Question 686988
In quadrilateral {{{ABCD}}}, 

given:

{{{AE = x + 2}}}, and 

{{{BE =3x -18}}} 

find: For what value of {{{x}}} is {{{ABCD}}} a {{{rectangle}}}?

 

{{{drawing( 600,600, -15, 15, -15, 15, 
         grid(0),locate( 0, 0, A( 0, 0 ) ),locate( 0, 10, D( 0, 10 ) ),locate( 10, 0, B( 10, 0 ) ),locate( 10, 10, C( 10, 10 ) ),locate( 5, 4, E ),locate( 2, 3, AE =( x + 2 ) ),locate( 7, 3, BE =( 3x -18 ) ),
         line( 10, 0,10, 8 ),line( 0, 0, 10, 0 ),line( 0, 0, 10, 8 ),line( 10, 0, 0, 8 ),line( 0, 8, 10, 8 ),line(0, 0, 0, 8 )
)}}}



The diagonals of a rectangle are congruent; so

{{{AC=BD}}}

{{{AC=2AE = 2(x + 2)=2x+4}}}

{{{BD=2BE =2(3x-18)=6x-36}}}


than

{{{2x+4=6x-36}}}......solve for {{{x}}}

{{{36+4=6x-2x}}}

{{{40=4x}}}

{{{highlight(10=x)}}}