We begin by recalling the formulas for the perimeter and area of a square respectively.

\(\displaystyle P=4s\)

\(\displaystyle A=s^2\)

Using these formulas and the fact that the perimeter is half the area, we can create an equation.

\(\displaystyle 4s=\frac{1}{2}(s^2)\)

We can multiply both sides by 2 to eliminate the fraction.

\(\displaystyle 8s=s^2\)

To get one side of the equation equal to zero, we will move everything to the right side.

\(\displaystyle 0=s^2-8s\)

Next we can factor.

\(\displaystyle 0=s(s-8)\)

Setting each factor equal to zero provides two potential solutions.

\(\displaystyle s=0\)       or        \(\displaystyle s-8=0\)

                             \(\displaystyle s=8\)

However, since a square cannot have a side of length 0, 8 is our only answer.

\(\displaystyle Area = Length \times Length\)

\(\displaystyle 100 = (Length)^2\)

\(\displaystyle Length = \sqrt{100}=10\)

If diagonal \(\displaystyle \overline{SU}\) of Square \(\displaystyle SQUA\) is constructed, then \(\displaystyle \bigtriangleup SQU\) is a 45-45-90 triangle with hypotenuse \(\displaystyle SU = \sqrt{2x}\). By the 45-45-90 Theorem, the sidelength \(\displaystyle SQ\) can be calculated as follows:

\(\displaystyle SQ = \frac{SU}{\sqrt{2}} = \frac{\sqrt{2x}}{\sqrt{2}} = \sqrt{\frac{2x}{2}} = \sqrt{x}\).

The diameter of a circle with circumference 20 is

\(\displaystyle \frac{20}{\pi } \approx \frac{20 }{3.1416} \approx 6.3662\)

The diameter of a circle that circumscribes a square is equal to the length of the diagonals of the square.

If diagonal \(\displaystyle \overline{SU}\) of Square \(\displaystyle SQUA\) is constructed, then \(\displaystyle \bigtriangleup SQU\) is a 45-45-90 triangle with hypotenuse approximately 6.3662. By the 45-45-90 Theorem, divide this by \(\displaystyle \sqrt{2} \approx 1.4142\) to get the sidelength of the square:

\(\displaystyle 6.3662 \div 1.4142 \approx 4.5\)

The area of Square \(\displaystyle SQUA\) is the square of sidelength \(\displaystyle SQ\), or \(\displaystyle (SQ)^{2}\).

The area of Rectangle \(\displaystyle RECT\) is \(\displaystyle RE \cdot EC\). Rectangle \(\displaystyle RECT\) has area 90% of that of Square \(\displaystyle SQUA\), which is \(\displaystyle \frac{9}{10} (SQ)^{2}\);  \(\displaystyle RE\) is 80% of \(\displaystyle SQ\), so \(\displaystyle RE = \frac{4}{5}SQ\). We can set up the following equation: 

\(\displaystyle \frac{9}{10} (SQ)^{2} = RE \cdot EC\)

\(\displaystyle \frac{9}{10} (SQ)^{2} = \frac{4}{5} SQ \cdot EC\)

\(\displaystyle \frac{9}{10} \cdot SQ = \frac{4}{5} \cdot EC\)

\(\displaystyle \frac{10} {9} \cdot \frac{9}{10} \cdot SQ = \frac{10} {9} \cdot \frac{4}{5} \cdot EC\)

\(\displaystyle SQ = \frac{8} {9} \cdot EC\)

As a percent, \(\displaystyle \frac{8} {9}\) of \(\displaystyle EC\) is \(\displaystyle \frac{8} {9} \times 100 \%= 88 \frac{8} {9} \%\)

The area of the square was originally 

\(\displaystyle A = s^{2}\), 

\(\displaystyle s\) being the sidelength.

Reducing the area by 12% means that the new area is 88% of the original area, or \(\displaystyle 0.88A\); the square root of this is the new sidelength, so

\(\displaystyle 0.88A = 0.88s^{2}\)

\(\displaystyle \sqrt{0.88A} = \sqrt{0.88s^{2}} = \sqrt{0.88} \cdot \sqrt{s^{2}} \approx 0.94 s\)

Each side of the new square will measure 94% of the length of the old measure - a reduction by 6%.

The diameter of a circle that is inscribed inside a square is equal to its sidelength \(\displaystyle SQ\), so all we need to do is find the diameter of the circle - which is circumference 16 divided by \(\displaystyle \pi\):

\(\displaystyle \frac{16}{\pi } \approx \frac{16}{3.1416 } \approx 5.1\).

Let \(\displaystyle s = CD\), the sidelength shared by the square and the equilateral triangle.

The area of \(\displaystyle \bigtriangleup CDE\) is

\(\displaystyle \frac{s^{2}\sqrt{3}}{4} = \frac{\sqrt{3}}{4} \cdot s^{2} \approx \frac{1.7321}{4} \cdot s^{2} \approx 0.4330 s^{2}\)

The area of Square \(\displaystyle ABCD\) is \(\displaystyle s^{2}\).

By symmetry, \(\displaystyle \overline{EF}\) bisects the portion of the square not in the triangle, so the area of Quadrilateral \(\displaystyle BCEF\) is half the difference of those of the square and the triangle. Since the area of Quadrilateral \(\displaystyle BCEF\)is 100, we can set up an equation:

\(\displaystyle \frac{1}{2}\left (s ^{2} - 0.4330 s ^{2} \right ) = 100\)

\(\displaystyle \frac{1}{2}\left ( 0.5670 s ^{2} \right ) = 100\)

\(\displaystyle 0.2835 s ^{2} = 100\)

\(\displaystyle s ^{2} \approx 100 \div 0.2835\)

\(\displaystyle s ^{2} \approx 352.73\)

\(\displaystyle s \approx\sqrt{ 352.73} \approx 18.7\)

Of the five choices, 20 comes closest.

Recall how to find the perimeter of a square:

\(\displaystyle \text{Perimeter}=4(\text{side length})\)

By dividing both sides by \(\displaystyle 4\), we can write the following:

\(\displaystyle \text{Side length}=\frac{\text{Perimeter}}{4}\)

For the square in question,

\(\displaystyle \text{Side length}=\frac{444}{4}\)

\(\displaystyle \text{Side length}=111\)

Recall how to find the perimeter of a square:

\(\displaystyle \text{Perimeter}=4(\text{side length})\)

By dividing both sides by \(\displaystyle 4\), we can write the following:

\(\displaystyle \text{Side length}=\frac{\text{Perimeter}}{4}\)

For the square in question,

\(\displaystyle \text{Side length}=\frac{96}{4}\)

\(\displaystyle \text{Side length}=24\)