The length of one side of the square is $\displaystyle 1,200 \div 4 = 300$ feet, or $\displaystyle 300 \div 3 = 100$ yards. Square this to get the area in square yards:

$\displaystyle A =100^{2} =10,000$ square yards.

The square, the pentagon, and the hexagon have a total of 15 sides, all of which are of equal length; the sum of the lengths is one mile, or 5,280 feet, so the length of one side of any of these polygons is 

$\displaystyle 5,280 \div 15 = 352$ feet.

The square has area equal to the square of this sidelength:

$\displaystyle 352^{2} = 123,904 \textup{ ft} ^{2}$

The regular perimeter has sidelength 60 centimeters and therefore perimeter $\displaystyle 5 \times 60 = 300$ centimeters. The square has as its sidelength $\displaystyle 300 \div 4 = 75$ centimeters and area $\displaystyle 75 ^{2} = 5,625$ square centimeters.

The areas of the squares are the squares of the sidelengths, so add the squares of the sidelengths. Since 1 foot is equal to 12 inches and 2 feet are equal to 24 inches, the sum of the areas is:

$\displaystyle 8^{2} + 12 ^{2} + 15 ^{2} + 20 ^{2} + 24^{2} + 25^{2}$

$\displaystyle = 64 + 144 + 225 + 400 + 576 + 625$

$\displaystyle =2,034$ square inches

As a square, $\displaystyle ABCD$ is also a rhombus. The area of a rhombus is half the product of the lengths of its diagonals, one of which is $\displaystyle \overline{AC }$. Since the diagonals are congruent, this is equal to half the square of $\displaystyle AC$ :

$\displaystyle \frac{1}{2} \left (AC \right) ^{2}$

$\displaystyle = \frac{1}{2} \left (2t + 5 \right) ^{2}$

$\displaystyle = \frac{1}{2}\left [ \left (2t \right) ^{2} + 2\left (2t \right) \cdot 5 + 5^{2} \right ]$

$\displaystyle = \frac{1}{2}\left ( 4t ^{2} +20t + 25 \right )$

$\displaystyle = 2t ^{2} +10t +\frac{ 25 }{2}$

This problem requires us to find the area of a square. Don't let the story behind it distract you, it is simply an area problem. Use the following equation to find our answer:

$\displaystyle Area=s^2=(15\:m)^2=225\:m^2$

$\displaystyle s$ is the length of one side of the square; in this case we are told that it is $\displaystyle 15\:m$, so we can solve accordingly!

Square $\displaystyle ABCD$ has area 49, so each of its sides has as its length the square root of 49, or 7. Each side of Square $\displaystyle WXYZ$ is therefore a hypotenuse of a right triangle with legs 1 and $\displaystyle 7-1 = 6$, so each sidelength, including $\displaystyle WX$, can be found using the Pythagorean Theorem:

$\displaystyle WX = \sqrt{(WB)^{2}+(BX)^{2}}$

$\displaystyle WX = \sqrt{1^{2}+6^{2}} = \sqrt{1 +36} = \sqrt{37}$

The square of this, which is 37, is the area of Square $\displaystyle WXYZ$.

Square $\displaystyle WXYZ$ has area 25, so each side has length the square root of 25, or 5. 

Specifically, $\displaystyle WX = 5$, and, as given, $\displaystyle BW = 1$.

Since $\displaystyle \bigtriangleup WBX$ is a right triangle with hypotenuse $\displaystyle \overline{WX}$ and legs $\displaystyle \overline{BW}$ and $\displaystyle \overline{BX}$, $\displaystyle BX$ can be found using the Pythagorean Theorem:

$\displaystyle BX = \sqrt{(WX)^{2}-(BW)^{2}}$

$\displaystyle = \sqrt{5^{2}-1^{2}}$

$\displaystyle = \sqrt{25-1}$

$\displaystyle = \sqrt{24}$

$\displaystyle =\sqrt{4} \cdot \sqrt{6}$

$\displaystyle = 2\sqrt{6}$

The area of $\displaystyle \bigtriangleup WBX$ is 

$\displaystyle \frac{1}{2} \cdot BW \cdot BX = \frac{1}{2} \cdot 1 \cdot 2 \sqrt{6} = \sqrt{6}$

Since all four triangles, by symmetry, are congruent, all have this area. the area of Square $\displaystyle ABCD$ is the area of Square $\displaystyle WXYZ$ plus the areas of the four triangles, or $\displaystyle 25 + 4\sqrt{6}$.

To make this easier, assume that $\displaystyle BX = 13$ and $\displaystyle BW=CX = 2$ - the reasoning generalizes. Then Square $\displaystyle ABCD$ has sidelength 15 and area $\displaystyle 15^{2}= 225$. The sidelength of Square $\displaystyle WXYZ$, each side being a hypotenuse of a right triangle with legs 2 and 13, is 

$\displaystyle \sqrt{2^{2}+13^{2}}= \sqrt{4+169} = \sqrt{173}$.

The square of this, 173, is the area of Square $\displaystyle WXYZ$.

The ratio is $\displaystyle 225:173$.

To make this easier, assume that $\displaystyle BX = 7$ and $\displaystyle BW = XC = 1$; the results generalize. 

Each side of Square $\displaystyle ABCD$ has length 8, so the area of Square $\displaystyle ABCD$ is 64. 

Each of the four right triangles has legs 7 and 1, so each has area $\displaystyle \frac{1}{2} \cdot 1 \cdot 7 = 3\frac{1}{2}$; Square $\displaystyle WXYZ$ has area four times this subtracted from the area of Square $\displaystyle ABCD$, or

$\displaystyle 64 - 4 \cdot 3 \frac{1}{2} = 50$.

The area of Square $\displaystyle WXYZ$ is

$\displaystyle \frac{50}{64} \times 100 \% = 78 \frac{1}{8} \%$

of that of Square $\displaystyle ABCD$.

Of the five choices, 80% comes closest.