tag:blogger.com,1999:blog-6933544261975483399.post1746140039641184908..comments2024-03-26T19:10:02.918-07:00Comments on Go Geometry (Problem Solutions): Geometry Problem 971: Equilateral Triangle, Rectangle, Common Vertex, Sum of Right Triangles AreasAntonio Gutierrezhttp://www.blogger.com/profile/04521650748152459860noreply@blogger.comBlogger3125tag:blogger.com,1999:blog-6933544261975483399.post-58493493915423319442015-09-13T19:47:52.767-07:002015-09-13T19:47:52.767-07:00Sir, can you give me the idea how to get Z(F) Z(B)...Sir, can you give me the idea how to get Z(F) Z(B) and Z(D)<br />thx for a lot<br />here is my Gmail account : bukolight@yahoo.comAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-2721966142757481512014-01-29T18:11:46.491-08:002014-01-29T18:11:46.491-08:00Let z(P) be the complex number representing P.
Le...Let z(P) be the complex number representing P. <br />Let z(C)=0, z(B)=−a, z(E)=−a+bi. <br /><br />Then<br />z(F) = (−a+bi)(1/2 + √3/2 i) = (−1/2 a − √3/2 b) + (−√3/2 a + 1/2 b)i<br />z(A) = −a + (−√3/2 a + 1/2 b)i<br />z(D) = (−√3/2 a + 1/2 b)i<br /><br />Thus <br />BE = b, BC = a<br />AF = −1/2 a + √3/2 b, AE = √3/2 a + 1/2 b<br />DC = √3/2 a − 1/2 b, DF = 1/2 a + √3/2 b<br /><br />Area of ΔEBC = 1/2 ab<br />Area of ΔFAE = 1/8 (−√3 a² + 2 ab + √3 b²)<br />Area of ΔFDC = 1/8 (√3 a² + 2 ab − √3 b²)<br /><br />Hence, <br />Area of ΔEBC = Area of ΔFAE + Area of ΔFDCJacob HA (EMK2000)https://www.blogger.com/profile/17238561555526381028noreply@blogger.comtag:blogger.com,1999:blog-6933544261975483399.post-65283164943661967792014-01-29T08:49:03.730-08:002014-01-29T08:49:03.730-08:00http://imagizer.imageshack.us/v2/800x600q90/835/8n...http://imagizer.imageshack.us/v2/800x600q90/835/8nxu.png<br /><br />Draw rectangular FDNM as per sketch<br />Note that area ( EBC) = area(ENC)<br />And Area(EAF)= area(FME)<br />This problem becomes problem 969.<br />Peter Tranhttps://www.blogger.com/profile/02320555389429344028noreply@blogger.com