Is the perimeter of equilateral triangle T greater than the perimeter

\(P_{equilateral}=3t\) and \(P_{square}=4s\), where \(t\) and \(s\) are the sides of triangle and square respectively. Question: is \(P_{equilateral}>P_{square}\). You can notice that if we knew the ratio of the side \(t\) to the side \(s\) then we would be able to answer the question.

(1) The ratio of the area of T to area of S is root (3) : 1 --> both the area of the equilateral triangle (\(area_{equilateral}=t^2\frac{\sqrt{3}}{4}\)) and the area of a square (\(area_{square}=s^2\)) can be expressed with their sides, so we could get the ratio of the sides from the ratio of the areas. Sufficient.

(2) The ratio of length of a side of T to a side of S is 2 : 1 --> directly gives us the ratio of the sides. Sufficient.

Answer: D.
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What I'm unclear about is the formula t squared=root 3 divided by four. Does that formula give the ratio of the sides? Like if the sides of an equilateral were 2 : 2 : 2 then would 2^2 times root divided by four equal the ratio of the side of a square- does that mean the side of a square is 2 root 3 then? Two squared= 4 times root 3 divided by four then cancels out to root 3 then multiply 2 by root 3 and that's the length of the side of the square?

First of all please use math formulas (check here: https://gmatclub.com/forum/rules-for-po ... l#p1096628).

Next, \(t^2*\frac{\sqrt{3}}{4}\) is a formula for an area of an equilateral triangle with the length of a side equal to t.
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We can find the sides of triangle and square using a and b separately. Hence it is D.

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Let the side for the triangle be t and square be s

St 1:\sqrt{3}/4 *t^2/s^2 = \sqrt{3}

or t/s = 2
or t = 2s

therefroe the perimeter for the triangle = 3t = 6s
area of suare = 4s
hence 6s>4s
ANSWER

St 2: t/s = 2 or t = 2s
therefroe the perimeter for the triangle = 3t = 6s
area of suare = 4s
hence 6s>4s
ANSWER

Option D

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