouseOut=\"GTTabsShowLinks();\"  class='GTTabsLinks'>Enunciado<\/a><\/li>\n<li id='GTTabs_li_1_13420' ><a  id=\"13420_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_13420'>\n<span class='GTTabs_titles'><b>Enunciado<\/b><\/span><\/p>\n<p><a href=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/02\/9V1Pag144-6.png\"><img loading=\"lazy\" decoding=\"async\" data-attachment-id=\"13421\" data-permalink=\"https:\/\/www.acasinhadamatematica.pt\/?attachment_id=13421\" data-orig-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/02\/9V1Pag144-6.png\" data-orig-size=\"230,225\" 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=\"Segmento de c\u00edrculo\" data-image-description=\"\" data-image-caption=\"\" data-large-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/02\/9V1Pag144-6.png\" class=\"alignright size-full wp-image-13421\" src=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/02\/9V1Pag144-6.png\" alt=\"\" width=\"230\" height=\"225\" \/><\/a>Na figura, o c\u00edrculo de centro O tem \\(400\\pi \\) cm<sup>2<\/sup> de \u00e1rea e os raios tra\u00e7ados s\u00e3o perpendiculares.<\/p>\n<p>Determina a \u00e1rea, arredondada \u00e0s d\u00e9cimas, do segmento circular colorido.<\/p>\n<p><div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_next'><a href='#GTTabs_ul_13420' onClick='GTTabs_show(1,13420)'>Resolu\u00e7\u00e3o &gt;&gt;<\/a><\/span><\/div><\/div>\n\n<div class='GTTabs_divs' id='GTTabs_1_13420'>\n<span class='GTTabs_titles'><b>Resolu\u00e7\u00e3o<\/b><\/span><!--more--><\/p>\n<p><a href=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/02\/9V1Pag144-6.png\"><img loading=\"lazy\" decoding=\"async\" data-attachment-id=\"13421\" data-permalink=\"https:\/\/www.acasinhadamatematica.pt\/?attachment_id=13421\" data-orig-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/02\/9V1Pag144-6.png\" data-orig-size=\"230,225\" 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=\"Segmento de c\u00edrculo\" data-image-description=\"\" data-image-caption=\"\" data-large-file=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/02\/9V1Pag144-6.png\" class=\"alignright size-full wp-image-13421\" src=\"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/02\/9V1Pag144-6.png\" alt=\"\" width=\"230\" height=\"225\" \/><\/a>Comecemos por determinar o comprimento do raio do c\u00edrculo:\u00a0\\(r = \\sqrt {\\frac{{400\\pi }}{\\pi }} = \\sqrt {400} = 20\\) cm.<\/p>\n<p>A \u00e1rea, em cm<sup>2<\/sup>, do segmento circular colorido \u00e9:<\/p>\n<p>\\[\\begin{array}{*{20}{l}}{{A_{Colorida}}}&amp; = &amp;{{A_{Setor}} &#8211; {A_{\\left[ {OAB} \\right]}}}\\\\{}&amp; = &amp;{\\frac{{90^\\circ }}{{360^\\circ }} \\times 400\\pi &#8211; \\frac{{\\overline {OA} \\times \\overline {OB} }}{2}}\\\\{}&amp; = &amp;{100\\pi &#8211; \\frac{{20 \\times 20}}{2}}\\\\{}&amp; = &amp;{100\\pi &#8211; 200}\\\\{}&amp; \\approx &amp;{114,2}\\end{array}\\]<\/p><\/p>\n<div class='GTTabsNavigation' style='display:none'><span class='GTTabs_nav_prev'><a href='#GTTabs_ul_13420' onClick='GTTabs_show(0,13420)'>&lt;&lt; Enunciado<\/a><\/span><\/div><\/div>\n\n","protected":false},"excerpt":{"rendered":"<p>Enunciado Resolu\u00e7\u00e3o Enunciado Na figura, o c\u00edrculo de centro O tem \\(400\\pi \\) cm2 de \u00e1rea e os raios tra\u00e7ados s\u00e3o perpendiculares. Determina a \u00e1rea, arredondada \u00e0s d\u00e9cimas, do segmento circular colorido. Resolu\u00e7\u00e3o &gt;&gt;&#46;&#46;&#46;<\/p>\n","protected":false},"author":1,"featured_media":20493,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"_jetpack_newsletter_access":"","_jetpack_dont_email_post_to_subs":false,"_jetpack_newsletter_tier_id":0,"_jetpack_memberships_contains_paywalled_content":false,"_jetpack_feature_clip_id":0,"_jetpack_memberships_contains_paid_content":false,"footnotes":"","jetpack_post_was_ever_published":false},"categories":[213,97,278],"tags":[426,188,458,284],"series":[],"class_list":["post-13420","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-9--ano","category-aplicando","category-circunferencia-e-poligonos","tag-9-o-ano","tag-circunferencia","tag-segmento-de-circulo","tag-setor-circular"],"views":2290,"jetpack_featured_media_url":"https:\/\/www.acasinhadamatematica.pt\/wp-content\/uploads\/2018\/02\/9V1Pag144-6_520x245.png","jetpack_sharing_enabled":true,"jetpack_likes_enabled":true,"_links":{"self":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/13420","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=13420"}],"version-history":[{"count":0,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/posts\/13420\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=\/wp\/v2\/media\/20493"}],"wp:attachment":[{"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=13420"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=13420"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=13420"},{"taxonomy":"series","embeddable":true,"href":"https:\/\/www.acasinhadamatematica.pt\/index.php?rest_route=%2Fwp%2Fv2%2Fseries&post=13420"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}