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: Ladder Alley ( 4015 )
Ladder Alley
«
:
November 23, 2007, 12:35:26 PM From
Rajesh
»
Ladder Alley
In an alley two ladders are placed crosswise. The lengths of these ladders are resp. 2 and 3 meters. They cross one another at one meter above the ground.
What is the width of the alley?
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Re: Ladder Alley
«
#1 :
November 28, 2007, 05:35:57 PM From
Ria
»
Looks like insufficient data. Or is there any relation between the 2 triangles formed with the hypotenuse 2 & 3??
Re: Ladder Alley
«
#2 :
November 29, 2007, 12:28:54 PM From
Rajesh
»
@ Ashish
Data is sufficient dear.....!!
Re: Ladder Alley
«
#3 :
December 23, 2007, 08:56:52 PM From
preeps
»
is it 4.560....??
Re: Ladder Alley
«
#4 :
December 25, 2007, 11:52:19 PM From
spazinvader
»
How you have done it preeps?
Could you explain it?
Re: Ladder Alley
«
#5 :
December 31, 2007, 03:24:05 PM From
Rajesh
»
Hi guys....here is the solution...!!!
Define the following variables (see the figure in attachment):
x is the height at which the ladder of 2 meters touches the wall of the alley;
y is the height at which the ladder of 3 meters touches the wall of the alley;
a is the horizontal distance between the point where the ladders cross and the wall to which the ladder of 2 meters is standing;
b is the horizontal distance between the point where the ladders cross and the wall to which the ladder of 3 meters is standing;
w is the width of the alley (equals a+b);
h is the height at which the ladders cross (1 meter).
Because of the similarity of triangles,
x / w = h / b
and
y / w = h / a
so
b = (w ï¿½ h) / x
and
a = (w ï¿½ h) / y.
Combining this with
a + b = w
gives
(w ï¿½ h) / y+(w ï¿½ h) / x = w
from which it follows that
h ï¿½ x + h ï¿½ y = x ï¿½ y
from which we conclude that
y = (h ï¿½ x) / (x  h).
Because of Pythagoras' theorem,
w = sqrt(32  y2) = sqrt(9  y2)
and
w = sqrt(22  x2) = sqrt(4  x2).
Combining these two equations gives
9  y2 = 4  x2
so
y2  x2 = 5.
Combining this with
y = (h ï¿½ x) / (x  h)
and h = 1, results in:
(x / (x  1))2  x2 = 5.
Solving this equation gives:
x = 1/2 + (sqrt(c) + sqrt((24 ï¿½ sqrt(2) / sqrt(c))  c 14)) / (2 ï¿½ sqrt(2))
where
c = 2 ï¿½ (d + (25 / d)  7) / 3
and
d = (395 + 60 ï¿½ sqrt(39))1/3.
And since w = sqrt(4  x2), this gives 1.231185724... meters for the width of the alley.
:
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