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Wire lines

by Rastis » Thu Nov 13, 2014 10:32 am
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?

a) (sqrt3/64)A^2

b) (sqrt3/4)A

c) (9/16)A

d) (3/4)A

e) (4sqrt3/9)A

I see how to chose smart numbers (36) but the answer explanation talks about the xsqrt3 side of the triangle (height) being (sqrt3/2)8. Where does the 2 come from?
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by Brent@GMATPrepNow » Thu Nov 13, 2014 10:52 am
Rastis wrote: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?

a) (√3/64)A^2

b) (√3/4)A

c) (9/16)A

d) (3/4)A

e) (4√3/9)A
SQUARE
Area of square = (side length)²
So we know that A = (side length)²
Solve to get: side length = √A
So, perimeter = 4√A
IMPORTANT: So, the length of the wire is 4√A

EQUILATERAL TRIANGLE
The wire has length 4√A
So, EACH of the 3 sides of the triangle will have length (4√A)/3

USEFUL FORMULA: Area of equilateral triangle = (√3/4)(side)²
So, area = (√3/4)(4√A/3)²
= (√3/4)(16A/9)
= (4A√3)/9
= E

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by GMATGuruNY » Thu Nov 13, 2014 10:57 am
Rastis wrote: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?

a) (sqrt3/64)A^2

b) (sqrt3/4)A

c) (9/16)A

d) (3/4)A

e) (4sqrt3/9)A
Since a square has 4 equal sides and an equilateral triangle has 3 equal sides, let the length of the wire = 4*3 = 12.

Square:
Since the perimeter of the square = 12, each side = 3.
Thus, A = 3² = 9.

Equilateral triangle;
Area of an equilateral triangle = (s²/4)√3.
Since the perimeter of the triangle = 12, s=4.
Area = (s²/4)√3 = (4²/4)√3 = 4√3. This is our target.

Now plug A=9 into the answers to see which yields our target of 4√3.
Only E works:
(4√3)/9 * A = (4√3)/9 * 9 = (4√3).

The correct answer is E.
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by [email protected] » Thu Nov 13, 2014 11:06 am
Hi Rastis,

In your example, you mention using the number 36. This number is a fine choice, but are you using it for the perimeter of the square OR the area of the square?

If you use it as the AREA of the square, then

Area = 36
Each side = 6
Perimeter = 24

So the Equilateral triangle has a perimeter of 24
Each side = 8

The Equilateral triangle can be split into two 30/60/90 right triangles with sides of 4/4(sqrt3)/8 respectively.

The 4(sqrt3) is the height.

If the explanation references ((sqrt3)/2)8 as the height, then that's technically correct here since the "8" divided by "2" is "4"... but I will admit that that's a strange way to present that information.

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by Rastis » Thu Nov 13, 2014 11:19 am
Rich,

In your previous message what you state "The Equilateral triangle can be split into two 30/60/90 right triangles with sides of 4/4(sqrt3)/8 respectively.", shouldn't it be 8(sqrt3) for the 60 degree angle, not 4(sqrt3)? Since X is 8?

Jesse
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by [email protected] » Thu Nov 13, 2014 3:11 pm
Hi Jesse,

In the 30/60/90 triangle, since the "8" is across from the 90 degree angle, 8 = 2X (thus X = 4).

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by Mathsbuddy » Fri Nov 14, 2014 6:52 am
Square with Area A, has side length SQRT(A) and perimeter = 4 x SQRT(A)

So, equilateral triangle has side length 4 x SQRT(A)/3

Area of any triangle = Base x side a x sinA / 2
So, area of this equilateral triangle = [4 x SQRT(A)/3] x [4 x SQRT(A)/3] x [sin 60 ]/ 2
= [4 x SQRT(A)/3] x [4 x SQRT(A)/3] x [SQRT(3/4)] / 2
= 16 x A x (SQRT(3)/2) / 18
= 8 x SQRT(3)/2 x A/9
= (4SQRT(3)/9) x A

ANSWER E
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