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 306090 triangle and 306090 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.
Can anyone help me figure this one out?
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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
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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Perfect solution, just to clarify a few steps: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
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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I guess your equation read like: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 306090 triangle and 306090 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.
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
 GMATGuruNY
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Another approach is to plug in the answers, which represent the length of each side of the triangle.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 306090 triangle and 306090 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.
It is helpful to know the formula for the area of an equilateral triangle:
A = (sÂ²âˆš3)/4.
Since p=3s, we get:
3s = (sÂ²âˆš3)/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 âˆš3.
Answer choice E: 4âˆš3
Plugging this value into 3s = (sÂ²âˆš3)/4, we get:
3(4âˆš3) = ((4âˆš3)Â²*âˆš3)/4
12âˆš3 = ((16*3)âˆš3)/4
12âˆš3 = 12âˆš3.
Success!
The correct answer is E.
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