Can anyone help me figure this one out?

Question

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.

Mr Smith · Answer

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

Kindly Use Thanks Button as liberally as you receive replies.

Mr. Smith

Stuart@KaplanGMAT · Answer

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

Stuart Kovinsky | Kaplan GMAT Faculty | Toronto

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nisagl750 · Answer

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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GMATGuruNY · Answer

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.

Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.

As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.

For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3

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