Proposed Problem
Click the figure below to see the complete problem 306.
See also:
Complete Problem 306
Collection of Geometry Problems
Level: High School, SAT Prep, College geometry
Saturday, June 20, 2009
Problem 306: Square, Triangles, Angle, Side
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From triangle BCE, x^2 = b^2 + BC^2. Now triangles BCE & FAB are similar; hence, a/AB =BC/b or BC^2 = ab since AB=BC. Thus, x^2 = b^2 +ab. QED.
ReplyDeleteLet length of square ABCD be y
ReplyDeleteTriangle BCE ~ Triangle FDE
BC/CE=FD/DE
y/b=FD/(y-b)
FD=y(y-b)/b
AF=AD+DF
a=y+y(y-b)/b=y^2/b
y^2=ab
x^2=y^2+b^2=ab+b^2