Ratio between area of equilateral triangle and Square

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ELYAC Realty: Senior | Next Rank: 100 Posts; Posts: 30; Joined: Sat Aug 30, 2008 12:23 am; Location: Los Angeles CA

by ELYAC Realty » Tue Sep 02, 2008 9:18 pm

Can anyone help with this?

Thanks

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Source: Beat The GMAT — Problem Solving | Tue Sep 02, 2008 9:18 pm

Nycgrl: Master | Next Rank: 500 Posts; Posts: 124; Joined: Mon May 19, 2008 1:12 pm; Thanked: 3 times

by Nycgrl » Tue Sep 02, 2008 9:51 pm

Area of square = s^2
Area of trainge = 1/2*t* {(sqt3)/4}t

1/2*t* {(sqt3)/4}t = s^2

t^2/S^2 = 4/sqt3

t/s = 2/4rt 3

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reza: Junior | Next Rank: 30 Posts; Posts: 15; Joined: Wed Aug 27, 2008 7:57 pm; GMAT Score:760

Two solutions

by reza » Tue Sep 02, 2008 9:54 pm

Plug-in some random numbers : if t = 4 then area of height of triangle will be 2V3 (2 times root 3)
and its area will be 2 * V12 / 2 = V12

Now for the square to have an area of V12,
S * S = 12 ^ (1/4)
then S is root of root 12.

Now the ratio of t/s is :

4 / (12 ^(1/4))

Once found, your weakness is your strength. https://www.gmat760.com

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loki.gmat: Master | Next Rank: 500 Posts; Posts: 187; Joined: Sat May 24, 2008 5:44 amFollowed by: 1 member; Thanked: 6 times

by loki.gmat » Tue Sep 02, 2008 9:56 pm

from data given we have
[3^(1/2)/4]*t^2 = s^2

(t/s)^2 = 4/(3^1/2)

t/s = 2/(3^1/4)

hence IMO D.

Thanks!

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ELYAC Realty: Senior | Next Rank: 100 Posts; Posts: 30; Joined: Sat Aug 30, 2008 12:23 am; Location: Los Angeles CA

by ELYAC Realty » Tue Sep 02, 2008 10:04 pm

Nycgrl wrote:Area of square = s^2
Area of trainge = 1/2*t* {(sqt3)/4}t

1/2*t* {(sqt3)/4}t = s^2

t^2/S^2 = 4/sqt3

t/s = 2/4rt 3

I was under the impression that the area of an equilateral triangle is: s^2(sqrt3)/4.

And when you use that equation: s^2(sqrt3)/4=s^2(area of square), yo will eventually get sqrt3=4...what am I missing?

[/img]https://www.mathwords.com/a/area_equilat ... iangle.htm

Last edited by ELYAC Realty on Tue Sep 02, 2008 10:12 pm, edited 1 time in total.

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Nycgrl: Master | Next Rank: 500 Posts; Posts: 124; Joined: Mon May 19, 2008 1:12 pm; Thanked: 3 times

by Nycgrl » Tue Sep 02, 2008 10:12 pm

If you look at the question ,side of an equilateral triangle is T and that of square is S.

Question does not say that T= S so we take t and s as different values.

therefore area of traingle is {(srt3)/4}*t

If you still have any doubt let me know

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ELYAC Realty: Senior | Next Rank: 100 Posts; Posts: 30; Joined: Sat Aug 30, 2008 12:23 am; Location: Los Angeles CA

by ELYAC Realty » Tue Sep 02, 2008 10:24 pm

Can you tell me the algebra of how you went from:

1/2*t* {(sqt3)/4}t = s^2

to

t^2/S^2 = 4/sqt3

to finally

t/s = 2/4rt 3

I still dont know why the area of the triangle isnt: s^2(sqrt3)/4 like it says here: https://www.mathwords.com/a/area_equilat ... iangle.htm

Thanks

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gmat2010: Junior | Next Rank: 30 Posts; Posts: 29; Joined: Mon Sep 01, 2008 3:24 pm; Thanked: 3 times

by gmat2010 » Wed Sep 03, 2008 5:24 pm

ELYAC Realty wrote:Can you tell me the algebra of how you went from:

1/2*t* {(sqt3)/4}t = s^2

to

t^2/S^2 = 4/sqt3

to finally

t/s = 2/4rt 3

I still dont know why the area of the triangle isnt: s^2(sqrt3)/4 like it says here: https://www.mathwords.com/a/area_equilat ... iangle.htm

Thanks

It's is indeed t^2(sqrt3)/4.

Hence, t^2(sqrt3)/4 = s^2

t^2/s^2 = 4/sqrt3

square root both sides. t/s = 2/(3^(1/4))

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