Statement 1 is unhelpful in that it gives the volume, not the surface area, of the tank. The volume of a cylinder depends on two independent values, the height and the area of a base; neither can be determined, so neither can the surface area.

From Statement 2 alone, we can find the surface area. One gallon of paint covers 350 square feet, so, since \(\displaystyle 10 \frac{3}{4}\) gallons of this paint will cover about

\(\displaystyle 10 \frac{3}{4} \cdot 350 \approx 3,800\) square feet, the surface area of the tank.

The surface area of the cylinder can be calculated from radius \(\displaystyle r\) and height \(\displaystyle h\) using the formula:

\(\displaystyle a = 2 \pi rh + 2 \pi r^{2}\).

Let the radius of the sphere be \(\displaystyle R\). Then its surface area can be calculated to be 

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

From Statement 1 alone, \(\displaystyle h = R\), so the surface area of the cylinder can be expressed as

\(\displaystyle a = 2 \pi r R + 2 \pi r^{2}\).

We cannot compare this to the surface area of the sphere without knowing anything about the radius of a base. 

Likewise, from Statement 2 alone, we know that \(\displaystyle r > R\), but without knowing anything about the height, we cannot compare the surface areas.

Now assume both statements. Again, from Statement 1,

\(\displaystyle a = 2 \pi r R + 2 \pi r^{2}\).

Since, from Statement 2, \(\displaystyle r > R\), if follows that

\(\displaystyle 2 \pi R \cdot r >2 \pi R \cdot R\), or, \(\displaystyle 2 \pi r R > 2 \pi R ^{2}\).

It also follows that

\(\displaystyle r^{2} > R^{2}\)

and

\(\displaystyle 2 \pi r^{2} > 2 \pi R^{2}\)

Therefore, we can add both sides of the inequalities:

\(\displaystyle 2 \pi r R + 2 \pi r^{2} > 2 \pi R^{2}+ 2 \pi R ^{2}\)

\(\displaystyle 2 \pi r R + 2 \pi r^{2} > 4 \pi R^{2}\)

and 

\(\displaystyle a > A\),

so the cylinder has the greater surface area of the two solids.

Assume Statement 1 alone. Since \(\displaystyle \widehat{XY}\) has length one fourth the circumference of a base, then each base has circumference \(\displaystyle 5 \pi \cdot 4 = 20 \pi\), and radius \(\displaystyle r = C \div \left (2\pi \right )= 20 \pi \div \left (2 \pi \right )= 10\). It follows that each base has area \(\displaystyle A = \pi r^{2}= \pi \cdot 10^{2}= 100 \pi\)

Also, the diameter is \(\displaystyle C \div \pi = 20 \pi \div \pi = 20\); it is also the length of each edge, and it is therefore the height. The lateral area is the product of height 20 and circumference \(\displaystyle 20 \pi\), or \(\displaystyle 20 \cdot 20 \pi = 400 \pi\).

The surface area can now be calculated as the sum of the areas:

\(\displaystyle 400 \pi + 100 \pi+ 100 \pi= 600 \pi\).

Statement 2 is actually a redundant statement; since each base is inscribed inside a square, it already follows that \(\displaystyle \widehat{XY}\) is one fourth of a circle - that is, a \(\displaystyle 90^{\circ }\) arc.

The surface area of a cylinder, given height \(\displaystyle h\) and radius of the bases \(\displaystyle r\), is given by the formula

\(\displaystyle A =2 \pi r (r+h)\)

The surface area of a cube, given the length \(\displaystyle s\) of each edge, is given by the formula

\(\displaystyle A ' = 6s^{2}\).

Assume Statement 1 alone. Then \(\displaystyle h = s\) and, since the diameter of a base is \(\displaystyle s\), the radius is half this, or \(\displaystyle r = \frac{1}{2}s\). The surface area of the cylinder is, in terms of \(\displaystyle s\), equal to 

\(\displaystyle A =2 \pi s \left ( s+ \frac{1}{2}s \right )= 2 \pi s \cdot \frac{3}{2}s = 3 \pi s^{2}\).

Since \(\displaystyle 3 \pi > 6\), the cylinder has the greater area regardless of the actual measurements.

Assume Statement 2 alone. The cube has six faces with area \(\displaystyle 4r^{2}\), so its surface area is six times this, or \(\displaystyle 6 \cdot 4r^{2} = 24 r^{2}\). The surface area of the cylinder is \(\displaystyle A =2 \pi r (r+h) = 2 \pi r^{2}+ 2 \pi rh\); however, without knowing anything aobout the height of the cylinder, we cannot compare the two surface areas.

The surface area of the cylinder can be calculated from radius \(\displaystyle r\) and height \(\displaystyle h\) using the formula:

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

We show that both statements together provide insufficient information by first noting that if the two cylinders have the same height, and their bases have the same radius, their surface areas will be the same.

Now we explore the case in which Cylinder 1 has height 6 and bases with radius 8, and Cylinder 2 has height 8 and bases of radius 6.

The surface area of Cylinder 1 is

\(\displaystyle A = 2 \pi \cdot 8 \cdot 6 + 2 \pi \cdot 8^{2}\)

\(\displaystyle = 2 \pi \cdot 8 \cdot 6 + 2 \pi \cdot 8 \cdot 8\)

\(\displaystyle =96 \pi + 128 \pi\)

\(\displaystyle =224 \pi\)

The surface area of Cylinder 2 is

\(\displaystyle A = 2 \pi \cdot 6 \cdot 8 + 2 \pi \cdot 6^{2}\)

\(\displaystyle = 2 \pi \cdot 6 \cdot 8 + 2 \pi \cdot 6 \cdot 6\)

\(\displaystyle =96 \pi + 72\pi\)

\(\displaystyle =168\pi\)

In this scenario, Cylinder 1 has the greater surface area.

The surface area of the cylinder can be calculated from radius \(\displaystyle r\) and height \(\displaystyle h\) using the formula:

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

It can be seen from the diagram that if we let \(\displaystyle s\) be the length of one edge of the cube, then \(\displaystyle h = s\) and \(\displaystyle r = \frac{1}{2}s\). The surface area formula can be rewritten as

\(\displaystyle A = 2 \pi \cdot \frac{1}{2}s \cdot s + 2 \pi \left ( \frac{1}{2}s \right ) ^{2}\)

\(\displaystyle A = \pi s ^{2}+ 2 \pi \cdot \frac{1}{4}s ^{2}\)

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

\(\displaystyle A = \frac{3}{2} \pi s ^{2}\)

Subsequently, the length of one edge of the cube is sufficient to calculate the surface area of the cylinder.

From Statement 1 alone, the length of an edge of the cube can be calculated using the volume formula:

\(\displaystyle s^{3} = V = 729\)

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

From Statement 2 alone, the length of an edge of the cube can be calculated using the surface area formula:

\(\displaystyle 6s^{2} = A = 486\)

\(\displaystyle s^{2}= 486 \div 6 = 81\)

\(\displaystyle s = \sqrt{81}= 9\)

Since \(\displaystyle s\) can be calculated from either statement alone, so can the surface area of the cylinder:

\(\displaystyle A = \frac{3}{2} \pi s ^{2} = \frac{3}{2} \pi \cdot 9 ^{2} = \frac{243}{2} \pi\)

Statement 1 alone provides insufficient information. The surface area of a cylinder with bases of radius \(\displaystyle r\) and height \(\displaystyle h\) is 

\(\displaystyle a = 2 \pi r ^{2}+ 2 \pi r h\).

The surface area of a regular triangular prism with height \(\displaystyle H\) and whose (equilateral triangular) bases have common sidelength \(\displaystyle S\) - and perimeter \(\displaystyle 3S\) - is

\(\displaystyle A = 2 \cdot \frac{S^{2}\sqrt{3}}{4} + 3S \cdot H\), or

\(\displaystyle A = \frac{S^{2}\sqrt{3}}{2} + 3S H\).

Statement 1 alone tells us that \(\displaystyle S = \frac{1}{3} \cdot 2 \pi r\) , or \(\displaystyle S = \frac{2 \pi }{3} r\). However, without any information about the heights, we cannot compare \(\displaystyle a\) and \(\displaystyle A\). Similarly, Statement 2 alone tells us that \(\displaystyle h = H\), but nothing about \(\displaystyle r\) or \(\displaystyle S\).

However, if we combine what we know from Statement 2 with the information from Statement 1, we can answer the question. The surface area of a cylinder or a prism is equal to its lateral area plus the areas of its two congruent bases.

The lateral area of a cylinder or a prism is the height multiplied by the perimeter or circumference of a base. From Statement 1, the circumference of the bases of the cylinder is equal to the perimeter of the bases of the prism, and from Statement 2, the heights are equal. It follows that the lateral areas are equal, and that the figure with the bases that are greater in area has the greater surface area. 

For simplicity's sake, we will assume that the circumference of the base of the cylinder is \(\displaystyle 6 \pi\), for reasons that will be apparent later; this reasoning works for any circumference. The radius of this base is \(\displaystyle 6 \pi \div 2 \pi = 3\), and the area is \(\displaystyle \pi \cdot 3^{2} = 9 \pi\). The perimeter of the triangular base of the prism is also \(\displaystyle 6 \pi\), so its sidelength is \(\displaystyle 2 \pi\), making its area \(\displaystyle \frac{(2 \pi)^{2}\sqrt{3}}{2} = 2 \pi^{2}\sqrt{3}\), which is greater than \(\displaystyle 9 \pi\). This makes the prism the greater in surface area as well.