Online Geometry theorems, problems, solutions, and related topics.
Geometry Problem. Post your solution in the comment box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the diagram below to enlarge it.
http://s15.postimg.org/vfo5l4wgr/Pro_1082.png1. In triangle BFD with median FO we haveBF^2+FD^2=2(BO^2+FO^2)replace BF= a/sqrt(2), FD= a, BO= a/sqrt(2) in above we get FO= a/2Circles centered A ,B, C and D have symmetry axis Ox so OE=OF’Circles centered B and D have symmetry axis BD so OF=OF’ => OE=OF= a/22. Due to symmetric property as shown above we have triangles OFC congruent to OF’A and OEB ( case ASA)Triangle OEB is the image of OFC in the rotational transformation centered O , rotational angle = 90so angle ( EOF)=903. Area S1= ½. a/2 . a/2= a^2/8= S/8