tag:blogger.com,1999:blog-6933544261975483399.post4182947732796746913..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 1078: Square, Inscribed Circle, Tangent, Triangle, AreaAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-6933544261975483399.post-57002608598953383182019-05-06T02:57:59.446-07:002019-05-06T02:57:59.446-07:00Drop a perpendicular CX from C to GT3. Let CX = h
...Drop a perpendicular CX from C to GT3. Let CX = h<br /><br />Tr.s T1FT3 & CXT3 are similar right triangles. Since CT3 = T1T3/2, h = a/2<br /><br />So S = h(a+b)/2 = a(a+b)/4<br /><br />Sumith Peiris<br />Moratuwa<br />Sri LankaSumith Peirishttps://www.blogger.com/profile/06211995240466447227noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-90146137556825940382015-02-05T09:17:52.158-08:002015-02-05T09:17:52.158-08:00https://docs.google.com/file/d/0Bx5WrBkYmaBHRFdEVT...https://docs.google.com/file/d/0Bx5WrBkYmaBHRFdEVTh4a1ExcGc/edit<br /><br />Complete rectangle GCDE. Extend T3 OT1 to meet GE at H.<br />G,F,T1,H are concyclic.<br />So a(a + b) = T3F.T3G = T3T1. T3H = CD. GC = 2 CT3. GC = 4 S<br />Pravinhttps://www.blogger.com/profile/05947303919973968861noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-86941498178309184982015-01-31T17:45:48.630-08:002015-01-31T17:45:48.630-08:00http://s13.postimg.org/z1zzlirbr/pro_1078.png
Draw...http://s13.postimg.org/z1zzlirbr/pro_1078.png<br />Draw lines per sketch<br />Triangle T1T3F similar to triangle T3CH ( case AA)<br />Ratio of similarity= FT3/CH=T1T3/T3C=2<br />So FT3=2. CH<br />S= ½. GT3.CH= (a+b).1/2. FT3<br />So S=1/4.a.(a+b)<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com