A cubical box has a side-length of 1m. It is placed next to a wall which is perpendicular to the ground. A ladder 4.9 meters long is placed so that it rests against the ground, the and an edge of the box as shown in the diagram.
Let xm and ym be the lengths of AB and AC respectively a. find BD in tems of x b. find CF in terms of y c. prove that /\ DBE is similar to /\ FEC. Hence show that xy=x+y d. prove that x2 + y2=24 e. use parts c and d to prove that (y+y)2 - 2(x+y)-24=0 f. at what height does the ladder rest against the wall? g. does f have more than one solution?
Okay. a. find BD in terms of x AD + BD = AB right?
Okay. We know what AD and AB are from what the problem gives us, right?
So write the equation for BD in terms of x?
yes
no, not necessarily....we don't know what BD is at this point. AB = AD + BD =1 + BD = xm rearrange and solve for BD BD=xm-1
would what be 7 meters?
I don't know how you are getting these numbers? Can you show me your work?
Oh. Yeah, I was a little thrown by that as well. I don't think we can use those boxes, because the box is 1 m but 9 of the little boxes, so what does that mean? So I am ignoring them for now.
a. they only seem to want the length of BD in terms of x so what is that?
I think you might be getting ahead of yourself. They just want you to find BD and CF in terms of x and y for now.
I showed you above that for a. BD = xm - 1, remember?
I don't know yet. The first question is Let xm and ym be the lengths of AB and AC respectively a. find BD in tems of x BD = xm - 1 the second question is b. find CF in terms of y Can you do this one?
I could. I have already shown you. AC = AF + CF =1 + CF = ym rearrange and solve for CF CF=ym-1
I know there are comments from another person but can anyone explain it in a better way to where i can understand it easier?
|dw:1336774065283:dw| so you have the expression for DB and FC, and FE=DE=1 put in the values and solve to prove xy=x+y
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