An equilateral triangle has three sides of equal length, so  and .

Assume Statement 1 alone. Since , then, by substitution, .

Assume Statement 2 alone. Since , it follows that , and again by substitution, .

Assume Statement 1 alone, and let  be the common perimeter of the triangles. Since an equilateral triangle has three sides of equal length,  and , so .

Assume Statement 2 alone, and let  be the common area of the triangles. Using the area formula for an equilateral triangle, we can note that:

 and ,  

so

.

Assume Statement 1 alone. Since all three sides of  are congruent - specifically,  - and , it follows by transitivity that . However, no information is given as to whether  has length greater than, equal to, or less than , so it cannot be determined which of  and , if either, is the longer. By a similar argument, Statement 2 yields insufficient information.

Now assume both statements are true.  and  are each congruent to one of the congruent sides of equilateral  and are therefore congruent to each other. However, the hypotenuse of a right triangle must be longer than both legs, so the hypotenuse of   is .  is also longer than any segment congruent to one of the legs, which includes all three sides of  - specificially,  is longer than .

Assume Statement 1 alone.  is equilateral, so . Also, by the Triangle Inequality, the sum of the lengths of two sides of a triangle must exceed the third, so . From Statement 1, , so by substitution, , and .

Statement 2 alone provides insufficient information. For example, assume  is an equilateral triangle with sidelength 9. If  is an equilateral triangle with sidelength 8, the conditions of the statement hold, and . However, if  is a right triangle in which , , and , the conditions of the statement still hold, but .

All sides of an equilateral triangle have the same measure, so we can let  be the common sidelength of , and  be that of .

Statement 1 can be rewritten as ; Statement 2 can be rewritten as . The equivalent question is whether we can determine which, if either, is greater,  or . 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   and , or vice versa. Therefore, we do not know for sure whether a side of  is longer than a side of  - specifically, which of  or  is longer.

All sides of an equilateral triangle have the same measure, so we can let  be the common sidelength of , and  be that of .

Statement 1 can be rewritten as ; Statement 2 can be rewritten as . The equivalent question is whether we can determine which, if either, is greater,  or .

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  or  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  or  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, , and  and  have the same sidelengths. Specifically,  and  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.

From there we can find the area.

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  is 2; since a Square is a rhombus, the area is one half the product of the lengths of the diagonals, or .

Now examine the figure below, which shows , its altitudes, and its circumscribed circle:

The three altitudes of an equilateral triangle meet at the center of the circumscribed circle, or circumcenter, so ; they also divide one another into segments such that one has twice the length of the other, so . Therefore, . Also, the six smaller triangles are all 30-60-90 by symmetry, so by the 30-60-90 Theorem, , and .

Therefore, the area of the triangle is 

,

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 , so we can keep the area of the equilateral triangle as before, ; this argument works regardless of the dimensions. The length of a diagonal of the square will as before be , and its area will again be 2.