Question 203426
{{{drawing( 400, 400, 0, 40, 0, 40,
  locate( 5, 10, A(0,0) ),
  locate( 29, 10, B(8,0) ),
  locate( 29, 14, C(8,1) ),
  locate( 27.5, 19, D(p,q) ),
  locate( 35.5, 21.5, E ),
  locate( 18, 15, 5x-2y=0 ),
  blue(locate( 28, 16.5, 2x+5y+k=0 )),
  line( 5, 10, 35, 22 ),
  green(line( 29, 10, 29, 13 )),
  blue(line( 29, 13, 26.7242, 18.6895 )),
  red(line( 0, 10, 37, 10 ))
  )}}}

Ground is represented by {{{red(red)}}} line.
Ramp is represented by {{{AE}}}, support by {{{green(BC)}}} and beam by {{{blue(CD)}}}.
According to given conditions, support BC is perpendicular to ground at B and beam CD is perpendicular to ramp at D.


The coordinates of all the points are mentioned in the diagram.
Let the coordinates of the point D be (p,q).
Since, the slope of the ramp is 2/5 so equation to ramp AE is 5x - 2y = 0.
The equation to beam CD, perpendicular to AE, is of the form 2x + 5y + k = 0 where k = constant. 
To find k, we note that this line passes through C whose coordinates are (8,1).
So equation for CD must be satisfied by substituting x=8 and y=1.
Thus, we have 2*8 + 5*1 + k = 0 i.e. k = -21.
So, equation for CD is 2x + 5y - 21 = 0


D is the point of intersection of lines AE and CD. To find its coordinates we must solve the equations of those lines.


Solve:
5x - 2y = 0 and
2x + 5y -21 = 0


Solution: x = 7.2414, y = 2.8965
Hence, p = 7.2414, q = 2.8965


Now,length of the CD = {{{sqrt((8-7.2414)^2 + (1 - 2.8965)^2)}}} = 2.0426.


Thus, the length of the beam is 2.0426 feet.