Question 1193117

 Rectangle {{{ABCD}}} is given

Given:
{{{AB = x + y}}}
{{{BC = x + 2y}}}
{{{CD = 2x -y - 2}}}
{{{DA = 3x -3y + 3}}}

Find: {{{x}}} and {{{y}}}


recall that opposite sides of a rectangle are equal in length, so

{{{AB = CD}}} and {{{BC = DA }}}

then we have

{{{x + y= 2x - y - 2}}}........solve for {{{x}}}

{{{y + y+2= 2x - x }}}

{{{x=2y+2 }}}.........eq.1


{{{x + 2y=3x- 3y + 3}}}......solve for {{{x}}}

{{{ 3y + 2y-3=3x -x }}}

{{{ 5y-3=2x }}}

{{{x=(5y-3)/2............eq.2


from eq.1 and eq.2 we have

{{{2y+2= (5y-3)/2}}}......multiply by {{{2}}}

{{{4y+4= 5y-3}}}.......solve for {{{y}}}

{{{3+4= 5y-4y}}}

{{{y=7}}}


go to

{{{x=2y+2 }}}.........eq.1, substitute {{{y}}}

{{{x=2*7+2 }}}

{{{x=16}}}



now we can find the sides length:


{{{AB = x + y=16+7=23}}}
{{{BC = x + 2y=16+2*7=30}}}
{{{CD = 2x -y -2=2*16-7-2=32-9=23}}}
{{{DA = 3x -3y + 3=3*16-3*7+3=30}}}