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# PS: Similar Triangles

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Intern
Joined: 03 Feb 2009
Posts: 27
Schools: Duke, UNC, Emory, Ga Tech, Vanderbilt, Indiana, Wash U, Texas, Rollins

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05 Mar 2009, 17:42
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Please see attached. I am stuck on how to approach this problem.
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PracTest1_Geo_SimTri.doc [70.5 KiB]

Senior Manager
Joined: 30 Nov 2008
Posts: 483
Schools: Fuqua

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05 Mar 2009, 20:20
One of the properties of Similar Triangles worth remembering: -

If two triangles ABC and DEF are similar, then ratio of their areas = square of the (ratio of the corresponding sides).

Coming back to the original question, Area of the left side triable be a. Then Area of the right side triable will be 2a.

Based on the above rule, we have $$2a / a = (S/s)^2$$ ==> $$sqrt(2) = S/s. ==> S = s sqrt(2).$$
Manager
Joined: 17 Dec 2008
Posts: 171

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06 Mar 2009, 08:27
Area of any polygon is proportional to square of the side...
The above is not any special rule. Let minimize the number of special rules to learn

$$A1 = 1/2 (sh) => 1/2 * k1* s1^^2 (h = k1s1) A2 = 1/2 (s1h1) => 1/2 *k2 * s2^^2 (h= k2s2)$$

Since they are similar triangles k1=k2 since the angles are the same, h will be same function of s.

$$A1/A2 = (s1/s2)^^2$$
Re: PS: Similar Triangles   [#permalink] 06 Mar 2009, 08:27
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