Online Geometry theorems, problems, solutions, and related topics.
Geometry Problem. Post your solution in the comments box below.Level: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the complete problem 891.
Distance between DE & AH = 6+5 = 7+xx = FG = 4
Like Jacob Ha, I extended DE, BC, AH and GFDE line meet BC line at WDE line meet GF line at TAH and FG at Uand AH and BC at KWTUK is a rectangle [easy to prove]WK=TUWK=d/sqrt(2)+5+a/sqrt(2)TU=c/sqrt(2)+d/sqrt(2)+x6+5=x+7x=4
that was really helpful