(a) One meter is equal to 100 centimeters; a square with this sidelength has perimeter $\displaystyle 100 \times 4 = 400$ centimeters.

(b) A regular pentagon has five congruent sides; since the sidelength is 75 centimeters, the perimeter is $\displaystyle 75 \times 5 = 375$ centimeters.

This makes (a) greater.

Let $\displaystyle s$ be the sidelength of Square 1. Then the length of a diagonal of this square - which is $\displaystyle \sqrt{2}$ times this sidelength, or $\displaystyle s \sqrt{2}$ by the $\displaystyle 45^{\circ }-45^{\circ }-90^{\circ }$ Theorem - is the same as the diameter of this circle, which, in turn, is equal to the sidelength of Square 2. 

Since the perimeter of a square is four times its sidelength, Square 1 has perimeter $\displaystyle 4s$; Square 2 has perimeter $\displaystyle 4 s \sqrt{2}$, which is $\displaystyle \sqrt{2}$ times the perimeter of Square 1. $\displaystyle \sqrt{2} < 2$, making the perimeter of Square 2 less than twice than the perimeter of Square 1.

The perimeters of the squares are 

$\displaystyle 4 \times 1 = 4$ feet

$\displaystyle 4 \times 2 = 8$ feet

$\displaystyle 4 \times 3 = 12$ feet

$\displaystyle 4 \times 4 = 16$ feet

$\displaystyle 4 \times 5 = 20$ feet

The mean of the perimeters is their sum divided by five;

$\displaystyle (4+8+12+16+20) \div 5 = 60 \div 5 = 12$ feet.

The median of the perimeters is the value in the middle when they are arranged in ascending order; this can be seen to also be 12 feet.

The quantities are equal.

First find the perimeters of the squares:

$\displaystyle 4 \times 100 = 400$ centimeters (one meter being 100 centimeters)

$\displaystyle 4 \times 100 = 400$ centimeters

$\displaystyle 4 \times 120 = 480$ centimeters

$\displaystyle 4 \times 140 = 560$ centimeters

The mean of the perimeters is their sum divided by four:

$\displaystyle (400+400+480+560) \div 4 = 1,840 \div 4 = 460$ feet.

The median of the perimeters is the mean of the two values in the middle, assuming the values are in numerical order:

$\displaystyle (400 + 480) \div 2 = 880 \div 2 = 440$

The mean, (A), is greater.

The length of one side of a square is the square root of its area. The polynomial representing the area of the square can be recognized as a perfect square trinomial:

$\displaystyle 36t ^{2} + 84t + 49$

$\displaystyle =( 6t )^{2} + 2\cdot 6t \cdot 7 + 7^{2}$

$\displaystyle = (6t+7)^{2}$

Therefore, the square root of the area is

$\displaystyle \sqrt{36t ^{2} + 84t + 49} = \sqrt{ (6t+7)^{2}} = |6t+7|$,

which is the length of one side.

The perimeter of the square is four times this length, or

$\displaystyle 4 |6t+7| = |24t+28|$.

Let the length of one side of the first square be $\displaystyle s_{1}$. Then the length of one side of the second-smallest square is $\displaystyle s_{2}= s_{1}+ 2$, and the perimeters of the squares are

$\displaystyle P_{1}= 4s_{1}$

and 

$\displaystyle P_{2} = 4 s_{2}= 4 (s_{1}+ 2) = 4s_{1} + 8 = P_{1}+ 8$

This makes the common difference of the perimeters 8 units.

The perimeters of the squares being in arithmetic progression, the perimeter of the $\displaystyle n$th-smallest square is

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

Since $\displaystyle d = 8$, this becomes 

$\displaystyle P_{n} = P_{1} + 8 (n-1)$

The perimeter of the third-smallest square is 

$\displaystyle P_{3} = P_{1} + 8 (3-1) = P_{1}+ 8 \cdot 2 = P_{1} + 16$

The perimeter of the largest, or sixth-smallest, square is

$\displaystyle P_{6} = P_{1} + 8 (6-1) = P_{1}+ 8 \cdot 5= P_{1} + 40$

The length of one side of this square is one fourth of this, or

$\displaystyle s_{6} = \frac{ P_{6} }{4}= \frac{ P_{1} + 40}{4} = \frac{1}{4} P_{1} + 10$

Therefore, we are comparing $\displaystyle P_{1} + 16$ and $\displaystyle \frac{1}{4} P_{1} + 10$.

Since a perimeter must be positive,

$\displaystyle \frac{1}{4} P_{1} < P_{1}$;

also, $\displaystyle 10 < 16$.

Therefore, regardless of the value of $\displaystyle P_{1}$,

$\displaystyle P_{1} + 16 > \frac{1}{4} P_{1} + 10$,

and 

$\displaystyle P_{3}> s_{6}$,

making (a) the greater quantity.

The perimeter of a square is four times the length of a side, which here is $\displaystyle 2 ^{x }$:

$\displaystyle P = 4 \cdot 2^{x} = 2 ^{2} \cdot 2^{x} =2 ^{x+2}$

The length of a side of a square can be determined by dividing the length of a diagonal by $\displaystyle \sqrt{2}$ - that is, $\displaystyle 2 ^{\frac{1}{2}}$. A diagonal has length $\displaystyle 2^{x}$, so the sidelength is

$\displaystyle 2^{x} \div 2 ^{\frac{1}{2}}= 2 ^{x-\frac{1}{2}}$

Multiply this by four to get the perimeter:

$\displaystyle 4 \cdot 2 ^{x-\frac{1}{2}} = 2^{2} \cdot 2 ^{x-\frac{1}{2}} = 2 ^{x-\frac{1}{2}+2} = 2 ^{x+\frac{3}{2}}$

The two smallest squares have areas 9 and 25, so their sidelengths are the square roots of these, or, respectively, 3 and 5. Their perimeters are the sidelengths multiplied by four, or, respectively, 12 and 20. Therefore, in the arithmetic sequence,

$\displaystyle P_{1} = 12$

$\displaystyle P_{2} = 20$

and the common difference is $\displaystyle d = 20- 12 = 8$.

The perimeter of the $\displaystyle n$th smallest square is

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

Setting $\displaystyle P_{1} = 12, n = 6, d=8$, the perimeter of the largest (or sixth-smallest) square is

$\displaystyle P_{6}=P_{1}+ (6-1)8 = 12+ 5 \cdot 8 = 12 + 40 = 52$.

One yard is equal to three feet, so the length of one side of a square with this perimeter is $\displaystyle \frac{3}{4}$ feet. The area of the square is $\displaystyle \frac{3}{4}\times \frac{3}{4} = \frac{9}{16}$ square feet. $\displaystyle \frac{9}{16} > \frac{1}{2}$, making (a) greater.