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If the two regions above have the same area, what is the [#permalink ]
11 Oct 2009, 09:03

Question Stats:

85% (01:50) correct

15% (01:24) wrong

based on 73 sessions
Attachment:

Triangle and square.gif [ 4.55 KiB | Viewed 1123 times ]
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)

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 ]
11 Oct 2009, 09:25
Last edited by

Bunuel on 11 Oct 2009, 14:59, edited 1 time in total.

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Re: Good Geometry Question.... [#permalink ]
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 ]
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 ]
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 ]
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 ]
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 ]
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 ]
16 Aug 2010, 14:19

Simple question I got it wrong

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Re: PS - Same Area, Ratio? [#permalink ]
13 Dec 2010, 20:45
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, 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 ]
13 Dec 2010, 21:26
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, 21:31

Thank you very much Karishma. Kudos! +1!

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