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Problem Solving — algebra and arithmetic (GMAT Focus Edition)
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Quant problem from prep test

by diegocuenca » Thu Aug 04, 2011 10:27 am
A fly flies from one corner of a rectangular room to the opposite corner using the shortest possible route. The room is x meters long, y meters wide and z meters high. What is the distance in meters that the fly travels?

A. sq rt (z*x*y)
B. sq rt (z^2 + x^2 + y^2)
C. (z + x + y)
D. sq rt (z^2*x^2*y^2)
E. (z^2 + x^2 + y^2)

I had some issues with this problem, can anyone offer assistance?
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by diegocuenca » Thu Aug 04, 2011 12:25 pm
Answer is B
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by gmatboost » Thu Aug 04, 2011 7:18 pm
First, find the distance from one corner of the floor to the opposite corner of the floor.
Since the room is x by y, using the Pythagorean Theorem, that distance is sqrt(x^2 + y^2).

Now, imagine a right triangle with the following characteristics:
1. One leg is this line of length sqrt(x^2 + y^2) that goes from one corner of the floor to the other
2. The other leg is the height of the room, extending up from one of the end points of Leg 1. This height is given to us, it is z.
3. The hypotenuse is then the distance from one corner of the floor to the opposite corner on the ceiling. This is the distance the question is asking us for.

Using the Pythagorean Theorem on this right triangle:
c^2 = a^2 + b^2
c^2 = sqrt(x^2 + y^2)^2 + z^2

The square root of something squared gets you back to the original number, so sqrt(x^2 + y^2)^2 = x^2 + y^2

c^2 = (x^2 + y^2) + z^2
c = sqrt(x^2 + y^2 + z^2)

Let me know what you think.
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by MrR » Thu Aug 04, 2011 9:36 pm
I good geometry fact to know is that the length of the diagonal across a rectangular prism (floor corner to opposite ceiling corner as described) is the squareroot of the quantity L^2+W^2+H^2. You can see the previous post for how it is derived but it might just be something you want to add to your tool box of tricks.
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by diegocuenca » Sat Aug 06, 2011 6:27 pm
gmatboost wrote:First, find the distance from one corner of the floor to the opposite corner of the floor.
Since the room is x by y, using the Pythagorean Theorem, that distance is sqrt(x^2 + y^2).

Now, imagine a right triangle with the following characteristics:
1. One leg is this line of length sqrt(x^2 + y^2) that goes from one corner of the floor to the other
2. The other leg is the height of the room, extending up from one of the end points of Leg 1. This height is given to us, it is z.
3. The hypotenuse is then the distance from one corner of the floor to the opposite corner on the ceiling. This is the distance the question is asking us for.

Using the Pythagorean Theorem on this right triangle:
c^2 = a^2 + b^2
c^2 = sqrt(x^2 + y^2)^2 + z^2

The square root of something squared gets you back to the original number, so sqrt(x^2 + y^2)^2 = x^2 + y^2

c^2 = (x^2 + y^2) + z^2
c = sqrt(x^2 + y^2 + z^2)

Let me know what you think.
Thanks for the explanation. I was able to understand clearer than the explanation in the book. Also props to MrR. I'll make a flash card for the formula. Do you guys think this is something that will be tested on the gmat?
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by MrR » Sat Aug 06, 2011 11:06 pm
Who knows? It seems a fair possibility. But the concepts and formula are easy. Its just pythagorean theorem with an added dimension. But remember. like pythagorean theorem only applies to right triangles the formula I gave is contingent on right angles being present in the geometric figure. It won't work for any old parallelpiped. THis is why it is good to understand the other explanation,so that you can see how the pythagorean theorem leads to this shortcut formula result and understanding that, hopefully, will guard you against misapplying it.

The same idea can also be used to find the length of a diagonal edge going up a rectangular pyramid from corner to tip but you can't take the whole length and width. You should just make up a some dimensions for a pyramid and see if you can use a similar thought process to calculate the distance from any corner to its apex. Post it after, if you'd like and we'll make sure you've done it correctly.

And thanks for posting. I love to help.
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by diegocuenca » Sun Aug 07, 2011 3:36 pm
I'm trying to find the distance from corner to tip of a pyramid. Dimensions are x*x*y, and I have attached a picture to help.

Take the base first: x^2 + x^2 = c^2 yields sqrt(x^2+x^2) = c
then use the height: (sqrt(x^2+x^2))^2 + y^2 = c^2 yields x^2+x^2+y^2 = c^2 yields sqrt(x^2+x^2+y^2) = c

I believe I did this wrong because I should not take the entire diagonal from the base, so how would this be done? Half it? I think this is what you meant when you said that you can't take the whole length and width.
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by gmatboost » Sun Aug 07, 2011 5:56 pm
Hi,

First, I agree that this absolutely could be asked on the GMAT.

Second, you are right that you should use half of the hypotenuse rather than the whole hypotenuse of the base of the pyramid.

The reason is that you are ultimately trying to construct a right triangle whose hypotenuse is the length you are looking for. In this case, the triangle whose hypotenuse is the distance from a corner of the base to the top has two legs: a) half the diagonal of the base; b) the height.

So your instincts were right on. Definitely continue to let us know if you have more questions.
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by diegocuenca » Sun Aug 07, 2011 6:02 pm
What would the math on that look like?
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by gmatboost » Sun Aug 07, 2011 7:02 pm
The diagonal of the base is
sqrt(x^2 + x^2) =
sqrt(2x^2) =
sqrt(x^2) * sqrt(2) =
x*sqrt(2)

This should look familiar, since the diagonal of square base creates a 45-45-90 triangle, and sides of a 45-45-90 are x, x, x*sqrt(2). Note that this true because the base of your pyramid is a square. If it were a rectangle, we would just say the diagonal was sqrt(L^2 + W^2).

So, if the diagonal of the square base is x*sqrt(2), then half of it, the distance from the center of the base to a corner of the base, is x*sqrt(2)/2.

Then, the right triangle that connects the center of the base, the corner of the base, and the top of the pyramid has legs x*sqrt(2)/2 and y.

Using the Pythagorean Theorem:
c^2 = (x*sqrt(2)/2)^2 + y^2
c^2 = x^2*[sqrt(2)]^2/[2^2] + y^2
c^2 = x^2*[2]/[4] + y^2
c^2 = x^2/2 + y^2
c = sqrt(x^2/2 + y^2)
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by MrR » Mon Aug 08, 2011 11:41 am
Sorry for my late response, especially as I posed the question. I was at a wedding.

I looked at your work and concur with you thoughts on using half the diagonal of the base, as well as GMAT Booost's final answer.

Going along with my short cut formula, here is an alternative and equivalent way of solving it.

(I did it on paint, and its pretty, but how I wish i had a scanner hooked up after seeing your quickwork)

The Basic Idea: if you divide the base in to four quadrants, the apex is above the center where they meet. Then you are trying to find the diagonal of a x/2 by x/2 by y rectangualr prism and the short cut formula applies.

"a^2+b^2+c^2=d^2" or as presented before;
"Diagonal = SQRT(L^2+W^2+H^2)"

Work and pictures below:
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by diegocuenca » Tue Aug 09, 2011 4:53 pm
No worries, thanks for the help. It is much easier to understand the problem with a visual diagram. I believe I have seen something like this on a veritas HW problem too, the pyramid hypo.
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