tag:blogger.com,1999:blog-6933544261975483399.post4200732774096083722..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1420: Regular Hexagon, Inscribed Circle, Area, TangentAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-6933544261975483399.post-86882063163299425352019-03-05T07:27:24.597-08:002019-03-05T07:27:24.597-08:00Got the geometry right and the arithmetic wrong!
...Got the geometry right and the arithmetic wrong!<br /><br />6.75 X 4 is indeed 27 not 18Sumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-70579116582370960782019-03-05T03:06:27.332-08:002019-03-05T03:06:27.332-08:00Let the side of the regular hexagon be a. We note ...Let the side of the regular hexagon be a. We note that B,H,O,G,E are collinear.<br /><br />BE = 2a and BH = a/2 so EG = 3a/4 and r(Q)^2 = (9/16).a^2<br /><br />S(G)/S(Q) = r(G)^2/r(Q)^2 = (a^2*(9/16)) / (a^2/12) = 6.75<br />So S(G) = 18<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-11652560609193454072019-03-04T11:59:36.788-08:002019-03-04T11:59:36.788-08:00From ∆CHE => R(G) = (3r√3)/2 r = r(Q)
=>...From ∆CHE => R(G) = (3r√3)/2 r = r(Q)<br />=> S = 27c.t.e.onoreply@blogger.com