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This problem shows a diagram but can be just as easily

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This problem shows a diagram but can be just as easily [#permalink] New post 08 Oct 2007, 18:03
(This problem shows a diagram but can be just as easily understood without one).

Two triangles exist each with the respective angles of x, y, and z. Triangle one has a base of 's', while triangle 2 has a base of 'S'. The triangle with base 'S' has twice the area of triangle 1. In terms of 's' what is 'S'?
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juaz [#permalink] New post 08 Oct 2007, 18:34
thats correct, but i knew that was the answer obviously. can you provide explanation/
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 [#permalink] New post 08 Oct 2007, 18:48
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1/2SH=2*1/2sh
SH=2sh

s/h=S/H
Sh=sH
H=Sh/s

S*S*h/s=2sh
S^2=2s^2
S=sqrt(2) * s

does anyone know a quicker way of solving this??
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Re: juaz [#permalink] New post 08 Oct 2007, 19:05
dr908 wrote:
thats correct, but i knew that was the answer obviously. can you provide explanation/


Area is proportional to side*side if angles are same.
e.g. area of a triangle = 1/2*s^2*(sin x)

If angles are same, equation can be written as A = k * s^2
A1 = 2A2
=> S^2 = 2*s^2
=> S = sqrt2 * s

HTH
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 [#permalink] New post 08 Oct 2007, 19:24
young_gun wrote:
1/2SH=2*1/2sh
SH=2sh

s/h=S/H
Sh=sH
H=Sh/s

S*S*h/s=2sh
S^2=2s^2
S=sqrt(2) * s

does anyone know a quicker way of solving this??


Correct.

Both Triangles are "Similar Triangles".
So the ratio of the height to the base of both triangles will be same.

- Brajesh
  [#permalink] 08 Oct 2007, 19:24
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