Walking Under Ladders-- I

Geometry Level 3

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 (in Feet) from the point of intersection to the ground?


The answer is 4.52.

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3 solutions

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

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

4 3 x + 8 = 3 x 8 = x ( 4 3 + 3 ) x = 24 4 + 3 3 -\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 = 3 24 4 + 3 3 = 72 9 + 4 3 = 4.52 y = \sqrt{3} * \dfrac{24}{4 + 3\sqrt{3}} = \dfrac{72}{9 + 4\sqrt{3}} = \boxed{4.52} ft. to 2 2 decimal places.

This is the way I did it, but I really hope there's a better way than analytic geometry. :P

Finn Hulse - 6 years, 8 months ago

A F E \triangle AFE and A C D \triangle ACD are similar, so are D F E \triangle DFE and D B A \triangle DBA . Let A E = a AE = a , then:

a x = A D C D = 1 3 ( A C D = 3 0 o ) \dfrac {a}{x} = \dfrac {AD}{CD} = \dfrac {1}{\sqrt{3}}\quad (\angle ACD = 30^o)

Similarly.

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

Adding the two equations together:

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

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

By pythagorean theorem, we have

C D = 1 2 2 6 2 = 108 CD=\sqrt{12^2-6^2}=\sqrt{108}

A B = 1 0 2 6 2 = 64 = 8 AB=\sqrt{10^2-6^2}=\sqrt{64}=8

Since F E A C D A \triangle FEA \sim \triangle CDA , A E F E = A D C D = 6 108 \dfrac{AE}{FE}=\dfrac{AD}{CD}=\dfrac{6}{\sqrt{108}}

Since F E D B A D \triangle FED \sim \triangle BAD , E D F E = A D A B = 6 8 = 3 4 \dfrac{ED}{FE}=\dfrac{AD}{AB}=\dfrac{6}{8}=\dfrac{3}{4}

Now we add

A E F E + E D F E = 6 108 + 3 4 \dfrac{AE}{FE}+\dfrac{ED}{FE}=\dfrac{6}{\sqrt{108}}+\dfrac{3}{4}

But A E + E D = A D = 6 AE+ED= AD=6 , so

A D F E = 6 108 + 3 4 \dfrac{AD}{FE}=\dfrac{6}{\sqrt{108}}+\dfrac{3}{4}

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

F E FE \approx 4.52 \boxed{4.52}

Note:

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

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

x x\approx 4.52 \boxed{4.52}

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