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 the 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?
Answer
The answer is sqrt(108)/(1+(sqrt(108)/8)) =~ 4.52 feet.
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,sqrt(108)). 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 has the equation y=(4/3)x. The longer ladder has the equation y=(-sqrt(108)/6)x + sqrt(108).
The lines meet where the ladders cross.
Use substitution to solve for y.
Shopping Cart
