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: . Where  is the area and  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.

We need I) and II) to find the perimeter

The area of an equilateral triangle is given by the formula

,

where  is its common sidelength. It follows that the triangle with the greater sidelength has the greater area.

We will let  and  stand for the common sidelength of  and , respectively. The question becomes which, if either, of  and  is the greater.

Statement 1 alone can be rewritten by multiplying:

Therefore, .

Therefore, , the length of one side of  is less than , the length of one side of .

Statement 2 alone can be rewritten as . Again, it follows that .

From either statement alone, it follows that .  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  is longer than one side of , it follows that  has the longer perimeter.

Statement 2 gives that  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, , must also have the greater sidelength and, consequently, the greater perimeter.

Assume Statement 1 alone. To find the radius of a circle with area , use the area formula:

This is also the length of each side of , so its perimeter is three times this, or 24.

Assume Statement 2 alone. If we let  be the length of , then, since this the hypotenuse of a 30-60-90 triangle, by the 30-60-90 Theorem, the legs measure  and . Half the product of their lengths is equal to area , so

.

As before, the sidelength of  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  is the segment that connects the endpoints of two sides of , it is a midsegment of the triangle, whose length is half the length of the side of  to which it is parallel. Therefore, the sidelength of  is half that of , and their perimeters are similarly related. This makes  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  and  be the common sidelengths of  and , respectively, Statement 1 can be rewritten as the equation . This can be expressed as follows:

Therefore, .

Statement 2 can be rewritten as 

Once again,

Since either statement alone establishes that , it follows that  has the longer sides and, consequently, the greater perimeter of the triangles.

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

We will also need the circumference formula .

Assume Statement 1 alone. Since the midpoint of , which we will call , is inside the circle, the radius of the circle must be greater than  . This makes the circumference at least  times this, or , which is greater than 3.

Assume Statment 2 alone. Since the circle has as a radius the segment from  to the midpoint of the opposite side, it is an altitude of , and the radius is the height of the triangle. By way of the 30-60-90 Theorem, this height is , and the circumference of the circle is  times this, or . This is greater than 3.

Either statement alone establishes that the circumference of the circle is greater than 3, the perimeter of .

Assume both statements to be true. From Statement 1, since , it follows that ; since the perimeter of an equilateral triangle is three times the length of one side, it follows that  has perimeter greater than . Similarly, from Statement 2, it follows that  has perimeter greater than . However, there is no way to determine whether  or  has the greater perimeter of the two.

Assume Statement 1 alone, and examine the diagram below, which shows a circle circumscribed about :

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

If Statement 2 alone is assumed, then, similarly,  can be shown to have perimeter greater than that of . But nothing can be determined about .

From the two statements together, however,  has a perimeter greater than those of the other two triangles.