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If two regions have same area what is ratio t : s

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Where problem came from

by galinaphillips » Sun Jun 03, 2007 11:43 am
Sorry I forgot to mention my question comes from the MBA.com practice test.
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by rajatmehta » Sun Jun 03, 2007 1:02 pm
Area of equilateral triangle = (Sq. Root(3)/4) * Square(Side of Triangle)
Therefore, Area of triangle = (Sq. Root(3)/4) * Square(t)

Area can also be calculated by drawing perpendicular from vertex to the opposite side. Perpendicular will bisect the other side, and will give 2 right angled triangles.
Using Pythagorus theorem, length of perpendicular = (Sq. Root(3)/2) * t Area of One Right Angled Triangle = (1/2) * {(Sq. Root(3)/2) * t} * (t/2)

Total Area of Triangle = (Sq. Root(3)/4) * Square(t) ... (I)
Area of Square = Square(s) ... (II)

Eqaute (I) and (II):
(Sq. Root(3)/4) * Square(t) = Square(s)
Or, Square(t)/Square(s) = 4/Sq. Root(3)
Or, t/s = 2/(4th root of 3).
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Thanks

by galinaphillips » Sun Jun 03, 2007 6:49 pm
Thanks a lot!!!! :)
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by baf2010 » Thu Dec 23, 2010 9:22 pm
Does anyone else have a different way of explaining this problem? I got stuck on it too, and still do not understand. Thank you for your help!
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by anshumishra » Thu Dec 23, 2010 9:26 pm
baf2010 wrote:Does anyone else have a different way of explaining this problem? I got stuck on it too, and still do not understand. Thank you for your help!
Area of triangle (which is equilateral) = Area of square

=> sqrt3/4 * t^2 = s^2
=> t^2/s^2 = 4/sqrt3 = 4/(3)^1/2
=> t/s = [4/(3)^1/2]^1/2 = 2/(3)^1/4
Thanks
Anshu

(Every mistake is a lesson learned )
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by N:Dure » Thu Dec 23, 2010 9:58 pm
Both areas are equal so

T^2 root 3 /4 = S^2

T^2 root 3 = 4 S^2

T 3^1/4 = 2 S

T:S 2: 3^1/4
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