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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, 08:29

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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21 Mar 2010, 11:10

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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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.

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

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05 May 2016, 06:06

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