Crossed+Ladders

A narrow street is lined with tall buildings. The base of a 20 foot long ladder is resting at the base of the building on the right side of the street and leans on the building on the left side. The base of the 30 foot long ladder is resting at the base of the building on the left side of the street and leans on the building on the right side. The point where the two ladders cross is exactly 8 feet from the ground. Approximately, how wide is the street (to the nearest tenth)? Justify your answer.

First, we can tell there are similar triangles. math \angle AFD \text{is similar to} \angle CFE.

\text{Since} \angle ADF \text{and} \angle CEF \text{are both right angles, triangles} ADF \text{and} CEF \text{are similar.}

\text{Similarly,} \angle BDF \text{is similar to} \angle CDE \text{and}

\angle BFD \text{and} \angle CED \text{are right angles.}

\therefore \text{triangles} CED \text{and} BFD \text{are similar.} math Using the pythagorean theorem and proportions, we can find the missing sides necessary to find the width of the street. math x^2+ad^2=20^2 \to ad= \sqrt{20^2-x^2}

x^2+bf^2=30^2 \to bf= \sqrt{30^2-x^2} \therefore \dfrac{\sqrt{30^2-x^2}}{x} = \dfrac{8}{de} \Rightarrow de= \dfrac{8*x}{\sqrt{30^2-x^2}}

\text{and then} \dfrac{\sqrt{20^2-x^2}}{x} = \dfrac{8}{ef} \Rightarrow ef= \dfrac{8x}{\sqrt{20^2-x^2}}

\text{Since we know that} x=de+ef \text{we have} x= \dfrac{8x}{ \sqrt{30^2-x^2}} + \dfrac{8x}{\sqrt{20^2-x^2}}

math

Using Wolframalpha, I solved for x and got x=16.2, so the street is approximately 16.2 feet wide Solve for x