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PS - Geometry

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Manager
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PS - Geometry [#permalink] New post 02 Jan 2004, 11:40
00:00
A
B
C
D
E

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(N/A)

Question Stats:

0% (00:00) correct 0% (00:00) wrong based on 0 sessions
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Manager
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Great [#permalink] New post 02 Jan 2004, 12:00
Right answer, is there anyway you can explain? Thanks?
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Re: Great [#permalink] New post 02 Jan 2004, 12:07
ftoor wrote:
Right answer, is there anyway you can explain? Thanks?


half of the rectangle, the length is 10 feet = 120 inches.

X + A.A1 + X = 120 inches

angle A = 45 deg, thus, AA1 = 6 inches

2X + 6 = 120

X=57

AB = X + AA1 = 63 inches or 5 feets 3 inches
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 [#permalink] New post 02 Jan 2004, 12:14
1. B has to be the middle point.
2. The base of Triangle at A is common to both AB and segement before it. Base is 6 inches. (isos tirangle). So AB is 10/2 by 1/2 of the base of triangle. To is is 5+3 inches.

It took me a while to figure it out more than 2 mins for sure. One mistake is to not see that 6 was inches. Then it was more of logic. In GMAT try to think about the question rather than solve it with brute calculation. I am trying to follow this myself too.
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 [#permalink] New post 02 Jan 2004, 12:17
bat_car wrote:
Thanks dj for the figure.


figure is a little inconsistent though.. anyway, I think, it potrayed the desired
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here [#permalink] New post 02 Jan 2004, 12:28
I got the 6 inches. The part I was stuck on was that I did not see that the distance from A.1 to B is the same distance from the first "X" on your diagram to A.

Is there is concept I am missing?
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 [#permalink] New post 02 Jan 2004, 12:33
Think like this. For all parts to be equal, the B has to be in center and ideally A should have been the 1/4 mark. But to make the question interesting they have offset A to a side. Lookaing at dj's diagram you can see that the base of the 45 degree tirangle is common to both segement 1 and segment 2 on left side of B. And these segements are equal in length. So we get

(base of triangle) + 2x(length of each partial segement) = 10.

Hope this helps to explan that the triangle base is common area and 1/4 mark on the board is also the mid point of the base of the triangle.

Hope my logic is not too twisted in terms of using the result as the cause. Picked up the line from some critical reason ans choise. :lol:
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Re: here [#permalink] New post 02 Jan 2004, 12:37
ftoor wrote:
I got the 6 inches. The part I was stuck on was that I did not see that the distance from A.1 to B is the same distance from the first "X" on your diagram to A.

Is there is concept I am missing?


see, the question states that we have 4 identical figures. thus, the left half of the figure has 2 identical figures. consider this as a trapezium. except the angled part, two opposite sides are equal i.e. the smaller side of the trapezium is equal to longer side + side under angled part.

you know the angled one is 6 incehes. the remaining (remember, identical figure) part will be: whole length - 6 inches.

is that clear?? I still have hangover :drunk :wink: :drunk
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 [#permalink] New post 10 Jan 2004, 16:31
The cuts are made at 3 points A,B and C to divide the board into identical 4 pieces. then AB has to be 5 ft.
Am I not understanding the problem here ?
  [#permalink] 10 Jan 2004, 16:31
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