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GMAT Prep Similar Triangles

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GMAT Prep Similar Triangles [#permalink] New post 26 Jul 2006, 08:59
How can I afront this question?

Thanks
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 [#permalink] New post 26 Jul 2006, 09:11
It is evident that the two triangles are similar. In which case the ratio of the base to the height of one is equal to the ratio of the base to height of the other

i.e. S/H = s/h => s/S = h/H

Given SxH/2 = 2xsxh/2

or SH = 2sh

S= 2sx h/H=2s x s/S

S^2 = 2s^2

or S = s x sqrt(s)
Answer: C
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 [#permalink] New post 26 Jul 2006, 12:21
C

Since the triangles are similar. Ratios of their corresponding sides are equal.

SH = 2sh
Let S = ks
so H = kh

i.e k^2 * sh = 2sh
k^2 = 2
k = SQRT(2)
S = s* (SQRT(2)
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 [#permalink] New post 26 Jul 2006, 19:06
I am not good at solving via equations on these types of problems so I tried picking numbers. I assumed both triangles were a 45-45-90 triangle to make things easy...

I picked 4 for a "side" on the smaller one which would make the area =8.
For the larger once, since it was twice the area, I said each side was sqrt32 which would make the area 16...

When I do that and solve, the answer doesn't come out to choice c...because the answer is not sqrt2*4

Can't I pick numbers for this problem?
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 [#permalink] New post 29 Jul 2006, 12:14
ps_dahiya :

SH = 2sh
Let S = ks
so H = kh

i.e k^2 * sh = 2sh
k^2 = 2
k = SQRT(2)
S = s* (SQRT(2)

could you explain s = ks concept. I'm not getting it. thanks
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 [#permalink] New post 29 Jul 2006, 18:53
alimad wrote:
ps_dahiya :

SH = 2sh
Let S = ks
so H = kh

i.e k^2 * sh = 2sh
k^2 = 2
k = SQRT(2)
S = s* (SQRT(2)

could you explain s = ks concept. I'm not getting it. thanks

Both triangles are similar triangles because their angles are same.
So the ratio of their corresponding sides will be same
So I assumed S/s = k
Read more about similar triangles here:

http://regentsprep.org/Regents/math/sim ... imilar.htm
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  [#permalink] 29 Jul 2006, 18:53
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