To find the perimeter we should be able to calculate each sides of the triangle.

Statement 1 tells us the area of the triangle. From this we can't calculate anything else, since we don't know whether the triangle is of a special type.

Statement 2 tells us that the triangle is equilateral. Again This information alone is not sufficient.

Taken together these statements allow us to find the sides of the equilateral triangle ABC. Indeed, the area of an equilateral triangle is given by the following formula: \(\displaystyle a=\frac{s^{2}\sqrt{3}}{4}\). Where \(\displaystyle a\) is the area and \(\displaystyle s\) the length of the side.

Therefore both statements are sufficient.

To find perimeter, we need the side lengths.

I) Gives us the measure of two angles. The given measurement is equal to 60 degrees. This means the last angle is also 60 degrees.

II) Gives us one side length, but because we know from I) that this is an equilateral triangle, we know that all the sides have the same length. 

Add up all the sides to get the perimeter.

\(\displaystyle P=16+16+16\rightarrow P=48m\)

We need I) and II) to find the perimeter

The area of an equilateral triangle is given by the formula

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

where \(\displaystyle s\) is its common sidelength. It follows that the triangle with the greater sidelength has the greater area.

We will let \(\displaystyle x\) and \(\displaystyle y\!\) stand for the common sidelength of \(\displaystyle \bigtriangleup ABC\) and \(\displaystyle \bigtriangleup DEF\), respectively. The question becomes which, if either, of \(\displaystyle x\) and \(\displaystyle y\:\) is the greater.

Statement 1 alone can be rewritten by multiplying:

\(\displaystyle \frac{x}{y} = \frac{5}{6}\)

\(\displaystyle \frac{x}{y} \cdot y = \frac{5}{6} \cdot y\)

\(\displaystyle x= \frac{5}{6} \cdot y\)

Therefore, \(\displaystyle x< y\).

Therefore, \(\displaystyle x\), the length of one side of \(\displaystyle \bigtriangleup ABC\) is less than \(\displaystyle y\!\), the length of one side of \(\displaystyle \bigtriangleup DEF\).

Statement 2 alone can be rewritten as \(\displaystyle x = y -4\). Again, it follows that \(\displaystyle x< y\).

From either statement alone, it follows that \(\displaystyle x< y\). \(\displaystyle \bigtriangleup DEF\) has the greater sidelength, and, consequently, the greater area.

Since an equilateral triangle has three sides of equal measure, the perimeter of an equilateral triangle is three times its sidelength, so the triangle with the greater common sidelength has the greater perimeter.

Statement 1 gives precisely this information; since one side of \(\displaystyle \bigtriangleup ABC\) is longer than one side of \(\displaystyle \bigtriangleup DEF\), it follows that \(\displaystyle \bigtriangleup ABC\) has the longer perimeter.

Statement 2 gives that \(\displaystyle \bigtriangleup ABC\) has the greater area. Since the area of an equilateral triangle depends only on the common length of its sides, the triangle with the greater area, \(\displaystyle \bigtriangleup ABC\), must also have the greater sidelength and, consequently, the greater perimeter.

Assume Statement 1 alone. To find the radius of a circle with area \(\displaystyle 6 4 \pi\), use the area formula:

\(\displaystyle \pi r ^{2} = A\)

\(\displaystyle \pi r ^{2} = 6 4 \pi\)

\(\displaystyle r ^{2} = 6 4\)

\(\displaystyle r = 8\)

This is also the length of each side of \(\displaystyle \bigtriangleup ABC\), so its perimeter is three times this, or 24.

Assume Statement 2 alone. If we let \(\displaystyle s\) be the length of \(\displaystyle \overline{BC}\), then, since this the hypotenuse of a 30-60-90 triangle, by the 30-60-90 Theorem, the legs measure \(\displaystyle \frac{1}{2}s\) and \(\displaystyle \frac{1}{2}s \sqrt{3}\). Half the product of their lengths is equal to area \(\displaystyle 8 \sqrt{3}\), so

\(\displaystyle \frac{1}{2} \cdot \frac{1}{2}s \cdot \frac{1}{2}s \sqrt{3} = 8\sqrt{3}\)

\(\displaystyle \frac{1}{8}s^{2} \sqrt{3} = 8\sqrt{3}\)

\(\displaystyle \frac{1}{8}s^{2} = 8\)

\(\displaystyle s^{2} = 64\)

\(\displaystyle s = 8\).

As before, the sidelength of \(\displaystyle \bigtriangleup ABC\) is 8 and the perimeter is 24.

Neither statement alone is enough to determine which triangle has the greater perimeter, as each statement gives information about only one point. 

Assume both statements to be true. Since \(\displaystyle \overline{AB}\) is the segment that connects the endpoints of two sides of \(\displaystyle \bigtriangleup DEF\), it is a midsegment of the triangle, whose length is half the length of the side of \(\displaystyle \bigtriangleup DEF\) to which it is parallel. Therefore, the sidelength of \(\displaystyle \bigtriangleup ABC\) is half that of \(\displaystyle \bigtriangleup DEF\), and their perimeters are similarly related. This makes \(\displaystyle \bigtriangleup DEF\) the triangle with the greater perimeter.

Since the perimeter of an equilateral triangle is three times its common sidelength, comparison of the lengths of the sides is all that is necessary to determine which triangle, has the greater perimeter.

If we let \(\displaystyle x\) and \(\displaystyle y\:\) be the common sidelengths of \(\displaystyle \bigtriangleup ABC\) and \(\displaystyle \bigtriangleup DEF\), respectively, Statement 1 can be rewritten as the equation \(\displaystyle x - y = 1\). This can be expressed as follows:

\(\displaystyle x - y+ y = 1 + y\)

\(\displaystyle x = y+1\)

Therefore, \(\displaystyle x > y\).

Statement 2 can be rewritten as 

\(\displaystyle y - x = -1\)

\(\displaystyle y - x + x= -1 + x\)

\(\displaystyle y = x-1\)

Once again, \(\displaystyle x > y\)

Since either statement alone establishes that \(\displaystyle x > y\), it follows that \(\displaystyle \bigtriangleup ABC\) has the longer sides and, consequently, the greater perimeter of the triangles.

For simplicity's sake, we will assume that \(\displaystyle \bigtriangleup ABC\) has sidelength 1, and, consequently, perimeter 3; these arguments work regardless of the size of the triangle.

We will also need the circumference formula \(\displaystyle C = 2 \pi r\).

Assume Statement 1 alone. Since the midpoint of \(\displaystyle \overline{AC}\), which we will call \(\displaystyle M\), is inside the circle, the radius of the circle must be greater than  \(\displaystyle AM= \frac{1}{2}\). This makes the circumference at least \(\displaystyle 2 \pi\) times this, or \(\displaystyle 2 \pi \cdot \frac{1}{2} = \pi\), which is greater than 3.

Assume Statment 2 alone. Since the circle has as a radius the segment from \(\displaystyle C\) to the midpoint of the opposite side, it is an altitude of \(\displaystyle \bigtriangleup ABC\), and the radius is the height of the triangle. By way of the 30-60-90 Theorem, this height is \(\displaystyle \frac{\sqrt{3}}{2}\), and the circumference of the circle is \(\displaystyle 2 \pi\) times this, or \(\displaystyle \frac{\sqrt{3}}{2}\cdot 2 \pi = \pi \sqrt{3}\). This is greater than 3.

Either statement alone establishes that the circumference of the circle is greater than 3, the perimeter of \(\displaystyle \bigtriangleup ABC\).

Assume both statements to be true. From Statement 1, since \(\displaystyle GH > AC\), it follows that \(\displaystyle 3 \cdot GH > 3 \cdot AC\); since the perimeter of an equilateral triangle is three times the length of one side, it follows that \(\displaystyle \bigtriangleup GHJ\) has perimeter greater than \(\displaystyle \bigtriangleup ABC\). Similarly, from Statement 2, it follows that \(\displaystyle \bigtriangleup DEF\) has perimeter greater than \(\displaystyle \bigtriangleup ABC\). However, there is no way to determine whether \(\displaystyle \bigtriangleup DEF\) or \(\displaystyle \bigtriangleup GHJ\) has the greater perimeter of the two.

Assume Statement 1 alone, and examine the diagram below, which shows a circle circumscribed about \(\displaystyle \bigtriangleup GHJ\):

The diameter, which is equal to \(\displaystyle AB\) as given by Statement 1, is greater in length than any chord which is not a diameter - and all sides of \(\displaystyle \bigtriangleup GHJ\) are non-diameter chords. Therefore, \(\displaystyle \bigtriangleup ABC\) has sides of greater length than \(\displaystyle \bigtriangleup GHJ\), and its perimeter is therefore greater. However, nothing is given about \(\displaystyle \bigtriangleup DEF\). 

If Statement 2 alone is assumed, then, similarly, \(\displaystyle \bigtriangleup GHJ\) can be shown to have perimeter greater than that of \(\displaystyle \bigtriangleup DEF\). But nothing can be determined about \(\displaystyle \bigtriangleup ABC\).

From the two statements together, however, \(\displaystyle \bigtriangleup ABC\) has a perimeter greater than those of the other two triangles.