There is a 6 foot wide alley. Both walls of the alley are perpendicular to the ground. Two ladders, one 10 feet long, the other 12 feet, are propped up from opposite corners to the adjacent wall, forming an X shape. All four feet of each ladder are firmly touching either the corner or the wall. The two ladders are also touching each other at the intersection of the X shape. What is the distance from the point of intersection to the ground?

Consider the two ladders two lines on a graph. Let the shorter ladder extend from (0,0) to (6,8). Let the longer ladder extend from (6,0) to (0,108^1/2). The square root of 108 can be found using the Pythagorean formula. Then solve for the slope and y intercept to find the equations of the two lines:

The shorter ladder have the equation y=(4/3)*x.

The longer ladder has the equation y=(-108^1/2/6)*x + 108^1/2.

The lines meet where the ladders cross.

Use substitution to solve for y.

Pointed out that if x and y are the heights where the ladders touch the wall, and p is the height of the intersection then 1/p = 1/x + 1/y.

To prove this let x be the height on the right wall and y the height on the left wall. Call a the distance from the left edge of the alley to the point on the alley directly below the intersection point of the ladders. Call b the distance from the right edge of the alley to the point on the alley directly below the intersection point of the ladders. From similar triangles we get:

p/x = a/(a+b)
p/y = b/(a+b)

Add the two equations:

p/x + p/y = a/(a+b) + b/(a+b)
py/xy + px/xy = (a+b)/(a+b)
p(x+y)/xy = 1
p = xy/x+y
1/p = x+y/xy
1/p = x/xy + y/xy
1/p = 1/y + 1/x

The answer is 108^1/2/(1+(108^1/2/8)) =~ 4.52 feet.