In the Diagram, ABCD is a rectangle, and BEFG is a square. Find AD
Link to picture: https://latex.artofproblemsolving.com/a/1/9/a195ebdd79f1dd3386154bbe48a56bfbed7268e4.png
+129845 AG = sqrt (18^2 - 12^2) = sqrt (324 - 144) = sqrt (180) = 6sqrt (5)
Call the intersection of GF and DC = H
Let DG = x DH = y
Triangles DGH and ABG are similar
So
DG/ DH = AB / AG
x / y = 12 / 6 sqrt 5
x / y = 2 /sqrt 5
y = [ sqrt (5) x / 2 ] = DH
And triangles FCH and DGH and ABG are similar
CH = 12 - y = 12 - [sqrt(5) x / 2]
FH / CH = AG / BG
FH / CH = 6sqrt (5) / 18
FH / CH = sqrt (5) / 3
FH = CH * sqrt (5) /3
FH = (12 - sqrt (5)x / 2 ) (sqrt (5) /3) = (24sqrt (5) - 5x) / 6
DG^2 + DH^2 = FH^2
x^2 + (sqrt(5)x /2)^2 = [ (24sqrt (5) - 5x) / 6 ]^2
Solving for positive x produces x = (12/7)sqrt (5) = DG
AD =
DG + AG =
(12/7)sqrt (5) + 6sqrt (5) =
(12/7 + 6)sqrt (5) =
(54/7) sqrt (5)
