tag:blogger.com,1999:blog-6933544261975483399.post7597085327051604378..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Elearn Geometry Problem 69Antonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger8125tag:blogger.com,1999:blog-6933544261975483399.post-41220040590001068342022-12-12T01:35:22.223-08:002022-12-12T01:35:22.223-08:00Similar triangleSimilar triangleMarcohttps://www.blogger.com/profile/04632526355171968456noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-28472074754641753722018-08-15T06:15:33.424-07:002018-08-15T06:15:33.424-07:00the area of the trapezium ACRS is (1/2)(b+s)*s, si...the area of the trapezium ACRS is (1/2)(b+s)*s, since it's height is sGewürtztraminer68https://www.blogger.com/profile/00329980114313015297noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-88891354039833256352016-02-10T06:47:47.824-08:002016-02-10T06:47:47.824-08:00Or from similar triangles
(h-s)/h .= s/b and the...Or from similar triangles <br /><br />(h-s)/h .= s/b and the result follows Sumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-79522904725281615082015-10-05T00:09:20.419-07:002015-10-05T00:09:20.419-07:00Area of Tr. ABC is comprised of the areas of 3 Tr....Area of Tr. ABC is comprised of the areas of 3 Tr.s and the square. <br /><br />So s^2 + 1/2s(h-s) + 1/2s(b-s) = 1/2 bh<br /><br />From which the result follows<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-87656325356806242262009-09-10T21:41:33.694-07:002009-09-10T21:41:33.694-07:00Let PQRS be the square.
Area of triangle ABC= Area...Let PQRS be the square.<br />Area of triangle ABC= Area of triangle BSR + Area of Trapezium ACRS.<br />(1/2)bh = (1/2)s*(h-s) + (1/2)(b+s)* (h-s)<br />bh = (h-s)(2s+b)<br />on simplification we get the resultsuzynoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-69756656661168172552009-09-09T09:51:26.163-07:002009-09-09T09:51:26.163-07:00With P, Q, R and S vertices of the square and k be...With P, Q, R and S vertices of the square and k be the height of the triangle ΔSBR, we have ΔABC ~ ΔSBR. Therefore,<br /><br />From h/k = b/s, we have:<br />s = bk/h = b(h-s)/h = (bh-bs)/h<br />and sh + bs = bh or s = bh/(b+h)<br /><br />QEDAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-58216700371905536662008-11-08T23:27:00.000-08:002008-11-08T23:27:00.000-08:00Let the vertices of square are PQRS and BM perpend...Let the vertices of square are PQRS and BM perpendicular is drawn to BC. area triangle APM=(1/2).s.AM<BR/>area of triangle SMC=(1/2).s.MC<BR/>area of triangle PMS=(1/2).s.s<BR/>area of triangle BPS=(1/2).s.(h-s)<BR/>area of whole triangle=(1/2).b.h<BR/>now<BR/>1/2).s.AM+(1/2).s.MC+(1/2).s.s+(1/2).s.(h-s)=(1/2).b.h<BR/>and AM + MC=b<BR/>which on simplifying prove the relationgyanhttps://www.blogger.com/profile/02290100629801225136noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-21995784753247314992008-07-29T04:45:00.000-07:002008-07-29T04:45:00.000-07:00Thalès twiceThalès twiceAnonymousnoreply@blogger.com