danago
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Two ladders, one red, and the other green, are 2 and 3m long respectively. The base of the red ladder is resting on the side of a narrow hallway and leaning on the wall of the other side of the hallway. The green ladder is doing the same, but on the opposite side, such that the two ladders cross each other and form an 'X' shape. The point of intersection is 1m above the floor. Show how the following equation is derived, and thus, find the width of the hallway:
<br /> \frac{1}{{\sqrt {4 - x^2 } }} + \frac{1}{{\sqrt {9 - x^2 } }} = 1<br />
Ive been working on this for quite a while, and can't seem to get anywhere. Using pythagoras' theorem, i can show that the red ladder rests <br /> {\sqrt {4 - x^2 } }<br /> meters up the wall, and the green ladder rests <br /> {\sqrt {9 - x^2 } }<br /> meters up the wall.
Ive been trying to find expressions for other lengths so that i can create an equation, but i have been unsuccessfull in doing so.
If anybody is able to shed some light on the problem, id be very thankful.
Thanks in advance,
Dan.
<br /> \frac{1}{{\sqrt {4 - x^2 } }} + \frac{1}{{\sqrt {9 - x^2 } }} = 1<br />
Ive been working on this for quite a while, and can't seem to get anywhere. Using pythagoras' theorem, i can show that the red ladder rests <br /> {\sqrt {4 - x^2 } }<br /> meters up the wall, and the green ladder rests <br /> {\sqrt {9 - x^2 } }<br /> meters up the wall.
Ive been trying to find expressions for other lengths so that i can create an equation, but i have been unsuccessfull in doing so.
If anybody is able to shed some light on the problem, id be very thankful.
Thanks in advance,
Dan.