tag:blogger.com,1999:blog-6933544261975483399.post5807980093663633202..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1331: Two Squares Side by Side, Parallel, Perpendicular, 90 Degrees, MeasurementAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger4125tag:blogger.com,1999:blog-6933544261975483399.post-61881517367960801622017-04-14T21:07:23.427-07:002017-04-14T21:07:23.427-07:00AB = 8, BH = 6 so AH = 10
Tr. HAK is right isosce...AB = 8, BH = 6 so AH = 10<br /><br />Tr. HAK is right isosceles <br /><br />Hence HK = 10/sqrt2 = 5sqrt2<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-43198524860354847082017-04-13T22:30:53.286-07:002017-04-13T22:30:53.286-07:00Problem 1331
Suppose that the BC intersects GM in...Problem 1331<br />Suppose that the BC intersects GM in point N then NM=FG=6, or MG=8-6=2.<br />AM^2=AG^2+MG^2=14^2+2^2=200, or AM=10√2 .So HK=AM/2=5√(2 ).<br />APOSTOLIS MANOLOUDIShttps://www.blogger.com/profile/15561495997090211148noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-89402253232049897032017-04-13T13:00:22.174-07:002017-04-13T13:00:22.174-07:00Applying the knowledge from the previous problems,...Applying the knowledge from the previous problems, we have<br />AH=HM=GC=10 ----------(1)<br />and AHMG is cyclic with m(AHM) = 90-----------(2)<br />=> AM=10Sqrt(2) ---------(3)<br />=> K is the center of the circle on which A,H,M,G lie<br />=> HK=AM/2 = 5sQRT(2)Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-4772565867259139192017-04-13T12:29:28.446-07:002017-04-13T12:29:28.446-07:00HC=2, => BH=6 => AH²=100, HK²+AK²=100 =>...HC=2, => BH=6 => AH²=100, HK²+AK²=100 => HK=5√2c.t.e.onoreply@blogger.com