+0  
 
0
103
1
avatar

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

 Jul 30, 2022
edited by Guest  Jul 30, 2022
 #1
avatar+124676 
+1

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)

 

cool cool cool

 Jul 30, 2022

25 Online Users