Assume Statement 1 alone. Since all three sides of \(\displaystyle \bigtriangleup ABC\) are congruent - specifically, \(\displaystyle \overline{AB} \cong \overline{BC}\) - and \(\displaystyle \overline{BC} \cong \overline{EF}\), it follows by transitivity that \(\displaystyle \overline{AB} \cong \overline{EF}\). However, no information is given as to whether \(\displaystyle \overline{DE}\) has length greater than, equal to, or less than \(\displaystyle \overline{EF}\), so it cannot be determined which of \(\displaystyle \overline{AB}\) and \(\displaystyle \overline{DE}\), if either, is the longer. By a similar argument, Statement 2 yields insufficient information.

Now assume both statements are true. \(\displaystyle \overline{DF}\) and \(\displaystyle \overline{EF}\) are each congruent to one of the congruent sides of equilateral \(\displaystyle \bigtriangleup ABC\) and are therefore congruent to each other. However, the hypotenuse of a right triangle must be longer than both legs, so the hypotenuse of  \(\displaystyle \bigtriangleup DEF\) is \(\displaystyle \overline{DE}\). \(\displaystyle \overline{DE}\) is also longer than any segment congruent to one of the legs, which includes all three sides of \(\displaystyle \bigtriangleup ABC\) - specificially, \(\displaystyle \overline{DE}\) is longer than \(\displaystyle \overline{AB}\).

Assume Statement 1 alone. \(\displaystyle \bigtriangleup ABC\) is equilateral, so \(\displaystyle AB = BC\). Also, by the Triangle Inequality, the sum of the lengths of two sides of a triangle must exceed the third, so \(\displaystyle DF + EF > DE\). From Statement 1, \(\displaystyle BC= DF + EF\), so by substitution, \(\displaystyle BC > DE\), and \(\displaystyle AB> DE\).

Statement 2 alone provides insufficient information. For example, assume \(\displaystyle \bigtriangleup ABC\) is an equilateral triangle with sidelength 9. If \(\displaystyle \bigtriangleup DEF\) is an equilateral triangle with sidelength 8, the conditions of the statement hold, and \(\displaystyle AB > DE\). However, if \(\displaystyle \bigtriangleup DEF\) is a right triangle in which \(\displaystyle DF = 6\), \(\displaystyle EF = 8\), and \(\displaystyle DE = 10\), the conditions of the statement still hold, but \(\displaystyle AB < DE\).

All sides of an equilateral triangle have the same measure, so we can let \(\displaystyle x\) be the common sidelength of \(\displaystyle \bigtriangleup ABC\), and \(\displaystyle y\) be that of \(\displaystyle \bigtriangleup DEF\).

Statement 1 can be rewritten as \(\displaystyle x+ y = 15\); Statement 2 can be rewritten as \(\displaystyle xy = 50\). The equivalent question is whether we can determine which, if either, is greater, \(\displaystyle x\) or \(\displaystyle y\:\). The two statements together are insufficient to answer the question, however; 5 and 10 have sum 15 and product 50, but we cannot determine without further information whether  \(\displaystyle x = 5\) and \(\displaystyle y = 10\), or vice versa. Therefore, we do not know for sure whether a side of \(\displaystyle \bigtriangleup ABC\) is longer than a side of \(\displaystyle \bigtriangleup DEF\) - specifically, which of \(\displaystyle \overline{AB}\) or \(\displaystyle \overline{DE}\) is longer.

All sides of an equilateral triangle have the same measure, so we can let \(\displaystyle x\) be the common sidelength of \(\displaystyle \bigtriangleup ABC\), and \(\displaystyle y\:\) be that of \(\displaystyle \bigtriangleup DEF\).

Statement 1 can be rewritten as \(\displaystyle x+ y = 24\); Statement 2 can be rewritten as \(\displaystyle xy = 144\). The equivalent question is whether we can determine which, if either, is greater, \(\displaystyle x\) or \(\displaystyle y\:\).

Statement 1 alone yields insufficient information; for example, the two numbers added together could be 10 and 14, but it is impossible to determine whether \(\displaystyle x\) or \(\displaystyle y\) is the greater of the two. Statement 2 alone is also insufficient, for a similar reason; for example, the two numbers could be 9 and 16, but again, either \(\displaystyle x\) or \(\displaystyle y\) could be the greater.

Now assume both statements. The only two numbers that can be added to yield a sum of 24 and multiplied to yield a product of 144 are 12 and 12; therefore, \(\displaystyle x=y\), and \(\displaystyle \bigtriangleup ABC\) and \(\displaystyle \bigtriangleup DEF\) have the same sidelengths. Specifically, \(\displaystyle \overline{AB}\) and \(\displaystyle \overline{DE}\) have the same length.

To find an area of a triangle we need the length of the height and the length of the corresponding basis.

Each statement 1 and 2 alone is not sufficient, since we don't know whether the triangle is equilateral. Indeed, we need to take both statements to be able to calculate the area.

Hence, both statements together are sufficient.

To answer this question, we should be able to have information about any lengths in the circle or in the triangle, or also about the areas of the different regions.

Statement 1 alone is sufficient, since from the area of the circle we can find the length of the radius, which allows us to calculate the length of the height of the equilateral triangle. This would allow us to find the length of a side, giving us all the necessary information to calculate the area of the equilateral triangle.

Statement 2, although giving us information about the difference of the circle and the triangle, we cannot conclude anything because we don't know the proportions of the respective geometric figures. Therefore statement 2 is insufficient.

To conclude, statement one alone is sufficient.

To find the area of a triangle, we need the base and the height. Taking both statements together we can solve this problem.

I) tells us that OHT is an equilateral triangle.

II) gives us one side length. Which means we really know all the side lengths. 

We can use either Pythagorean Theorem or 30/60/90 triangle ratios to find the height of OHT.

\(\displaystyle a^2+b^2=c^2\)

\(\displaystyle 64^2-32^2=h^2\)

\(\displaystyle h=55.4\)

From there we can find the area.

\(\displaystyle A=\frac{1}{2}bh \rightarrow A=\frac{1}{2}(64)(55.4)=1773.6\)

Thus, both are needed. I) Tells us we have an equilateral, II) Lets us find the height. Both allow us to find the area.

Assume Statement 1 alone. Since the vertices of both the triangle and the square are on the same circle, then the same circle is circumscribed about both polygons. For the sake of simplicity, we will assume the radius of that circle to be 1 (and the diameter to be 2); this argument will work regardless of the size of the circle. 

A diagonal of a square doubles as a diameter of the circumscribed circle, so each diagonal of Square \(\displaystyle WXYZ\) is 2; since a Square is a rhombus, the area is one half the product of the lengths of the diagonals, or \(\displaystyle \frac{1}{2} \cdot 2\cdot 2= 2\).

Now examine the figure below, which shows \(\displaystyle \bigtriangleup ABC\), its altitudes, and its circumscribed circle:

The three altitudes of an equilateral triangle meet at the center of the circumscribed circle, or circumcenter, so \(\displaystyle CO = 1\); they also divide one another into segments such that one has twice the length of the other, so \(\displaystyle OM = \frac{1}{2}\). Therefore, \(\displaystyle CM = \frac{3}{2}\). Also, the six smaller triangles are all 30-60-90 by symmetry, so by the 30-60-90 Theorem, \(\displaystyle AM = OM \sqrt{3} = \frac{1}{2} \sqrt{3}\), and \(\displaystyle AB = 2\cdot AM = 2 \cdot\frac{1}{2} \sqrt{3} = \sqrt{3}\).

Therefore, the area of the triangle is 

\(\displaystyle \frac{1}{2}\cdot AB \cdot CM = \frac{1}{2}\cdot \sqrt{3}\cdot \frac{3}{2} = \frac{3\sqrt{3}}{4}\),

which can be shown to be less than the square's area of 2.

Assume Statement 2 alone. Again, for simplicity's sake, we will use \(\displaystyle CM = \frac{3}{2}\), so we can keep the area of the equilateral triangle as before, \(\displaystyle \frac{3\sqrt{3}}{4}\); this argument works regardless of the dimensions. The length of a diagonal of the square will as before be \(\displaystyle \frac{4}{3} \cdot \frac{3}{2} = 2\), and its area will again be 2.

Since the area of an equilateral triangle is wholly dependent on its common sidelength, comparison of the lengths of the sides is all that is necessary to determine which triangle, if either, has the greater area.

If we let \(\displaystyle x\) and \(\displaystyle y\:\) be the common sidelengths of \(\displaystyle \bigtriangleup ABC\) and \(\displaystyle \bigtriangleup DEF\), respectively, Statements 1 and 2, respectively, can be rewritten as:

Statement 1: \(\displaystyle x+y = 10\)

Statement 2: \(\displaystyle x - y = 2\). 

The question is now whether \(\displaystyle x < y\), \(\displaystyle x> y\), or \(\displaystyle x= y\).

Statement 1 alone is insufficient to determine which sidelength is greater; both \(\displaystyle (1,9)\) and \(\displaystyle (9,1)\) are easily seen to be solutions, with \(\displaystyle x < y\) in the first case, and \(\displaystyle x> y\) in the second. Consequently, it is possible for either triangle to have the greater sidelength and the greater area.

Statement 2 alone, however, is sufficient. If \(\displaystyle x - y = 2\), if follows that 

\(\displaystyle x - y + y = 2 + y\)

and 

\(\displaystyle x = y+2\)

This means that \(\displaystyle x > y\). \(\displaystyle \bigtriangleup ABC\) has the greater sidelength and, consequently, the greater area.

The area formula for an equilateral triangle is \(\displaystyle A =\frac{s^{2}\sqrt{3}}{4}\); that of a circle is \(\displaystyle A = \pi r^{2}\). Both will be used here.

Assume Statement 1 alone. If we let \(\displaystyle r\) be the radius of the circle, then, since the points on the circle include the midpoint of \(\displaystyle \overline{AB}\), the distance from center \(\displaystyle A\) to that midpoint is \(\displaystyle r\), and the length of \(\displaystyle \overline{AB}\) is \(\displaystyle 2r\). The area of the circle is \(\displaystyle \pi r^{2}\), and that of the triangle is 

\(\displaystyle \frac{(2r)^{2}\sqrt{3}}{4} = \frac{4r^{2}\sqrt{3}}{4} = r^{2} \sqrt{3}\).

Since \(\displaystyle \pi > \sqrt{3}\), the circle has the greater area.

Assume Statement 2 alone. For simplicity's sake, assume that the triangle has sidelength 1. Then its area is \(\displaystyle \frac{1^{2}\sqrt{3}}{4} = \frac{\sqrt{3}}{4}\). Since we only know that \(\displaystyle A\) is the center of the circle and \(\displaystyle C\) is outside it, it follows that the radius must be less than 1. This means that the area of the circle must be less than 

\(\displaystyle \pi \cdot 1^{2} = \pi\)

Since the area falls in the range \(\displaystyle \left ( 0, \pi\right )\), and \(\displaystyle 0 < \frac{\sqrt{3}}{4} < \pi\), we cannot tell whether the circle or the triangle has the greater area.