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If the two regions, an equilateral triangle with side t and [#permalink]
 09 Apr 2008, 23:44
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
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Director
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Re: GMATprep: Geom, ratios [#permalink]
09 Apr 2008, 23:56
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) )
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Manager
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Re: GMATprep: Geom, ratios [#permalink]
10 Apr 2008, 00:50
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 
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Manager
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Re: GMATprep: Geom, ratios [#permalink]
10 Apr 2008, 08:28
U ASSUME THAT IT IS A 30-60-90 DEGREE TRINALGE
WHY CANNOT IT CAN BE A 45-45-90
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Manager
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Re: GMATprep: Geom, ratios [#permalink]
10 Apr 2008, 08:36
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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Senior Manager
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Re: GMATprep: Geom, ratios [#permalink]
10 Apr 2008, 08:43
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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Director
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Re: GMATprep: Geom, ratios [#permalink]
10 Apr 2008, 09:41
I've got B on this one.
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CEO
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Re: GMATprep: Geom, ratios [#permalink]
10 Apr 2008, 09:45
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)
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Director
Joined: 01 May 2007
Posts: 809
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Re: GMATprep: Geom, ratios [#permalink]
10 Apr 2008, 09:48
GMATBLACKBELT, can you walk me through your logic?
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CEO
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Re: GMATprep: Geom, ratios [#permalink]
10 Apr 2008, 10:02
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.
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Re: GMATprep: Geom, ratios
[#permalink]
10 Apr 2008, 10:02
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