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Can anyone help me figure this one out?

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by debonairdrz » Wed Apr 25, 2012 12:01 pm
If the area of an equilateral triangle is x square meters and the perimeter is x meters, then
what is the length of one side of the triangle in meters?

(A)6
(B)8
(C)4 sqrt(2)
(D)2 sqrt(3)
(E)4 sqrt(3)


I apologize, this is my first post, and formatting's not my strong suit. Answers C,D,E are "four times the square root of two", "two times the square root of three", and "four times the square root of three", respectively.


I can't figure out what I'm doing wrong with this one. If the perimeter is the same as the area, I can set up an equation as follows:

x = length of one side (I know that in the wording of the problem x is used differently, but I like x. Please humor me.)

3x=(1/2)xh

Since it is an equilateral triangle, the height will be (sqrt(3)/2)X (I think this is necessarily the case since when we draw in the height we create a 30-60-90 triangle and 30-60-90 triangles have side lengths with the ratio of 1 - sqrt(3) - 2.

Therefore I now have:

3x=(1/2)x sqrt(3)(x/2)


That's where I get stuck. Whenever I solve this my answer matches none of the possible answers. Am I messing up the algebra or do I have the problem set up incorrectly?

Any and all help is greatly appreciated.
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by Mr Smith » Wed Apr 25, 2012 12:28 pm
hey debonairdrz,

U have got it right till the equation.
U are doing something wrong while simplyfying it.

3x = sqrt(3)*(x^2)/4
Thus x/4 = 3/sqrt(3)
Thus x = 4*sqrt(3)

Note: I am following the same nomenclature as you are n the solution. as in x is the side of the triangle :)
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by Stuart@KaplanGMAT » Wed Apr 25, 2012 7:42 pm
Mr Smith wrote:hey debonairdrz,

U have got it right till the equation.
U are doing something wrong while simplyfying it.

3x = sqrt(3)*(x^2)/4
Thus x/4 = 3/sqrt(3)
Thus x = 4*sqrt(3)

Note: I am following the same nomenclature as you are n the solution. as in x is the side of the triangle :)
Perfect solution, just to clarify a few steps:

3x = sqrt(3)*(x^2)/4

We know that x is positive (since it's the side of a triangle), so we can safely divide both sides by x to get:

3 = sqrt3 * x/4

Multiplying both sides by 4:

12 = sqrt 3 * x

Dividing both sides by sqrt3:

12/sqrt3 = x

It's mathematically inelegant to leave a root in the denominator of a fraction, so we "rationalize the denominator" by multiplying the fraction by (root we want to get rid of / root we want to get rid of). Accordingly:

12*sqrt3/sqrt3*sqrt3 = x
12(sqrt3)/3 = x
4(sqrt3) = x
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by nisagl750 » Thu Apr 26, 2012 12:20 am
debonairdrz wrote:If the area of an equilateral triangle is x square meters and the perimeter is x meters, then
what is the length of one side of the triangle in meters?

(A)6
(B)8
(C)4 sqrt(2)
(D)2 sqrt(3)
(E)4 sqrt(3)


I apologize, this is my first post, and formatting's not my strong suit. Answers C,D,E are "four times the square root of two", "two times the square root of three", and "four times the square root of three", respectively.


I can't figure out what I'm doing wrong with this one. If the perimeter is the same as the area, I can set up an equation as follows:

x = length of one side (I know that in the wording of the problem x is used differently, but I like x. Please humor me.)

3x=(1/2)xh

Since it is an equilateral triangle, the height will be (sqrt(3)/2)X (I think this is necessarily the case since when we draw in the height we create a 30-60-90 triangle and 30-60-90 triangles have side lengths with the ratio of 1 - sqrt(3) - 2.

Therefore I now have:

3x=(1/2)x sqrt(3)(x/2)


That's where I get stuck. Whenever I solve this my answer matches none of the possible answers. Am I messing up the algebra or do I have the problem set up incorrectly?

Any and all help is greatly appreciated.
I guess your equation read like:
3x = (1/2)x*sqrt(3)(x/2)
Simplifying it we get
3x = sqrt(3)/4*x^2

Solving this we get x=12/sqrt(3)

Hope you got this answer.
Now all you hacve to do is rationalise it.

i.e. multiply and divide it by sqrt(3)
So you'll get X=4*sqrt(3)

Hope it helps ;)
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by GMATGuruNY » Thu Apr 26, 2012 12:32 am
debonairdrz wrote:If the area of an equilateral triangle is x square meters and the perimeter is x meters, then
what is the length of one side of the triangle in meters?

(A)6
(B)8
(C)4 sqrt(2)
(D)2 sqrt(3)
(E)4 sqrt(3)


I apologize, this is my first post, and formatting's not my strong suit. Answers C,D,E are "four times the square root of two", "two times the square root of three", and "four times the square root of three", respectively.


I can't figure out what I'm doing wrong with this one. If the perimeter is the same as the area, I can set up an equation as follows:

x = length of one side (I know that in the wording of the problem x is used differently, but I like x. Please humor me.)

3x=(1/2)xh

Since it is an equilateral triangle, the height will be (sqrt(3)/2)X (I think this is necessarily the case since when we draw in the height we create a 30-60-90 triangle and 30-60-90 triangles have side lengths with the ratio of 1 - sqrt(3) - 2.

Therefore I now have:

3x=(1/2)x sqrt(3)(x/2)


That's where I get stuck. Whenever I solve this my answer matches none of the possible answers. Am I messing up the algebra or do I have the problem set up incorrectly?

Any and all help is greatly appreciated.
Another approach is to plug in the answers, which represent the length of each side of the triangle.
It is helpful to know the formula for the area of an equilateral triangle:
A = (s²√3)/4.

Since p=3s, we get:

3s = (s²√3)/4.

For the lefthand side to be equal to the righthand side, the value of s almost certainly will include a factor 4 and a factor of √3.

Answer choice E: 4√3
Plugging this value into 3s = (s²√3)/4, we get:
3(4√3) = ((4√3)²*√3)/4
12√3 = ((16*3)√3)/4
12√3 = 12√3.
Success!

The correct answer is E.
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