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# Good Geometry Question....

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Good Geometry Question.... [#permalink]  11 Oct 2009, 10:03
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Question Stats:

40% (01:25) correct 59% (01:20) wrong based on 22 sessions
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D) 2 : (3)^(1/4)
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Re: Good Geometry Question.... [#permalink]  11 Oct 2009, 10:25
robertrdzak wrote:
If the two objects 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)

D) 2 : (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}}.

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Last edited by Bunuel on 11 Oct 2009, 15:59, edited 1 time in total.
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Re: Good Geometry Question.... [#permalink]  11 Oct 2009, 10:46
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]  14 Oct 2009, 05:27
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]  15 Aug 2010, 23:31
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, 23:38, edited 1 time in total.
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Re: Geometry problem - Equal areas between triangle and square [#permalink]  15 Aug 2010, 23:32
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]  15 Aug 2010, 23:56
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)

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Re: Good Geometry Question.... [#permalink]  16 Aug 2010, 09:08
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]  16 Aug 2010, 15:19
Simple question I got it wrong
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geometry [#permalink]  13 Dec 2010, 21:34
(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

I can't really get my arms around this one. Can anyone break it down a bit further? thank you
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Re: PS - Same Area, Ratio? [#permalink]  13 Dec 2010, 21:45
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]  13 Dec 2010, 22:18
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]  13 Dec 2010, 22:26
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]  13 Dec 2010, 22:31
Thank you very much Karishma. Kudos! +1!
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Re: geometry [#permalink]  14 Dec 2010, 01:19
Merging similar topics.
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Re: geometry   [#permalink] 14 Dec 2010, 01:19
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