Walking Under Ladders-- I

Let point $A$ be the origin and line segment $AD$ lie along the positive $x$ -axis.

Now by Pythagoras $AB = 8$ ft. and $CD = 6\sqrt{3}$ ft.. So $DB$ lies on the line $y = -\frac{4}{3} x + 8$ , and $AC$ lies along the line $y = \sqrt{3}x$ . These two lines intersect when

$-\frac{4}{3}x + 8 = \sqrt{3}x \Longrightarrow 8 = x(\frac{4}{3} + \sqrt{3}) \Longrightarrow x = \dfrac{24}{4 + 3\sqrt{3}}$ ft..

This gives us the height of the point of intersection as

$y = \sqrt{3} * \dfrac{24}{4 + 3\sqrt{3}} = \dfrac{72}{9 + 4\sqrt{3}} = \boxed{4.52}$ ft. to $2$ decimal places.

$\triangle AFE$ and $\triangle ACD$ are similar, so are $\triangle DFE$ and $\triangle DBA$ . Let $AE = a$ , then:

$\dfrac {a}{x} = \dfrac {AD}{CD} = \dfrac {1}{\sqrt{3}}\quad (\angle ACD = 30^o)$

Similarly.

$\dfrac {6-a}{x} = \dfrac {AD}{BA} = \dfrac {3}{4} \quad (\triangle DBA =$ 3-4-5 $\triangle )$

Adding the two equations together:

$\Rightarrow \dfrac {6}{x} = \dfrac {1}{\sqrt{3}} + \dfrac {3}{4}$

$\Rightarrow x = \dfrac {6} {\frac {1}{\sqrt{3}} + \frac {3}{4}} = \boxed {4.52}$

By pythagorean theorem, we have

$CD=\sqrt{12^2-6^2}=\sqrt{108}$

$AB=\sqrt{10^2-6^2}=\sqrt{64}=8$

Since $\triangle FEA \sim \triangle CDA$ , $\dfrac{AE}{FE}=\dfrac{AD}{CD}=\dfrac{6}{\sqrt{108}}$

Since $\triangle FED \sim \triangle BAD$ , $\dfrac{ED}{FE}=\dfrac{AD}{AB}=\dfrac{6}{8}=\dfrac{3}{4}$

Now we add

$\dfrac{AE}{FE}+\dfrac{ED}{FE}=\dfrac{6}{\sqrt{108}}+\dfrac{3}{4}$

But $AE+ED= AD=6$ , so

$\dfrac{AD}{FE}=\dfrac{6}{\sqrt{108}}+\dfrac{3}{4}$

$\dfrac{6}{FE}=\dfrac{6}{\sqrt{108}}+\dfrac{3}{4}$

$FE \approx$ $\boxed{4.52}$

Note:

We can use the formula $\dfrac{1}{x}=\dfrac{1}{a}+\dfrac{1}{b}$ . Now substitute,

$\dfrac{1}{x}=\dfrac{1}{8}+\dfrac{1}{\sqrt{108}}$

$x\approx$ $\boxed{4.52}$