To find the area of an equilateral triangle, notice that it can be divided into two \(\displaystyle 30-60-90\) triangles:

The ratio of sides in a \(\displaystyle 30-60-90\) triangle is \(\displaystyle 1-\sqrt{3}-2\), and since the triangle is bisected such that the \(\displaystyle 30\) degree side is \(\displaystyle \frac{s}{2}\), the \(\displaystyle 60\) degree side, the height of the triangle, must have a length of \(\displaystyle \frac{\sqrt{3}}{2}s\).

The formula for the area of the triangle is given as:

\(\displaystyle Area =\frac{1}{2}base*height\)

So the area of an equilateral triangle can be written in term of the lengths of its sides as:

\(\displaystyle \frac{1}{2}s\frac{\sqrt{3}s}{2}=\frac{\sqrt{3}}{4}s^2\)

For this particular triangle, since \(\displaystyle s=1\), its area is equal to \(\displaystyle \frac{\sqrt{3}}{4}\).

\(\displaystyle \frac{\sqrt{3}}{4}< \frac{1}{2}\)

If the relation between ratios is hard to visualize, realize that \(\displaystyle \frac{\sqrt{4}}{4}=\frac{2}{4}=\frac{1}{2}\)

The area of an equilateral triangle is \(\displaystyle \frac{s^2\sqrt{3}}{4}\).

So let's set-up an equation to solve for \(\displaystyle s\). 

\(\displaystyle \frac{s^2\sqrt{3}}{4}=6\sqrt{3}\) Cross multiply.

\(\displaystyle s^2\sqrt{3}=24\sqrt{3}\) 

The \(\displaystyle \sqrt{3}\) cancels out and we get \(\displaystyle s^2=24\).

Then take square root on both sides and we get \(\displaystyle 2\sqrt{6}\). To find height, we need to realize by drawing a height we create \(\displaystyle 2\) \(\displaystyle 30-60-90\) triangles. 

The height is opposite the angle \(\displaystyle 60\). We can set-up a proportion. Side opposite \(\displaystyle 60\) is \(\displaystyle \sqrt{3}\) and the side of equilateral triangle which is opposite \(\displaystyle 90\) is \(\displaystyle 2\).

\(\displaystyle \frac{h}{\sqrt{3}}=\frac{2\sqrt{6}}{2}\) Cross multiply.

\(\displaystyle 2\sqrt{18}=2h\) Divide both sides by \(\displaystyle 2\)

\(\displaystyle h=\sqrt{18}\) 

We can simplify this by factoring out a \(\displaystyle \sqrt{9}\) to get a final answer of \(\displaystyle 3\sqrt{2}\). 

This problem requires a bit of creative thinking (unless you have memorized the fact that an equilateral triangle always has an area equal to its side length times \(\displaystyle \sqrt{3}\).

Consider the equilateral triangle:

Since this kind of triangle is a species of isoceles triangle, we know that we can drop down a height from the top vertex. This will create two equivalent triangles, one of which will look like:

This gives us a 30-60-90 triangle. We know that for such a triangle, the ratio of the side across from the 30-degree angle to the side across from the 60-degree angle is:

\(\displaystyle \frac{1}{\sqrt{3}}\)

We can also say, given our figure, that the following equivalence must hold:

\(\displaystyle \frac{1}{\sqrt{3}} = \frac{4}{x}\)

Solving for \(\displaystyle x\), we get:

\(\displaystyle x = 4\sqrt{3}\)

Now, since \(\displaystyle \sqrt{3} < \sqrt{4}\), we know that \(\displaystyle \sqrt{3}\) must be smaller than \(\displaystyle 2\). This means that \(\displaystyle 4\sqrt{3} < 4 * 2\) or \(\displaystyle 4\sqrt{3} < 8\). Quantity B is larger than quantity A.

If the perimeter of our equilateral triangle is \(\displaystyle 27\), each of its sides must be \(\displaystyle \frac{27}{3}\) or \(\displaystyle 9\). This gives us the following figure:

Since this kind of triangle is a species of isoceles triangle, we know that we can drop down a height from the top vertex. This will create two equivalent triangles, one of which will look like:

This gives us a 30-60-90 triangle. We know that for such a triangle, the ratio of the side across from the 30-degree angle to the side across from the 60-degree angle is:

\(\displaystyle \frac{1}{\sqrt{3}}\)

Therefore, we can also say, given our figure, that the following equivalence must hold:

\(\displaystyle \frac{1}{\sqrt{3}} = \frac{4.5}{x}\)

Solving for \(\displaystyle x\), we get:

\(\displaystyle x = 4.5\sqrt{3}\)

Now, since \(\displaystyle \sqrt{3} < \sqrt{4}\), we know that \(\displaystyle \sqrt{3}\) must be smaller than \(\displaystyle 2\). This means that \(\displaystyle 4.5\sqrt{3} < 4.5 * 2\) or \(\displaystyle 4.5\sqrt{3} < 9\)

Therefore, quantity B is larger than quantity A.