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Piece of wire is bent so as to form the boundary of a square

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by gmattesttaker2 » Sat Sep 14, 2013 3:47 pm
Hello,

Can you please assist with this:

A piece of wire is bent so as to form the boundary of a square with area A. If the wire is then bent into the shape of an equilateral triangle, what will be the area of the triangle thus bounded in terms of A?

Sorry, I don't have the OA.
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by [email protected] » Sat Sep 14, 2013 8:59 pm
Hi gmattesttaker2,

Do you have the answer choices for this question? The algebra/geometry involved to solve this question is rather long-winded; you'd be better off TESTing a value and using the answer choices.

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by gmattesttaker2 » Sat Sep 14, 2013 9:33 pm
[email protected] wrote:Hi gmattesttaker2,

Do you have the answer choices for this question? The algebra/geometry involved to solve this question is rather long-winded; you'd be better off TESTing a value and using the answer choices.

GMAT assassins aren't born, they're made,
Rich
Hello Rich,

I came across this question in the MGMAT CAT. However, when I was reviewing this question later the answers weren't being displayed. Sorry about that. I noticed this happening for quite a few questions.

Best Regards,
Sri
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by Yaj » Sat Sep 14, 2013 11:50 pm
Here it is -

A piece of wire is bent so as to form the boundary of a square with area A. If the wire is then bent into the shape of an equilateral triangle, what will be the area of the triangle thus bounded in terms of A?

The options are attached!

[spoiler]OA:E[/spoiler]

My method (please tell me where I am going wrong) -

Square:
Area: A
Each Side √A
Perimeter: 4√A

Triangle:
Perimeter: 4√A
Since it is an equilateral triangle, length of each side: 4√A/3
Area of an equilateral triangle: side^2 √3/4

Therefore =>
Area = 8A/9*√3/4
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by Brent@GMATPrepNow » Sun Sep 15, 2013 6:21 am
Yaj wrote:Here it is -

A piece of wire is bent so as to form the boundary of a square with area A. If the wire is then bent into the shape of an equilateral triangle, what will be the area of the triangle thus bounded in terms of A?

The options are attached!

[spoiler]OA:E[/spoiler]

My method (please tell me where I am going wrong) -

Square:
Area: A
Each Side √A
Perimeter: 4√A

Triangle:
Perimeter: 4√A
Since it is an equilateral triangle, length of each side: 4√A/3
Area of an equilateral triangle: side^2 √3/4

Therefore =>
Area = 8A/9*√3/4
Hey Yaj,

Your calculations are perfect, right up until the last step.

You have:
- Each side of equilateral triangle has length (4√A)/3 [great]
- Area formula for equilateral triangle: Area = (side length²)(√3/4) [great]

When we plug in 4√A/3 for side length, we get:
Area = (4√A/3)²(√3/4)
= (16A/9)(√3/4) [you have 8A/9]
= (4A√3)/9
= E

Cheers,
Brent
Brent Hanneson - Creator of GMATPrepNow.com
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by [email protected] » Sun Sep 15, 2013 11:50 am
Hi All,

Now that we have the answer choices to work with, you can try TESTing a value.

1) Draw a square with a side of 3. It's perimeter = 12 and area = 9 = A.

2) Create an equilateral triangle with perimeter = 12; each side = 4

3) This triangle has a base of 4 and a height of 2(root3)

4) Area of this triangle = (1/2)(4)(2(root3)) = 4(root3) when A = 9

5) Plug A = 9 into the answers and find the match....

Final Answer: E

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by Zach.J.Dragone » Mon Dec 02, 2013 10:42 am
I am a little confused.

I solved this problem by assigning a length of 10 to the square meaning the area is 100. I then solved the area of the equilateral triangle by cutting it into a 30:60:90 triangle, figuring out the height (in this case, with a base of 10 the triangle has a height of 5rt.3) and then solving the area which came out to be 1/2 * 10 * 5rt.3 = 25rt.3.

I think my calculations are correct but how do I get the answer that this problem is looking for?
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by Zach.J.Dragone » Mon Dec 02, 2013 10:42 am
I am a little confused.

I solved this problem by assigning a length of 10 to the square meaning the area is 100. I then solved the area of the equilateral triangle by cutting it into a 30:60:90 triangle, figuring out the height (in this case, with a base of 10 the triangle has a height of 5rt.3) and then solving the area which came out to be 1/2 * 10 * 5rt.3 = 25rt.3.

I think my calculations are correct but how do I get the answer that this problem is looking for?
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by [email protected] » Mon Dec 02, 2013 2:04 pm
Hi Zach,

I think that you made a mistake in your "setup." By making each side of the square = 10, you DO have an area of 100, but your perimeter is now 40. This means that your equilateral triangle must have a perimeter of 40 (so each side is 13 1/3). Doing this math is probably not going to be fun, so you might consider picking a different number for each side of the square (try picking a multiple of 3; that number will then be divisible by 3 and your triangle sides will be integers).

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