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)?

__**Solution**__

From the image above, we can identify two pairs of similar triangles. Two triangles are similar if they have at least two congruent corresponding angles. First let’s prove similarity between ΔABC and ΔFEC. The first corresponding congruent angle is angle C, which is shared by both ΔABC and ΔFED. The second pair of corresponding congruent angles is angle B and angle E. Since angle B is a right angle formed by one of the walls we can assume angle B is 90°. Likewise, angle E is also formed by a wall and the ground, which means we can assume angle E is also 90°. All right angles are congruent, which provides us with the second corresponding congruent angle; therefore ΔABC and ΔFEC are similar. In a likewise manner,we can prove ΔHCB similar to ΔFEB.

To simplify our variables from the image above, I will let BC=x, EC=a, BE=b. We want to know the width of the street, x. By the Segment Addition postulate, we have the equation x=a+b. Using our variables, x,a, and b, the Pythagorean theorem allows us to solve for AB and HC. math AB=\sqrt{20^{2}-x^{2}} math math HC=\sqrt{30^{2}-x^{2}} math

We can create equal proportions with the segments of triangles ABC and FEC to solve for a. math \frac{a}{x} = \frac{FE}{AB} \rightarrow \frac{a}{x} = \frac{8}{\sqrt{900-x^{2}}} \rightarrow a = \frac{8x}{\sqrt{900-x^{2}}}\\ math

We can create equal proportions with the segments of triangles HBC and FBE to solve for b. math \frac{b}{x} = \frac{FE}{HC} \rightarrow \frac{b}{x} = \frac{8}{\sqrt{400-x^{2}}} \rightarrow b = \frac{8x}{\sqrt{400-x^{2}}}\\ math Since we have solved for both a and b in terms of x, we can substitute this information into our x=a+b equation. math \hspace{2in} x= \frac{8x}{\sqrt{900-x^{2}}} + \frac{8x}{\sqrt{400-x^{2}}} math

Then we can set this equation to equal zero by subtracting x from both sides. math 0= -x + \frac{8x}{\sqrt{400-x^{2}}} + \frac{8x}{\sqrt{900-x^{2}}} math Since this is not a pretty equation, I let Wolfram Alpha solve for x. math x= -16.21 \hspace{.25in} \text{and} \hspace{.25in} x=16.21\\ \text{Since distance cannot be negative, the width of the street is 16.21 feet.}

math [|Wolfram Alpha Solution]

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