If two regions have same area what is ratio t : s? The answer is the fourth but I don't know which rule makes this true.
Thanks!
If two regions have same area what is ratio t : s
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Sorry I forgot to mention my question comes from the MBA.com practice test.
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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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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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- anshumishra
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Area of triangle (which is equilateral) = Area of squarebaf2010 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!
=> 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 )
Anshu
(Every mistake is a lesson learned )