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If the two regions, an equilateral triangle with side t and

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Manager
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If the two regions, an equilateral triangle with side t and [#permalink] New post 09 Apr 2008, 22:44
If the two regions, an equilateral triangle with side t and square with side s, have the same area, what is the ratio t:s?
A. 2:3
B. 16:3
C. 4:sqrt(3)
D. 2:sqrt(3)^4
E. 4:sqrt(3)^4


This the way I solved the question:

The area of the triangle can be calculated as height*base*1/2
(30-60-90 >> X : X*sqrt(3) : 2X >> t/2 : t*sqrt(3)/2 : t )
Height = sqrt(3)t/2
Base = t
Area = sqrt(3)*t^2/4

The area of square is s^2
sqrt(3)*t^2/4 = s^2 >> s = 3*t/2

The ratio: t/s = t/(3*t/2)

Completely blind here: what did I do wrong??
Please, point my mistake... Thank you
Director
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Re: GMATprep: Geom, ratios [#permalink] New post 09 Apr 2008, 22:56
chica wrote:
If the two regions, an equilateral triangle with side t and square with side s, have the same area, what is the ratio t:s?
A. 2:3
B. 16:3
C. 4:sqrt(3)
D. 2:sqrt(3)^4
E. 4:sqrt(3)^4


This the way I solved the question:

The area of the triangle can be calculated as height*base*1/2
(30-60-90 >> X : X*sqrt(3) : 2X >> t/2 : t*sqrt(3)/2 : t )
Height = sqrt(3)t/2
Base = t
Area = sqrt(3)*t^2/4

The area of square is s^2
sqrt(3)*t^2/4 = s^2



t^2 /s^2 = 4 / sqrt(3)

t /s = 2 / sqrt(sqrt(3) )
Manager
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Re: GMATprep: Geom, ratios [#permalink] New post 09 Apr 2008, 23:50
alpha_plus_gamma wrote:
chica wrote:
If the two regions, an equilateral triangle with side t and square with side s, have the same area, what is the ratio t:s?
A. 2:3
B. 16:3
C. 4:sqrt(3)
D. 2:sqrt(3)^4
E. 4:sqrt(3)^4


This the way I solved the question:

The area of the triangle can be calculated as height*base*1/2
(30-60-90 >> X : X*sqrt(3) : 2X >> t/2 : t*sqrt(3)/2 : t )
Height = sqrt(3)t/2
Base = t
Area = sqrt(3)*t^2/4

The area of square is s^2
sqrt(3)*t^2/4 = s^2



t^2 /s^2 = 4 / sqrt(3)

t /s = 2 / sqrt(sqrt(3) )


oooooooooooooops, I see now

Thank you :)
Manager
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Re: GMATprep: Geom, ratios [#permalink] New post 10 Apr 2008, 07:28
U ASSUME THAT IT IS A 30-60-90 DEGREE TRINALGE

WHY CANNOT IT CAN BE A 45-45-90
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Re: GMATprep: Geom, ratios [#permalink] New post 10 Apr 2008, 07:36
shobuj wrote:
U ASSUME THAT IT IS A 30-60-90 DEGREE TRINALGE

WHY CANNOT IT CAN BE A 45-45-90


Well, we are given equilateral triangle 60 - 60 - 60
A height bisects the angle of the equilateral triangle. Therefore, when we draw a height we will get 2 similar triangles: 30-60-90 and 30-60-90.
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Re: GMATprep: Geom, ratios [#permalink] New post 10 Apr 2008, 07:43
D

Isn't the perpendicular bisector of the base of an equilateral triangle the bisector of the opposite ange to the base as well?
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Re: GMATprep: Geom, ratios [#permalink] New post 10 Apr 2008, 08:41
I've got B on this one.
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Re: GMATprep: Geom, ratios [#permalink] New post 10 Apr 2008, 08:45
chica wrote:
If the two regions, an equilateral triangle with side t and square with side s, have the same area, what is the ratio t:s?
A. 2:3
B. 16:3
C. 4:sqrt(3)
D. 2:sqrt(3)^4
E. 4:sqrt(3)^4


This the way I solved the question:

The area of the triangle can be calculated as height*base*1/2
(30-60-90 >> X : X*sqrt(3) : 2X >> t/2 : t*sqrt(3)/2 : t )
Height = sqrt(3)t/2
Base = t
Area = sqrt(3)*t^2/4

The area of square is s^2
sqrt(3)*t^2/4 = s^2 >> s = 3*t/2

The ratio: t/s = t/(3*t/2)

Completely blind here: what did I do wrong??
Please, point my mistake... Thank you


s^2=(t^2*sqrt(3))/4

t/s=2/4thrt(3)
Director
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Re: GMATprep: Geom, ratios [#permalink] New post 10 Apr 2008, 08:48
GMATBLACKBELT, can you walk me through your logic?
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Re: GMATprep: Geom, ratios [#permalink] New post 10 Apr 2008, 09:02
jimmyjamesdonkey wrote:
GMATBLACKBELT, can you walk me through your logic?


We have side s and side t.

Area of square is s*s or s^2.

we have t for the side of the triangle. t/2 is the 30* angle side, thus the 60* angle is t/2*sqrt3

area is (t*t/2*sqrt3)4

thus s^2=t^2sqrt3/4

Now just simplify.
Re: GMATprep: Geom, ratios   [#permalink] 10 Apr 2008, 09:02
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