Doubling The Scales

Perimeter of an equilateral triangle with side length $a$ is equal to $3a$ .
Area of an equilateral triangle with side length $a$ is equal to $\dfrac{\sqrt{3}}4a^2$ .

If we have an equilateral triangle with side length $b$ and with area twice as large as in an equilateral triangle with side length $a$ , we have $\dfrac{\sqrt{3}}4b^2 = 2\dfrac{\sqrt{3}}4a^2 \implies b = \sqrt{2}a$ .

Thus, the perimeter is not doubled because it is enlargened by a factor of $\sqrt{2} \neq 2$ .

Hence the answer is $\boxed{\text{No}}$ .

The area of a polygon is given in square units, whereas the perimeter is expressed in linear units. Doubling the area is achieved by doubling just one dimension. Consider a rectangle, whose area is doubled by multiplying just one dimension by two. Multiplying both dimensions by two would result in an area four times as large as the original, whereas its perimeter would just be doubled.

$\text{Side in units = }a\\ \text{Area, }A = \dfrac{\sqrt[]{3}}{4}a^2 \\ A \propto a^2 \\ \sqrt[]{A} \propto a \\ \text{Perimeter, }P = 3a \\ P \propto a \\ \implies A \propto P^2 \\ \text{As A increases Perimeter drastically increases. So it's squared rather than doubled.}$

Area of a Triangle = $\dfrac{1}{2}bh$

Perimeter of a Triangle = $a+b+c$ , where $a,b,c$ being the sides of the triangle.

With an equilateral triangle, we can simplify these equations since $a=b=c$ .

WLOG, Perimeter = $3a$

Area = $\dfrac{1}{2}a(a\sqrt{3})=\dfrac{a^2\sqrt{3}}{2}$ ( $h=a\sqrt{3}$ from a $30,60,90$ triangle).

Now if we doulbe the Area, we are doubling a factor of $a^2$ . Thus, $a$ is being increased by a factor of $\sqrt{2}$ . Implying the perimeter is increased by a factor of $\sqrt{2}$ .

Proportional polygons have a constant of proportionality K in a way that, if a and b are the side lengths, and a' and b' are the side lengths of the Proportional polygon, then a/a' = b/b' = K. Supposing the proportional polygons as rectangles, the perimeter is given by 2a + 2b. We must realize that (2a + 2b)/(2a' + 2b') = K. The constant of proportionality for the area, however, is given by a b/a' b' = (a/a') * (b/b') = K*K = K^2. So, the proportion that relates the area of a polygon is equal to the square of the proportion that relates the side lengths and the perimeters.

Work backwards, Perimeter of an equilateral triangle is 3a, if you square the Perimeters (Area has dimension of 2) you'll get factor of 3 to 1, which is not double, and so Area does not double as a result from squaring the perimeter. (Area would go up by a factor of square root 2)