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600-700 level triangles and diagonals sol. provided ?

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600-700 level triangles and diagonals sol. provided ?

by c210 » Fri Jan 20, 2012 3:43 pm
Cant figure this out from the MGmat cat even with the sol. provided!

if the box pictured to the right is a cube, then the difference in length between line segment BC and line segment AB is approximately what fraction of the distance from A to C?

Image


solutions is :


Image

Let x be the length of an edge of the cube. We can find the length of BC by first finding the length of CD. CD must be x
Image
since it is the hypotenuse of a 45-45-90 triangle with legs of length x.

Using the Pythagorean theorem, BC can be calculated:

Image


THIS PART: doenst make sense either since they are assuming that CD and BD are different lengths???

Now, we need to subtract AB from BC to find their difference in length. The approximate values are 1.7x and 1.4x, so the difference is 0.3x. Dividing by the length of AC (which is an edge of the cube, so its length is x), we find that the answer is 0.3, or 30%.


THIS PART?: where did the 1.7x and 1.4x come from


My inability to comprehend this makes me sad :(
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by neelgandham » Fri Jan 20, 2012 4:48 pm
First up don't ever be sad, it is just a matter of time before you are equipped to crack these problems.
THIS PART: doenst make sense either since they are assuming that CD and BD are different lengths???
In fact CD and BD are of different lengths. Please refer to the figure in the attachment BTG33.jpg.
Let the length of CE(one side of the base)= X
Then length of DE(another side of the base) is also X
Length of CD = Square root(2)*X (Pythagorean theorem)
Since the figure is a cube, where all sides are equal. CE = DE = BD = X
So, we can clearly say that CD = Square root(2)*X and BD = X
THIS PART?: where did the 1.7x and 1.4x come from
From the above explanation we know that the value of CD = Square root(2)*X = 1.414 X. That explains the value, 1.4x present in the explanation. Refer to the figure BTG34.jpg attached

Triangle BDE is a right angled triangle, where the two perpendicular sides are BD and CD and are of lengths X and 1.414X respectively. As provided in the explanation.

BC^2 = CD^2 + BD^2 = Sqaure root(3) * X = 1.732*X

Question :The difference in length between line segment BC and line segment AB is approximately what fraction of the distance from A to C?

((BC-AB)/AC) * 100 = ?
= ((1.732X - CD)/X)*100 (CD = AB, both are diagonals of one of the surfaces)
= ((1.732X - 1.414X)/X)8100
= 0.318/100 = ~32%.

Drop me a PM or a question here if you still need help !
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by Anurag@Gurome » Fri Jan 20, 2012 8:20 pm
Let AC be of length x.
The line segment BC is sqrt(x^2 + x^2 + x^2) = x*sqrt3.
Line segment AB is sqrt(x^2 + x^2) = x*sqrt2.
Difference in length of BC and AB is x*(sqrt3 - sqrt2) = x*(1.732 - 1.414) = x*(0.3)
So, required % is 0.3x/x * 100 = [spoiler]30%[/spoiler].
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by c210 » Sat Jan 21, 2012 12:32 am
thank you for the responses! the problem is definitely a lot clearer, however, one more point that is confusing is the line



" The approximate values are 1.7x and 1.4x, so the difference is 0.3x."

that line is provided in the solution , not the question itself..

so how did they come up with those values in the solution?
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by c210 » Sat Jan 21, 2012 12:39 am
scratch that, i figured it out that those are set values..... thank you so muchneelgandham and anurag@gurome!! :)
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by ArunangsuSahu » Sat Jan 21, 2012 12:47 am
Let the side of the cube=1 unit
BC=3^1/2
AB=2^1/2

Difference = (3^1/2-2^1/2)=1.732-1.41=0.3(approx)
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by GMATGuruNY » Sat Jan 21, 2012 4:04 am
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