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# In the figures above, if the area of the triangle on the right is twic

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In the figures above, if the area of the triangle on the right is twic [#permalink]

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08 Mar 2010, 11:02
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In the figures above, if the area of the triangle on the right is twice the area of the triangle on the left, then in terms of s, S=

A. $$\frac{\sqrt{2}}{2}s$$

B. $$\frac{\sqrt{3}}{2}s$$

C. $$\sqrt{2}s$$

D. $$\sqrt{3}s$$

E. $$2s$$

OPEN DISCUSSION OF THIS QUESTION IS HERE: in-the-figures-above-if-the-area-of-the-triangle-on-the-128929.html

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Re: In the figures above, if the area of the triangle on the right is twic [#permalink]

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08 Mar 2010, 13:02
seems easy enough....

for similar triangles (all angles the same) the following holds:

(area A)/(area b) = ( x/y )^2 (where x and y are corresponding lengths)

in this case, let A be the area of the smaller triangle. We have:

2A/A = (S/s)^2 => S/s = 2^0.5
S = s*2^0.5
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Re: In the figures above, if the area of the triangle on the right is twic [#permalink]

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08 Mar 2010, 13:18
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For Small triangle, lets say h is the height. And for the big triangle, H is the height.
Area = 1/2 (s*h)
Area = 1/2 (S*H)

According to the info in the question -->
1/2 (S*H) = 2 * [1/2 (s*h)]
= s*h
=> s = S*H/2h
=> h/H = S/2s ----------------------1

For similar triangles
h/H = s/S -----------------------2

put 2 in 1 ==>
s/S = S/2s
=> s^2 = S^2/2
= sqrt(2).s = S

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Re: In the figures above, if the area of the triangle on the right is twic [#permalink]

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15 Mar 2010, 09:29
With the property ratio of area in similar triangle = (ratio of any similar side)^2
so s/S = sqrt(1/2) => S = s*sqrt(2) hence C is the answer
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Re: In the figures above, if the area of the triangle on the right is twic [#permalink]

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20 Mar 2010, 09:01
nickk wrote:
seems easy enough....

for similar triangles (all angles the same) the following holds:

(area A)/(area b) = ( x/y )^2 (where x and y are corresponding lengths)

in this case, let A be the area of the smaller triangle. We have:

2A/A = (S/s)^2 => S/s = 2^0.5
S = s*2^0.5

Where does this property comes from??
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Re: In the figures above, if the area of the triangle on the right is twic [#permalink]

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21 Mar 2010, 12:10
mustdoit wrote:
nickk wrote:
seems easy enough....

for similar triangles (all angles the same) the following holds:

(area A)/(area b) = ( x/y )^2 (where x and y are corresponding lengths)

in this case, let A be the area of the smaller triangle. We have:

2A/A = (S/s)^2 => S/s = 2^0.5
S = s*2^0.5

Where does this property comes from??

This is the similar trianlges property. Any two triangles are similar if they have same corresponding angles or if the ratio of their corresponding sides are same.
In case two triangles are similar, ratio of their areas = (ratio of side)^2
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GMAT PREP (PS) [#permalink]

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06 May 2010, 12:34
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Re: GMAT PREP (PS) [#permalink]

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06 May 2010, 13:05
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LM wrote:
In the figures above, if the area of the triangle on the right is twice the area of the triangle on the left, then in terms of s, S=

Property of similar triangles: In two similar triangles, the ratio of their areas is the square of the ratio of their sides.

Hence $$\frac{AREA}{area}=\frac{S^2}{s^2}=2$$ --> $$S=s\sqrt{2}$$.

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Re: GMAT PREP (PS) [#permalink]

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06 May 2010, 13:34
Bunuel wrote:
LM wrote:

Property of similar triangles: In two similar triangles, the ratio of their areas is the square of the ratio of their sides.

Hence $$\frac{AREA}{area}=\frac{S^2}{s^2}=2$$ --> $$S=s\sqrt{2}$$.

Thanks I did not know this! the ratio of their areas is the square of the ratio of their sides.
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Re: GMAT PREP (PS) [#permalink]

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25 Aug 2010, 02:24
Bunuel that was awesome...although read all concepts in your Polygon blog..but did not click to me. +1 to you as always
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Re: GMAT PREP (PS) [#permalink]

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08 Sep 2010, 06:39
This is certainly not a 700 Level question Bunuel can be moved
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06 Oct 2010, 06:29
I eliminated A and B for this because I knew that S had to be greater than s, but I did not know how to find the exact answer. I also eliminated 2s because that excludes the height when calculating area, but again I could not figure out how to get the exact answer. Any help please?? Thanks!!! The file is attached.

Jeremiah
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Geo PS1.docx [57.39 KiB]

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Re: Tough Geometry PS- Help Please! [#permalink]

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06 Oct 2010, 06:39
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Merging similar topics.
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Re: GMAT PREP (PS) [#permalink]

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23 Oct 2010, 10:08
If you don't know the property:
take H=height of the larger triangle
and h=height of the smaller triangle

As the area is twice ,so HS=2hs
because triangles are similar h/s=H/S or h/H=s/S

Now after solving we get S=(2^1/2)s
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Re: In the figures above, if the area of the triangle on the right is twic [#permalink]

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05 May 2016, 07:06
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Re: In the figures above, if the area of the triangle on the right is twic [#permalink]

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05 May 2016, 07:50
Expert's post
ronny13 wrote:

In the figures above, if the area of the triangle on the right is twice the area of the triangle on the left, then in terms of s, S=

A. $$\frac{\sqrt{2}}{2}s$$

B. $$\frac{\sqrt{3}}{2}s$$

C. $$\sqrt{2}s$$

D. $$\sqrt{3}s$$

E. $$2s$$

Since in the given triangles the three angles are identical then they are similar triangles.

Useful property of similar triangles: in two similar triangles, the ratio of their areas is the square of the ratio of their sides.

Hence $$\frac{AREA}{area}=\frac{S^2}{s^2}=2$$ --> $$S=s\sqrt{2}$$.

For more on this subject check Triangles chapter of Math Book: math-triangles-87197.html

Hope it helps.

OPEN DISCUSSION OF THIS QUESTION IS HERE: in-the-figures-above-if-the-area-of-the-triangle-on-the-128929.html
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Re: In the figures above, if the area of the triangle on the right is twic   [#permalink] 05 May 2016, 07:50
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