Since the ratio is the same regardless of the sidelengths, then for simplicity's sake, assume the sidelength of Square B is 7. The area of Square B is therefore the square of this, or 49.

Then the sidelength of Square A is three-sevenths of 7, or 3. Its area is the square of 3, or 9. 

The ratio of the area of Square B to that of Square A is therefore 49 to 9.

The areas of the squares are:

\(\displaystyle 1^{2} = 1\) square foot

\(\displaystyle 2^{2} = 4\) square feet

\(\displaystyle 3^{2} = 9\) square feet

\(\displaystyle 4^{2}= 16\) square feet

\(\displaystyle 5^{2}= 25\) square feet

Therefore, we are comparing the mean and the median of the set \(\displaystyle \left \{ 1, 4, 9, 16, 25\right \}\).

The mean of this set is the sum divided by 5:

 \(\displaystyle (1 + 4 + 9 + 16 + 25) \div 5 = 55 \div 5 = 11\) 

The median is the middle element after arrangement in ascending order, which is 9.

This makes (A), the mean, greater.

The areas of the squares are:

\(\displaystyle 100 \cdot 100 = 10,000\) square centimeters (one meter being 100 centimeters)

\(\displaystyle 120 \cdot 120 = 14,400\) square centimeters

\(\displaystyle 140 \cdot 140 = 19,600\) square centimeters

\(\displaystyle 140 \cdot 140 = 19,600\) square centimeters

The mean of these four areas is their sum divided by four:

\(\displaystyle (10,000+14,400+19,600+19,600) \div 4\)

\(\displaystyle = 63,600 \div 4 =15,900\) square centimeters.

The median is the mean of the two middle values, or

\(\displaystyle ( 14,400+19,600 ) \div 2 = 34,000 \div 2 = 17,000\) square centimeters.

The median, (B), is greater.

The length of one side of a square is one fourth its perimeter. Since the perimeter of the square is \(\displaystyle 12x+16\), the length of one side is

\(\displaystyle (12x+16) \div 4 = 3x+ 4\)

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

\(\displaystyle ( 3x+ 4)^{2}= (3x)^{2}+ 2 \cdot 3x \cdot 4 + 4 ^{2} = 9x^{2}+24x + 16\)

The area of a square is the square of its sidelength. Therefore, square \(\displaystyle 2^{x}\):

\(\displaystyle \left (2^{x} \right ) ^{2} = 2 ^{2 x}\)

A square being a rhombus, its area can be determined by taking half the product of the lengths of its (congruent) diagonals:

\(\displaystyle A = \frac{1}{2} \cdot 2^{x} \cdot 2^{x} = 2 ^{-1} \cdot 2^{x} \cdot 2^{x} = 2^{-1 + x + x } = 2^{2x-1}\)

Let \(\displaystyle a_{n}\) be the lengths of the sides of the squares in meters. \(\displaystyle a_{1} = 60 \div 100 = 0.6\) and \(\displaystyle a_{2} = 1\), so their common difference is

\(\displaystyle d =1 - 0.6 = 0.4\)

The arithmetic sequence formula is 

\(\displaystyle a_{n} = a_{1} + (n-1)d\)

The length of a side of the largest square - square 10 - can be found by substituting \(\displaystyle a_{1} = 0.6, n= 10, d = 0.4\):

\(\displaystyle a_{10} =0.6+ (10-1) 0.4 = 0.6 + 9 (0.4) = 0.6 + 3.6 = 4.2\) 

The largest square has sides of length 4.2 meters, so its area is the square of this, or \(\displaystyle A = 4.2^{2} = 17.64\) square meters.

Of the choices, 18 square meters is closest.

The two smallest squares have perimeters 16 and 20, so their sidelengths are one fourth of these, or, respectively, 4 and 5. Their areas are the squares of these, or, respectively, 16 and 25. Therefore, in the arithmetic sequence,

\(\displaystyle A_{1} = 16\)

\(\displaystyle A_{2} = 25\)

and the common difference is \(\displaystyle d = 25 - 16 = 9\).

The area of the \(\displaystyle n\)th smallest square is

\(\displaystyle A_{n}= A_{1}+ (n-1)d\)

Setting \(\displaystyle A_{1} = 16, n = 6, d=9\), the area of the largest (or sixth-smallest) square is

\(\displaystyle A_{n}= 16+ (6-1)9 = 16 + 5 \cdot 9 = 16 + 45 = 61\)

A square with perimeter \(\displaystyle 160 X\) centimeters has sides of length one-fourth of this, or \(\displaystyle 160 X \div 4 = 40 X\) centimeters. Since one meter is equal to 100 centimeters, divide by 100 to get the equivalent in meters - this is 

\(\displaystyle 40 X \div 100 = 0.4X\)

meters.

The square in (b) has sidelength less than that of the square in (a), so its area is also less than that in (a).

The length of a segment with endpoints \(\displaystyle (0,0)\) and \(\displaystyle (A, B)\) can be found using the distance formula with \(\displaystyle x_{1} = y_{1} = 0\), \(\displaystyle x_{2} = A\), \(\displaystyle y_{2} = B\):

\(\displaystyle d = \sqrt{(x_{2}- x_{1})^{2}+(y_{2}-y_{1})^{2}}\)

\(\displaystyle d = \sqrt{(A-0)^{2}+(B-0)^{2}}\)

\(\displaystyle d = \sqrt{A ^{2}+B^{2}}\)

The length of a segment with endpoints \(\displaystyle (0,0)\) and \(\displaystyle (-B, -A)\) can be found using the distance formula with \(\displaystyle x_{1} = y_{1} = 0\), \(\displaystyle x_{2} = -B\), \(\displaystyle y_{2} = -A\):

\(\displaystyle d = \sqrt{(x_{2}- x_{1})^{2}+(y_{2}-y_{1})^{2}}\)

\(\displaystyle d = \sqrt{(-B-0)^{2}+(-A-0)^{2}}\)

\(\displaystyle d = \sqrt{(-B )^{2}+(-A )^{2}}\)

\(\displaystyle d = \sqrt{B^{2}+A ^{2}}\)

\(\displaystyle d = \sqrt{A ^{2}+B^{2}}\)

The sides are of equal length, so the squares have equal area. Note that the fact that \(\displaystyle B > A\) is irrelevant to the question.