Question 874920
Since {{{system(red(x-y-8=0),green(x-t-3=0))}}} has no solution,
the lines {{{red(x-y-8=0)}}} and {{{green(x-y-3=0)}}} are parallel.
What's more, the slope of the lines is {{{1}}} ,
since {{{red(x-y-8=0)}}} --> {{{red(y=x-8)}}}
and {{{green(x-y-3=0)}}} --> {{{green(y=x-3)}}} .
The slope of the lines is {{{1}}} .
Since {{{tan(45^o)=1}}} , the lines make a {{{45^o}}} with the positive x-axis.
The bases are part of those parallel lines, which make a {{{45^o}}} with the positive x-axis.
A perpendicular to those lines from the y-intercept of {{{green(y=x-3)}}} at (0,-3} forms an isosceles right triangle with {{{red(y=x-8)}}} and the y-axis, with a hypotenuse of {{{8-3=5}}}. So the length of the legs of that triangle is
{{{5sin(45^o)=5sqrt(2)/2}}} , and that is the distance between the lines and the height of the trapezoid.
{{{drawing(300,300,-1,9,-9,1,grid(0),
red(line(-1,-9,9,1)),
green(line(-1,-4,9,6)),
line(0,-3,2.5,-5.5),
line(2.1,-5.5,2.3,-5.3),line(2.1,-5.5,2.3,-5.7)
)}}}
 
The vertices of the trapezoid can be calculated as the intersections of the lines.
{{{system(red(y=x-8),3x+2y+1=0)}}} --> {{{system(x=3,y=-5)}}}
{{{system(red(y=x-8),2x+3y+24=0)}}} --> {{{system(x=0,y=-8)}}}
{{{system(green(y=x-3),3x+2y+1=0)}}} --> {{{system(x=1,y=-2)}}}
{{{system(green(y=x-3),2x+3y+24=0)}}} --> {{{system(x=-3,y=-6)}}}
The vertices are (3,-5) and (0,-8) for the base on {{{red(y=x-8)}}}
The vertices are  (1,-2) and (-3,-6) for the base on {{{green(y=x-3)}}}
The length of the bases is the distance between their vertices, so those lengths are
{{{sqrt((3-0)^2+(-5+8)^2)=sqrt(9+9)=sqrt(2*9)=3sqrt(2)}}}
and {{{sqrt((1+3)^2+(-2+6)^2)=sqrt(16+16)=sqrt(2*16)=4sqrt(2)}}}
 
The area of a trapezoid (isosceles or not) can be calculated as
{{{In this case the area is
(1/2)(base[1]+base[2])(height)}}}
{{{(1/2)(3sqrt(2)+4sqrt(2))(5sqrt(2)/2)=(1/2)(7sqrt(2))(5sqrt(2)/2)=35/2}}}
{{{drawing(300,300,-10,10,-16,4,grid(0),
green(line(-10,-13,10,7)),
red(line(-10,-18,10,2)),
line(-12,0,12,-16),line(-10,14.5,10,-15.5)
)}}} and in close-up {{{drawing(300,300,-6,4,-9,1,grid(1),
green(line(-3,-6,1,-2)),
red(line(0,-8,3,-5)),
line(-3,-6,0,-8),line(1,-2,3,-5)
)}}}
 
To prove that it is an isosceles trapezoid, we could easily prove that the non-parallel sides (the legs) are congruent.
The length of one of those sides is the distance between vertices on {{{2x+3y+24=0}}} , (0,-8) and(-3,-6).
That distance is
{{{sqrt((0+3)^2+(-8+6)^2)=sqrt(9+4)=sqrt(13)}}} .
The length of the other leg is the distance between vertices on 
{{{3x+2y+1=0)}}} , (1,-2) and (3,-5).
That distance is
{{{sqrt((1-3)^2+(-2+5)^2)=sqrt(4+9)=sqrt(13)}}} .