Thursday

April 24, 2014

April 24, 2014

Posted by **holly** on Monday, March 12, 2007 at 2:51am.

a ladder length 13m rests against a vertical wall with its foot on a horizontal floor at a distance of 5m from the wall. when the top of the ladder slips down a distance of x, the foot of the ladder moves out x. find the distance of x.

a^2 + b^2 = c^2

a^2 + 5^2 = 13^2 ==> 169-25=144=a^2

therefore a = 12

after adjustment;

(12-x)^2 +(5+x)^2 = 13^2

144 - 24x +x^2 + 25 + 10x +x^2 = 169

2x^2-14x=0

x(2x-14)=0

x = 7 or 0

Therefore the distance moved is 7m.

thanks!

but shouldn't it be 0 or -7?

oh, sorry, i get it

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