tag:blogger.com,1999:blog-2962428197530077805.post7371568623688537503..comments2024-03-18T22:57:59.502+00:00Comments on Clube de Matemática: EE12Susana Martin Tenreirohttp://www.blogger.com/profile/09348824451614416496noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-2962428197530077805.post-82223714320313396932017-03-17T15:10:29.765+00:002017-03-17T15:10:29.765+00:0013-R:
13.1- AÔC = 140º
CA = 140º
ABC...13-R:<br /><br />13.1- AÔC = 140º<br /> CA = 140º <br /> ABC = CA/2<br /> ABC = 140/2 <br /> ABC = 70(Opção B) <br /><br />Resposta: Opção B <br /><br />13.2-<br /><br /> CÔA = 140º <br /> CA = 360-140 <br /> CA = 220 <br /> ADE = 220/2<br /> ADE = 110 <br /><br />Resposta: ADE = 110 <br /> Anonymoushttps://www.blogger.com/profile/16761716326437808740noreply@blogger.comtag:blogger.com,1999:blog-2962428197530077805.post-28143918497489505852017-03-17T15:03:59.940+00:002017-03-17T15:03:59.940+00:00EE12 David Castanheira
1.
18.R:
=AC +CD =AD
= AC...EE12 David Castanheira<br /><br />1.<br />18.R:<br />=AC +CD =AD <br />= AC + 80 = 180 <br />= AC = 180−80 <br />= AC = 100º<br /><br />2.<br />13.<br />R: ABC = AOC/2=140/2= 70º<br /><br />13.1 <br />R: Opção B<br /><br />13.2R:<br />ADE + ADC = 180 <br />= ADE + 40 = 180 <br />= ADE = 180−40 <br />= ADE = 140º Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-2962428197530077805.post-69506440434768479762017-03-17T15:03:04.763+00:002017-03-17T15:03:04.763+00:0013.2
AC = 360 - 140 = 220º
ADC = CA - AC ÷ 2
ADC ...13.2<br />AC = 360 - 140 = 220º<br />ADC = CA - AC ÷ 2 <br />ADC = 140 - 220 ÷ 2<br />ADC = 80 ÷ 2 = 40º<br /><br />ADE + ADC = 180º<br />ADE = 180 - ADC<br />ADE = 180 - 40<br />ADE = 140ºEvelyn_Claudinohttps://www.blogger.com/profile/01796920613863899122noreply@blogger.comtag:blogger.com,1999:blog-2962428197530077805.post-23149193849880875232017-03-17T14:35:52.843+00:002017-03-17T14:35:52.843+00:0018.
CD = 40 x 2 = 80º
AC = 180 - 80 = 100º
13.
13...18.<br />CD = 40 x 2 = 80º<br />AC = 180 - 80 = 100º<br /><br />13.<br />13.1<br />(B) 70º<br />O ângulo COA é um ângulo ao centro, então:<br />COA = CA <br />140º = 140º<br /><br />O ângulo CBA é um ângulo inscrito, então:<br />CBA = CA ÷ 2<br />CBA = 140 ÷ 2<br />CBA = 70ºEvelyn_Claudinohttps://www.blogger.com/profile/01796920613863899122noreply@blogger.com