The figure described is the triangular pyramid, or tetrahedron, in the coordinate three-space below.

The base of the pyramid can be seen as a triangle with the three known coordinates \(\displaystyle (0,0,0)\), \(\displaystyle (10,0,0)\),  and \(\displaystyle (0,14,0)\), and the area of its base is half the product of the lengths of its legs, which is 

\(\displaystyle b = \frac{1}{2} \cdot 10 \cdot 14 = 70\).

The volume of the pyramid is one third the product of the area of its base, which is 70, and its height, which is the perpendicular distance from the unknown point to the base. Since the base is entirely within the \(\displaystyle xy\)-plane, this distance is the \(\displaystyle z\)-coordinate, which is \(\displaystyle C\). Therefore, the only thing that is needed to determine the volume of the pyramid is \(\displaystyle C\).

Neither statement alone is enough to gain this information. However, if both statements are assumed true, we can subtract each side of the latter equation from the former as follows:

\(\displaystyle A+B+C= 50\)

\(\displaystyle \underline{A+B-C= 30}\)

              \(\displaystyle 2C = 20\)

                 \(\displaystyle C= 10\)

The value of \(\displaystyle C\) is obtained and the volume of the pyramid can be calculated.

The volume of a pyramid is one third the product of its height and the area its base. 

Pyramid 1 is shown below:

The base of the pyramid is on the \(\displaystyle xy\)-plane, so the height of the pyramid  is the perpendicular distance from apex \(\displaystyle (D,E,F)\) to this plane; this is the \(\displaystyle z\)-coordinate, \(\displaystyle F\). The base of the pyramid is a square of sidelength 10, so its area is the square of 10, or 100. This makes the volume of Pyramid 1 

\(\displaystyle V_{1} = \frac{1}{3} \cdot 100 \cdot F = \frac{100}{3}F = \left ( 33 \frac{1}{3}\right )F\)

Similarly, the volume of Pyramid 2 is

\(\displaystyle V_{2} = \frac{1}{3} \cdot 12 ^{2}\cdot J= \frac{1}{3} \cdot 144 \cdot J = 36 J\)

Therefore, the problem asks us to determine which of \(\displaystyle \left ( 33 \frac{1}{3}\right )F\) and \(\displaystyle 36J\) is the greater.

We show that the two statements together provide insufficient information by examining two scenarios.

Case 1: \(\displaystyle F =12, J = 10\)

\(\displaystyle V_{1} = \left ( 33 \frac{1}{3}\right )F = \left ( 33 \frac{1}{3}\right ) \cdot 12 = 400\)

\(\displaystyle V_{2} = 36 J = 36 \cdot 10 = 360\)

\(\displaystyle V_{1} > V_{2}\)

Case 2: \(\displaystyle F =10 \frac{1}{6}, J = 10\)

\(\displaystyle V_{1} = \left ( 33 \frac{1}{3}\right )F = \left ( 33 \frac{1}{3}\right ) \cdot 10 \frac{1}{6} = 338 \frac{8}{9}\)

\(\displaystyle V_{2} = 36 J = 36 \cdot 10 = 360\)

\(\displaystyle V_{1}< V_{2}\)

Each case fits the conditions of the two statements and the main body of the question; in one case, Pyramid 1 has the greater volume and in the other case, Pyramid 2 does.

The volume of any pyramid is one third the product of its height and the area of its base. Since both pyramids have the same height, their volumes are equal if and only if the areas of their bases are equal as well. Also, since both pyramids are regular, their bases are regular polygons.

Statement 1 alone is insufficient. It is possible for two regular polygons to have the same perimeter but different areas. For example, an equilateral triangle with perimeter 60 has sidelength 20 and, consequently, area

\(\displaystyle A = \frac{s^{2}\sqrt{3}}{4} = \frac{20^{2}\sqrt{3}}{4} = 100\sqrt{3}\)

A square with perimeter 60 has sidelength 15, and, consequently, area equal to the square of this, 225.

Statement 2 only gives the number of sides, and no information about their measures.

Now assume both statements are true. If the common perimeter is \(\displaystyle P\), then the length of one side of the base of Pyramid 1, it being a regular hexagon, is \(\displaystyle \frac{1}{6}P\), likewise, the length of one side of the base of Pyramid 2 is \(\displaystyle \frac{1}{8}P\). The area of each in terms of \(\displaystyle P\) can be calculated, and the two can be compared.

The volume of a pyramid is one third the product of its height and the area its base. The two pyramids have the same base, so the pyramid with the greater height will have the greater volume (and if their heights are equal, their volumes are equal).

Pyramid 1 is shown below:

The base of the pyramid is on the \(\displaystyle xy\)-plane, so the height of the pyramid  is the perpendicular distance from apex \(\displaystyle (D,E,F)\) to this plane; this is the \(\displaystyle z\)-coordinate, \(\displaystyle F\). Pyramid 2 has the same base and apex \(\displaystyle (G,H,J)\), so its height is the \(\displaystyle z\)-coordinate of its apex, \(\displaystyle J\). 

Therefore, whichever is greater - \(\displaystyle F\) or \(\displaystyle J\) - determines which pyramid has the greater volume. However, the two statements to not give this information.

The volume of the pyramid is one third the product of height \(\displaystyle DE\) and the area of its base, which in turn, since here it is a right triangle, is half the product of the lengths \(\displaystyle HE\) and \(\displaystyle HG\) of its legs. Since the prism in the figure is a cube, the three lengths are equal, so we can set each to \(\displaystyle s\). The volume of the pyramid is

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

\(\displaystyle V = \frac{1}{6} s^{3}\)

Therefore, knowing the length of one edge of the cube is sufficient to determine the volume of the pyramid.

Assume Statement 1 alone. If the volume of the circumscribing sphere is known to be \(\displaystyle 972 \pi\), the radius can be calculated as follows:

\(\displaystyle \frac{4}{3} \pi r^{3} = V\)

\(\displaystyle \frac{4}{3} \pi r^{3} =972 \pi\)

\(\displaystyle \frac{3}{4\pi} \cdot \frac{4}{3} \pi r^{3} =\frac{3}{4\pi} \cdot 972 \pi\)

\(\displaystyle r^{3} =729\)

\(\displaystyle r = \sqrt[3]{729}= 9\)

The diameter, which is twice this, or 18, is the length of a diagonal of the cube. By the three-dimensional extension of the Pythagorean Theorem, the relationship of this length to the side length of the cube is

\(\displaystyle s^{2}+s^{2}+s^{2} = d^{2}\), or

\(\displaystyle 3s^{2} = d^{2}\)

so

\(\displaystyle 3s^{2} = 18^{2}\)

\(\displaystyle s^{2} = \frac{18^{2}}{3} = 108\)

\(\displaystyle s = \sqrt{108} = 6 \sqrt{3}\),

Assume Statement 2 alone. If the surface area of the inscribed sphere is known to be \(\displaystyle 108 \pi\), then its radius can be calculated as follows:

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

\(\displaystyle 4 \pi r^{2} = 108 \pi\)

\(\displaystyle \frac{4 \pi r^{2} }{4 \pi }=\frac{ 108 \pi}{4 \pi }\)

\(\displaystyle r^{2} = 27\)

\(\displaystyle r = \sqrt{27} = 3\sqrt{3}\).

The diameter of the inscribed sphere, which is twice this, or \(\displaystyle 6 \sqrt{3}\), is equal to the length \(\displaystyle s\) of one edge of the cube.

Either statement alone gives us the length of one side of the cube, which is enough to allow the volume of the pyramid to be calculated.

 \(\displaystyle \bigtriangleup ADC\), \(\displaystyle \bigtriangleup ADB\), and \(\displaystyle \bigtriangleup BDC\), all being right triangles with the same leg lengths, are congruent, and, consequently, their diagonals are congruent, making \(\displaystyle \bigtriangleup ABC\) equilateral. 

Assume Statement 1 alone. The sidelength of equilateral \(\displaystyle \bigtriangleup ABC\) is one third of its perimeter of 60, or 20. This is also the common length of the hypotenuses of isoseles right triangles  \(\displaystyle \bigtriangleup ADC\), \(\displaystyle \bigtriangleup ADB\), and \(\displaystyle \bigtriangleup BDC\), so by the 45-45-90 Theorem, each leg length can be computed by dividing 20 by \(\displaystyle \sqrt{2}\):

\(\displaystyle \frac{20}{\sqrt{2}} = \frac{20\cdot \sqrt{2}}{\sqrt{2}\cdot \sqrt{2}} = \frac{20 \sqrt{2}}{2} = 10 \sqrt{2}\)

Assume Statement 2 alone. \(\displaystyle \bigtriangleup ADC\) has area 100; since the area of a right triangle is half the product of the lengths of its legs, we can find the common leg length using this formula:

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

\(\displaystyle \frac{1}{2} s^{2} = 100\)

\(\displaystyle s^{2} = 200\)

\(\displaystyle s =\sqrt{200} = \sqrt{100}\cdot \sqrt{2}= 10 \sqrt{2}\)

This is the common length of the legs of the three right triangles; by the 45-45-90 Theorem, each hypotenuse - and each side of equilateral \(\displaystyle \bigtriangleup ADC\) - is \(\displaystyle \sqrt{2}\) times this, or \(\displaystyle 10 \sqrt{2} \cdot \sqrt{2} = 10 \cdot 2 = 20\).

Therefore, from either statement alone, the lengths of all sides of the tetrahedron can be found, and the area formulas for a right triangle and an equilateral can be applied to find the areas of all four faces.

The surface area of a tetrahedron is found by \(\displaystyle SA=a^2 \sqrt{3}\) where \(\displaystyle a\) represents the edge value.

Situation 1: We're given our \(\displaystyle a\) value so we just need to plug it into our equation.

\(\displaystyle SA=(3\sqrt{2})^2 \sqrt{3}\) 

\(\displaystyle SA=18\sqrt{3}\)

Situation 2: We use the given volume to solve for the length of the edge.

\(\displaystyle V=\frac{a^3}{6\sqrt{2}}=9\)

\(\displaystyle a=3\sqrt{2}\)

Now that we have a length, we can plug it into the equation for the surface area: 

\(\displaystyle SA=(3\sqrt{2})^2 \sqrt{3}=18\sqrt{2}\)

Thus, each statement alone is sufficient to answer the question.