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# GMATPrep2 - Geometry ps

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Director
Joined: 29 Aug 2005
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20 Jul 2009, 15:10
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Kudos [?]: 489 [0], given: 7

Intern
Joined: 11 Jul 2009
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Re: GMATPrep2 - Geometry ps [#permalink]

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20 Jul 2009, 16:35
2
KUDOS
This is a similar triangle problem.

raito of the areas = (ratio of the sides)^2

2 = (ratio of the sides)^2

ratio of the sides = sqrt(2)

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Manager
Joined: 03 Jul 2009
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Location: Brazil
Re: GMATPrep2 - Geometry ps [#permalink]

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20 Jul 2009, 16:41
Seofah

Is the answer 2s, or is just your option? in my calculations I got $$S = sqrt{2}*s$$, but I did it very fast, so I am not sure.

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Manager
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Posts: 106

Kudos [?]: 93 [0], given: 13

Location: Brazil
Re: GMATPrep2 - Geometry ps [#permalink]

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20 Jul 2009, 16:49
Ok, someone has also posted at the same time.

So here is what I did, although the first solution is faster if you assimilate things easily.

I have used similarity:

H = Highest height
h = lowest height

$$S/s = H/h$$
So $$h/H = s/S$$

The area of the second is two the area of the first, so
$$S*H/2 = s*h/2$$
$$S=2*s*h/H$$
$$S=2*s*s/S$$
$$S=sqrt{2}*s$$

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Senior Manager
Joined: 18 Jun 2009
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Re: GMATPrep2 - Geometry ps [#permalink]

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20 Jul 2009, 16:49
Let us take Area of right triangle as A1 and left as A2

A1 = 2A2
1/2*S*H = 2*(1/2*s*h )

1/2*S*H= s*h

S*H = 2*s*h .... (1)

Now note that angle are same for both the triangles therefor they are similar triangles

h/H = s/S

h=sH/S

Using the above equation in equation 1 we have

S*H=2*s(sH/S)

Solving it we get S=\sqrt{2} s

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Manager
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Re: GMATPrep2 - Geometry ps [#permalink]

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21 Jul 2009, 03:08
Nicely explained GMATQuantCoach3 +1 Kudo

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Director
Joined: 29 Aug 2005
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Re: GMATPrep2 - Geometry ps [#permalink]

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21 Jul 2009, 12:58
OA C.
Thanks for the effort!

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Re: GMATPrep2 - Geometry ps   [#permalink] 21 Jul 2009, 12:58
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