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If the two regions above have the same area, what is the

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If the two regions above have the same area, what is the [#permalink] New post 11 Oct 2009, 09:03
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If the two regions above have the same area, what is the ratio of t:s?

A. 2 : 3
B. 16 : 3
C. 4 : (3)^(1/2)
D. 2 : (3)^(1/4)
E. 4 : (3)^(1/4)
[Reveal] Spoiler: OA

Last edited by Bunuel on 09 Jul 2013, 09:10, edited 1 time in total.
Renamed the topic and edited the question.
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Re: Good Geometry Question.... [#permalink] New post 11 Oct 2009, 09:25
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If the two regions above have the same area, what is the ratio of t:s?

A. 2 : 3
B. 16 : 3
C. 4 : (3)^(1/2)
D. 2 : (3)^(1/4)
E. 4 : (3)^(1/4)

Area of equilateral triangle is \(area_{equilateral}=t^2*\frac{\sqrt{3}}{4}\);

Area of square is \(area_{square}=s^2\);

As areas are equal, then \(t^2*\frac{\sqrt{3}}{4}=s^2\) --> \(\frac{t^2}{s^2}=\frac{4}{\sqrt{3}}\) --> \(\frac{t}{s}=\frac{2}{\sqrt[4]{3}}\).

Answer: D.
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Last edited by Bunuel on 11 Oct 2009, 14:59, edited 1 time in total.
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Re: Good Geometry Question.... [#permalink] New post 11 Oct 2009, 09:46
thanks, I was curious, how did you get 1/4 as part of the solution? When i was doing the problem I kept ending up with 2^(1/2) : 3^(1/4)
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Re: Good Geometry Question.... [#permalink] New post 14 Oct 2009, 04:27
Sqr (S) = Sqrt (3) / 4 * sqr (T)
on simplification T/S = 2 : (3)^(1/4)
OA D
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Re: Geometry problem - Equal areas between triangle and square [#permalink] New post 15 Aug 2010, 22:31
A(triangle) = \(1/2 * t * (t/2)*sqrt{3} = (t^2sqrt{3}) / 4\)

A(square) = \(s^2\)

\((t^2 sqrt{3}) / 4 = s^2\) Areas are equal.

\(t^2 = 4s^2 / sqrt{3}\) Isolate t.

\(t = sqrt{4s^2 / 3^{1/2}}\) Take the square root of both sides.

\(t = sqrt{4s^2)} / sqrt{3^{1/2}}\) Square root of a fraction: \(sqrt{a/b} = sqrt{a} / sqrt{b}\)

\(t = 2s / 3^{1/4}\) Simplify.

\(t/s = 2 / 3^{1/4}\) Finally, the ratio.

Last edited by jpr200012 on 15 Aug 2010, 22:38, edited 1 time in total.
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Re: Geometry problem - Equal areas between triangle and square [#permalink] New post 15 Aug 2010, 22:32
I thought this was a good problem. I overlooked that the triangle was equilateral the first time. I was looking at the shape and not the labels. One reason to always redraw figures!
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Re: Geometry problem - Equal areas between triangle and square [#permalink] New post 15 Aug 2010, 22:56
I find this problem to be really easy if you just plug in numbers.

Let's find the area of the triangle first, since finding the area of a square is easier to do with a given value.

Say t =2

Area of equilateral triangle with side of 2 = \(\sqrt{3}\)

Set this area equal to \(s^2\) and take the square root of both sides

s = 3^(1/4)

Put t over s, and you have your answer!
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Re: Good Geometry Question.... [#permalink] New post 16 Aug 2010, 08:08
YourDreamTheater: That works really fast, too. I've been using plugging in numbers more lately for saving time.

Bunuel: How the heck do you keep track of all these topics? :) Can you add GMAT Prep tag to this topic?
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Re: Good Geometry Question.... [#permalink] New post 16 Aug 2010, 14:19
Simple question I got it wrong :(
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Re: PS - Same Area, Ratio? [#permalink] New post 13 Dec 2010, 20:45
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consultinghokie wrote:
(imagine a picture of an equilateral triangle with sides T and a square with sides S)

If the two regions above have the same area, what is the ratio of T:S?

2:3

16:3

4: sq root 3

2: fourth root 3

4: third root 3


Area of an equilateral triangle of side \(T = (\sqrt{3}/4)T^2\)

Area of square of side S = \(S^2\)

Given: \((\sqrt{3}/4)T^2\) = \(S^2\)

\(T^2/S^2 = 4/\sqrt{3}\)

\(T/S = 2/fourth root 3\)
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Re: PS - Same Area, Ratio? [#permalink] New post 13 Dec 2010, 21:18
one quick question where I am stumped. When you square root a square root is that where you are getting the 4th root?
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Re: PS - Same Area, Ratio? [#permalink] New post 13 Dec 2010, 21:26
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spyguy wrote:
one quick question where I am stumped. When you square root a square root is that where you are getting the 4th root?


Yes.

\(\sqrt{3} = 3^{\frac{1}{2}}\)

When you take the root again, you get \((3^{\frac{1}{2}})^{\frac{1}{2}}\) which is equal to \(3^{\frac{1}{4}}\)
In other words, it the fourth root of 3.
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Re: PS - Same Area, Ratio? [#permalink] New post 13 Dec 2010, 21:31
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Re: geometry [#permalink] New post 14 Dec 2010, 00:19
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If the two regions above have the same area, what is the [#permalink] New post 29 Jun 2015, 15:51
A. 2 : 3 formula of area for equilateral triangles includes irrational number and area of square is the sides squared, a result without irrational number. One side must have an irrational number and therefore 2:3 cannot not be correct.
B. 16 : 3 same reasoning as above.
C. 4 : (3)^(1/2) Trick to see whether the final root was taken
D. 2 : (3)^(1/4) True statement
E. 4 : (3)^(1/4) Trick to test whether you're precise enough when selecting answer choices.

IMO D
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If the two regions above have the same area, what is the   [#permalink] 29 Jun 2015, 15:51
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