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October 18, 2017, 03:00:17 PM
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: Ladder Alley  ( 3978 )
« : November 23, 2007, 12:35:26 PM From Rajesh»



Ladder Alley


In an alley two ladders are placed cross-wise. 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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« #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??
« #2 : November 29, 2007, 12:28:54 PM From Rajesh»

@ Ashish

 Data is sufficient dear.....!!  :)
« #3 : December 23, 2007, 08:56:52 PM From preeps»

is it 4.560....??
« #4 : December 25, 2007, 11:52:19 PM From spazinvader»

How you have done it preeps?
Could you explain it?
« #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 � h) / x
and
a = (w � h) / y.
Combining this with
a + b = w
gives
(w � h) / y+(w � h) / x = w
from which it follows that
h � x + h � y = x � y
from which we conclude that
y = (h � 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 � 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 � sqrt(2) / sqrt(c)) - c -14)) / (2 � sqrt(2))
where
c = 2 � (d + (25 / d) - 7) / 3
and
d = (395 + 60 � 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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