First, we need the radius \(\displaystyle r\) of the circle, which can be determined from the area of a circle formula by setting \(\displaystyle A = 4\):

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

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

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

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

\(\displaystyle r = \sqrt{\frac{4 }{\pi} }\)

Simplifying the expression by splitting the radicand and rationalizing the denominator:

\(\displaystyle r = \frac{\sqrt{4}}{\sqrt{\pi}}\)

\(\displaystyle r = \frac{2}{\sqrt{\pi}}\)

\(\displaystyle r = \frac{2 \cdot \sqrt{\pi}}{\sqrt{\pi} \cdot \sqrt{\pi}}\)

\(\displaystyle r= \frac{2 \sqrt{\pi}}{\pi}\)

The circumference of the circle is \(\displaystyle 2 \pi\) multiplied by the radius, or

\(\displaystyle C = 2 \pi r\)

\(\displaystyle C = 2 \pi \cdot \frac{2 \sqrt{\pi}}{\pi} = 4 \sqrt{\pi}\)

This is also the perimeter of the square, so the length of each side is one fourth of this perimeter, or

\(\displaystyle \frac{ 4 \sqrt{\pi}}{4} = \sqrt{\pi}\)

The area of the square is the square of this common sidelength, or 

\(\displaystyle \left ( \sqrt{\pi} \right )^{2} = \pi\).

A circle with diameter 10 has radius half this, or 5. The area of the circle can be found using the formula

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

setting \(\displaystyle r = 5\):

\(\displaystyle A = \pi \cdot 5^{2} = 25 \pi\).

The square also has this area. The length of one side of this square is equal to the square root of this, which is

\(\displaystyle s = \sqrt{A} = \sqrt{ 25 \pi}\)

Simplify this by breaking the radicand, as follows:

\(\displaystyle S = \sqrt{ 25 \pi} = \sqrt{ 25} \cdot \sqrt{ \pi} = 5\sqrt{ \pi}\)

The area of a right triangle is equal to half the product of the lengths of its legs. Since we know the lengths of one leg and the hypotenuse, we can calculate the length of the other leg using the Pythagorean Theorem. We can use this form:

\(\displaystyle b = \sqrt{ c^{2} - a ^{2}}\)

Setting \(\displaystyle c\) and \(\displaystyle a\) equal to the lengths of the hypotenuse and the known leg - 13 and 5, respectively:

\(\displaystyle b = \sqrt{ 13^{2} - 5 ^{2}} = \sqrt{169 - 25} = \sqrt{144} = 12\)

The area of the above triangle is

\(\displaystyle A = \frac{1}{2} ab = \frac{1}{2} \cdot 5 \cdot 12 = 30\)

The square also has this area. The length of one side of this square is equal to the square root of this, which is \(\displaystyle \sqrt{30 }\).

A rectangle has perimeter \(\displaystyle P = 2L + 2W\), \(\displaystyle L\) the length and \(\displaystyle W\) the width. Substitute \(\displaystyle L= x + 16\) and \(\displaystyle W = 2x-7\) in the perimeter formula, and simplify.

\(\displaystyle P = 2L + 2W\)

\(\displaystyle = 2 \left ( x+16\right ) + 2 \left ( 2x-7\right )\)

\(\displaystyle = 2 \cdot x+2 \cdot 16 +2 \cdot 2x-2 \cdot 7\)

\(\displaystyle = 2x+32 +4x-14\)

\(\displaystyle = 6x+18\)

In inches, the perimeter of the rectangle can be calculated by substituting \(\displaystyle L = 30, W = 25\) in the following formula:

\(\displaystyle P = 2 (L + W)\)

\(\displaystyle P = 2 (30 + 25) = 2\times 55 = 110\)

The perimeter is 110 inches.

Now divide by 12 to convert to feet:

\(\displaystyle 110 \div 12 = 9 \textrm{ R }2\)

This makes the perimeter 9 feet 2 inches, which is between 9 feet and 10 feet.

Let:

\(\displaystyle Length=x\Rightarrow Width=x-3\)

The perimeter of a rectangle is \(\displaystyle P=2L+2W\), where \(\displaystyle L\) is the length and \(\displaystyle W\) is the width of the rectangle. The perimeter is known so we can set up an equation in terms of \(\displaystyle x\) and solve it:

\(\displaystyle P=2L+2W=2(x)+2(x-3)=36\)

\(\displaystyle \Rightarrow2x+2x-6=36\)

\(\displaystyle \Rightarrow 4x-6=36\)

\(\displaystyle \Rightarrow 4x=42\)

\(\displaystyle \Rightarrow x=10.5\)

So we can get:

\(\displaystyle Length=x=10.5\) inches

\(\displaystyle Width=x-3=10.5-3=7.5\) inches

The perimeter of a rectangle is \(\displaystyle P=2L+2W\), where \(\displaystyle L\) is the length and \(\displaystyle W\) is the width of the rectangle. In order to find the perimeter we can substitute the \(\displaystyle L=t+1\) and \(\displaystyle W=t-1\) in the perimeter formula:

\(\displaystyle P=2L+2W\)

\(\displaystyle =2(t+1)+2(t-1)\)

\(\displaystyle =2t+2+2t-2\)

\(\displaystyle =4t\)

The length of the rectangle is known, so we can find the width in terms of \(\displaystyle x\):

 \(\displaystyle Width=Length-3=(2x+4)-3=2x+1\)

The perimeter of a rectangle is \(\displaystyle P=2L+2W\), where \(\displaystyle L\) is the length and \(\displaystyle W\) is the width of the rectangle.

In order to find the perimeter we can substitute the \(\displaystyle L=2x+4\) and \(\displaystyle W=2x+1\) in the perimeter formula:

\(\displaystyle P=2L+2W\)

\(\displaystyle =2(2x+4)+2(2x+1)\)

\(\displaystyle =4x+8+4x+2\)

\(\displaystyle =8x+10\)

Substitute \(\displaystyle t=5\) to get \(\displaystyle L\) and \(\displaystyle W\):

\(\displaystyle L=t^2=5^2=25\)

\(\displaystyle W=5t=5\times 4=20\)

The perimeter of a rectangle is \(\displaystyle P=2L+2W\) , where \(\displaystyle L\) is the length and \(\displaystyle W\) is the width of the rectangle. So we have:

\(\displaystyle P=2L+2W=2\times 25+2\times20=50+40=90\) inches

Now we should divide the perimeter by 12 in order to convert to feet:

\(\displaystyle Perimeter=90\div 12=7.5\) feet

So the perimeter is 7 feet and 6 inches, which is between 7 and 8 feet.

The perimeter of a rectangle is twice the sum of its length and its width; a rectangle with dimensions 55 inches and 15 inches has perimeter 

\(\displaystyle 2 (55+15) = 140\) inches.

All of the polygons in the choices are regular - that is, all have congruent sides - and all have sidelength two feet, or 24 inches, so we divide 140 by 24 to determine how many sides such a polygon would need to have a perimeter equal to the rectangle. However, 

\(\displaystyle 140 \div 24 = 5 \frac{5}{6}\),

so there cannot be a regular polygon with these characteristics. All of the choices fail, so the correct response is that none are correct.