A piece of wire is bent so as to form the boundary of a square with ar

Question

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.  \frac{\sqrt{3}}{64}A^2

B.  \frac{\sqrt{3}}{4}A

C.  \frac{9}{16}A

D.  \frac{3}{4}A

E.  \frac{4\sqrt{3}}{9}A

Kudos for a correct  solution .

PareshGmat · Accepted Answer

Answer = E  = \frac{4\sqrt{3}}{9}A

Area A is achieved when each side of square = \sqrt{A}

Perimeter  = 4\sqrt{A}

Each side of equilateral  	riangle = \frac{4\sqrt{A}}{3}

Area of equilateral  	riangle = \frac{\sqrt{3}}{4} * (\frac{4\sqrt{A}}{3})^2 = \frac{4\sqrt{3}}{9}A

chetan2u · Answer

E...let sides be x so A=x^2...... sides of triangle=4x/3 
 we get ans as E by substituting in formula

vedavyas9 · Answer

let side of square = a
perimeter = 4a && area = a^2 = A

let side of triangle = b
perimeter = 3b && area =(sqrt{3}/4) b^2

since length of wire is same. therefore perimeters must be same => 4a = 3b => b = (4/3)a

substituting in area of triangle = (4* sqrt{3}/9) *a^2 => (4*sqrt{3}/9) * A (since a^2 = A)

Cadaver · Answer

Let P be the side of the square....P^2 = A...P =A^1/2....

Now 4P = 3S , where s = side of equilateral triangle,

S=4A^(1/2)/3...

Area of equilateral triangle = 3^(1/2)/4.S^2 = 3^(1/2)/4.(16A/9) = (3)^1/2.4 A/9 = Option E

82vkgmat · Answer

Since area of square is A, each side is sqrt(A). So perimeter = 4*sqrt(A)
So each side of traingle = 4*sqrt(A)/3

So area of triange = 1/2 * sqrt(3)/2 * (side)^2 = sqrt(3)/4*16/9*A = 4*sqrt(3)/9 *A.

Answer = E.

rhine29388 · Answer

we can conclude from the initial statement that perimeter of both figures are equal . consider a number which is divisible by both 3 and 4 (side of triangle and side of square).a number such as 60.therefore length of wire = 60
side of square = 60/4 = 15
area of square = 225 sq units
side of triangle = 60/ 3 = 20
area of square = (sqrt 3/ 4)* 20 *20 = sqrt 3 * 100
therefore ratio = (100 *sqrt 3)/ (225) = {4 (sqrt 3)}/ 9 * A
correct answer option E

ajdse22 · Answer

one way to solve this problem :
we know that if the same wire is bent both ways,
length of wire = 4s , where s is the side of the square with area A
length of wire = 3x , where x is the side of the equi-triangle.
if we need to express area which is in terms of x to something which needs to be expressed in A, we need to express x in terms of s
x = 4/3s
so area of triangle = root(3)/4 *(16/9)s^2
=> root(3)*4/9*A -- this is choice E

AmritaSarkar89 · Answer

I will go for a detailed solution here. I had encountered this question during a mock and it came around the end when I was seriously out of patience.

Now gathering your patience and breaking the question into parts. Area of square is A then one side=sqrt(A). 
So the length of the entire wire becomes 4sqrt(A)

Twisting it into an equilateral triangle will require me to divide the entire wire into 3 segments so each side of the triangle is 4sqrt(A)/3

(you can draw the triangle for convenience) 
Now for area of eqilateral triangle we need the formula 1/2*b*h
base =4sqrt(A)/3
and height is sqrt(3)/2(side)
sqrt(s)/2*sqrt(a)4/3

Putting in the value to get the value of the area as 
2sqrt(3)/3*sqrt(A)

Nunuboy1994 · Answer

In order to solve this question we should know the area of square (we can plug in a value) and the area of the triangle

Assuming a 3 by 3 square the area of the square would be 9 and the perimeter would be 12- an equilateral triangle has 3 congruent sides so we can divide the perimeter of the square by the number of sides of our triangle which equals 4- a triangle with 3 sides each with a length of 4

Now I learned an equation from Bunuel that helps with relating the area of a square to an equilateral triangle- the area of a equilateral triangle can also calculated using the formula s^2 \sqrt{3}/4- in this formula s represents the number of sides

If we plug in the length of any of the sides of our equilateral triangle then the formula is

4^2 x \sqrt{3}/4 
16 x \sqrt{3} /4
16\sqrt{3}/4
4\sqrt{3}

If we plug the area of our square, 9 , into equation D then the result equals the area of our triangle

Thus, the correct answer is (D)

mbaapp1234 · Answer

Using smart numbers, you can set the area of the square equal to 36. So each side is 6.

Thus, the total length of the wire is 4*6 = 24 units.

Then, each side of the equilateral triangle is 24/3 = 8 units. So then you can find the area of the equilateral triangle either via

sqrt(3)/4 * (8^2) = sqrt(3)/4 * 64 = sqrt(3)*16

Now we can plug in A = 36 and see which one gives sqrt(3)*16.

Answer E matches.

BrentGMATPrepNow · Answer

Let's say 4x = the length of the wire
So when we create a square, each side has length x, which means the area of the square = (x)(x) = x²
This means  A = x²

Now take the same length of wire (length = 4x) and create an equilateral triangle
This means each side will have length  4x/3

Nice formula: Area of equilateral triangle = (√3/4)(side length)²
= (√3/4)( 4x/3 )²
= (√3/4)(16x²/9)
= (4√3/9) x² 
= (4√3/9) A

Answer: E

Cheers,
Brent

Farina · Answer

Hi Brent, Can you please tell me how did you get sqrt 3 / 4?

BrentGMATPrepNow · Answer

That's just the formula for the area of an equilateral triangle. It can save a lot of time on test day. 
 
So, for example, if an equilateral triangle has sides of length 12, then the triangle's area = (√3/4)(12)² = (√3/4)(144) = 36√3

Does that help?

Cheers, 
Brent

GmatPoint · Answer

Let the side of the square is X and that of the triangle be Y.
The perimeter of the square = 4*X

Since the same length of the wire was used to form equilateral triangle,
The perimeter of an equilateral triangle = Perimeter of a square
3*Y = 4*X
Y = (4/3)*X

Area of square(A) = X^2

Area of triangle = (√3/4)*Y^2
Area of triangle = (√3/4)* ^2
Area of triangle = (4*√3/9)*A^2

Thus, the correct option is E.

bumpbot · Answer

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A piece of wire is bent so as to form the boundary of a square with ar

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