How Wide is the Alley?

# Gayle Gilbert

Problem 1: A 6-foot wide alley has both walls perpendicular to the ground. Two ladders, one 10 feet long, the other 12 feet, are propped up from opposite bottom corners to the adjacent wall, forming an X shape. All four legs of each ladder are firmly touching either the bottom corner or the opposite wall. The two ladders are side by side and also touching each other at the intersection of the X shape. What is the distance from the point of intersection to the ground?

First, letŐs label all the points of intersection.  See the picture below:

Since  and  are right triangles, we can use the Pythagorean Theorem to find AB and DC.

Notice by AAA,  and , so we can use ratios to solve for x:

Since , then we can substitute  for  in the second equation.

Now, letŐs set the equations equal to each other and solve for x.

Thus, the ladders intersect at approximately 4.52 feet off the ground.

LetŐs change the situation a little bit and see if we are able to figure it out the missing information still.  LetŐs say we know the distance of the point of intersection to the ground.  This time we want to find how wide the alley is.  Look at the following situation.

Problem 2: Two buildings are separated by an alley. Two ladders are placed so that the base of each ladder is against one of the buildings and reaches the top of the other building. The two ladders are 40 feet and 30 feet long. Further, they cross at a point 10 feet from the ground. How wide is the alley?

First, letŐs label all the intersection points, like we did in the previous problem.  However, this time we will put it on a coordinate system.

LetŐs use the Pythagorean Theorem for  and  to solve for the heights of the two buildings.

For :

For :

Since we are putting this situation on a graph, then we can derive the formulas for the two ladders.

The equation for the 40-foot ladder is .

The equation for the 30-foot ladder is .

To find the point where the two ladders intersect, we can just set the two equations equal to each other.

LetŐs substitute this value of x into the second equation.

Since we know , , and , we can plug in these values to find b, or the width of the alley.

To solve for x, letŐs graph this equation.

When we trace the graph, we see that the two graphs intersect at (26.0329,10), so the alley is approximately 26.03 feet wide.

If we continue to solve, we can see that the height of the first building is approximately 30.3692 feet, the height of the second building is approximately 14.9093 feet, and the distance from B to C is approximately 30.2774.  These lengths yield a trapezoid with diagonals of lengths 30 and 40 feet such that the distance from the intersection of the diagonals to the base is 10 feet.