Intersection of two ladders

[BMAD]

Two ladders are placed cross-wise in an alley to form a lopsided X-shape. The walls of the alley are not vertical, but are parallel to each other. The ground is flat and horizontal. The bottom of each ladder is placed against the opposite wall. The top of the longer ladder touches the alley wall 18 feet vertically higher than the top of where the shorter ladder touches the opposite wall, which in turn is 6 feet vertically higher than the intersection of the two ladders. How high vertically above the ground is that intersection?

[k-man]

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12 feet.

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The intersection of the ladders divides them into 2 parts - a top part and a bottom part. Let's call the parts of the long ladder LT (top) and LB (bottom). Similarly, the short ladder is divided into ST and SB. From the congruency of the triangles formed by the ladders and the alley walls we can establish that LT/LB = SB/ST. Projecting this ratio onto the vertical and calling the height of the intersection point x we get the equation:

(18+6)/x = x/6.

Multiplying both sides by 6x we get

x²=24*6=144.

Therefore x=12.

[tammie]

12 ft? not entirely sure this is right. but i'm thinking that the intersection on the smaller ladder would be 2/3 of the ladders length. and the shorter ladderwould meet the top of the wall at the halfway point of the length of the point that the tall ladders top meets the walls. 18ft from the top of the tall to the top of the short and 18 from the top of the short to the ground minus the 6ft down to the intersection =12 ft. from the ground? and i think that my logic only makes sense in my own head.

Edited February 28, 2013 by tammie

[k-man]

Welcome to the Den, tammie!

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[tammie]

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Awesome thanks k-man i was wondering how everyone was doing that.